LECTURE 2, 18.155, 13 SEPTEMBER 2011
(1)
(1)
Theorem 1. The space S(Rn ) of Schwartz test functions is a
complete metric space with respect to
X
kφ − ψkN
.
d(φ, ψ) =
2−N
1 + kφ − ψkN
N ≥0
(2) First check that (1) is indeed a distance. This has nothing
to do with the nature of the norms involved. Only the triangle
inequality needs checking and it suffices to do this term by term,
i.e. to show that for any φ, ψ and η
kφ − ψk
kφ − ηk
kη − ψk
(2)
≤
+
1 + kφ − ψk
1 + kφ − ηk 1 + kη − ψk
for any norm. Multiplying out this is equivalent to checking
that
(3) kφ−ψk+kφ−ψkkφ−ηk+kφ−ψkkη −ψk+kφ−ψkkφ−ηkkη −ψk
is dominated by
(4)
kφ − ηk + kη − ψk + kφ − ηkkφ − ψk + kφ − ηkkη − ψk
+kη − ψkkφ − ψk + kη − ψkkφ − ηk + 2kη − ψkkφ − ηkkφ − ψk.
By the triangle inequality for the norm, the first term on the
left in (2) is dominated by the sum of the first two terms in (4).
The last, cubic, term in (2) appears as twice in (4) so it suffices
to observe that the two quadratic terms in (2) also appear in
(4).
(3) So, we need to show that a Cauchy sequence in S(Rn ) converges.
First observe a sequence is Cauchy with respect the distance if
it is Cauchy with respect to each of the norms k · kN – with no
uniformity of any sort in N. For φn to be Cauchy means that
given > 0 there exists k such that
(5)
n, m > k =⇒ d(φn , φm ) < .
Given an N and a δ > 0 choose < 2−N −1 min(δ, 12 )(> 0). Since
(6)
2−N
kφn − φm kN
< d(φn , φm ),
1 + kφn − φm kN
1
2
LECTURE 2, 18.155, 13 SEPTEMBER 2011
it then follows from (5) that
(7)
n, m > k =⇒ kφn − φm kN < δ/4 + kφn − φm kN )/2 =⇒ kφn − φm kN < δ.
Thus the sequence is Cauchy with respect to each norm.
(4) We don’t need the reverse implication here, but observe that
if a sequence is Cauchy with respect to each of the norms, so
given δ > 0 there exists k = k(N ) such that
n, m > k =⇒ kφn − φm kN < δ.
(8)
Then given > 0 choose N so large that
X
X
kφ − ψkM
2−M
≤
2−M = 2−N < /2
1 + kφ − ψkM
M >N
M >N
(9)
and having chosen N then choose k using (8) with δ = /N.
Then using (9), if n, n > k,
(10) d(φn , φm ) =
X
M
≤
X
M ≤N
2−M
2−M
kφ − ψkM
1 + kφ − ψkM
kφ − ψkM
+ /2 ≤ N kφ − ψkN + /2 < .
1 + kφ − ψkM
Here I use the fact that the norms are increasing with N, although it is not necessary for the argument to work.
(5) So, we now know that a Cauchy sequence is Cauchy with respect
to each of the norms, which are
X
(11)
kφkN =
sup |xβ ∂ α φ(x)|.
|α|+|β|≤N
x∈Rn
So this means that each of the sequences xβ ∂ α φn is Cauchy with
respect to the supremum norm on Rn . We use the completeness
of the space of bounded continuous functions on Rn to conclude that in fact they each converge uniformly to a continuous
functions
(12)
0
(Rn ).
xβ ∂ α φn → gα,β ∈ C∞
(6) We need an additional argument to conclude that φ = g0,0 ∈
S(Rn ) and that φn → φ in S(Rn ). Just consider the gα,0 first,
these are suppose to be the derivatives ∂ α φ. Why is this so?
Well, we concentrate on the ‘first derivatives’, gj being the limit
of ∂j φn since then we can use induction to handle the higher
derivatives (being derivatives of derivatives). One way to do
LECTURE 2, 18.155, 13 SEPTEMBER 2011
(13)
3
this is to use the Fundamental Theorem of Calculus in each
variable and observe that
Z t
∂j φn (x + sej )ds
φn (x + tej ) − φn (x) =
0
(14)
since dφn (x + sej )/ds = ∂j φ(x + sej ) essentially by definition.
This is a Riemann integral and these behave well under uniform
convergence so taking the limit we find that
Z t
gj (x + sej )ds.
φ(x + tej ) − φ(x) =
0
Now, we can reverse the argument using the Fundamental Theorem of Calculus again to conclude that dφ(x + tej )/dt) exists
and is equal to gj (x + tej ). Setting t = 0 this means that φ has
first partial derivatives everywhere and equal to the gj (which
are continuous). Now, repeat the argument ad infinitem to see
that ∂ α φ = gα,0 .
An even easier version of the argument then shows that for
every β, gα,β = xβ paα φ so indeed, φ ∈ S(Rn ).
(7) Don’t forget the last step, going back and letting m → ∞ in
(5) to conclude that d(φn , φ) → 0 so φn → φ in S(Rn ) and we
have shown completeness.
Definition 1. The space of tempered (or temperate) distributions is the dual of S(Rn ), denoted S 0 (Rn ) and so consists of all
continuous linear function(al)s
(15)
u : S(Rn ) −→ C.
(8) Of course, we need to flesh out this definition a good deal before
we see that these things are useful. First we need to check what
continuity means.
Proposition 1. A linear map (15), or more generally u :
S(Rn ) −→ B where B is a normed space, is continuous iff
there exists N and C such that
(16)
ku(φ)kB (= |u(φ)| for (15)) ≤ Ckφ|N ∀ φ ∈ S(Rn ).
A linear map T : S(Rn ) −→ S(Rn ) is continuous iff for each
N 0 there exists N and C such that
(17)
kT φkN 0 ≤ Ckφ|N ∀ φ ∈ S(Rn )
and a linear map from a normed space I : B −→ S(Rn ) is
continuous iff for each N there exists C such that
(18)
kI(b)kN ≤ CkbkB .
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LECTURE 2, 18.155, 13 SEPTEMBER 2011
(9) To check (16) observe that continuity for