18.155 Notes, Fall 2011
Michael Andrews
Department of Mathematics, MIT
December 5, 2011
Abstract
These notes summarise some of the material covered in 18.155 in the Fall of 2011. At present
there are still a few gaps. Some are filled in by the problems at the end.
1
1.1
Introduction
Schwartz Space
Definition: The Schwartz space on Rn is the function space
0
S(Rn ) = {f : Rn −→ C : ιβ ∂ α f ∈ C∞
(Rn ) for all α, β ∈ (N ∪ {0})n }.
For ϕ ∈ S(Rn ), let
X
kϕkN =
sup |xβ ∂ α ϕ(x)|.
|α|+|β|≤N
x∈Rn
Theorem: k · kN defines a norm for each N ∈ N ∪ {0}.
Theorem: S(Rn ) is a complete metric space with respect to
d(ϕ, ψ) =
∞
X
2−N
N =0
kϕ − ψkN
1 + kϕ − ψkN
and we have d-convergence if and only if we have k · kN -convergence for all N ∈ N ∪ {0}.
kϕ−ψk
Proof : Firstly, there’s a claim: whenever k · k is a norm, D(ϕ, ψ) = 1+kϕ−ψk
is a metric.
Now suppose (ϕn ) is Cauchy with respect to d. Let N ∈ N, > 0 and choose M ∈ N such that
n, m ≥ M =⇒ d(ϕn , ϕm ) < 2−N min{/2, 1/2}
Then
n, m ≥ M =⇒
kϕ − ψkN
< min{/2, 1/2} =⇒ kϕ − ψkN < .
1 + kϕ − ψkN
Thus (ϕn ) is Cauchy with respect to k · kN and we see that (ϕn ) converges to some ϕ uniformly
and (∂ α ϕ) converges to some ψ α uniformly. One uses induction, the fundamental theorem of
calculus
Z
t
ϕn (x + tej ) − ϕn (x) =
∂j ϕn (x + sej )ds
0
1
and uniform continuity to show that ∂ α ϕ = ψ α . One checks ϕn −→ ϕ with respect to each
norm k · kN and hence the metric d.
Note: Let SN denote S(Rn ) equipped with the norm k · kN ; let Sd denote S(Rn ) equipped
with the metric d. Since k · kN ≤ k · kN +1 the map SN +1 −→ SN is continuous. Since dconvergence implies k · kN -convergence the map Sd −→ SN is continuous. Thus we have a
commuting diagram:
Sd
...
"
/ SN +1
)/
+/
SN
/, . . .
SN −1
,/ S
0
Theorem:
1. Let X be a topological space. A map X −→ Sd is continuous if and only if X −→ SN is
continuous for all N . In particular,
Sd = lim(. . . −→ SN +1 −→ SN −→ SN −1 −→ . . .)
2. Let V be a normed vector space. A linear map T : Sd −→ V is continuous if and only if
T : SN −→ V is continuous for some N .
3. A linear map T : Sd −→ Sd is continuous if and only if for each N 0 , T : SN −→ SN 0 is
continuous for some N .
Proof :
1. Let τN denote the topology on SN and τd denote the topology on Sd . We have τN ⊂ τd
for all N . Let
∞
[
UN : UN ∈ τN for each N }.
τ ={
N =0
τ is a topology on S(R ) and we want to show τ = τd . Certainly τ ⊂ τd . Suppose > 0.
Choose N ∈ N such that
∞
X
2−K < /2.
N
K=N +1
Then kϕ − ψkN ≤ /4 gives
d(ϕ, ψ) ≤
N
X
∞
X
2−K kϕ − ψkN +
K=0
2−K < .
K=N +1
Thus
{ϕ : kϕ − ψkN < /4} ⊂ {ϕ : d(ϕ, ψ) < }
and we see τd ⊂ τ .
2. One implication is trivial. For the other suppose that T : Sd −→ V is linear and continuous.
Then T −1 {v : kvkV < 1} is open. Thus there exists > 0 such that
{ϕ : d(ϕ, 0) < } ⊂ T −1 {v : kvkV < 1}
Choose N ∈ N such that
∞
X
2−K < /2.
K=N +1
2
Then kϕkN ≤ /4 gives
d(ϕ, 0) ≤
N
X
∞
X
2−K kϕkN +
K=0
2−K < .
K=N +1
Thus given ϕ ∈ Sd \ {0} we have
< 1 =⇒ kT (ϕ)kV < 4 kϕkN .
T ϕ
4 kϕkN V
We conclude T : SN −→ V is continuous.
3. This follows immediately by applying 1 and then 2.
Examples:
1. Let δ : S(Rn ) −→ C be defined by δ(ϕ) = ϕ(0). Then |δ(ϕ)| ≤ kϕk0 and so δ is continuous.
R
2. Let 1 : S(Rn ) −→ C be defined by 1(ϕ) = Rn ϕ.
Z
2 n+1
2 −n−1
(1 + |x| )
ϕ(x)(1 + |x| )
dx ≤ Ckϕk2n+2 .
|1(ϕ)| = Rn
and so 1 is continuous.
3. Let F be an infinitely differentiable function of polynomial growth, i.e. for each α ∈
(N ∪ {0})n there exists an N (α) and a C(α) such that
|∂ α F | ≤ C(α)(1 + kxk)N (α)
and let f ∈ S(Rn ). Then
∂ α (f F ) =
X α
γ≤α
γ
∂ α−γ f ∂ γ F
so that f F is infinitely differentiable and
β α
X α β α−γ
x ∂ (f F )(x) ≤
x ∂
f (x) |∂ γ F (x)|
γ
γ≤α
X α xβ ∂ α−γ f (x) C(γ) (1 + kxk)N (γ) .
≤
γ
γ≤α
Hence, supx∈Rn |xβ ∂ α (f F )(x)| is finite, f F ∈ S(Rn ) and so we have a map
×F : S(Rn ) −→ S(Rn ).
From the above inequality one can also see that for each N 0 ∈ N ∪ {0}, there exists an
N ∈ N ∪ {0} and C ∈ N such that kf F kN 0 ≤ Ckf kN . Thus ×F is continuous.
4. ∂ α : S(Rn ) −→ S(Rn ) is continuous since k∂ α ϕkN ≤ kϕkN +|α| .
1.2
The Fourier Transform
Definition: Let ϕ ∈ S(Rn ). The Fourier transform ϕ̂ of ϕ is given by
Z
ϕ̂(ξ) =
e−ix·ξ ϕ(x)dx
Rn
3
where x · ξ = x1 ξ1 + . . . + xn ξn .
Theorem: F : S(Rn ) −→ S(Rn ), ϕ 7−→ ϕ̂ defines a continuous map.
Proof : We have the following relation
\
ξ β ∂ξα ϕ̂ = (−i)|α|+|β| ∂xβ (xα ϕ).
To see this proceed in two stages: differentiate under the integral sign and then integrate by
parts. By using the same technique as in Example 2 of the last subsection and the equivalence
of two norms proved in the exercises we see that
kϕ̂kN ≤ Ckϕk2n+2+N .
Theorem: F is an isomorphism with inverse G defined by
Z
Gϕ(ξ) = (2π)−n (Fϕ)(−ξ) = (2π)−n
eix·ξ ϕ(x)dx.
Rn
Proof : The main part of the proof is to show that
Z
ϕ̂(ξ)dξ = (2π)n ϕ(0)
Rn
for all ϕ ∈ S(Rn ). If ϕ ∈ S(Rn ) and y ∈ Rn define ψ(x) = ϕ(x + y). Then
Z
ψ̂(ξ)dξ = (2π)n ϕ(y).
Rn
Also
Z
ψ̂(ξ) =
e−ix·ξ ϕ(x + y)dx = eiy·ξ
Z
Rn
e−iζ·ξ ϕ(ζ)dζ = eiy·ξ ϕ̂(ξ)
Rn
giving
−n
Z
−n
ϕ(y) = (2π)
Z
eiy·ξ ϕ̂(ξ)dξ = GFϕ(y).
ψ̂(ξ)dξ = (2π)
Rn
Rn
Thus F is injective and G is surjective. Thus F is surjective too and we have proved the theorem.
We prove the result used using the following lemmas.
Lemma: We can construct a function χ ∈ Cc∞ (Rn ) ⊂ S(Rn ) such that
(
1 when kxk ≤ 21
χ(x) =
0 when kxk ≥ 1
Lemma: If ϕ ∈ S(Rn ) then there exist ϕj ∈ S(Rn ) such that
ϕ(x) = ϕ(0) exp(−|x|2 ) +
n
X
xj ϕj (x).
j=1
Corollary: For any ϕ ∈ S(Rn ).
Z
Z
ϕ̂(ξ)dξ = ϕ(0)
Rn
where F (ξ) = F(exp(−|x|2 )).
Lemma:
R
Rn
F (ξ)dξ
Rn
F (ξ)dξ = (2π)n .
4
1.3
Integration
Theorem: Suppose (ψj ) ⊂ Cc0 (Rn ) is a sequence such that
∞
X
kψj kLp = C < ∞.
j=1
Then
E = {x ∈ Rn :
∞
X
|ψj (x)| = ∞}
j=1
has measure zero.
Proof : Fix > 0 and choose T ∈ N so that T > C(2/)1/p . Let x ∈ E. There exists an
Mx ∈ N such that
Mx
X
|ψj (x)| ≥ T + 1
j=1
and there exists a rectangle R(x) containing x such that
Mx
X
|ψj (y)| ≥ T whenever y ∈ R(x).
j=1
By choosing smaller rectangles, if necessary, we may assume that the coordinates of the vertices
of the rectangles are rational. In this way we obtain a covering of E by a countable collection
of rectangles {Rj }∞
j=1 such that
0
for each k ∈ N, there exists an
Mk0
∈ N such that
Mk
X
|ψj (y)| ≥ T whenever y ∈ Rk .
j=1
We may also assume that the Rj are disjoint. Let Mk = max{M10 , . . . , Mk0 }. Then
p

p 
p
X
Z
Z
Mk
Mk
Mk
k
X
X
X


|ψj | ≤ Tp
vol(Rj ) = S
Tp ≤ S
|ψj |
kψj kLp  ≤ C p
≤
k
k
R
R
j
j
j=1
j=1
j=1
j=1
j=1
j=1
Lp
and so
k
X
j=1
vol(Rj ) ≤
C
T
p
<
∞
X
=⇒
vol(Rj ) < 2
j=1
and E has measure zero.
2
2.1
The Structure of S 0 (Rn )
S(Rn ) ⊂ L2 (R2 ) ⊂ S 0 (Rn ), S(Rn ) ⊂ L1 (R2 ) ⊂ S 0 (Rn )
Observation: We have an embedding S(Rn ) −→ S 0 (Rn ) given by ϕ 7−→ tϕ where
Z
tϕ (ψ) = ϕψ.
5
This is well-defined since | tϕ (ψ)| ≤ kϕkL2 kψkL2 and
2 n
kψkL2 ≤ sup |ψ(x)|(1 + |x| )
Z
2 −2n
12
(1 + |x| )
≤ Ckψk2n .
x∈Rn
Injectivity is seen by setting ψ = ϕ.
Theorem: We have an embedding L2 (Rn ) −→ S 0 (Rn ) given by f 7−→ tf .
Proof : For injectivity we use the density of Cc∞ (Rn ) in L2 (Rn ). Suppose tf = 0. Choose
ψn ∈ Cc∞ (Rn ) with ψn −→ f . Then 0 = hf, ψn i −→ kf k2 =⇒ f = 0.
Lemma (Continuity in the mean): Let f ∈ Lp (Rn ) and define ft (x) = f (x − t). Then ft −→ f
as t −→ 0 with respect to the Lp norm.
Proof : The result holds for Cc0 (Rn ) functions by uniform continuity. Use a 3-argument with
approximating sequences.
R
Lemma: Suppose χ ∈ S(Rn ) with χ = 1. Let χ (x) = −n χ(x/) whenever > 0. Then for
any f ∈ L1 (Rn ) we have f ∗ χ −→ f in L1 (Rn ).
R R
RR
Proof : kf − f ∗ χ kL1 = (f (y) − f (y − x))χ (x)dx dy ≤
|f (y) − f (y − x)| dy χ (x)dx.
Thus kf − f ∗ χ kL1 ≤ sup|t|≤ kf − ft kL1 and we are done by the previous lemma.
Theorem: We have an embedding L1 (Rn ) −→ S 0 (Rn ) given by f 7−→ tf .
Proof : Suppose ∪f = 0. Then f ∗ χ = 0 for all > 0. Thus f = 0.
2.2
Operations on S 0 (Rn )
Observation: For ϕ, ψ ∈ S(Rn ) the fundamental theorem of calculus gives t∂j ϕ (ψ) = tϕ (−∂j ψ).
Definition: For u ∈ S 0 (Rn ), ∂j u(ψ) = u(−∂j ψ).
This definition is chosen so that the following diagram commutes:
S(R n )
_
∂j
t
S 0 (Rn )
/ S(Rn )
_
t
∂j
/ S 0 (Rn )
Observation: For ϕ, ψ ∈ S(Rn ) interchanging the order of intergration gives tϕ̂ (ψ) = tϕ (ψ̂).
This is
Definition: The Fourier transform on S 0 (Rn ) is given by F : S 0 (Rn ) −→ S 0 (Rn ), u 7−→ u ◦ F.
i.e. û(ϕ) = u(ϕ̂). Since F : S(Rn ) −→ S(Rn ) is an isomorphism, F : S 0 (Rn ) −→ S 0 (Rn ) is a
bijection.
Observation: Let ϕ, ψ ∈ S 0 (Rn ) and let F be an infinitely differentiable function of polynomial growth. Then tF ϕ (ψ) = tϕ (F ψ).
6
Definition: Let F be an infinitely differentiable function of polynomial growth. Multiplication by F is given by ×F : S 0 (Rn ) −→ S 0 (Rn ), u 7−→ u ◦ (×F ).
\
\
Lemma: The relation ξ β ∂ξα ϕ̂ = (−i)|α|+|β| ∂xβ (xα ϕ) gives ξ β ∂ξα û = (−i)|α|+|β| ∂xβ (xα u).
Theorem: The Fourier transfrom extends by continuity (using the density of S(Rn ) in L2 (Rn ))
to an essentially isometric isomorphism
F : L2 (Rn ) −→ L2 (Rn );
kFukL2 = (2π)n/2 kukL2 and we have a commuting diagram:
S(R n )
_
F
∼
=
/ S(Rn )
_
L2 (R n )
_
F
∼
=
/ L2 (Rn )
_
S 0 (Rn )
Proof : From the identity
R
ϕ̂ψ =
R
F
∼
=
/ S 0 (Rn )
ϕψ̂, which holds for all ϕ, ψ ∈ S(Rn ), we obtain
Z
Z
ϕ̂µ̂ = (2π)n ϕµ
for all ϕ, µ ∈ S(Rn ). (Take µ = G(ψ).) Thus the stated identity holds on the dense subspace
S(Rn ). By general theory of Banach spaces the map extends to L2 (Rn ) and the stated equality
remains true. The top square automatically commutes. Commutativity of the bottom square
comes from a strong implies weak argument: given (ϕn ) ⊂ S(Rn ) converging to u ∈ L2 (Rn ),
(ϕ̂n ) ⊂ S(Rn ) converges to û ∈ L2 (Rn ). Thus for any ψ ∈ S(Rn ) considering the limit of the
cu (ψ). To see the map is an isomorsequence (tϕ̂n (ψ)) = (tϕn (ψ̂)) we see tû (ψ) = tu (ψ̂) = t
phism we apply the same argument to G.
\
Meta: Because of the formula ξ β ∂ξα û = (−i)|α|+|β| ∂xβ (xα u) we should think of the Fourier
transform as interchanging growth and regularity. Thus L2 (Rn ) has the the same growth as
regularity!
Lemma: The Fourier transform extends by continuity (using the density of S(Rn ) in L1 (Rn ))
to a map
F : L1 (Rn ) −→ C00 (Rn )
and we have a commuting diagram:
S(R n )
_
F
∼
=
/ S(Rn )
_
L1 (R n )
_
F
/ C00 (Rn )
_
S 0 (Rn )
F
∼
=
7
/ S 0 (Rn )
Proof : For f ∈ S(Rn )
|fˆ(ξ)| ≤
Z
|f (x)|dx = kf kL1 .
Thus we have a continuous function F : (S(Rn ), k·kL1 ) −→ (S(Rn ), k·k∞ ) ⊂ C00 (Rn ). A densely
defined bounded linear map from a normed space into a Banach space extends uniquely to a
continuous linear map and so we obtain F : L1 (Rn ) −→ C00 (Rn ). The commutativity of the
top square is immediate. The commutativity of the bottom square is given by the fact that for
f ∈ L1 (Rn ) and ϕ ∈ S(Rn )
Z
Z
f ϕ̂ = fˆϕ.
Notation: From now on we will omit the notation t and view elements of L1 and L2 as elements
of S 0 (Rn ).
2.3
Sobolev Spaces
Definition: For k ∈ N ∪ {0}
H k (Rn ) = {u ∈ L2 (Rn ) : ∂ α u ∈ L2 (Rn ) for all |α| ≤ k}.
Theorem: H k (Rn ) is a Hilbert space when equipped with the inner product
X
hu, vik =
h∂ α u, ∂ α vi.
|α|≤k
Proof : (uj ) is Cauchy on H k if and only if ∂ α uj is Cauchy in L2 for all |α| ≤ k. Thus (uj )
converges to some u ∈ L2 (Rn ) and (∂ α uj ) converges to some v α ∈ L2 (Rn ). Now
∂ α u(ϕ) = u((−1)|α| ∂ α ϕ) = lim uj ((−1)|α| ∂ α ϕ) = lim ∂ α uj (ϕ) = v α (ϕ).
j−→∞
j−→∞
Thus u −→ v ∈ H k .
Inequalities: (|x1 | + . . . + |xn |)/n ≤ |x| ≤ |x1 | + . . . + |xn |, (1 + |x|)2 /2 ≤ (1 + |x|2 ) ≤ (1 + |x|)2
show that the functions
(1 + |x|),
(1 + |x1 | + . . . + |xn |),
hxi = (1 + |x|2 )1/2
are all the same size on Rn .
Lemma: Let s ∈ R and define Fs : Rn −→ C, x 7−→ (1 + |x|2 )s/2 . Then Fs is an infinitely
differentiable function of polynomial growth.
Proof : We show that ∂ α Fs (x) = (1 + |x|2 )s/2−|α| P (x) where P is a polynomial of degree
less than or equal to |α|. By induction on |α| it is enough to note that if P is a polynomial of
degree less than or equal to n, then
∂i ((1 + |x|2 )s/2−n P (x)) = (1 + |x|2 )s/2−n−1 (2xi (s/2 − n)P (x) + (1 + |x|2 )∂i P (x))
and
2xi (s/2 − n)P (x) + (1 + |x|2 )∂i P (x)
is a polynomial of degree less than or equal to n + 1.
Now for each polynomial there exists a CP and an NP such that
P (x) ≤ CP (1 + |x|)NP .
8
Also, if s ≤ N , then (1 + |x|2 )s/2 ≤ (1 + |x|)s ≤ (1 + |x|)N .
Relevance: This lemma shows that hxis u makes sense whenever u ∈ S 0 (Rn ), an important
observation that will be used immediately.
Proposition: Let k ∈ N and u ∈ S 0 (Rn ). Then u ∈ H k (Rn ) if and only if hξik û ∈ L2 (Rn ).
Definition: For s ∈ R
H s (Rn ) = {u ∈ S 0 (Rn ) : hξis û ∈ L2 (Rn )}.
R
We have an inner product defined by hu, vis = hξi2s ûv̂ = hhξis û, hξis v̂iL2 .
This inner product is consistent with the one previously defined because F is essentially an
isometric isomorphism on L2 and
\
ξ β ∂ξα û = (−i)|α|+|β| ∂xβ (xα u).
0
Note: s ≥ s0 ⇐⇒ H s (Rn ) ⊂ H s (Rn ).
Theorem: S(Rn ) ⊂ H s (Rn ) is a dense inclusion.
Proof : Let u ∈ H s . Then f = hξis û ∈ L2 . S(Rn ) is dense in L2 (Rn ) and so there exists
a sequence (fj ) ⊂ S(Rn ) with fj −→ f . Let uj = G(hξi−s fj ) ∈ S(Rn ). Then
kuj − uks = khξis (ûj − û)kL2 = kfj − f kL2 −→ 0.
Corollary (Weak = Strong): Each u ∈ H s (Rn ) has strong derivatives of all order orders |α| ≤ s,
i.e. there exists a sequence (uj ) ⊂ S(Rn ) such that ∂ α uj −→ ∂ α u in L2 for all |α| ≤ s.
αu −∂
α uk 2 = kξ α û −ξ α ûk 2 ≤ Ckhξis (û − û)k 2 = Cku −uk −→ 0,
d
Proof : For |α| ≤ s, k∂[
j
j
j
j
s
L
L
L
which completes the proof because F is essentially an isometry on L2 .
Theorem (Sobolev Embedding): Suppose k ∈ N ∪ {0} and s > n/2 + k then H s (Rn ) ⊂ C0k (Rn ).
Proof : Using polar coordinates we see that hξis ∈ L1 (Rn ) if and only if s < −n. Suppose
s > n/2 and u ∈ H s (Rn ). Then hξi−s ∈ L2 (Rn ) and hξis û ∈ L2 (Rn ). Thus û ∈ L1 (R) which
implies u ∈ C00 (Rn ) and the theorem is proved for k = 0.
Suppose s > n/2 + k, u ∈ H s (Rn ) and |α| ≤ k. Then ∂ α u ∈ H s−|α| (Rn ) and s − |α| > n/2 so
that ∂ α u ∈ C00 (Rn ). We need to show this element is the strong derivative of u. Let (uj ) be a
sequence in S(Rn ) which converges to u in H s . Then
αu − ∂
α uk 1 ≤ Ckhξi−(s−|α|) k 2 khξis−|α| (∂
αu − ∂
α u)k 2
d
d
[
k∂ α uj − ∂ α uk∞ ≤ Ck∂[
j
j
L
L
L
≤ C 0 kuj − uks .
Thus, the convergence of (∂ α uj ) to ∂ α u is uniform and we are done. Notice this inequality
shows the embedding is continuous, too.
2.4
The Structure of S 0 (Rn )
Definition (Working): u ∈ S 0 (Rn ) has compact support if there exists R > 0 such that u(ϕ) = 0
whenever ϕ ∈ S(Rn ) and ϕ = 0 on {x : |x| < R}.
9
Lemma: u ∈ S 0 (Rn ) has compact support if and only if χu = u for some χ ∈ Cc∞ (Rn ).
Theorem (Paley-Wiener-Schwartz): If u ∈ S 0 (Rn ) has compact support then û ∈ C ∞ (Rn )
and there exists an N ∈ N and Cα > 0 such that
|∂ α û(ξ)| ≤ Cα hξiN .
Proof : First we will outline the idea. Suppose u ∈ S(Rn ) and χ ∈ Cc∞ (Rn ) and that χu = u.
Then
Z
Z
−ix·ξ
û(ξ) = u(x)e
dx = u(x)χ(x)e−ix·ξ dx = u(x 7−→ χ(x)e−ix·ξ ).
Thus we suppose u ∈ S 0 (Rn ) has compact support, that χu = u where χ ∈ Cc∞ (Rn ), and we
define F : Rn −→ C by
F (ξ) = u x 7−→ χ(x)e−ix·ξ .
We will check the requisite properties are satisfied by F and that û = F .
0
As ξ 0 −→ ξ, kχ(x)e−ix·ξ − χ(x)e−ix·ξ kN −→ 0 and so F is continuous. Since u is continuous there is an N with |F (ξ)| ≤ Ckχ(x)e−ix·ξ kN ≤ C 0 hξiN . One checks smoothness in a similar
way to continuity; the difference quotient converges in norm.
Suppose ϕ ∈ S(Rn ). Then û(ϕ) = u(ϕ̂) = χu(ϕ̂) = u(χϕ̂) and F (ϕ) =
so we must verify
Z
u(χϕ̂) = F (ξ)ϕ(ξ)dξ.
R
F (ξ)ϕ(ξ)dξ and
By continuity it is enough to check this for ϕ ∈ Cc∞ (Rn ).
Let σj = 2−j Zn ∩ supp ϕ. Then
Z
X
ϕ̂(x) = e−ix·ξ ϕ(ξ)dξ = lim 2−nj
e−ix·ξij ϕ(ξij )
j−→∞
ξij ∈σj
and this convergence is uniform in x and the same remains true after differentiating with respect
to xj . Since u is continuous and linear


X
u(χϕ̂) = u x 7−→ χ(x) lim 2−nj
e−ix·ξij ϕ(ξij )
j−→∞
= lim 2−nj
j−→∞
ξij ∈σj
u x 7−→ χ(x)e−ix·ξij ϕ(ξij )
ξij ∈σj
−nj
= lim 2
j−→∞
X
X
F (ξij )ϕ(ξij )
ξij ∈σj
Z
=
F (ξ)ϕ(ξ)dξ
which is what we wished to show.
Definition: A sequence (uj ) in S 0 (Rn ) is said to converge weakly to u ∈ S 0 (Rn ) if for all
ϕ ∈ S(Rn ) we have uj (ϕ) −→ u(ϕ).
10
Note: Whenever (uj ) ⊂ S 0 (Rn ) converges weakly to u, (ûj ) converges weakly to û.
Lemma: Compactly supported distributions are dense in S 0 (Rn ).
Proof : Suppose u ∈ S 0 (Rn ). Choose χ ∈ Cc∞ (Rn ) ⊂ S(Rn ) with 0 ≤ χ ≤ 1, χ = 1 whenever |x| ≤ 1/2 and χ = 0 whenever |x| ≥ 1. Let uj = χ(x/j)u. Then uj −→ u weakly.
Proposition: S(Rn ) is weakly dense in S 0 (Rn ).
Proof : Since weak convergence is characterized by a topology we have transitivity of density
and it is enough to show S(Rn ) is dense in Cc−∞ (Rn ), the compactly supported distributions.
Let u ∈ Cc−∞ (Rn ). Then û ∈ C ∞ (Rn ) is of slow growth. The previous proof shows we have
uk ∈ S(Rn ) with ûk ∈ Cc∞ (Rn ) and such that ûk −→ û weakly. Thus uk −→ u weakly.
Proposition (Reisz Representation): If u ∈ S 0 (Rn ), then |u(ϕ)| ≤ CkϕkL2 for all ϕ ∈ S(Rn ) if
and only if u ∈ L2 (Rn ).
Proof : The ‘if’ is Cauchy-Schwarz. Together with the density of S(Rn ) in L2 (Rn ) the inequality tells us that the map u extends by continuity to an element of L2 (Rn )0 . The Riesz
representation theorem for Hilbert spaces then gives the result. However, it is possible to see
another proof using the theory we have set up...
Theorem (Structure Theorem): If u ∈ S(Rn ) then there exists N ∈ N and w ∈ C00 (Rn )∩L2 (Rn )
such that u = hxi2N hDi2N w. Here hDi2N = (1 + |D|2 )N = (1 − ∂12 − . . . − ∂n2 )N .
Proof : Let u ∈ S 0 (Rn ) and N ∈ N. Pick v, w ∈ S 0 (Rn ) so that v = hxi−2N u and ŵ = hξi−2N v̂.
Then u = hxi2N hDi2N w. By the Sobolev embedding theorem it is enough to choose N ∈ N
so that w ∈ H n (Rn ). By definition we need to show that hξin ŵ ∈ L2 (Rn ) and by Riesz it is
enough to show that there exists a C > 0 with
|hξin ŵ(ϕ)| ≤ CkϕkL2 .
Since u is continuous there exists an M ∈ N and C 0 > 0 such that |u(ϕ)| ≤ C 0 kϕkM whenever
ϕ ∈ S(Rn ). Thus |v(ϕ)| ≤ C 0 khxi−2N ϕkM . Choose M 0 > n/2 + M and N > M 0 /2 + n/2. Then
|v(ϕ)| ≤ C 0 khxi−2N ϕkM ≤ C 00 kϕkC M ≤ C 000 kϕkH M 0 .
The second inequality holds because N > M/2 as one checks by applying the Leibniz rule. The
third inequality is given by the Sobolev embedding theorem because M 0 > n/2 + M . Finally,
0
|v̂(ϕ)| = |v(ϕ̂)| ≤ C 000 kϕ̂kH M 0 = CkhξiM ϕkL2
and
0
|hξin ŵ(ϕ)| = |v̂ hξin−2N ϕ | ≤ Ckhξin+M −2N ϕkL2 ≤ CkϕkL2 .
Note: In the proceeding proof we can give w more regularality and decay by increasing N .
Summary: The Sobolev embedding
theorem and the Schwartz structure theorem give S(Rn ) =
T
S
−s s
n
0
n
H (R ) and S (R ) = s hxis H −s (Rn ). We may give the first the limit topology and
s hxi
the second the colimit topology although we will not think about this now.
Exercise 1...
11
Proposition: u ∈ H −s (Rn ) if and only if there exists a C ∈ R such that for all ϕ ∈ S(Rn ) we
have |u(ϕ)| ≤ Ckϕks . Thus H −s (Rn ) = H s (Rn )0 .
2.5
Support
Ω will always denote an open subset of Rn .
Motivation: Let ϕ : Ω −→ C be a continuous function. Then the support of ϕ is
supp ϕ = {x ∈ Ω : ϕ(x) 6= 0}
Ω
One can check that this is the complement (in Ω) of the largest open set on which ϕ vanishes.
Definition: Let u ∈ S 0 (Rn ). u is said to be identically zero on an open subset U ⊂ Rn if
u(ϕ) = 0 for all ϕ ∈ Cc∞ (Rn ) with supp ϕ ⊂ U .
Definition: Let u ∈ S 0 (Rn ). The support of u, supp u is the complement of the maximal
open set on which u vanishes, i.e. the complement of the set
[
{U : U is open and u is identically zero on U }
To check this is well-defined we need a lemma.
Lemma: If u ∈ S 0 (Rn ) vanishes on a collection of open set {Uα }. Then u vanishes on
S
α
Uα .
Consistency: One checks that the working definition we used in 2.4 of a distribution having compact support agrees with the terminology we now have available to us.
Definition: The singular support of an element u ∈ S 0 (Rn ), sing supp u is the complement of
the maximal open set on which u restricts to a smooth function.
Explanation: For each open subset Ω ⊂ Rn we have a commutative diagram:
Cc∞ (Ω)
i
/ S(Rn )
∗
/ S 0 (Rn )
/ C ∞ (Ω)
∗
/ (Cc∞ (Ω))0
µ
t
λ
(C ∞ (Ω))0
r
r
i
An element u ∈ S 0 (Rn ) restricts to a smooth function on Ω if there exists a function v ∈ C ∞ (Ω)
such that i∗ u = µv. Said explicitly, an element
R u restricts to a smooth function on Ω if there
exists a function v ∈ C ∞ (Ω) such that u(ϕ) = vϕ for all ϕ ∈ Cc∞ (Ω). Since µ is injective the appropriate lemma is proved using naturality of µ with respect to inclusions and that C ∞ is a sheaf.
Note: ri, i, i∗ r∗ , r∗ , t, λ and µ are injective; ri and r have dense image; i∗ r∗ , i∗ , t, λ
and µ have weakly dense images.
Theorem: An element u ∈ S 0 (Rn ) has compact support contained in Ω if and only if it
extends by continuity to C ∞ (Ω).
Proof : Look at the top left square of the diagram above. An element u ∈ S 0 (Rn ) has compact
support contained in Ω if and only if there exists a χ ∈ Cc∞ (Ω) with χu = u. Suppose this holds.
12
Then any sequence (un ) ⊂ S(Rn ) weakly converging to u can be taken to lie in Cc∞ (Ω). Thus u
is in the image of r∗ , which means we can extend to C ∞ (Ω). Conversely, if we can extend u to
C ∞ (Ω) then it lies in the image of r∗ and there is a sequence (un ) ⊂ Cc∞ (Ω) converging weakly
to u. There exists a χ ∈ Cc∞ (Ω) with χun = un for all n ∈ N (by construction; see problems).
Then we are done since χu = u.
Theorem: If u ∈ S 0 (Rn ) and supp u ⊂ {0}. Then u =
P
|α|≤N
cα ∂ α δ.
Proof : Suppose u ∈ S 0 (Rn ) and supp u ⊂ {0}. Pick N ∈ N and C > 0 so that |u(ϕ)| ≤ CkϕkN
for all ϕ ∈ S(Rn ). We can write any ϕ ∈ S(Rn ) as
ϕ(x) =
X 1
(∂ α ϕ(0))xα + ψN (x)
α!
|α|≤N
where ψN ∈ C ∞ (Rn ) and ∂ α ψN (0) = 0 for all |α| ≤ N . Choose χ ∈ Cc∞ (Rn ) with 0 ≤ χ ≤ 1,
χ = 1 whenever |x| ≤ 1/2 and χ = 0 whenever |x| ≥ 1. Then u = χu and so
X 1
(∂ α ϕ(0))u(χxα ) + u(χψN )
α!
|α|≤N
X
=
cα (∂ α δ)(ϕ) + u(χψN )
u(ϕ) =
|α|≤N
for appropriate cα which do not depend on ϕ. Thus we just need to show u(ϕ) = 0 whenever
ϕ ∈ S(Rn ) and ∂ α ϕ(0) = 0 for all |α| ≤ N . Since
X
|u(ϕ)| = |u(χϕ)| ≤ CkχϕkN ≤ C 0
sup |∂ α ϕ(x)|
|α|≤N
|x|≤1
the following lemma will complete the proof.
Lemma: Suppose ϕ ∈ S(Rn ) and that ∂ α ϕ(0) = 0 for all |α| ≤ N . Then there is a sequence (ϕj ) ⊂ S(Rn ) such that 0 ∈
/ supp ϕj for each j and ∂ α ϕj −→ ∂ α ϕ uniformly on |x| ≤ 1
for all |α| ≤ N .
Proof : Let χ ∈ Cc∞ (Rn ) be equal to 1 on |x| ≤ 1, 0 outside |x| ≤ 2 and satisfy 0 ≤ χ ≤ 1. Let
ϕj (x) = (1 − χ(jx))ϕ(x). We just need to show that (χϕj ) converges to χϕ uniformly.
Taylor’s theorem with remainder gives
ϕ(x) =
X
xβ ϕβ (x)
|β|=N +1
for some smooth ϕβ so that for j > 1
(χϕ − χϕj )(x) =
X
(xβ χ(jx))(χ(x)ϕβ (x)).
|β|=N +1
We now apply the Leibniz rule to see that we really only need to analyse
X
ψj (x) =
xβ χ(jx).
|β|=N +1
13
3
P (D)Es
3.1
Homogeneous Distributions
Definition: A function f : R −→ C is said to be homogeneous of degree z ∈ C if
f (tx) = tz f (x) for all x ∈ R, t > 0.
Examples: The obvious examples are the powers themselves:
(
(
xz when x > 0
0
z
z
x− =
x+ =
0
when x ≤ 0
(−x)z
when x ≥ 0
when x < 0
Definition: u ∈ S 0 (Rn ) is homogeneous of degree z ∈ C if u(ϕt ) = tz+1 u(ϕ) for all ϕ ∈ S(Rn ).
Here we have ϕt (x) = t−n ϕ(x/t).
Example: The functions xz± are smooth and of polynomial growth away from the origin. They
are locally integrable at the origin if and only if Rez > −1. In this case they define elements of
S 0 (R) which are homogeneous of degree z ∈ C.
Proposition: For any z ∈ C the elements of S 0 (R) which are homogeneous of degree z form a
two-dimensional space.
3.2
Fundamental Solutions
Definition: Let P be a polynomial. E ∈ S 0 (Rn ) (or Cc∞ (Rn )) is a fundamental solution of P if
P (D)E = δ.
Motivation: We can define a convolution u ∗ v with the property that Dj (u ∗ v) = (Dj u) ∗ v =
u ∗ (Dj v). If E is a fundamental solution of P and u = E ∗ f , then P (D)(u) = (P (D)E) ∗ f =
δ ∗ f = f.
Definition: F is a (convolution) parametrix for P if P (D)F = δ + G for G ∈ C ∞ (Rn ).
Theorem: If P is elliptic (see next subsection), then there exists a parametrix F ∈ Cc∞ (Rn ) for
P.
P (D)F = δ + R
where R ∈ Cc∞ (Rn ). sing supp (F ) ⊂ {0} and F ∗ H s ⊂ H m+s for all s ∈ Rn .
Note: I think the next subsection makes this precise. Figure this out...
3.3
Elliptic Regularity
Definition: A polynomial P is elliptic of order m if P (ξ) =
following statements hold:
P
1. Pm (ξ) = |α|=m cα ξ α 6= 0 on Rn \ 0.
P
|α|≤m cα ξ
α
and any of the
2. |Pm (ξ)| ≥ C|ξ|m for some C > 0.
3. There exists > 0 such that |P (ξ)| ≥ |ξ|m in |ξ| ≥ 1/.
4. There exists Q(ξ) ∈ C ∞ (Rn ) such that Q(ξ)P (ξ) = 1 − χ(ξ) where χ ∈ Cc∞ (Rn ) and
|∂ α Q(ξ)| ≤ Cα hξi−m−|α| .
14
5. There exists Q̃(ξ) ∈ C ∞ (Rn ) such that Q̃(ξ)P (ξ) = hξim − χ̃(ξ) where χ̃ ∈ Cc∞ (Rn ) and
|∂ α Q̃(ξ)| ≤ Cα0 hξi−|α| .
Consistency Check: (1) ⇐⇒ (2) follows from compactness of S n−1 ⊂ Rn . (2) =⇒ (3)
follows because for large |ξ|
X
α
cα ξ ≥ C|ξ|m − C 0 (|ξ|m−1 + 1)
|P (ξ)| ≥ |Pm (ξ)| − |α|<m
C|ξ|m
C|ξ|m
C|ξ|m
0
m−1
0
=
+
− C |ξ|
−C ≥
.
2
2
2
(3) =⇒ (2) follows in identical fashion together with the fact that both sides of the equation
in (2) scale the same way under scalings of ξ.
(3) =⇒ (4): Choose χ ∈ Cc∞ (Rn ) such that χ = 1 on |ξ| ≤ 1/ and define
(
1−χ(ξ)
when kξk ≥ 1
P (ξ)
Q(ξ) =
0
when kξk ≤ 1
C
Then Q(ξ) ∈ C ∞ (Rn ), Q(ξ)P (ξ) = 1 − χ(ξ) and |Q(ξ)| ≤ |ξ|
m in |ξ| ≥ 1/. It suffices to check
the required condition on |ξ| > R where χ = 0. One checks using an induction that
|∂ α (1/P )(ξ)| ≤ C 0 hξi−m−|α| .
(4) =⇒ (5): We let Q̃(ξ) = hξim Q(ξ), use Leibniz and note that ∂ β hξim ≤ Chξim−|β| .
(5) =⇒ (3): Outside the support of χ̃ we have Q̃(ξ)P (ξ) = hξim and so hξim ≤ C00 |P (ξ)|.
Proposition: Let P be elliptic of order m. Suppose that u ∈ S 0 (Rn ) and that û ∈ L2loc (Rn );
an example of such a u is an element of H N (Rn ). Suppose, in addition, that P (D)u ∈ H s (Rn ).
Then u ∈ H s+m (Rn ).
Proof : By definition hξis P (ξ)û = hξis P\
(D)u ∈ L2 (Rn ). Choose Q̃(ξ) ∈ C ∞ (Rn ) as in (5)
m
so that hξi − χ̃(ξ) = Q̃(ξ)P (ξ) where χ̃ ∈ Cc∞ (Rn ) and Q̃ is bounded. Then
hξis+m û − hξis χ̃(ξ)û = hξis Q̃(ξ)P (ξ)û ∈ L2 (Rn ).
Since, hξis χ̃(ξ)û ∈ L2 (Rn ) we have hξis+m û ∈ L2 (Rn ) and so u ∈ H s+m (Rn ).
Theorem: Suppose Ω ⊂ Rn is open, u ∈ (Cc∞ (Ω))0 and P is an elliptic polynomial of ors+m
s
der m. If P (D)u ∈ Hloc
(Ω) then u ∈ Hloc
(Ω).
Explanation: There are two points to make. Firstly, if u ∈ (Cc∞ (Ω))0 then the usual definition of P (D)u returns another element of (Cc∞ (Ω))0 . For the second point let χ ∈ Cc∞ (Ω). We
obtain an element of (C ∞ (Rn ))0 by the formula ϕ 7−→ u(χϕ) and thus an element of S 0 (Rn ). If,
s
in fact, we obtain an element of H s (Rn ) for all choices of χ then we say u ∈ Hloc
(Ω).
Proof : Let χ ∈ Cc∞ (Ω). We wish to show χu ∈ H s+m (Rn ). Construct a sequence (χj ) ⊂ Cc∞ (Ω)
such that each χj is 1 on a neighborhood of supp χ and supp χj+1 ⊂ {χj = 1}. χ0 u is a compactly supported tempered distribution and so by the structure theorem (or rather, the proof of
15
it) χ0 u ∈ H N (Rn ) for some N ∈ R. We claim that χj u ∈ H pj (Rn ) where pj = min{N +j, m+s}.
To prove the claim we suppose inductively that χj u ∈ H pj (Rn ). Then we show that P (D)(χj+1 u)
is contained in H pj −m+1 (Rn ) + H s (Rn ). Global regularity allows us to deduce that χj+1 u ∈
H qj (Rn ) where qj = min{pj − m + 1, s} + m = pj+1 .
To fill in the details in the middle part of the argument we use the Leibniz rule together with
the facts that χj+1 χj = χj+1 and χj+1 χβj = 0 for β > 0:
P (D)(χj+1 u) = P (D)(χj+1 χj u) =
X
Dγ χj+1 Dβ (χj u) + χj+1 P (D)u
where in the sum |β| ≤ m − 1.
4
Hilbert Space
Throughout, H will denote an infinite dimensional seperable Hilbert space.
4.1
Basics
Definition: A sequence (un ) ⊂ H converges weakly to u ∈ H if hun , ϕi −→ hu, ϕi for all ϕ ∈ H.
Corollary (of the Uniform Boundedness Theorem): If (un ) ⊂ H converges weakly to u ∈ H,
then (kun k) is a bounded sequence.
Lemma: Suppose (ei )∞
i=1 is an orthonormal basis for H. If (un ) is a bounded sequence in
H then (un ) converges to u weakly if and only if hun , ei i −→ hu, ei i for all i ∈ N.
Lemma: Every bounded sequence (un ) ⊂ H has a weakly convergent subsequence.
Definition: A sequence of operators (An ) ⊂ B(H) converges strongly to A ∈ B(H) if for
each f ∈ H, AN f −→ Af .
PN
Remark: Given an orthonormal basis (ei ) of H, EN f =
i=1 hf, ei iei defines a bounded
operator EN : H −→ H for each N ∈ N. Since each EN is a non-trivial projection we have
∗
2
= EN and kEN k = k1H − EN k = 1 for all N ∈ N. (EN ) converges strongly to I
= EN , EN
EN
by Bessel’s theorem.
Theorem: Suppose (ei )∞
i=1 is an orthonormal basis for H. A set K ⊂ H is compact if and only
it is closed and bounded and for each > 0 there exists an N ∈ N such that k(1H − EN )uk < whenever u ∈ K. One might paraphrase this last condition by saying “elements of K have
uniformly small tails”.
To prove this result we use a lemma.
4.2
Self-Adjoint Operators
Lemma: Let A ∈ B(H) with A∗ = A. Then kAk = sup{|hAu, ui| : kuk ≤ 1}.
Proof : Let a = sup{|hAu, ui| : kuk ≤ 1}. We need kAk ≤ a. Let u, v ∈ H. Then
4hAu, vi = hA(u + v), u + vi − hA(u − v), u − vi + ihA(u + iv), u + ivi − ihA(u − iv), u − ivi.
16
We are interested in |hAu, vi| when kuk = kvk = 1. By replacing u by eiθ u for some θ we do
not change |hAu, vi| and we can assume hAu, vi = |hAu, vi|. In this case
4|hAu, vi| = hA(u + v), u + vi − hA(u − v), u − vi ≤ aku + vk2 + aku − vk2 = 4a
so that kAk = sup{hAu, vi : kuk = kvk = 1} ≤ a.
Proposition: Let A ∈ B(H) with A∗ = A. Then A − λI is invertible for all λ ∈ C \ R
and either A − kAkI or A + kAkI is not invertible.
Proof : Let λ = α + iβ where α, β ∈ R and β 6= 0. Then k(λI − T )vk ≥ |β|kvk as one
sees by expanding k(λI − T )vk2 . Similarly k(λI − T )vk ≥ |β|kvk. Let v ∈ (λI − T )(H))⊥ . Then
h(λI − T )u, vi = 0 for all u ∈ H giving (λI − T )v = 0 and thus v = 0. We conclude the image
of (λI − T ) is dense and so (λI − T ) is invertible (we have verified two conditions which are
sufficient to imply this). We now prove the second claim. The previous lemma gives a sequence
(un ) in H with kun k = 1 and hAun , un i −→ ∓kAk. Then
kAun ± kAkun k2 = kAun k2 ± 2kAkhAun , un i + kAk2
≤ 2kAk (kAk ± hAun , un i) −→ 0
and thus A ± kAkI is not invertible.
4.3
Finite Rank Operators
Definition: A ∈ B(H) is of finite rank if A(H) is finite dimensional.
Lemma: Let A be a finite rank operator. Then there exists elements e1 , . . . , eN , f1 , . . . , fN ∈ H
where N = dim A(H) such that
N
X
Aϕ =
hϕ, ei ifi
i=1
Note: Ran(A∗ ) ⊂ Null(A)⊥ . If A is of finite rank then Null(A)⊥ is finite dimensional and so
A∗ is of finite rank. With ei , fi as above
hϕ, A∗ ψi = hAϕ, ψi =
N
X
hhϕ, ei ifi , ψi =
i=1
which gives A∗ ψ =
N
X
hϕ, ei ihfi , ψi =
i=1
N
X
hϕ, hψ, fi iei i
i=1
PN
i=1 hψ, fi iei .
Definition: R(H) will denote the set of finite rank operators on H.
Note: R(H) is a two-sided ideal in B(H).
Example: Any operator A ∈ B(H) is the strong limit of a sequence of finite rank operators.
We can see this by choosing an orthonormal basis of H and letting AN = EN A.
4.4
Compact Operators
Definition: K(H) = R(H) where the closure is taken with respect to the norm on B(H).
Note: K(H) is a two-sided ideal in B(H).
Philosophy: Elements of K(H) are supposed to behave like infinite matrices.
Definition: An element A ∈ B(H) is said to be compact if A{u ∈ H : kuk ≤ 1} is compact.
17
Since A{u ∈ H : kuk ≤ 1} is always closed and bounded, this is equivalent to specifying that
elements of A{u ∈ H : kuk ≤ 1} have uniformly small tails.
Theorem: K(H) is equal to the set of compact operators.
Proof : If A is compact, let AN = EN A ∈ F(H). kA−AN k = supkf k≤1 k(1H −EN )Af kH −→ 0,
which gives A ∈ K. Conversely, if A ∈ K there exists a sequence (FM ) ⊂ F with kA−FM k −→ 0.
k(1H − EN )Af k ≤ k(1H − EN )(A − FM )f k + k(1H − EN )FM f
≤ kA − FM k + k(1H − EN )FM f k
It is now clear how to proceed.
Lemma: An operator K ∈ B(H) is compact if and only for any weakly convergent sequence
(un ) in H, (Kun ) is norm convergent in H.
Proof : Let (un ) converge weakly to u. Then (un ) is bounded. Thus any subsequence of
(Kun ) has a convergent subsequence and it must necessarily converge to Ku. Thus (Kun )
converges to Ku. The converse is easier.
Theorem (Spectral Theorem): If A ∈ K and A∗ = A, then H has an orthonormal basis
of eigenvectors of A. Eigenvectors with non-zero eigenvalues can be arranged into a sequence
such that (|λj |) is a non-increasing sequence with λj −→ 0 (in the case that Null(A)⊥ is finite
dimensional this sequence is finite).
The proof relies heavily on a lemma.
Lemma: Let A ∈ K(H) and A∗ = A. Then F : {u ∈ H : kuk = 1} −→ R, u 7−→ hAu, ui is a
continuous function which attains its supremum and infimum. If the maximum is strictly positive then it is attained at an eigenvector of A with this extremal value as eigenvalue; similarly,
if the minimum is strictly negative.
Both results are proved very cleanly in Professor Melrose’s notes which can be found at http:
//math.mit.edu/classes/18.102/Chapter3.pdf
4.5
The Functional Calculus
Suppose A ∈ K(H) and A∗ = A. Then we have a homomorphism C 0 (Spec A) −→ B(H),
A 7−→ f (A): choose an orthononormal basis
P of eigenvectors (ej ) for A; let λj be the eigenvalue
corresponding to ej and define f (A)u = j f (λj )hu, ej iej ; provided f (0) = 0, f (A) is compact.
The most general extension of this fact which I have seen proved is the following theorem.
Theorem: Suppose A ∈ B(H) and suppose A∗ A = AA∗ . Then there exists an isometric
∗-homomorphism C 0 (Spec A) −→ B(H), f 7−→ f (A) such that 1SpecA (A) = 1H , ιj (A) = Aj and
ι(A) = A∗ . The image of the mapping is the closure of the algebra generated by 1H , A and A∗ .
The mapping f 7−→ f (A) is called the functional calculus.
We will be content to prove the following theorem.
Theorem: Let A ∈ B(H) and suppose A∗ = A. There exists a isometric homomorphism
CR0 (Spec A) −→ B(H), f 7−→ f (A) such that 1SpecA (A) = 1H , ιj (A) = Aj .
18
Proof : The definition for real polynomials p is determined and one can check that Spec p(A) =
p(Spec A). In addition, p(A) is self adjoint and so
kp(A)k = sup{|t| : t ∈ Spec p(A)} = sup{|p(t)| : t ∈ (Spec A)} = kpkCR0 (SpecA) .
The result holds by density of the real polynomials in CR0 (SpecA) (Stone-Weierstrass).
Theorem (Polar Decomposition): If B ∈ B(H), there exists a unique self-adjoint A ∈ B(H)
such that hAu, ui ≥ 0 and U a partial isometry such that B = U A.
1
Proof : Take A = (B ∗ B) 2 , using the functional calculus. Note that U is uniquely determined
on Ran(A). If we demand that U be 0 on Ran(A)⊥ then U is uniquely determined.
4.6
Hilbert-Schmidt
Definition: A ∈ B(H) is said to be Hilbert-Schmidt if there exists an orthonormal basis (ei )
such that
∞
X
kAei k2 < ∞.
i=1
P
Proposition: Let A, B be Hilbert-Schmidt. Then i hAei , Bei i converges for any orthonormal
basis and its value does not depend on the choice of basis.
P∞
2
Proof : Let (ei ) be an orthonormal basis for
which i=1 kAe
Suppose (fj ) is
P
Pi k converges.
2
2
another orthonormal basis. Then kAei k = j |hAei , fj i| = j |hei , A∗ fj i|2 , which gives
X
kAei k2 =
i
XX
i
|hei , A∗ fj i|2 =
XX
j
j
|hei , A∗ fj i|2 =
i
X
kA∗ fj k2 .
j
2
Thus
argument with (fj ) in place of (ei )).
j kAfj k converges (use the sameP
P
kAe
kkBe
k
≤
kAk
kBk
so
that
i
i
HS
HS
i
i hAei , Bei i converges and
X
X
hAei , Bei i =
hAei , fj ihfj , Bei i =
hB ∗ fj , ei ihei , A∗ fj i
P
j
gives
P
i hAei , Bei i
=
P
j hB
∗
P
i
|hAei , Bei i| ≤
j
fj , A∗ fj i.
Proposition: The space HS(H)
P of Hilbert-Schmidt operators is a Hilbert Space with inner
product given by hA, BiHS = i hAei , Bei i. kAkHS = kA∗ kHS .
Proof : Need to check completeness.
Note: HS(H) is a two-sided ideal in B(H). For B ∈ B(H) and A ∈ HS(H), kBAkHS ≤
kBkB kAkHS and kABkHS ≤ kAkHS kBkB .
Note: Let B ∈ B(H) and > 0. There exists u ∈ H with kuk = 1 and kBkB ≤ kBuk + ≤
kBkHS + , the last inequality coming from extending u to an orthonormal basis of H. So
kBkB ≤ kBkHS .
Proposition: R(H) ⊂ HS(H) is a dense inclusion with respect to the norm on HS and
HS(H) ⊂ K(H).
19
Proof : Let A ∈ HS(H) and let (ei ) be an orthonormal basis of H. From (Aei ) construct
of H. With respect to this basis we have
an orthonormal basis of Ran(A) and extend to a basis P
projections EN . EN A ∈ R(H) and kA − EN Ak2HS = i k(1H − EN )Aei k2 −→ 0 as N −→ ∞.
Thus the density statement is proved and kA − EN AkB −→ 0 giving A ∈ K(H).
Note: HS(H) is in general not complete with respect to k · kB because K(H) \ HS(H) 6= ∅.
4.7
Trace Class
PN
Definition: T ∈ B(H) is said to be of trace class if T = i=1 Ai Bi where Ai , Bi ∈ HS. We
denote the set of trace class elements by T C(H). So T C(H) = HS(H)2 is a two-sided ideal,
closed under adjoints.
Proposition: kT kT C = supo.n.b
P
i
|hT ei , fi i| defines a norm on T C(H).
Proof : Let A, B ∈ HS(H), T = AB and (ei ) be an orthonormal basis for H. Then
X
X
X
|hT ei , fi i| =
|hBei , A∗ fi i| ≤
kBei kkA∗ fi k ≤ kAkHS kBkHS .
i
i
i
Thus kT kT C is well-defined. It is easy to check that it is a norm.
4.8
K = c0 , HS = l2 , T C = l1
Recall: Suppose B ∈ K(H) and B ∗ = B. Then λj −→ 0 where (λj ) are the eigenvalues of B.
1
Theorem: Let B ∈ K(H), (ej ) be an orthonormal basis of eigenvectors (B ∗ B) 2 and let (λj )
1
be the corresponding eigenvalues. Then kBej k2 = λ2j = k(B ∗ B) 2 ej k2 .
1
Proof : Let A = (B ∗ B) 2 and U be as in the polar decomposition theorem. Then kBej k2 =
1
kU Aej k2 = λ2j kU ej k2 = λ2j = k(B ∗ B) 2 ej k2 .
1
Corollary: Let B ∈ K(H). Then B ∈ HS(H) ⇐⇒ (B ∗ B) 2 ∈ HS(H) ⇐⇒
P
this case kBkHS = j λ2j .
P
j
Corollary: Suppose B ∈ K(H) and B ∗ = B. Then B ∈ HS(H) if and only if
where (λj ) are the eigenvalues of B.
λ2j < ∞. In
P
j
λ2j < ∞
1
Theorem: Suppose T ∈ B(H). Then T ∈ T C(H) ⇐⇒ (T ∗ T ) 4 ∈ HS(H).
1
Proof : Suppose D = (T ∗ T ) 4 ∈ HS(H); then T = (U D)D ∈ T C(H). Suppose T ∈ T C(H).
1
Let (ei ) be an orthonormal basis of Ran(T ∗ T ) 2 and fi = U ei so that (fi ) is P
an orthonor0 2
mal basis of Ran(T ). Extend (ei ) to an orthonormal basis (e0i ) of H. Then
i kDei k =
P
P
1
∗
2
i |h(T T ) ei , ei i| =
i |hT ei , fi i| < ∞.
Corollary: Suppose T ∈ K(H). Then T ∈ T C(H) ⇐⇒
1
eigenvalues of (T ∗ T ) 2 .
P
j
λj < ∞ where (λj ) are the
Corollary: Suppose T ∈ K(H) and T ∗ = T . Then T ∈ T C(H) if and only if
where (λj ) are the eigenvalues of T .
Corollary: R(H) ⊂ T C(H) ⊂ HS(H) ⊂ K(H).
20
P
j
λj < ∞
4.9
Trace
Proposition:
Let (ei ) be an orthonormal basis of H. Define T r : T C(H) −→ C by T r(T ) =
P
hT
e
,
e
i.
T
r
is independent of the choice of orthonormal basis and satisfies the following
i
i
i
properties: |T r(T )| ≤ kT kT C , T r(GAG−1 ) = T r(A) whenever A ∈ T C(H) and G ∈ B(H) is
invertible, T r([A, B]) = 0 whenever A ∈ T C(H) and B ∈ B(H).
Proof : Let A, B ∈ HS(H) and (ei ), (fj ) be orthonormal bases for H. Then
X
X
X
X
hABei , ei i =
hBei , A∗ ei i =
hAfj , B ∗ fj i =
hBAfj , fj i.
i
i
j
j
For T = AB and G ∈ B(H) invertible
T r(GT G−1 ) = T r(GABG−1 ) = T r(BG−1 GA) = T r(BA) = T r(AB) = T r(T ).
Note: For A ∈ HS(H) we have kAk2HS = T r(A∗ A).
1
Theorem: For A ∈ T C, kAkT C = T r((A∗ A) 2 ).
Proposition: R(H) ⊂ T C(H) is a dense inclusion with respect to the norm on T C.
4.10
Summary
Polar decomposition and the spectral theorem for compact self adjoint operators say that for
any B ∈ K(H) there exist orthonormal bases (ei ) and (fi ) of H such that
X
Bu =
λi hu, ei ifi
i
where λi ≥ 0 and αi −→ 0.
P
An elementP
B ∈ K(H) is Hilbert-Schmidt if and only if i λ2i < ∞ and is of trace class if
and only if i λi < ∞; the norms are defined accordingly. The operator norm becomes supi λi
and so
kBkB ≤ kBkHS ≤ kBkT C .
Accordingly we have R(H) ⊂ T C(H) ⊂ HS(H) ⊂ K(H) and the three objects on the right are
closure of R(H) with respect to the different norms.
4.11
Pairings
HS(H) is a Hilbert space space so that HS(H)0 ∼
= HS(H) via an antilinear isomorphism.
However, HS(H) has a real structure: BH(H) is the complexification of the real vector space
BH(H)R = {B ∈ HS(H) : B ∗ = B}.
Complexifications of real vector spaces are isomorphic to their conjugates and so HS(H)0 ∼
=
HS(H) via a linear isomorphism: we have a perfect pairing
X
HS(H) × HS(H), (A, B) 7−→ T r(AB) =
hBei , A∗ ei i
i
T r is like the integral on L1 : the integral on L1 gives a pairing between L2 functions; T r gives
a pairing between elements of HS(H).
21
4.12
Theorem
We have parings
T C(H) × K(H) −→ C, (A, B) 7−→ T r(AB)
B(H) × T C(H) −→ C, (A, B) 7−→ T r(AB)
and these give rise to isomorphisms K(H)0 ∼
= T C(H) and T C(H)0 ∼
= B(H).
The next two subsections will be devoted to a clear proof of this theorem.
4.13
K(H)0 ∼
= T C(H)
Throughout this section A will be an element of T C(H) and B an element of K(H).
1. Let ϕ(A)(B) = T r(AB). ϕ(A)(B) makes sense because T C(H) is an ideal in K(H).
ϕ(A)(B) is linear in A and B.
2. There exist orthonormal bases (ej ) and (fj ) of H such that
X
Bu =
λj hu, fj iej
j
where λj ≥ 0. αj −→ 0 as j −→ ∞ and kBkB = supj λj .
|ϕ(A)(B)| ≤
X
|hABfi , fi i| =
i
X
λi |hAei , fi i|
i
≤ kBkB
X
|hAei , fi i| ≤ kAkT C kBkB .
i
Thus ϕ(A) ∈ K(H)0 , kϕ(A)k ≤ kAkT C and we have a linear map ϕ : T C(H) −→ K(H)0 .
3. There exist orthonormal bases (ei ) and (fi ) of H such that
X
Au =
λi hu, ei ifi
i
where λi ≥ 0 and
P
i
λi < ∞. Define B ∈ R(H) by
Bu =
N
X
hu, fi iei .
i=1
kBkB = 1 and so
kϕ(A)k ≥ ϕ(A)(B) =
∞
N
N
N
X
X
X
X
hABfi , fi i =
hAei , fi i =
hλi fi , fi i =
λi
i=1
i=1
i=1
i=1
which gives kϕ(A)k ≥ kAkT C . We have shown that ϕ : T C(H) −→ K(H)0 is a linear
isometry.
4. Let Φ ∈ K(H)0 . For x, y ∈ H, define x⊗y : H −→ H, u 7−→ hu, yix. S = {x⊗y : x, y ∈ H}
is a spanning set for R(H) and R(H) is dense in K(H), so Φ is determined by its values
on S .
|Φ(x ⊗ y)| ≤ kΦkkx ⊗ ykB = kΦkkxkH kykH
22
so Φ(− ⊗ y) : H −→ C, x 7−→ Φ(x ⊗ y) is a bounded linear functional and kΦ(− ⊗
y)kB ≤ kΦkkykH . By the Riesz representation lemma there is an element T y ∈ H with
kT ykH ≤ kΦkkykH uniquely determined by the property that
Φ(x ⊗ y) = hx, T yi.
By uniqueness, varying y determines a bounded linear map T : H −→ H with kT kB ≤ kΦk.
1
1
Write T = U (T ∗ T ) 2 as in the polar decompostion theorem. We’ll show that D = (T ∗ T ) 4 ∈
1
HS(H). Let (fi ) be an orthonormal basis of the closure of Ran(T ∗ T ) 2 and ei = U fi so
that (ei ) is an orthonormal basis of Ran(T ). Extend (ei ) and (fi ) to orthonormal bases of
H. Then
!
N
N
N
N
N
X
X
X
X
X
1
hei , T fi i =
kDfi k2 =
Φ(ei ⊗ fi ) = Φ
ei ⊗ fi ≤ kΦk
hfi , (T ∗ T ) 2 fi i =
i=1
because k
i=1
i=1
PN
i=1 ei
i=1
i=1
⊗ fi kB = 1.
Thus T is of trace class. Choose an orthongonal basis (gi ) with g1 = y/kykH . Then
X
X
ϕ(T ∗ )(x ⊗ y) =
h(x ⊗ y)gi , T gi i =
hhgi , yix, T gi i = hx, T yi
i
i
= Φ(x ⊗ y).
Thus ϕ(T ∗ ) = Φ, which means ϕ is surjective.
5. We conclude that ϕ : T C(H) −→ K(H)0 is an isometric isomorphism.
4.14
T C(H)0 ∼
= B(H)
Throughout this section A will be an element of B(H) and B an element of T C(H).
1. Let ϕ(A)(B) = T r(AB). ϕ(A)(B) makes sense because T C(H) is an ideal in B(H).
ϕ(A)(B) is linear in A and B.
2. There exist orthonormal bases (ej ) and (fj ) of H such that
X
Bu =
λj hu, fj iej
j
where λj ≥ 0.
P
j
αj < ∞ and kBkT C =
|ϕ(A)(B)| ≤
X
P
j
αj .
|hABfi , fi i| =
i
≤ kAkB
X
λi |hAei , fi i|
i
X
λi = kAkB kBkT C .
i
Thus ϕ(A) ∈ T C(H)0 , kϕ(A)k ≤ kAkB and we have a linear map ϕ : B(H) −→ T C(H)0 .
3. Assume A 6= 0 and let ∈ (0, kAkB ). Then there exists x ∈ H with kxkH = 1 and
kAxkH ≥ kAkB − > 0. Let
Ax
Ax
: H −→ H, u 7−→ u,
x
B =x⊗
kAxkH
kAxkH
23
Then kBkT C = 1. Choosing an orthonormal basis (ei ) with e1 = Ax/kAxkH we obtain
X
kϕ(A)k ≥ ϕ(A)(B) =
hABei , ei i = kAxkH ≥ kAkB − .
i
We have shown that ϕ : B(H) −→ T C(H)0 is a linear isometry.
4. Let Φ ∈ T C(H)0 . S = {x ⊗ y : x, y ∈ H} is a spanning set for R(H) and R(H) is dense
in T C(H), so Φ is determined by its values on S .
|Φ(x ⊗ y)| ≤ kΦkkx ⊗ ykT C = kΦkkxkH kykH
so Φ(− ⊗ y) : H −→ C, x 7−→ Φ(x ⊗ y) is a bounded linear functional and kΦ(− ⊗ y)kB ≤
kΦkkykH . As in section 1 we obtain a bounded linear map T : H −→ H with kT kB ≤ kΦk
and the property that
Φ(x ⊗ y) = hx, T yi.
Choose an orthongonal basis (ei ) with e1 = y/kykH . Then
X
X
ϕ(T ∗ )(x ⊗ y) =
h(x ⊗ y)ei , T ei i =
hhei , yix, T ei i = hx, T yi
i
i
= Φ(x ⊗ y).
Thus ϕ(T ∗ ) = Φ, which means ϕ is surjective.
5. We conclude that ϕ : B(H) −→ T C(H)0 is an isometric isomorphism.
4.15
ˆ
HS(H) ∼
= H ⊗H
Let H1 and H2 be two Hilbert spaces with inner products h· , ·i1 and h· , ·i2 , respectively. We
make H1 ⊗C H2 into an inner product space by defining
hϕ1 ⊗ ϕ2 , ψ1 ⊗ ψ2 i = hϕ1 , ψ1 i1 hϕ2 , ψ2 i2
ˆ 2 for the completion of H1 ⊗C H2 under this inner
and extending by linearity. Write H1 ⊗H
product.
We have an isomorphism
H ⊗C H −→ (R(H), h· , ·iHS ), x ⊗ y 7−→ x ⊗ y
ˆ −→ HS(H).
which induces an isomorphism H ⊗H
4.16
Fredholm Operators
Definition: Let
CR(H) = {A ∈ B(H) : Ran(A) is closed}
F̃(H) = {A ∈ CR(H) : either Null(A) or Ran(A)⊥ is finite-dimensional}
F(H) = {A ∈ CR(H) : both Null(A) and Ran(A)⊥ are finite-dimensional}
F(H) consists of the Fredholm operators and F̃(H) consists of the semi-Fredholm operators.
Observation: Any operator A ∈ B(H) defines a bounded bijection à : Null(A)⊥ −→ Ran(A).
24
If A ∈ CR(H) then the open mapping theorem gives a bounded inverse G̃ : Ran(A) −→
Null(A)⊥ . Extending G̃ to be zero on Ran(A)⊥ we obtain G ∈ B(H) with
GA = I − πNull(A) , AG = I − πRan(A)⊥ .
Conversely, if these relations hold Ran(G) ⊃ Null(A)⊥ and so Ran(A) = Ran(AG) = Ran(A).
Lemma: Let A ∈ B(H). Then A ∈ CR(H) if and only if the range of A has a closed complement, i.e. there exists a closed subspace F ⊂ H such that H = F ⊕ Ran(A).
Proof : The ‘only if’ is clear. Suppose F exists then F ⊕ H −→ H, (f, u) 7−→ f + Au is
bounded surjection and thus there exists a bounded G : H −→ F ⊕ H such that
GA = I − πNull(A) , AG = I.
Suppose (un ) ⊂ H and Aun −→ v ∈ H. Since Aun = AGAun , v = AGv ∈ Ran(A).
Corollary: If a bounded linear map has range with a finite-dimensional complement then
it is semi-Fredholm. If a bounded linear map has a finite dimensional null space and range with
a finite-dimensional complement then it is Fredholm.
Theorem: Suppose K ∈ K(H). Then there exists a decomposition
I − K = (I − B1 )(I − D)(I − B2 ), kB1 k < 1, kB2 k < 1
and D = πN DπN where πN denotes the projection onto the span of the first N elements of an
orthonormal basis.
Lemma: Suppose K ∈ K(H). Then I − K ∈ F(H).
Proof : Use the decomposition above. I − D ∈ F(H) and I − B1 , I − B2 are invertible.
Theorem: An operator A ∈ B(H) is Fredholm if and only if it has both a left and right
parametrix, in the sense that there are operators BR and BL in B(H) such that
ABR = I − KR , BL A = I − KL ,
where KR , KL ∈ K(H).
Note: By considering the quotient ring B(H)/K(H), we see that in any case BL − BR ∈ K(H),
and we can take BL = BR .
Proof of Theorem: The ‘only if’ is clear since πNull(A) and πRan(A)⊥ lie in R(H) ⊂ K(H)
when A ∈ F(H). Suppose we have a left and right parametrix. Then Null(A) ⊂ Null(BL A) =
Null(I − KL ) and Ran(A) ⊃ Ran(ABR ) = Ran(I − KR ) =⇒ Ran(A)⊥ ⊂ Ran(I − KR )⊥ . The
lemma gives the result.
Proposition: The Fredholm operators form an open set in B(H) which is closed under composition and adjoints; the image under a Fredholm operator of a closed subspace is closed. The
semi-Fredholm operators form and open and dense subset of B(H) closed under composition
with Fredholm operators and adjoints.
Definition: If A ∈ F(H) the index of A is
Ind(A) = dim Null(A) − dim Ran(A)⊥ .
25
Proposition: If A ∈ F(H) and B is a two-sided parametrix modulo trace class operators, then
[A, B] ∈ T C(H) and
Ind(A) = Tr([A, B]).
Proof : AB = I − TR , BA = I − TL where TR , TL ∈ T C(H). Thus [A, B] = AB − BA =
TL − TR ∈ T C(H). If G is a generalised inverse then
Tr([A, G]) = Tr πNull(A) − πRan(A)⊥ = Ind(A).
In general by considering the quotient ring B(H)/T C(H) we see B − G = T ∈ T C(H).
[A, B] = [A, G] + [A, T ] and Tr([A, T ]) = 0.
Proposition: The index is constant on components of F(H) and log-multiplicative in the
sense that
Ind(A1 A2 ) = Ind(A1 ) + Ind(A2 ), Ind(A∗ ) = −Ind(A)
If B is a two-sided parametrix modulo compact operators for A ∈ F(H) then B ∈ F(H) and
Ind(B) = −Ind(A).
5
5.1
Problems
The Leibniz formula
For α = (α1 , . . . , αn ) and β = (β1 , . . . , βn ) define
α
β
to be the product
α1
β1
···
αn
βn
.
Lemma: For sufficently differentiable functions f, g : Rn −→ C and α = (α1 , . . . , αn ) we
have the Leibniz formula
X α ∂ α (f g) =
∂ β f ∂ α−β g.
β
β≤α
Proof : The formula clearly holds when |α| = 0 and the formula holds when |α| = 1 by the
usual Leibniz rule. Assume inductively that the formula holds whenever |α| = N and suppose
we are given α with |α| = N + 1. We can write α = α0 + ei for some i ∈ {1, . . . , n}, where
|α0 | = N . Thus,
X α 0 0
α
α0
∂ (f g) = ∂i ∂ (f g) = ∂i
∂ β f ∂ α −β g
β
β≤α0
X α0
0
0
=
[(∂i ∂ β f ) ∂ α −β g + ∂ β f (∂i ∂ α −β g)]
β
β≤α0
X α − ei =
[∂ β+ei f ∂ α−(β+ei ) g + ∂ β f ∂ α−β g]
β
β≤α−ei
X α − ei X α − ei =
∂ β f ∂ α−β g +
∂ β f ∂ α−β g.
β − ei
β
ei ≤β≤α
β≤α−ei
αi > 0 and for 0 < βi < αi we have
αi − 1
αi − 1
αi
+
=
.
βi − 1
βi
βi
This gives
α − ei
β − ei
+
α − ei
β
26
=
α
β
whenever ei ≤ β ≤ α − ei . We also have
αi − 1
αi
=
αi − 1
αi
and
αi − 1
αi
=
0
0
giving
α − ei
β − ei
=
α
α − ei
α
when βi = αi and
=
when βi = 0.
β
β
β
Thus
α
∂ (f g) =
X α
β
β≤α
∂ β f ∂ α−β g
and the result is proved by induction.
5.2
An equivalent norm
α
Let ι : Rn −→ C, x 7−→ xα and 1 : Rn −→ R, x 7−→ 1.
For α = (α1 , . . . , αn ) define α! to be the product α1 ! · · · αn !.
Theorem: kf k0N =
P
|α|+|β|≤N
supx∈Rn |∂ α (ιβ f )(x)| defines an equivalent norm to kf kN .
Proof : Suppose |α| + |β| ≤ N . Then for f ∈ S(Rn ),
α β X α γ β α−γ ∂ (ι f ) ≤
∂ ι ∂
f
γ
γ≤α


X α X 0 
β!
≤
ιβ ∂ α−γ f 
γ
γ≤α
β 0 ≤β
X α ≤
β! kf kN 1.
γ
γ≤α
0
Let Cα,β
=
P
γ≤α
α
γ
0
β! and CN
=
P
|α|+|β|≤N
0
0
Cα,β
. Then |f k0N ≤ CN
kf kN .
Suppose |α| + |β| ≤ N and f ∈ S(Rn ). We have
ιβ ∂ α f = ∂ α (ιβ f ) −
X α ∂ γ ιβ ∂ α−γ f
γ
06=γ≤α
and so
X α β α α β ι ∂ f ≤ ∂ (ι f ) +
∂ γ ιβ ∂ α−γ f γ
06=γ≤α


X α X 0 
≤ kf k0N 1 +
β!
ιβ ∂ α−γ f  .
γ
0
β ≤β
06=γ≤α
If |α| + |β| = 0 then |ιβ ∂ α f | ≤ kf k0N 1. Let M < N and suppose inductively that if |α| + |β| ≤ M
then |ιβ ∂ α f | ≤ C̃M kf k0N 1. Then if |α| + |β| = M + 1


X α
β α ι ∂ f ≤ 1 +
β! |{β 0 : β 0 ≤ β}| C̃M  kf k0M 1.
γ
06=γ≤α
27
Letting
C̃M +1




X α
= max 1 +
β! |{β 0 : β ≤ β}| C̃M : |α| + |β| = M + 1 ∪ {C̃M }


γ
06=γ≤α
we have |ιβ ∂ α f | ≤ C̃M +1 kf k0N 1 whenever |α| + |β| ≤ M + 1. By induction we have C̃N so that
|ιβ ∂ α f | ≤ C̃N kf k0N 1 whenever |α| + |β| ≤ N . Letting
CN = |{(α, β) : |α| + |β| ≤ N }| C̃N
we have kf kN ≤ CN kf k0N .
5.3
Some vaguely useful facts
Lemma: Suppose f : R\{0} −→ R is continuously differentiable. Suppose also, that limx−→0 f (x)
and limx−→0 f 0 (x) exist and are finite. Then f extends to a continuously differentiable fuction
F : R −→ R such that F 0 (0) = limx−→0 f 0 (x).
Proof : Define
(
F (x) =
(
G(x) =
f (x)
limx−→0 f (x),
when x 6= 0
when x = 0
f 0 (x)
limx−→0 f 0 (x),
when x 6= 0
when x = 0
Then F, G : R −→ R are continuous.
For x > y > 0 we have
Z
F (x) − F (y) = f (x) − f (y) =
x
f 0 (t)dt
y
x
Z
=
G(t)dt.
y
Letting y −→ 0 gives
x
Z
F (x) − F (0) =
G(t)dt for x > 0.
0
Similarly, the result holds for x < 0 and so we have
Z x
F (x) − F (0) =
G(t)dt for all x ∈ R.
0
0
Differentiating we obtain F (0) = G(0), which completes the proof.
By induction we easily obtain the following lemma.
Lemma: Suppose f : Rn \ {0} −→ C is infinitely differentiable and limx−→0 ∂ α f (x) = 0
for all α ∈ (N ∪ {0})n . Then f extends to an infinitely differentiable function F : Rn −→ C such
that ∂ α F (0) = 0 for all α ∈ (N ∪ {0})n .
28
Observation: Suppose f : Rn \{0} −→ C is infinitely differentiable and limx−→0 ∂ α f (x) = 0 for
all α ∈ (N∪{0})n . Let k ∈ N∪{0}. Then by the lemma above and Taylors theorem we may write
X
∂ α f (x) =
xβ hβ (x)
|β|=k
where limx−→0 hβ (x) = 0. Since |xβ | ≤ kxk|β| this gives
∂ α F (x)
= 0.
x−→0 kxkk
lim
5.4
Another description of S(Rn )
Lemma: Suppose f ∈ S(Rn ). Then lim|x|−→∞ |xβ ∂ α f (x)| = 0.
Proof : By definition supx∈R |xβ ∂ α f (x)| is finite for each α and β.
Suppose the result does not hold. Then we can find α, β, > 0 and a sequence (xj ) ⊂ Rn
such that
|xβj ∂ α f (xj )| ≥ for each j, and kxj k −→ ∞ as j −→ ∞.
For some i, the absolute value of the ith coordinate of xj tends to ∞ as j −→ ∞; thus
i α
|xβ+e
∂ f (xj )| −→ ∞ as j −→ ∞
j
giving a contradiction.
Conversely, the following lemma is immediate by the boundedness of continuous functions on
compact sets.
Lemma: Suppose that f is infinitely differentiable and lim|x|−→∞ |xβ ∂ α f (x)| = 0 for all α, β.
Then f ∈ S(Rn ).
Thus we have proven
Theorem:
S(Rn ) = {f : Rn −→ C : f is infinitely differentiable and
5.5
lim
kxk−→∞
|xβ ∂ α f (x)| = 0 for all α, β}.
Cc∞ (Rn ) ⊂ S(Rn ) is dense
Let χ : Rn −→ R denote the bump function constructed in class so that
(
1 when kxk ≤ 21
χ(x) =
0 when kxk ≥ 1
Let ϕ ∈ S(Rn ) and for each k ∈ N define ϕk ∈ Cc∞ (Rn ) by
x
ϕ(x).
ϕk (x) = χ
k
Lemma: For all α, β ∈ (N ∪ {0})n
sup |xβ ∂ α ϕk (x)| −→ 0 as k −→ ∞.
kxk≥ k
2
29
Proof : By the Leibniz rule and the chain rule
x
X α xβ
α−γ
∂ γ ϕ(x).
∂
xβ ∂ α ϕk (x) =
χ
k
γ k |α−γ|
γ≤α
∂ γ χ is a bounded function for all γ and so there exists a C ∈ N such that for all x ∈ Rn
X
β α
x ∂ ϕk (x) ≤ C
xβ ∂ γ ϕ(x) .
γ≤α
By a previous lemma, for each γ,
sup |xβ ∂ γ ϕ(x)| −→ 0 as k −→ ∞.
kxk≥ k
2
Thus the lemma is proven.
Let α, β ∈ (N ∪ {0})n . Then
sup xβ ∂ α (ϕ − ϕk )(x) ≤ sup xβ ∂ α (ϕ − ϕk )(x) + sup xβ ∂ α ϕ(x) + sup xβ ∂ α ϕk (x) .
x∈Rn
kxk< k
2
kxk≥ k
2
kxk≥ k
2
The first term on the right hand side is zero; we have just argued that the last term tends to 0
as k −→ ∞; using the lemma that I proved last time, once more, the second term tends to 0 as
k −→ ∞. Thus
sup xβ ∂ α (ϕ − ϕk ) −→ 0 as k −→ ∞.
x∈Rn
This is easily seen to give ϕk −→ ϕ in S(Rn ) and so the inclusion Cc∞ (Rn ) ⊂ S(Rn ) is dense.
5.6
5.6.1
A criterion for compactness in S(Rn )
Motivated by the hint
First we note that for any f ∈ S(Rn ) and x, y ∈ Rn we have
f (x) − f (y) =
n Z
X
i=1
1
∂i f (tx + (1 − t)y) dt (xi − yi )
0
by the fundamental theorem of calculus. Hence,
|f (x) − f (y)| ≤ kf k1
n
X
|xi − yi | .
i=1
Also,
kxk2 |f (x)| = |x21 f (x)| + . . . + |x2n f (x)| ≤ kf k2
so that
|f (x)| ≤
kf k2
whenever x 6= 0.
kxk2
We now proof a lemma, similar in theme, to the Arzelà-Ascoli Theorem.
Lemma: Let C ∈ N. Suppose that D ⊂ S(Rn ) and that kf k2 ≤ C for all f ∈ D. Then
D is totally bounded with respect to the k · k0 norm.
30
Proof : Let > 0. When f ∈ D, kf k1 ≤ kf k2 ≤ C and the remark above gives
n
X
|xi − yi | <
i=1
Choose N ∈ N such that N >
q
8C
.
=⇒ |f (x) − f (y)| < .
4C
4
The other remark says that for f ∈ D
kxk ≥ N =⇒ |f (x)| <
.
8
Thus
4
and we can cover Rn by finitely many open sets U1 , . . . , Ur with the property that
kxk, kyk ≥ N =⇒ |f (x) − f (y)| <
x, y ∈ Ui =⇒ |f (x) − f (y)| <
for all f ∈ D.
4
The rest of the proof is identical to that of the Arzelà-Ascoli Theorem. Choose xi ∈ Ui for each
i ∈ {1, . . . , r} and consider the set
B = {(f (x1 ), . . . , f (xr )) : f ∈ D} ⊂ Cn .
Since kf k0 ≤ kf k2 ≤ C for f ∈ D, B is bounded, and hence totally bounded. Thus, there exist
f1 , . . . , fs ∈ D with the property that for any f ∈ D, there exists a j ∈ {1, . . . , s} such that
|f (xi ) − fj (xi )| <
for all i ∈ {1, . . . , r}.
4
Given f ∈ D, choose such a j ∈ {1, . . . , s}. Then
kf − fj k0 ≤
3
<
4
so that f1 , . . . fs give a -net for D.
We actually need stronger statements than those above; I decided against scrapping them because I felt it was good motivation for what follows.
5.6.2
Answer to the Question
First we note that for any f ∈ S(Rn ) and x, y ∈ Rn we have
n
n
X
X
β α
x ∂ f (x) − y β ∂ α f (y) ≤ kιβ ∂ α f k1
|xi − yi | ≤ |β|kf k|α|+|β|+1
|xi − yi |
i=1
i=1
Also,
kxk2 |xβ ∂ α f (x)| = |x21 xβ ∂ α f (x)| + . . . + |x2n xβ ∂ α f (x)| ≤ kf k|α|+|β|+2
so that
|xβ ∂ α f (x)| ≤
kf k|α|+|β|+2
whenever x 6= 0.
kxk2
We now prove a stronger version of the lemma from the last section.
31
Lemma: Let C ∈ N. Suppose that D ⊂ S(Rn ) and that kf kN +2 ≤ C for all f ∈ D. Then D is
totally bounded with respect to the k · kN norm.
Proof : Let > 0. When f ∈ D, the first remark above allows us to find a δ > 0 such
that
n
X
X
|xi − yi | < δ =⇒
|xβ ∂ α f (x) − y β ∂ α f (y)| < .
4
i=1
|α|+|β|≤N
The other remark allows us to choose M ∈ N such that for f ∈ D
X
kxk ≥ M =⇒
|xβ ∂ α f (x)| < .
8
|α|+|β|≤N
Thus
X
kxk, kyk ≥ M =⇒
|xβ ∂ α f (x) − y β ∂ α f (y)| <
|α|+|β|≤N
4
n
and we can cover R by finitely many open sets U1 , . . . , Ur with the property that
X
for all f ∈ D.
x, y ∈ Ui =⇒
|xβ ∂ α f (x) − y β ∂ α f (y)| <
4
|α|+|β|≤N
Choose xi ∈ Ui for each i ∈ {1, . . . , r} and consider the set
n
o
1≤i≤r
B = (xβi ∂ α f (xi ))|α|+|β|≤N
: f ∈ D ⊂ Cn .
B is bounded, and hence totally bounded. Thus, there exist f1 , . . . , fs ∈ D with the property
that for any f ∈ D, there exists a j ∈ {1, . . . , s} such that
X
|xβi ∂ α f (xi ) − xβi ∂ α fj (xi )| <
|α|+|β|≤N
for all i ∈ {1, . . . , r}.
4
Given f ∈ D, choose such a j ∈ {1, . . . , s}. Then
kf − fj kN ≤
3
<
4
so that f1 , . . . fs give a -net for D.
We are now ready to prove the theorem.
Theorem: Let (CN ) ⊂ N be a sequence of natural numbers. Suppose that D ⊂ S(Rn ) is
closed and that kf kN ≤ CN for all f ∈ D. Then D is compact.
Proof : Since S(Rn ) is a complete metric space and D is closed it is enough to show that
any sequence in D has a Cauchy subsequence; let (fj ) ⊂ D be a sequence. By the lemma, D
is totally bounded with respect to each norm k · kN . In particular, for a fixed N we can find a
Cauchy subsequence with respect to k · kN . By a diagonal argument we can find a subsequence
which is Cauchy with respect to all norms k · kN simultaneously. A sequence which is Cauchy
with respect to all norms k · |kN is Cauchy with respect to the metric d, and we are done.
32
5.6.3
Remark
I now realise that I did not need the stronger lemma. I could have used the weaker one to
obtain a subsequence of (fj ), such that for each α and β, (ιβ ∂ α fjk ) is Cauchy with respect to
the k · k0 norm; this sequence is Cauchy with respect to the metric d. I am sorry you had to
read a slightly longer answer.
I have also realised that I could have used the previous subsection in order to apply the ArzelàAscoli theorem, c.f. the answer that I think Alex Moll is going to give.
5.7
Rectangles
In all of the following R will denote a rectangle, i.e. a set of the form
[a1 , b1 ) × . . . × [an , bn )
Qn
where aj , bj ∈ R. Since (a1 , . . . , an , b1 . . . , bn ) 7−→ j=1 (bj − aj ) is a continuous function, given
> 0, there exists a δ > 0 such that
vol ([a1 − δ, b1 + δ) × . . . × [an − δ, bn + δ)) ≤ vol ([a1 , b1 ) × . . . × [an , bn )) + .
SN
PN
Lemma: Suppose R ⊂ j=1 Rj . Then vol(R) ≤ j=1 vol(Rj ).
Proof : We consider the grid formed by indefinitely extending the sides of all the rectangles
R, R1 , . . . , RN . This construction yields finitely many disjoint rectangles R̃1 , . . . , R̃M , and a
collection of subsets J, J1 , . . . , JN of {1, . . . , M }, such that
[
[
R̃i for each j ∈ {1, . . . , N }.
R=
R̃i and Rj =
i∈J
i∈Jj
vol(R) =
X
vol(R̃i ) and vol(Rj ) =
i∈J
SN
j=1
SM
Rj ⊂ R, and the Rj are disjoint. Then
S∞
Lemma: Suppose R ⊂
Jj , the result is clear.
j=1
Proof : We proceed as above but in this case
Corollary: Suppose
j=1
vol(R̃i )
i∈Jj
by basic arithmetic, so observing that J ⊂
Lemma: Suppose
X
`M
j=1
PN
j=1
Rj . Then vol(R) ≤
vol(Rj ) ≤ vol(R).
Jj ⊂ J.
Rj ⊂ R, and the Rj are disjoint. Then
S∞
j=1
P∞
j=1
P∞
j=1
vol(Rj ) ≤ vol(R).
vol(Rj ).
Proof : Let > 0. Replace
Rj = [a1,j , b1,j ) × . . . × [an,j , bn,j )
with
R̃j = [a1,j − δj , b1,j + δj ) × . . . × [an,j − δj , bn,j + δj )
S∞
where δj > 0 is chosen so that vol(R̃j ) ≤ vol(Rj ) + 2−j . Then R ⊂ j=1 R̃j◦ . By Heine-Borel,
SN
SN
R is compact so there exists an N ∈ N such that R ⊂ j=1 R̃j◦ . Thus R ⊂ j=1 R̃j and by the
first lemma
N
∞
X
X
vol(R) ≤
vol(R̃j ) ≤ +
vol(Rj ).
j=1
j=1
33
Since > 0 was arbitary we are done.
Corollary: Suppose R =
Rj are disjoint.
S∞
j=1
Rj . Then vol(R) ≤
P∞
j=1
vol(Rj ) and equality holds if the
Lemma: A countable union of rectangles can be written as a countable union of disjoint
rectangles.
Proof : Let {Rj }∞
j=1 be a countable collection of rectangles. Proceeding with the grid construction above for the rectangles R1 , . . . , Rk we can find disjoint rectangles R̃1,k , . . . , R̃ik ,k such
that
ik
k−1
[
[
R̃j,k = Rk \
Rj .
j=1
j=1
The collection {R̃j,k : 1 ≤ j ≤ ik , k ∈ N} is the required countable collection of disjoint
rectangles.
5.8
A countable union of measure zero sets
n
Let {Ej }∞
j=1 be a countable collection of measure zero subsets of R ; fix > 0. We can find a
∞
countable collection of rectangles {Rij }i,j=1 such that for each j ∈ N,
∞
[
Rij ⊃ Ej and
∞
[
i,j=1
so that
5.9
S∞
j=1
Rij ⊃
vol(Rij ) < 2−j .
i=1
i=1
Then
∞
X
∞
[
Ej and
j=1
∞
X
vol(Rij ) <
i,j=1
∞
X
2−j = j=1
Ej has measure zero.
L1loc (Rn )
Given f : Rn −→ C and R > 0 write fR for the function
(
f (x) when kxk < R
n
R −→ C, x 7−→
0
when kxk ≥ R
Then
L1loc (Rn ) = {f : Rn −→ C : fN ∈ L1 (Rn ) for all N ∈ N}
and
L1loc (Rn ) = L1loc (Rn )/a.e.
Since (−)N : L1loc (Rn ) −→ L1 (Rn ) is linear we can define a seminorm by kf kN = kfN kL1 . For
f, g ∈ L1loc (Rn ) let
∞
X
kf − gkN
.
d(f, g) =
2−N
1 + kf − gkN
N =1
Suppose d(f, g) = 0. Then kf − gkN = 0 for all N ∈ N and so fN = gN a.e. for each N ∈ N.
But this proves f = g a.e. since we proved last time that a countable union of measure zero sets
has measure zero. d induces a function
d : L1loc (Rn ) × L1loc (Rn ) −→ R.
34
It was proved in class that d satisfies the triangle inequality and it is clear that d(f, g) = d(g, f ),
d(f, g) ≥ 0 and d(f, f ) = 0. The above argument shows d(f, g) = 0 implies f = g, so d is a
metric.
Let (fn ) ⊂ L1loc (Rn ) be a Cauchy sequence. Then for each N ∈ N, ((fn )N ) ⊂ L1 (Rn ) is a Cauchy
sequence and thus has a convergent subsequence. By a diagonal argument we can choose a subsequence (fnk ) of (fn ) so that ((fnk )N ) is convergent in L1 (R) for each N . Let f N denote the
limit of ((fnk )N ). (−)N : L1 (Rn ) −→ L1 (Rn ) is continuous and so (fnk )N +1 −→ f N +1 implies
(fnk )N −→ (f N +1 )N and thus (f N +1 )N = f N . Hence, we can define an element f ∈ L1loc (Rn )
so that fN = f N ; a representative for f is given by
f˜ : Rn −→ C, x 7−→ f˜N (x) whenever N − 1 ≤ kxk < N.
For each N ∈ N, kfnk − f kN −→ 0, so d(fnk , f ) −→ 0 and L1loc (Rn ) is complete.
5.10
Montone Convergence
P
P
Lemma: Suppose (fj ) is a sequence in L1 (Rn ) and j kfj kL1 < ∞. Then j fj (x) is convergent off a set, E, of measure zero. Moreover
(P
when x ∈
/E
n
j fj (x)
f : R −→ C, x 7−→
0
when x ∈ E
defines a function in L1 (Rn ) and
Z
f=
Rn
XZ
fj .
Rn
j
Proof : Let (ϕj,k )∞
k=1 be an approximating series for fj . Choose Nj ∈ N such that
kϕj,Nj +1 − fj kL1 < 2−j and
∞
X
kϕj,Nj +k − ϕj,Nj +k−1 kL1 < 2−j
k=2
Set
ψj,1 = ϕj,Nj +1 and ψj,k = ϕj,Nj +k − ϕj,Nj +k−1 for k > 1.
Then for any finite subset A ⊂ N2
X
(j,k)∈A
kψj,k kL1 ≤
X
kfj kL1 +kϕj,Nj +1 −fj kL1 +
(j,k)∈A
with k=1
∞
X
kfj kL1
kϕj,Nj +k −ϕj,Nj +k kL1 ≤ 2+
X
j=1
(j,k)∈A
with k>1
Thus
∞
X
kψj,k kL1
j,k=1
is convergent. Thus by the theorem in the integration chapter
X
|ψj,k (x)| < ∞
j,k
off a set, E, of measure zero. The last two statements allow us to sum all subsequent series
involving the ψj,k in any order. Thus for x ∈ Rn \ E,
X
j
fj (x) =
X
j
lim ϕj,Nj +k (x) =
k−→∞
∞ X
∞
X
j=1 k=1
35
ψj,k (x)
is convergent. Define f : Rn −→ C as in the statement of the lemma. Let I : N −→ N2 be some
bijection and define
j
X
ξj =
ψI(k) .
k=1
Then for x ∈ Rn \ E, ξj (x) −→ f (x) and
∞
X
∞
X
kξj − ξj−1 kL1 =
j=2
kψI(j) k < ∞.
j=2
By definition f ∈ L1 (Rn ). Also,
Z
def
f =
Z
lim
j−→∞
ξj = lim
Z X
j
j−→∞
ψI(k) = lim
j−→∞
k=1
j Z
X
ψI(k) =
∞ X
∞ Z
X
def
ψj,k =
j=1 k=1
k=1
∞ Z
X
fj .
j=1
Theorem: Suppose (fj ) is a sequence of real-valued
functions in L1 (Rn ), that fj (x) is monoR n
tone increasing for each x ∈ R and that R fj is Rbounded. Then there exists an f ∈ L1 (Rn )
such that fj (x) −→ f (x) a.e. and limj−→∞ fj = f .
Proof : Choose K ∈ N such that
N
X
kfj+1 − fj kL1 =
j=1
R
fj ≤ K for all j ∈ N. Note that for all N ∈ N
N Z
X
|fj+1 − fj | =
j=1
N Z
X
Z
(fj+1 − fj ) ≤ K −
f1 .
j=1
Let g1 = f1 and gj = fj − fj−1 for j > 1. Then the first lemma applies to the sequence (gj ) to
give the result.
5.11
Lebesgue Measurable Sets
Lemma: The characteristic function χ of {x ∈ Rn : kxk ≤ R} is an element of L1 (Rn ).
Proof : When R < R0 define
ϕR,R0 : Rn −→ C, x 7−→



1
0

 R0 −kxk
R0 −R
when kxk < R
when kxk ≥ R0
when R ≤ kxk ≤ R0
Choose an increasing sequence (nj ) ⊂ N so that whenever R0 = R +
Z
R0
R
1
nj
we have
R0 − r n−1
r
dr ≤ 2−j .
R0 − R
Then defining
ϕj = ϕR,R+ n1
j
P
0
n
we have (ϕj ) ⊂ Cc (R ), ϕj −→ χ pointwise and j kϕj+1 − ϕj kL1 < ∞. Actually one does
not have to estimate so much: if we define
ϕj = ϕR,R+ 1j
P
then ϕj −→ χ and we still have j kϕj+1 − ϕj kL1 < ∞ as one sees by considering a surface of
revolution. Alex Moll spotted this nicer argument.
36
We conclude that the constant function
1 : Rn −→ C, x 7−→ 1
lies in L1loc (Rn ). Thus Rn is measurable. We should have remarked before that L1loc (Rn ) is a
vector space; this is because L1 (Rn ) is a vector space and (−)N : {f : Rn −→ C} −→ {f :
Rn −→ C} is linear. So given any set A ⊂ Rn
χA ∈ L1loc (Rn ) ⇐⇒ χRn \A = 1 − χA ∈ L1loc (Rn )
and the complements of measurable sets are measurable.
Now
χU + χV + |χU − χV |
.
2
Since L1 (Rn ) is closed under | · | and (−)N commutes with | · | this shows that whenever U and V
are measurable, so is χU ∪V . By induction the set of measurable sets is closed under finite unions.
χU ∪V =
Let {Uj }∞
j=1 be a countable collection of measurable sets. We wish to show
able.
Fix N ∈ N. Then the sequence
S∞
j=1
Uj is measur-
∞
χSj
i=1
Ui
N
j=1
satisfies the hypotheses of the Monotone convergence theorem; the bound is given by
using the monotone property of the integral. Clearly we have pointwise convergence
χSj Ui
−→ χS∞
i=1 Ui N
i=1
π n/2 N n
Γ(n/2+1)
N
so we deduce from the theorem that
χS∞
i=1 Ui
N
∈ L1 (Rn ).
Since this holds for all N ∈ N we see that
χS∞
∈ L1loc (Rn )
i=1 Ui
and so
and
S∞
j=1
Uj is measurable. Thus, the set of measurable sets is closed under countable unions
∞
\
Ui is measurable ⇐⇒ Rn \
i=1
∞
\
Ui =
i=1
∞
[
Rn \ Ui is measurable
i=1
⇐= each Rn \ Ui is measurable ⇐⇒ each Ui is measurable
so that it is closed under countable intersections too.
5.12
Random example
Define
v : R −→ C,
x 7−→ exp(iex )
and let
u = v 0 : R −→ C.
37
Lemma: Given ϕ ∈ Cc∞ (R) the following inequality holds:
Z
uϕ ≤ πkϕk3 .
R
Proof : First note that
(1 + x2 )|ϕ0 (x)| ≤ kϕk3
for all x ∈ R. Choose R > 0 such that ϕ = 0 outside of (−R, R). Then
Z R
Z R
Z
Z R
R
0
vϕ0
vϕ = −
uϕ = [vϕ]−R −
uϕ =
and so
Z
Z
uϕ ≤
R
0
Z
|vϕ | ≤
−R
R
−R
−R
−R
R
R
kϕk3
dx = πkϕk3 .
1 + x2
Lemma: Suppose ϕ ∈ S(R) and that (ϕj ), (ψj ) are sequences in Cc∞ (Rn ) which converge to ϕ
with respect to the metric on S(R). Then the sequences
Z
Z
uϕj
and
uψj
R
R
converge in R with the same limit.
R
Proof : The sequence R uϕj converges since it is Cauchy:
Z
Z
Z
uϕj − uϕk = u(ϕj − ϕk ) ≤ πkϕj − ϕk k3 −→ 0 as j, k −→ ∞
R
R
R
and similarly R uψj converges. The limits are equal because
Z
Z
Z
uϕj − uψj = u(ϕj − ψj ) ≤ πkϕj − ψj k3 −→ 0 as j −→ ∞.
R
R
Define
Z
U : S(R) −→ C, ϕ 7−→ lim
j−→∞
uϕj
R
where (ϕj ) ⊂ Cc∞ (R) and ϕj −→ ϕ. The funcion is well-defined by the proceeding lemma.
Lemma: The function U is continuous.
Proof : Let ϕ ∈ S(R) and (ϕj ) ⊂ S(R) with ϕj −→ ϕ. Choose (ψj ) ⊂ Cc∞ (R) such that
Z
−j
d(ϕj , ψj ) < 2
and U (ϕj ) −
uψj < 2−j .
R
Then ψj −→ ϕ; so
R
uψj −→ U (ϕ) and
Z
Z
|U (ϕ) − U (ϕj )| ≤ U (ϕ) − uψj + U (ϕj ) −
uψj −→ 0 as j −→ ∞.
R
R
Remark: If ϕ ∈ Cc∞ (R) then U (ϕ) =
R
R
R
uϕ. Also,
|u(x)| = ex
38
so that u ∈ C 0 (R) is not polynomially bounded.
The argument above is easily adapted when R is replaced by Rn , v is replaced by
v : Rn −→ C,
x 7−→ exp(iex1 )
and u is replaced by
u = ∂1 v : Rn −→ C.
The only subtlety is that when proving the appropriate version of the first lemma we need to
n+1
replace 1 + x2 with 1 + kxk 2 and thus 3 by a larger value.
5.13
Complex Borel Measures
Let B(Rn ) = (C00 (Rn ))0 , the space of complex Borel measures.
5.13.1
Define µ : L1 (Rn ) −→ B(Rn ) by µf (ψ) =
We note that µ is continuous:
R
f ψ and t : L1 (Rn ) −→ S 0 (Rn ) by tf (ψ) =
R
f ψ.
|µf (ψ)| ≤ kf kL1 kψk∞ =⇒ kµf kB ≤ kf kL1
n
Let i : (S(R ), k · k∞ ) −→ C00 (Rn ) be the inclusion and j : (S(Rn ), d) −→ (S(Rn ), k · k∞ ) be the
continuous map, which is the identity on underlying sets. Precomposing with these maps gives
a restriction map r = j ∗ i∗ : B(Rn ) −→ S 0 (Rn ). We don’t know about the topology on S 0 (Rn )
yet, but I’d put money on r being continuous.
The following diagram commutes
L1 (Rn )
t
µ
B(Rn )
r
$
/ S 0 (Rn )
We know from class that t is injective and so a purely formal argument shows that µ is injective:
µf = µg =⇒ r(µf ) = r(µg ) =⇒ tf = tg =⇒ f = g.
We have a commuting diagram of inclusions
(S(Rn ), k · k∞ )
O
/ C00 (Rn )
O
Cc∞ (Rn )
/ Cc0 (Rn )
Although we won’t need it, we note that the left inclusion is dense; we proved a stronger
statement previously. The bottom inclusion is dense as one sees by using the convolution or the
Weierstrass approximation theorem and a bump function; this argument was given in class. The
right inclusion is dense: if > 0 and ϕ ∈ C00 (Rn ), then there exists R > 0 such that |ϕ(x)| < /2
whenever |x| ≥ R; defining



ϕ(x)
when kxk ≤ R
ϕ0 : Rn −→ C, x 7−→
(1 − t)ϕ


0
Rx
kxk
when kxk = R + t, t ∈ [0, 1]
when kxk ≥ R + 1
39
we have ϕ0 ∈ Cc0 (Rn ) and kϕ − ϕ0 k∞ ≤ . Thus the top inclusion is dense.
Suppose ϕ ∈ B(Rn ) and r(ϕ) = 0. Since j is the identity on underlying sets, j ∗ is injective and so i∗ (ϕ) = 0. This means ϕ is zero on (S(Rn ), k · k∞ ), a dense subset of C00 (R), and we
conclude by continuity of ϕ that ϕ = 0. Hence, r is injective.
5.13.2
Given a function ψ : Rn −→ C and y ∈ Rn , let ψy : Rn −→ C be defined by ψy (x) = ψ(y − x).
For µ ∈ B(Rn ) and ψ ∈ Cc0 (Rn ) let
µ ∗ ψ : Rn −→ C, y 7−→ µ(ψy ).
Firstly, we argue that this funtion is continuous. Because µ is continuous it is enough to show
the function
Rn −→ C00 (Rn ), y 7−→ ψy
is continous. Let > 0. Since ψ has compact support, it is uniformly continuous and we can
choose δ > 0 such that |ψ(x) − ψ(x0 )| < whenever |x − x0 | < δ. Then |y − y 0 | < δ implies
kψy − ψy0 k∞ ≤ and we are done. The function µ ∗ ψ is bounded since
|µ ∗ ψ(y)| = |µ(ψy )| ≤ kµkB kψy k∞ = kµkB kψk∞ .
From this inequality we obtain
kµ ∗ ψk∞ ≤ kµkB kψk∞
which, together with bilinearity of the convolution, shows
0
B(Rn ) × Cc0 (Rn ) −→ C∞
(Rn ),
(µ, ψ) 7−→ µ ∗ ψ
is continuous where B(Rn ) × Cc0 (Rn ) has the product topology.
Let µ ∈ B(Rn ) and ψ, ϕ ∈ Cc0 (Rn ). We wish to show
Z
µ(ψ̆ ∗ ϕ) = (µ ∗ ψ)ϕ
Since ϕ has compact support we can compute the convolution on the left and the integral on
the right using a limit of Riemann sums.
Let σj = 2−j Zn ∩ supp ϕ. Then
Z
X
ψ̆ ∗ ϕ(x) = ψy (x)ϕ(y)dy = lim 2−nj
ψyij (x)ϕ(yij )
j−→∞
yij ∈σj
and this convergence is uniform in x since ψ is bounded. Since µ is continuous and linear


X
X
µ(ψ̆ ∗ ϕ) = µ  lim 2−nj
ψyij ϕ(yij ) = lim 2−nj
µ(ψyij )ϕ(yij )
j−→∞
=
j−→∞
yij ∈σj
Z
Z
µ(ψy )ϕ(y)dy =
(µ ∗ ψ)ϕ
40
yij ∈σj
which is what we wished to show.
Now µ ∗ ψ can be interpreted as a linear functional
Cc0 (Rn ) −→ C, ϕ 7−→
Z
(µ ∗ ψ)ϕ
but this does not necessary make sense when ϕ ∈ C00 (Rn ). We claim that
C00 (Rn ) −→ C, ϕ 7−→ µ(ψ̆ ∗ ϕ)
is a well-defined continuous linear functional and this is how we interpret µ ∗ ψ as an element
of B(Rn ). Let’s check the details.
Suppose ψ ∈ Cc0 (Rn ) and ϕ ∈ C00 (Rn ). The following inequalities together with continuity
in the mean show that ψ ∗ ϕ is continuous.
Z
Z
|ψ ∗ ϕ(x) − ψ ∗ ϕ(x0 )| = ψ(x − y)ϕ(y)dy − ψ(x0 − y)ϕ(y)dy Z
≤ |ψ(x − y) − ψ(x0 − y)| |ϕ(y)| dy
Z
≤ kϕk∞ |ψx − ψx0 |
Also,
Z
|ψ ∗ ϕ(x)| ≤
|ψ(y)ϕ(x − y)| dy ≤ kψkL1
sup
|ϕ(x − y)|.
y∈supp ψ
There exists R > 0 such that supp ψ ⊂ [−R, R] and given > 0, there exists R0 > 0 such
that kψkL1 |ϕ(x)| < whenever |x| ≥ R0 . Thus |x| ≥ R + R0 implies |ψ ∗ ϕ(x)| ≤ and so
ψ ∗ ϕ ∈ C00 (Rn ).
Finally the inequality, kψ ∗ ϕk∞ ≤ kψkL1 kϕk∞ , shows that for fixed ψ ∈ Cc0 (Rn ) the function C00 (Rn ) −→ C00 (Rn ), ϕ 7−→ ψ ∗ ϕ is continuous. Because ψ ∈ Cc0 (Rn ) implies ψ̆ ∈ Cc0 (Rn )
(which we used above without mention) and µ is continuous the claim follows.
5.13.3
R
Let γ ∈ Cc∞ (Rn ). Suppose, in addition, that γ is real-valued, non-negative and that Rn γ = 1.
n
∞
n
For
R each k ∈ N define γk (x) = k γ(kx). Then for each k ∈ N we have γk ∈ Cc (R ) and
γ = 1, and supp γk −→ {0}.
Rn k
Fix µ ∈ B(Rn ) and ψ ∈ C00 (Rn ). We wish to show (µ ∗ γk )(ψ) −→ µ(ψ) as k −→ ∞. By
the definition in question 2 we must show
µ(γ̆k ∗ ψ) −→ µ(ψ) as k −→ ∞
and since µ is continuous and it is enough to show
kγ̆k ∗ ψ − ψk∞ −→ 0 as k −→ ∞.
We have
Z
Z
|(γ̆k ∗ ψ)(x) − ψ(x)| = γ̆k (y)ψ(x − y)dy − ψ(x) = γ̆k (y) [ψ(x − y) − ψ(x)] dy Z
≤ |γ̆k (y)| |ψ(x − y) − ψ(x)| dy ≤ sup |ψ(x − y) − ψ(x)|
y∈ supp γ̆k
41
Let > 0. Since ψ ∈ C00 (Rn ) there exists N ∈ N such that |ψ(x)| < /2 whenever |x| ≥ N − 1.
ψ is uniformly continuous on {x ∈ Rn : |x| ≤ N + 1} so there exists a δ ∈ (0, 1) such that
|ψ(x) − ψ(x0 )| < whenever both |x − x0 | < δ and |x|, |x0 | ≤ N + 1. Choose K ∈ N such that
supp γ̆K ⊂ (−δ, δ). One checks, by considering the cases |x| ≤ N and |x| ≥ N separately, that
k ≥ K =⇒ kγ̆k ∗ ψ − ψk∞ ≤ and so we are done.
5.14
H s (Rn ) = (H −s (Rn ))0
Preliminaries
First, we make note of a result from class.
Theorem (Riesz Representation): Let u ∈ S 0 (Rn ) and suppose there exists a C ∈ R such
that
|u(ϕ)| ≤ CkϕkL2 for all ϕ ∈ S(Rn ).
Then u ∈ L2 (Rn ).
If u ∈ L2 (Rn ) and ϕ ∈ S(Rn ) then the Cauchy-Schwarz inequality gives
|u(ϕ)| ≤ kukL2 kϕkL2
and so. . .
Theorem: Let u ∈ S 0 (Rn ). Then u ∈ L2 (Rn ) if and only if there exists a C ∈ R such that
|u(ϕ)| ≤ CkϕkL2 for all ϕ ∈ S(Rn ).
We proved on a previous problem set that ×hξis : S(Rn ) −→ S(Rn ) is an isomorphism with
inverse ×hξi−s and we know from class that F : S(Rn ) −→ S(Rn ) is an isomorphism with
inverse (2π)−n F.
Finally, we note that for s ∈ R
H s (Rn ) = {u ∈ S 0 (Rn ) : hξis û ∈ L2 (Rn )}
and kukH s = khξis ûkL2 , by definition.
Conclusion
Let u ∈ S 0 (Rn ) and ϕ = ϕ(ξ) ∈ S(Rn ). Then
−s ϕ).
\
hξi−s û (ϕ) = û hξi−s ϕ = u(hξi
Thus, applying the results noted above we obtain
u ∈ H −s (Rn ) ⇐⇒ ∃C ∈ R : | hξi−s û (ϕ)| ≤ CkϕkL2 , ∀ϕ ∈ S(Rn )
−s ϕ)| ≤ Ckϕk 2 , ∀ϕ ∈ S(Rn )
\
⇐⇒ ∃C ∈ R : |u(hξi
L
⇐⇒ ∃C ∈ R : |u(ϕ)| ≤ (2π)−n Ckhξis ϕ̂kL2 , ∀ϕ ∈ S(Rn )
⇐⇒ ∃C ∈ R : |u(ϕ)| ≤ CkϕkH s , ∀ϕ ∈ S(Rn ).
42
5.15
δ ∈ H −s (Rn ) whenever s > n/2
We showed above that
−s ϕ)| ≤ Ckϕk 2 , ∀ϕ ∈ S(Rn ).
\
δ ∈ H −s (Rn ) ⇐⇒ ∃C ∈ R : |δ(hξi
L
Let ϕ ∈ S(Rn ). hξi−s ∈ L2 (Rn ) if and only if s > n/2 and in this case we have
Z
−s ϕ) = hξi
−s ϕ(0) =
\
\
δ(hξi
hξi−s ϕ(ξ)dξ
Rn
so that
−s ϕ)| ≤ khξi−s k 2 kϕk 2 .
\
|δ(hξi
L
L
Taking C = khξi−s kL2 we see that
δ ∈ H −s (Rn ) whenever s > n/2.
5.16
A criterion to be in H m+k (Rn ), k ∈ N
Suppose m ∈ R and u ∈ H m (Rn ). Let k ∈ N and suppose that Djk u ∈ H m (Rn ) for all
j ∈ {1, . . . , n}.
d
2
n
k
Then hξim û ∈ L2 (Rn ) and hξim ξjk û = hξim D
j u ∈ L (R ).
If we can show hξim+k û ∈ L2 (Rn ) then u ∈ H m+k (Rn ).
Note that for l ≤ k and x1 , . . . , xl ≥ 0
x1 · · · xl ≤ max {xi }l ≤ 1 + max {xi }k
1≤i≤l
1≤i≤l
and so there exists some constant C ∈ N such that
k X
k
2k
2 k
(ξ12 + . . . + ξn2 )l ≤ C(1 + ξ12k + . . . + ξn2k )
hξi = (1 + |ξ| ) =
l
l=0
2
2
Thus hξim+k û ≤ C(1 + ξ12k + . . . + ξn2k ) |hξim û| . The right hand side is in L1 (Rn ). The
dominated convergence theorem tells us that the left hand side is in L1 (Rn ), as it is easily seen
to be measurable. Thus hξim+k û ∈ L2 (Rn ) as required.
5.17
Hölder-type characterisation of H s (Rn ), 0 < s < 1
Let 0 < s < 1 and consider the integral
Z
|u(x) − u(y)|2
dxdy.
n+2s
R2n |x − y|
I
In 0
0 Let (x, y) = (t, y)
. Because n
= 1 the change of variables formula gives
In In
In In Z
Z
|u(x) − u(y)|2
dxdy
=
|u(y + t) − u(y)|2 |t|−n−2s dtdy.
n+2s
|x
−
y|
2n
2n
R
R
Fix t and let g(y) = u(y + t) − u(y). Plancherel’s theorem tells us
Z
Z
|g(y)|2 dy =
|ĝ(η)|2 dη
Rn
Rn
43
But ĝ(η) = û(η)(eiη·t − 1). So by swapping the order of integration we see that
Z
Z
|u(y + t) − u(y)|2 |t|−n−2s dtdy =
|û(η)|2 |eiη·t − 1|2 |t|−n−2s dtdη
R2n
R2n
For fixed η, |eiη·t − 1|2 |t|−n−2s is integrable:
Fix η. Using the Taylor expansion for exp and Cauchy Schwarz we see that on {t ∈ Rn : |t| ≤ 1}
there exists a C ∈ N such that
|eiη·t − 1|2 ≤ C|t|2
so that
|eiη·t − 1|2 |t|−n−2s ≤ C|t|−n+2(1−s) .
Because s < 1, |eiη·t − 1|2 |t|−n−2s is integrable on {t ∈ Rn : |t| ≤ 1}.
|eiη·t − 1|2 |t|−n−2s ≤ 4|t|−n−2s
and so, because s > 0, |eiη·t − 1|2 |t|−n−2s is integrable on {t ∈ Rn : |t| ≥ 1}. Thus |eiη·t −
1|2 |t|−n−2s is integrable.
Because
Z
|eiη·t − 1|2 |t|−n−2s dt
{η : |η| = 1} −→ R, η 7−→
Rn
is a continuous function and {η : |η| = 1} is compact, there exist K, K 0 > 0 such that
Z
K≤
|eiη·t − 1|2 |t|−n−2s dt ≤ K 0 whenever |η| = 1.
Rn
For η 6= 0, by making the substitution τ = |η|t we obtain
Z Z
2
2 −n−2s
iη·t
η
i |η|
·τ
2s
e
dt = |η|
− 1 |t|
− 1 |τ |−n−2s dτ
e
Rn
Rn
so that
K|η|2s ≤
Z
2
iη·t
e
− 1 |t|−n−2s dt ≤ K 0 |η|2s .
Rn
This clearly remains true when η = 0. Mulitplying by |û(η)|2 and integrating with respect to η
gives
Z
Z
Z
|u(x) − u(y)|2
0
K
|η|2s |û(η)|2 dη ≤
dxdy
≤
K
|η|2s |û(η)|2 dη.
n+2s
Rn
R2n |x − y|
Rn
Theorem: Let 0 < s < 1. Then u ∈ H s (Rn ) if and only if u ∈ L2 (Rn ) and I(u) < ∞ where
Z
|u(x) − u(y)|2
I(u) =
dxdy.
n+2s
R2n |x − y|
Proof : Suppose u ∈ H s (Rn ). Then by definition u ∈ L2 (Rn ) and hξis û ∈ L2 (Rn ). Thus
Z
Z
I(u) ≤ K 0
|η|2s |û(η)|2 dη ≤ K 0
hηi2s |û(η)|2 dη < ∞.
Rn
2
Rn
n
Conversely, suppose u ∈ L (R ) and I(u) < ∞. Then
Z
Z
Z
Z
I(u)
1
hηi2s |û(η)|2 dη ≤
|û(η)|2 dη +
|η|2s |û(η)|2 dη ≤
|û(η)|2 dη +
< ∞.
2 Rn
K
n
n
n
R
R
R
Remark: Once we know that one integral exists all the others exist by applying the theorems
we have used; when we compare the integrals above we are really applying the dominated
convergence theorem.
44
5.18
Restriction and extension between Rn and Rn−1
Redefine Dn : S(Rn ) −→ S(Rn ) by Dn ϕ =
S(Rn ) −→ S(Rn−1 ) by
Z
In ϕ(x1 , . . . , xn−1 ) =
∂ϕ
∂xn
(to avoid i’s everywhere) and define In :
∞
ϕ(x1 , . . . , xn−1 , t)dt
−∞
Let χ ∈ Cc∞ (R) be chosen so that χ ≥ 0, supp χ ⊂ [−1, 1] and
R
R
χ = 1.
Define Jn : S(Rn−1 ) −→ S(Rn ) by
Jn (ψ)(x1 , . . . , xn ) = ψ(x1 , . . . , xn−1 )χ(xn )
and define Kn : S(Rn ) −→ S(Rn ) by
Z
Z xn
ϕ(x1 , . . . , xn−1 , t)dt −
Kn ϕ(x1 , . . . , xn ) =
∞
Z
χ(t)dt
−∞
−∞
−∞
xn
ϕ(x1 , . . . , xn−1 , t)dt
We will check later that all these maps are well-defined and continuous.
It is trivial that In ◦ Jn = idS(Rn−1 ) and so In is surjective. The fundamental theorem of
calculus shows that Kn ◦ Dn = idS(Rn ) , so Dn is injective, and that
Z ∞
In Dn ϕ(x1 , . . . , xn−1 ) =
Dn ϕ(x1 , . . . , xn−1 , t)dt = lim f (x1 , . . . , xn ) = 0.
xn −→∞
−∞
Suppose In ϕ = 0. Then the fundamental theorem of calculus gives Dn Kn ϕ = ϕ (∗).
All of this shows that
D
I
n
n
0 −→ S(Rn ) −−→
S(Rn ) −→
S(Rn−1 ) −→ 0
is a split short exact sequence.
We have a dual sequence
I∗
D∗
n
n
0 −→ S 0 (Rn−1 ) −→
S 0 (Rn ) −−→
S 0 (Rn ) −→ 0
Jn∗ ◦In∗ = idS 0 (Rn−1 ) so that In∗ is injective; Dn∗ ◦Kn∗ = idS 0 (Rn ) so that Dn∗ is surjective; Dn∗ ◦In∗ = 0.
Also, Dn∗ u = 0 implies that In∗ Jn∗ u = 0 (∗). It follows that this is split short exact sequence, too.
Remark: I’ve omitted checking the starred statements because they are easy to check and
also because they follow formally from the algebra involved, i.e. split short exact sequences.
Lemma: In is well-defined and continuous.
Proof : Write x for (x1 , . . . , xn−1 ). Then
Z ∞
Z
xβ ∂ α In ϕ(x) = xβ ∂ α
ϕ(x, t)dt =
−∞
so
β α
x ∂ In ϕ(x) ≤
∞
xβ ∂ (α,0) ϕ(x, t)dt
−∞
Z
∞
−∞
45
β (α,0)
ϕ(x, t) dt.
x ∂
Now
2 β (α,0)
ϕ(x, t) ≤ kϕk|α|+|β|+2
t x ∂
which implies
kϕk|α|+|β|+2
β (α,0)
≤
min
kϕk
,
∂
ϕ(x,
t)
.
x
|α|+|β|
t2
Thus
1
kϕk|α|+|β|+2
dt
t2
−1
R\(−1,1)
= 2 kϕk|α|+|β| + kϕk|α|+|β|+2 ≤ 4kϕk|α|+|β|+2
β α
x ∂ In ϕ(x) ≤
Z
Z
kϕk|α|+|β| dt +
This is easily seen to complete the proof.
Lemma: Jn is well-defined and continuous.
Proof : It is clear that Jn is well-defined. Also, kJn ψkN ≤ kψkN kχkN , which shows Jn is
continuous.
Lemma: Kn is well-defined and continuous.
Proof : Write x for (x1 , . . . , xn−1 ) so that
Z xn
Z
Kn ϕ(x, xn ) =
ϕ(x, t)dt −
−∞
∞
Z
xn
ϕ(x, t)dt
χ(t)dt
−∞
−∞
Let α, β ∈ (N ∪ {0})n and write β 0 for (β1 , . . . , βn−1 ).
Case 1: Suppose αn > 0. Then
(x, xn )β ∂ α Kn ϕ(x, xn ) = (x, xn )β ∂ α−en ϕ(x, xn ) −
Z
∞
ϕ(x, t)dt χ(αn −1) (xn ).
−∞
Thus
(x, xn )β ∂ α Kn ϕ(x, xn ) ≤ kϕk|α|+|β| + 4kϕk2 kχk|α| ≤ Ckϕk|α|+|β|+2 .
Case 2: Suppose αn = 0. For xn ≤ −1
(x, xn )β ∂ α Kn ϕ(x, xn ) = xβnn
Z
xn
0
xβ ∂ α ϕ(x, t)dt
−∞
and for xn ≥ 1
β α
(x, xn ) ∂ Kn ϕ(x, xn ) =
xβnn
Z
∞
0
xβ ∂ α ϕ(x, t)dt.
xn
For |xn | ≤ 1
(x, xn )β ∂ α Kn ϕ(x, xn ) = xβnn
Z
xn
0
xβ ∂ α ϕ(x, t)dt −
−∞
We have
Z
∞
ϕ(x, t)dt (x, xn )β
−∞
Z
xn
χ(t)dt.
−∞
βn +2 β 0 α
x ∂ ϕ(x, t) ≤ kϕk|α|+|β|+2 .
t
Thus for xn ≤ −1
(x, xn )β ∂ α Kn ϕ(x, xn ) ≤ |xn |βn
Z
xn
∞
kϕk|α|+|β|+2
kϕk|α|+|β|+2
dt =
≤ kϕk|α|+|β|+2
β
+2
n
|t|
|xn |(βn + 1)
46
and for xn ≥ 1
(x, xn )β ∂ α Kn ϕ(x, xn ) ≤ |xn |βn
Z
∞
xn
kϕk|α|+|β|+2
kϕk|α|+|β|+2
dt =
≤ kϕk|α|+|β|+2 .
β
+2
n
|t|
|xn |(βn + 1)
For |xn | ≤ 1
(x, xn )β ∂ α Kn ϕ(x, xn ) ≤
Z
∞
−∞
Z
0
β α
x ∂ ϕ(x, t) dt +
∞
0
|x|β |ϕ(x, t)|dt ≤ 2kϕk|α|+|β|+2 .
−∞
This is easily seen to complete the proof.
5.19
Support
Cc∞ (Ω)
i
/ S(Rn )
r∗
/ S 0 (Rn )
5.19.1
/ C ∞ (Ω)
i∗
/ (Cc∞ (Ω))0
µ
t
λ
(C ∞ (Ω))0
r
Definitions
Objects
In the following, by the limit and colimit topology I mean the categorical limit and colimit in
the category of topological spaces. Ω is an open subset of Rn .
1. S(Rn ) is the Schwartz space with the limit topology obtained from the spaces (S(Rn ), k ·
kN ). By definition of the categorical limit we know what it means to map continuously
from any topological space into S(Rn ). Of course, we also have a simple description of
what it means to map continuously out of S(Rn ) to a normed vector space.
2. S 0 (Rn ) is the dual of S(Rn ), the space consisting of all continuous linear functionals from
S(Rn ) to C. We have not yet been enlightened as to what the strong topology on S 0 (Rn )
is. However, we do know what it means for a sequence of linear functionals to converge
weakly.
3. Cc∞ (Ω) consists of the smooth functions Rn −→ C with compact support contained in
Ω. Let {Kj : j ∈ N} be an exhaustion of Ω by compact sets. Each Cc∞ (Kj ) is a closed
subspace of S(Rn ) and
∞
[
Cc∞ (Ω) =
Cc∞ (Kj ).
j=1
We give
4.
(Cc∞ (Ω))
∞
Cc∞ (Ω)
0
the colimit topology.
is the dual of Cc∞ (Ω). As in 2 we only know about weak convergence.
5. C (Ω) consists of the smooth functions Ω −→ C. For N ∈ N ∪ {0} and j ∈ N we have a
seminorm defined by
X
kϕkN,j = sup
|Dα ϕ(x)|,
x∈Kj
|α|≤N
where the Kj are as above. Let f : N −→ (N ∪ {0}) × N be a bijection and define
d(ϕ, ψ) =
∞
X
2−k
k=1
47
kϕ − ψkf (k)
.
1 + kϕ − ψkf (k)
One can show, exactly as was done in class, all the metric properties except “d(ϕ, ψ) =
0 =⇒ ϕ = ψ”. However, this final property is evident because any two functions which
differ on a point of Ω must differ on a point of Kj for some j. The metric defines a
topology on C ∞ (Ω). In fact, the seminorms k · kN,j define (non-Hausdorff) topologies on
C ∞ (Ω). The same proofs as for the metric on S(Rn ) show that the topology is the limit
topology. Thus, as in 1 we know what continuity means when mapping into C ∞ (Ω). The
same criterion for continuity when mapping out of C ∞ (Ω) also holds; the proof is identical.
6. (C ∞ (Ω))0 is the dual of C ∞ (Ω). As in 2, we only know about weak convergence.
Morphisms
1.
2.
3.
4.
5.
i : Cc∞ (Ω) −→ S(Rn ) is the inclusion.
r : S(Rn ) −→ C ∞ (Ω) is given by restriction.
Once we show i is continuous we will obtain i∗ : S 0 (Rn ) −→ (Cc∞ (Ω))0 via pullback.
Once we show r is continuous we will obtain r∗ : (C ∞ (Ω))0 −→ S 0 (Rn ) via pullback.
λ, t and µ are essentially defined by the same formula:
Z
λϕ (ψ) = µψ (ϕ) =
ϕψ, when ϕ ∈ Cc∞ (Ω), ψ ∈ C ∞ (Ω)
Ω
Z
ϕψ, when ϕ, ψ ∈ S(Rn ).
and tϕ (ψ) =
Rn
5.19.2
Well-definedness and Continuity
1. i is continuous:
i
Each composite Cc∞ (Kj ) ,→ Cc∞ (Ω) −→ S(Rn ) is continuous and so by definition of the
colimit topology we are done.
2. Thus i∗ is well-defined.
3. r is continous:
Let N ∈ N ∪ {0} and j ∈ N. Then krϕkN,j ≤ kϕkN . Thus
r : S(Rn ) −→ (C ∞ (Ω), k · kN,j )
is continuous and by definition of the limit topology we are done.
4. Thus r∗ is well-defined.
5. λ is well-defined:
Let ϕ ∈ Cc∞ (Ω). Because the {Kj } are an exhaustion of Ω by compact sets there exists a
j ∈ N such that supp ϕ ⊂ Kj . Then for any ψ ∈ C ∞ (Ω) we have
Z
|λϕ (ψ)| ≤
|ϕ| kψk0,j .
Ω
∞
Thus µϕ : C (Ω) −→ C is continuous and we are done.
6. µ is well-defined:
Let ψ ∈ C ∞ (Ω), j ∈ N and ϕ ∈ Cc∞ (Kj ). Then
!
Z
|µψ (ϕ)| ≤
|ψ| kϕk0 .
Kj
Thus
µψ
Cc∞ (Kj ) ,→ Cc∞ (Ω) −→ C
is continuous. By definition of the colimit topology µψ is continuous and we are done.
7. We know from class that t is well-defined.
48
5.19.3
Commutativity
1. When ϕ ∈ Cc∞ (Ω) and ψ ∈ S(Rn ) we have
Z
Z
ti(ϕ) (ψ) =
ϕψ =
ϕψ = λϕ (rψ) = r∗ (λϕ )(ψ).
Rn
Ω
Thus the first square commutes.
2. When ψ ∈ S(Rn ) and ϕ ∈ Cc∞ (Ω) we have
Z
Z
µr(ψ) (ϕ) =
ϕψ =
ϕψ = tψ (iϕ) = i∗ (tψ )(ϕ).
Rn
Ω
Thus the second square commutes.
5.19.4
Injectivity and Density
1. For ϕ ∈ Cc∞ (Ω), ri(ϕ) = ϕ. In particular, ri is injective and thus i is too.
2. ri has dense image:
A lemma from class tells us that for r ∈ N, we can find χr ∈ Cc∞ (Ω) such that 0 ≤ χr ≤ 1,
χr (x) = 1 when x ∈ Kr (and supp χr ⊂ Kr+1 ). Given f ∈ C ∞ (Ω) and > 0, choose
N ∈ N such that
∞
X
2−k < .
k=N +1
Let r = max{π2 f (1), . . . , π2 f (N )}, where f is as in section 1.1.5 and π2 : (N∪{0})×N −→
N is the projection onto the second factor. Then f χr ∈ Cc∞ (Ω) and d(f, f χr ) < .
3. Since ri has dense image, (ri)∗ = i∗ r∗ is injective. Thus r∗ is injective.
4. We know from class that t is injective.
5. λ is injective: i is injective and t is injective, so r∗ λ = ti is injective and thus λ is injective.
6. µ is injective:
Suppose ψ ∈ C ∞ (Ω) \ {0} and choose ω 0 ∈ Ω such that ψ(ω 0 ) 6= 0. Since µ is linear, for
the purpose of showing µψ 6= 0 we can assume ψ(ω 0 ) = 1. There exists an open set U ⊂ Ω
such that Re ψ(ω) > 0 for all ω ∈ U . Choose χ ∈ Cc∞ (Ω) such that χ(ω 0 ) = 1, 0 ≤ χ ≤ 1
and χ = 0 outside of U . Then µψ (χ) 6= 0 and thus µψ 6= 0.
7. We know from class that t has a weakly dense image.
8. i∗ r∗ has a weakly dense image:
Suppose u ∈ (Cc∞ (Ω))0 . Choose χr ∈ Cc∞ (Ω) such that 0 ≤ χr ≤ 1, χr (x) = 1 when x ∈ Kr
and supp χr ⊂ Kr+1 . Define
χj u : C ∞ (Ω) −→ C, ψ 7−→ u(χj ψ).
χj u ∈ (C ∞ (Ω))0 because u is continuous, the inclusion Cc∞ (Kj+1 ) −→ Cc∞ (Ω) is continuous,
and the function
C ∞ (Ω) −→ Cc∞ (Kj+1 ), ψ 7−→ χj ψ
is continuous: apply the Liebniz rule to see that kχj ψkN ≤ CkψkN,j+1 for some C ∈ N.
For each ϕ ∈ Cc∞ (Ω) it is clear that χj ϕ −→ ϕ; in fact for large enough j we have equality.
Thus by continuity of u we have
u(χj ϕ) −→ u(ϕ).
But u(χj ϕ) = χj u(ϕ) = i∗ r∗ (χj u)(ϕ) and so the line above tells us that i∗ r∗ (χj u) converges weakly to u.
49
9. µ has a weakly dense image:
Since the image of i∗ r∗ is weakly dense, so is the image of i∗ . By 7 the image of t is
weakly dense. Thus the image of µr = i∗ t has weakly dense image. In particular µ has
weakly dense image.
10. λ has a weakly dense image:
Let u ∈ (C ∞ (Ω))0 . Then there exists N ∈ N ∪ {0} and j, C ∈ N such that
|u(ψ)| ≤ CkψkN,j for all ψ ∈ C ∞ (Ω).
Let χ ∈ Cc∞ (Ω) be identically 1 on Kj . Then for ψ ∈ C ∞ (Ω)
|u(ψ − χψ)| ≤ Ckψ − χψkN,j = 0 =⇒ u(χψ) = u(ψ).
Choose (ψn ) ⊂ C ∞ (Ω) such that (µψn ) converges weakly to i∗ r∗ u and let ϕn = χψn ∈
Cc∞ (Ω). Then for ψ ∈ C ∞ (Ω)
Z
Z
λϕn (ψ) =
ϕn ψ =
ψn χψ = µψn (χψ) −→ i∗ r∗ u(χψ) = u(χψ) = u(ψ)
Ω
Ω
which means (λψn ) converges weakly to u.
5.20
5.20.1
Tempered distributional De Rham cohomology
Introduction
De Rham cohomology is a fundamental tool in geometry and the classical contruction uses the
notion of smooth forms. At each point x of a manifold M we may form the pth exterior power
of the cotangent space Tx∗ M , which we denote by Λp (Tx∗ M ). Using suitable charts we may glue
these vector spaces together to form a bundle over M ; the smooth sections of this bundle form
the vector space Ωp (M ) of p-forms on M . One has at hand a linear map d : Ωp (M ) −→ Ωp+1
with the property that d2 = 0. This makes a Ω∗ (M ) into a cochain complex. Taking the cohomology of the complex gives a diffeomorphism invariant of the manifold M .
On a manifold M one has local diffeomorphisms with Rn and so another way one might approach constructing smooth p-forms is to start of defining them locally. On Rn we may use
the formula Ω∗ (Rn ) = C ∞ (Rn ) ⊗R Ω∗n . Here Ω∗n is the exterior algebra Λ∗ (Vn ) where Vn is the
free R-vector space on the symbols {dx1 , . . . , dxn }. One may then define the forms globally by
working locally subject to compatibility conditions. In other words we have a sheaf of forms. We
know what value this sheaf should take on open sets diffeomorphic to Rn and this determines
what the global sections should be.
The advantage of this second point of view is that we have a clear dependence on C ∞ (Rn ).
When analysing solutions to differential equations we are naturally lead to the notion of (tempered) distributions, which should be thought of as an extension of the notion of a function.
A natural question to ask is whether we can replace C ∞ (Rn ) with S 0 (Rn ) in the above. Of
course we can. However, it may not be so clear that we get a theory having the same properties as the classical one. The object of this short project is to start investigating to what exend
the results from the classical theory carry over. In particular, we will prove the Poincaré Lemma.
5.20.2
Recall
Firstly,
let us fix once and for all χ ∈ Cc∞ (R), chosen so that χ ≥ 0, supp χ ⊂ [−1, 1] and
R
χ = 1.
R
50
Define Dn : S(Rn ) −→ S(Rn ) by Dn ϕ =
∂ϕ
∂xn
and In : S(Rn ) −→ S(Rn−1 ) by
Z
∞
ϕ(x1 , . . . , xn−1 , t)dt.
In ϕ(x1 , . . . , xn−1 ) =
−∞
Define Jn : S(Rn−1 ) −→ S(Rn ) by Jn (ψ)(x1 , . . . , xn ) = ψ(x1 , . . . , xn−1 )χ(xn ) and Kn :
S(Rn ) −→ S(Rn ) by
Z xn
Z ∞
Z xn
χ(t)dt.
ϕ(x1 , . . . , xn−1 , t)dt
ϕ(x1 , . . . , xn−1 , t)dt −
Kn ϕ(x1 , . . . , xn ) =
−∞
−∞
−∞
We saw that these maps are well-defined and continuous and that
I
D
n
n
S(Rn−1 ) −→ 0
S(Rn ) −→
0 −→ S(Rn ) −−→
is a split short exact sequence with the splittings given by Kn and Jn . From this we deduced
that we have a dual split short exact sequence
I∗
D∗
n
n
0 −→ S 0 (Rn−1 ) −→
S 0 (Rn ) −−→
S 0 (Rn ) −→ 0
with splittings given by Jn∗ and Kn∗ .
In particular we saw that the following relations hold:
Kn Dn = id, In Jn = id, Jn In + Dn Kn = id
and dually
Dn∗ Kn∗ = id, Jn∗ In∗ = id, In∗ Jn∗ + Kn∗ Dn∗ = id.
5.20.3
The De Rham Complex of Distributional Forms on Rn
Let Vn be the free R-vector space on the symbols {dx1 , . . . , dxn } and let Ω∗n be the the exterior
algebra Λ∗ (Vn ). The distributional forms on Rn are elements of
Ω∗S 0 (Rn ) = S 0 (Rn ) ⊗R Ω∗n .
n
We make Ω∗S 0 (Rn ) into a chain complex by defining d : ΩpS 0 (Rn ) −→ Ωp+1
S 0 (R ) as follows:
P
∂u
1. if u ∈ Ω0S 0 (Rn ), then du =
∂x dxi ;
P
P i
2. if ω =
uI dxI , then dω =
duI dxI .
Throughout this exposition we will take as definition
∂ϕ
∂u
(ϕ) = u
for ϕ ∈ S(Rn ),
∂xi
∂xi
i.e.
∂u
=
∂xi
∂
∂xi
∗
u.
This differs from the usual definition by a sign and I apologise to those in the field who are upset
by my convention; the original minus sign is there for a very important reason. However, the
subsequent computations will be more pleasing to the eye as a result. Referring back to problem
∂u
set 5, we also omitted signs for convenience; with our current conventions we have ∂x
= Dn∗ u.
n
Since
∂2u
∂xi ∂xj
=
∂2u
∂xj ∂xi
the classical proof that d2 = 0 still holds.
51
5.20.4
The Poincaré Lemma
We are now in a position to state the result we wish to prove.
Let Ω∗S 0 (Rn ) denote the complex of distributional forms on Rn , defined as above, and let HS∗ 0 (Rn )
denote the cohomology of this complex.
Theorem: HSp 0 (Rn ) = R when p = 0 and 0 otherwise.
In order to prove this theorem, known classically as the Poincaré Lemma, we will proceed
by using the following lemma:
p
Lemma: HSp 0 (Rn ) ∼
= HS 0 (Rn−1 ).
5.20.5
Proof of the Lemma
We will adapt the classical proof and construct chain homotopy equivalences between Ω∗S 0 (Rn )
and Ω∗S 0 (Rn−1 ). With this as our goal let π : Rn−1 × R −→ Rn−1 be the projection and
s : Rn−1 −→ Rn−1 × R be the zero section. In the classical context we immediately obtain chain
maps between Ω∗ (Rn ) and Ω∗ (Rn−1 ). In order for us to obtain such maps here we utilise what
we recalled in section 2. Define
X
X
π ∗ : Ω∗S 0 (Rn−1 ) −→ Ω∗S 0 (Rn−1 × R),
uI dxI 7−→
In∗ (uI )π ∗ (dxI )
and
s∗ : Ω∗S 0 (Rn−1 × R) −→ Ω∗S 0 (Rn−1 ),
X
uI dxI 7−→
X
Jn∗ (uI )s∗ (dxI ).
We immediately obtain s∗ ◦ π ∗ = id. This might be viewed as unsurprising since π ◦ s = id.
However, there is some unnaturality in our definition since we had to choose χ in section 2. We
should check that π ∗ and s∗ are chain maps. That π ∗ is a chain map comes down to proving
that for dxI ∈ Ω∗n−1 and uI ∈ S 0 (Rn−1 ),
n−1
X
In∗
j=1
∂uI
∂xj
π ∗ (dxj ∧ dxI ) =
n
X
∂I ∗ (uI )
n
j=1
∂xj
dxj ∧ π ∗ (dxI ).
∂
commutes with In and Dn∗ In∗ = 0. Thus the requisite identity is seen to hold.
For j < n, ∂x
j
∗
That s is a chain map comes down to proving that for dxI ∈ Ω∗n and uI ∈ S 0 (Rn ),
n
n−1
X
X ∂J ∗ (uI )
∂uI
n
Jn∗
s∗ (dxj ∧ dxI ) =
dxj ∧ s∗ (dxI ).
∂x
∂x
j
j
j=1
j=1
∂
For j < n, ∂x
commutes with Jn and s∗ (dxn ) = 0. Thus both maps are indeed chain maps.
j
We now define a chain homotopy between id and π ∗ ◦ s∗ . In what follows
I = {i1 , . . . , ip } where 0 < i1 < . . . < ip < n,
J = {j1 , . . . , jp−1 } where 0 < j1 < . . . < jp−1 < n
and there are obvious conventions when p or p − 1 is zero.
n−1
Define K : ΩpS 0 (Rn−1 × R) −→ Ωp−1
× R) by requiring both linearity and that
S 0 (R
K:
uI dxI
uJ dxJ ∧ dxn
7−→
7−→
52
0
p−1
(−1)
Kn∗ (uJ )dxJ .
Let ω = uI dxI . Then
(id − π ∗ s∗ )ω = (uI − In∗ Jn∗ uI )dxI and dKω = 0.
Also,
Kdω = K(Dn∗ (uI )dxn ∧ dxI ) = Kn∗ Dn∗ (uI )dxI .
Since In∗ Jn∗ + Kn∗ Dn∗ = id we have
(id − π ∗ s∗ )ω = (Kd + dK)ω.
Let ω = uJ dxJ ∧ dxn . Then (id − π ∗ s∗ )ω = ω,
dKω = (−1)p−1 dKn∗ (uJ )dxJ = Dn∗ Kn∗ (uJ )dxJ ∧ dxn +
X ∂K ∗ (uJ )
n
dxJ ∧ dxj
∂xj
j6=n
and


X ∂uJ
X
∂uJ
∗


Kn
dxj ∧ dxJ ∧ dxn = −
dxJ ∧ dxj .
Kdω = K
∂xj
∂xj
j6=n
For j < n,
∂
∂xj
j6=n
commutes with Kn and Dn∗ Kn∗ = id. Thus
(id − π ∗ s∗ )ω = (Kd + dK)ω.
We have shown id − π ∗ s∗ = (Kd + dK) and so, by basic homological algebra, the lemma is
proved.
5.20.6
Proof of the Theorem
When n = 1 the exact sequences in section 2 reduce to
D
I
1
1
0 −→ S(R) −−→
S(R) −→
R −→ 0
and
I∗
D∗
1
1
0 −→ R −→
S 0 (R) −−→
S 0 (R) −→ 0.
The second exact sequence tells us that HSp 0 (R) = R when p = 0 and 0 otherwise. The theorem
is proved in an obvious manner using induction together with the lemma.
5.20.7
Coefficients in S(Rn )
One might work with Schwartz-forms. In this context we will have a Poincaré Lemma analogous
to the one for compactly supported De Rham cohomology.
Let Ω∗S (Rn ) denote the complex of Schwartz-forms on Rn , defined in the obvious manner, and
let HS∗ (Rn ) denote the cohomology of this complex.
Lemma: HS∗ (Rn ) ∼
= HS∗−1 (Rn−1 ).
(Sketch) Proof: First we define two chain maps
n−1
π∗ : Ω∗S (Rn ) −→ Ω∗−1
),
S (R
53
∗−1
e∗ : Ω∗−1
) −→ ΩpS (Rn ).
S (R
With the same notation as before, we define π∗ by
π∗ :
We define e∗ by
ϕI dxI
7−→
0
ϕJ dxJ ∧ dxn
7−→
In (ϕJ )dxJ
ω 7−→ (π ∗ ω) ∧ χ(xn )dxn
One can check that these chain maps induce isomorphisms on cohomology.
5.20.8
Final Offerings
The extension of the distributional De Rham complex to manifolds takes a bit of work. One
would like to check the behaviour Ω∗S 0 (Rn ) under diffeomorphisms of Rn . I believe it is necessary
to look at a restricted class of diffeomorphism for this to make sense. For instance, we would
like there to be an induced map on S(Rn ) but not every diffeomorphism will give rise to such
a map. Can we guarantee that the transition maps on our manifold lie in this restricted class
of diffeomorphisms? Questions like this make the extension to arbitary sommoth manifolds a
more subtle point than one might initially think.
One might like to answer such questions directly. However, I am an algebraic topologist and
thus a lover of building big powerful mathematical machines; I am happy to agree with Professor Melrose that it is probably better to introduce some more machinery and kick out this
computational aspect.
Supposing one sets things up properly, the classical proof that de Rham cohomology agrees
with C̆ech cohomology should go through without too much trouble in this new setting. In
fact, there is a paper on the Arχiv written by Professor Melrose where it is shown that the
distributional De Rham complex can be retracted directly to the simplicial complex!
5.21
Weak compactness of the unit ball
Let H be a seperable Hilbert space and suppose (un ) is a bounded sequence in H.
H has a complete orthonormal basis (ej )∞
j=1 and Bessel’s ‘inequality’ tells us that
kun k2 =
∞
X
|hun , ej i|2 .
j=1
Thus for each j ∈ N, (hun , ej i)∞
n=1 is a bounded sequence in C. Using the Heine-Borel theorem
and a diagonal argument we can find a subsequence (unk ) of (un ) such that for each j ∈ N,
∞
(hunk , ej i)∞
k=1 converges. Let wj be the limit of (hunk , ej i)k=1 . Given > 0 and n ∈ N, for large
k we have
n
n
X
X
2
|wj | ≤
|hunk , ej i|2 + ≤ sup kun k2 + .
j=1
Thus
n∈N
j=1
∞
X
j=1
|wj |2 ≤ sup kun k2
n∈N
P∞
and we may set w = i=1 wi ei ∈ H. Then hw, ej i = wj so that hunk , ej i −→ hw, ej i for each j.
Since (ej )∞
j=1 is a complete orthonormal basis, (unk ) converges to w weakly.
54
5.22
A criterion for weak convergence
Let H be a seperable Hilbert space and suppose (un ) is a sequence in H such that hun , ϕi
converges in C for each ϕ ∈ H.
By the uniform boundedness theorem we know that (un ) is bounded. Also, H has a com∞
plete orthonormal basis (ej )∞
j=1 . Let wj be the limit of (hun , ej i)n=1 . Using the argument in 1
we see that
∞
X
|wj |2 ≤ sup kun k2
j=1
n∈N
P∞
and we may set w = i=1 wi ei ∈ H. Then hw, ej i = wj so that hun , ej i −→ hw, ej i for each j.
Since (ej )∞
j=1 is a complete orthonormal basis, (un ) converges to w weakly.
5.23
A strong limit of bounded operators is bounded
Let (An ) ⊂ B(H) be a sequence of bounded operators on a Hilbert space H which converges
strongly.
Define Au := limn−→∞ An u. By the uniform boundedness theorem the sequence (kAn k) is
bounded. For each n and u ∈ H we have
kAn uk ≤ kAn kkuk ≤ sup kAn kkuk =⇒ kAuk ≤ sup kAn kkuk.
n∈N
n∈N
A : H −→ H is linear by the algebra of limits and because each An is linear. Thus A ∈ B(H).
5.24
Open mapping theorem
Let X and Y be Banach spaces and let T : X −→ Y be a bounded linear map of X onto Y .
S∞
Let Un = {x ∈ X : kxk < n} and V = {y ∈ Y : kyk < }. Then n=1 T (Un ) = T (X) = Y .
By Baire’s theorem (T (Un ))◦ 6= ∅ for some n. Since y 7−→ ny is a homeomorphism of Y taking T (U1 ) to T (Un ) we see that (T (U1 ))◦ 6= ∅ and there exist y0 ∈ Y and > 0 such that
B(y0 , 2) ⊂ T (U1 ). We claim V ⊂ T (U2 ).
Suppose this for now and let U ⊂ X be open and x0 ∈ U . Choose δ > 0 such that B(x0 , δ) ⊂ U .
Let y ∈ B(T x0 , δ/2). Then k2(y − T x0 )/δ)k < . Thus there exists x ∈ U2 such that
T x = 2(y − T x0 )/δ and y = T (x0 + δx/2). x0 + δx/2 ∈ B(x0 , δ) ⊂ U which implies y ∈ T (U ).
Thus B(T x0 , δ/2) ⊂ T (U ). This shows T is open.
Proof of claim: y0 ∈ B(y0 , 2) ⊂ T (U1 ) implies −y0 ∈ T (U1 ). Now y ∈ V implies z =
y0 + 2y ∈ B(y0 , 2) ⊂ T (U1 ) which implies y = (z − y0 )/2 ∈ T (U1 ) since T (U1 ) is convex. Thus
V ⊂ T (U1 ). Take y ∈ V . There exists x1 ∈ U1 such that ky − T x1 k < /2. There exists
x2 ∈ P
U1 such that k2(y − T x1 ) − T x2 k < /2. Continue
P∞ finding xj ∈ X such that kxj k < 1,
n
ky − j=1 21−j T xj k < 2−n for all n ∈ N. Let x = j=1 21−j xj which converges since X is
complete. Then kxk < 2 and T x = y which implies y ∈ T (U2 ).
5.25
∆ : H 2 (Rn ) −→ L2 (Rn ) is a closed unbounded operator
We know that H 2 (Rn ) is dense in L2 (Rn ) and so we wish to show that {(u, ∆u) : u ∈ H 2 (Rn )}
is a closed subspace of L2 (Rn ) × L2 (Rn ). For this it is enough to take a sequence (un ) ⊂ H 2 (Rn )
such that (un ) converges to u in L2 (Rn ) and (∆un ) converges to v in L2 (Rn ) and show that
55
u ∈ H 2 (Rn ) and v = ∆u.
Since (un ) converges to u in L2 (Rn ), (un ) converges to u in S 0 (Rn ) weakly. In particular
for all ϕ ∈ S(Rn ), un (∆ϕ) −→ u(∆ϕ), i.e. ∆un (ϕ) −→ ∆u(ϕ). Thus (∆un ) converges weakly
to ∆u. On the other hand, (∆un ) converges to v in L2 (Rn ) and thus weakly in S 0 (Rn ). We
conclude that v = ∆u. Since v ∈ L2 (Rn ) = H 0 (Rn ) and ∆ is elliptic of order 2 we conclude
from the global regularity result for elliptic operators that u ∈ H 2 (Rn ). This completes the
proof.
5.26
An unbounded self-adjoint operator is closed
Let A be an unbounded self-adjoint operator on a Hilbert space H. Then we have a linear
subspace D ⊂ H and a linear map A : D −→ H such that
1. D ⊂ H is dense,
2. hAu, vi = hu, Avi for all u, v ∈ D,
3. If v ∈ H and D −→ C, u 7−→ hAu, vi extends to an element of H ∗ , then v ∈ D.
We will prove that Γ = {(u, Au) : u ∈ D} is closed in H × H.
Let (vn ) be a sequence in D such that
1. vn −→ v ∈ H,
2. Avn −→ w ∈ H.
We wish to show v ∈ D and w = Av. For any u ∈ D and n ∈ N we have hAu, vn i = hu, Avn i.
Taking limits as n −→ ∞ gives hAu, vi = hu, wi. Thus the linear map
D −→ C, u 7−→ hAu, vi
extends to the element h−, wi ∈ H ∗ and we deduce v ∈ D. For all u ∈ D we have hu, wi =
hAu, vi = hu, Avi. Since D ⊂ H is dense we obtain h−, wi = h−, Avi which implies w = Av by
the Riesz representation lemma.
5.27
Closed graph theorem
Theorem: Let X and Y be Banach spaces and T : X −→ Y be a linear map such that
Γ = {(x, T x) : x ∈ X} is closed in X × Y . Then T is continuous.
Proof : X × Y is a Banach space and so Γ is a Banach space. S : Γ −→ X, (x, T x) 7−→ x is
a bounded linear bijection. By the open mapping theorem S −1 : X −→ Γ is continuous and
so bounded. Thus for x ∈ X, kT xk ≤ k(x, T x)k = kS −1 xk ≤ kS −1 kkxk which implies T is
continuous.
5.28
Invertibility of A + itI, A is an unbounded self-adjoint operator
Let A be an unbounded self-adjoint operator and let t ∈ R \ {0}.
First we note that A + itI : D −→ H has a closed graph in H × H. Let (vn ) be a sequence in
D such that
1. vn −→ v ∈ H,
2. Avn + itvn −→ w ∈ H.
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We immediately obtain Avn −→ w −itv ∈ H and it follows from 1 that v ∈ D and w −itv = Av.
Next we make note of a useful identity. For u ∈ D, hitu, Aui+hAu, itui = ithu, Aui−ithAu, ui =
0. Thus
kAu + ituk2 = kAuk2 + kituk2 = kAuk2 + t2 kuk2 =⇒ kuk2 ≤ kAu + ituk2 /t2 .
From this it is clear that A + itI : D −→ H is injective.
We show that (A + itI)(D) is closed. Let (vn ) be a sequence in D such that (Avn + itvn )
converges to some w ∈ H. Then (Avn + itvn ) is Cauchy. The above inequality shows that (vn )
is Cauchy and so it converges to some v ∈ H. We are now in the position of the above so that
w ∈ (A + itI)(D) as required.
We show that A + itI is surjective. Since (A + itI)(D) is closed it is enough to prove that
((A + itI)(D))⊥ = {0}.
Let v ∈ ((A + itI)(D))⊥ . Then hAu + itu, vi = 0 for all u ∈ D. Thus D −→ C, u 7−→ hAu, vi
extends to h−, itvi ∈ H ∗ , and this shows v ∈ D. Because hu, Av − itvi = hu, Avi − hu, itvi =
hAu, vi + hitu, vi = hAu + itu, vi = 0 for all u ∈ D and D is dense, we see that Av − itv = 0.
We obtain v = 0 using the injectivity proved above (replacing t by −t).
Let B̃ : H −→ D be the inverse of A + itI. What we proved above shows that the composite
B̃
B : H −→ D ,→ H
has a closed graph. By the closed graph theorem this composite is continuous and hence
bounded.
5.29
((A + i)−1 )∗ = (A − i)−1
For all u, v ∈ D we have
hu, (A − i)vi = hu, Avi − hu, ivi = hAu, vi + hiu, vi = h(A + i)u, vi.
We wish to show (A + i)−1 : H −→ H and (A − i)−1 : H −→ H are adjoints of one another. By
definition, we must show that for all u0 , v 0 ∈ H
h(A + i)−1 u0 , v 0 i = hu0 , (A − i)−1 v 0 i.
By writing u0 = (A + i)u and v 0 = (A − i)v for unique u, v ∈ D we are reduced to the statement
above.
5.30
Define
U :H
(A+i)−1
/D
A−i
/H
so that
U −1 : H
(A−i)−1
/D
A+i
/ H.
We wish to prove that U ∗ = U −1 and so we must go about showing that
h(A − i)(A + i)−1 u0 , v 0 i = hu0 , (A + i)(A − i)−1 v 0 i
57
for all u0 , v 0 ∈ H. By writing u0 = (A + i)u and v 0 = (A − i)v for unique u, v ∈ D we are reduced
to showing
h(A − i)u, (A − i)vi = h(A + i)u, (A + i)vi
for all u, v ∈ D, which is true because
h(A − i)u, (A − i)vi = hAu, Avi + hu, vi + ihAu, vi − ihu, Avi = hAu, Avi + hu, vi
and
h(A + i)u, (A + i)vi = hAu, Avi + hu, vi − ihAu, vi + ihu, Avi = hAu, Avi + hu, vi.
This, of course, implies that U is bounded.
Using the theory of Möbius transformations we have a bijection
R −→ S 1 \ {1},
λ 7−→ s =
with inverse
S 1 \ {1} −→ R,
s 7−→ λ =
λ−i
λ+i
i(1 + s)
.
1−s
With this relationship we would like to show (1 − s)(A − λ) = (U − s)(A + i). This follows from
the following calculation:
i(1 + s)
(1 − s)(A − λ) = (1 − s) A −
= (A − i) − s(A + i)
1−s
and
(U − s)(A + i) = ((A − i)(A + i)−1 − s)(A + i) = (A − i) − s(A + i).
Thus, if (λ − i)(λ + i)−1 is not in the spectrum of U then (A − λ) : D −→ H is a bijection
with inverse (1 − s)(A + i)−1 (U − s)−1 . Conversely, if (A − λ) : D −→ H is a bijection, then
(U − s) : H −→ H is a bijection with inverse (1 − s)−1 (A + i)(A − λ)−1 and so (λ − i)(λ + i)−1
is not in the spectrum of U .
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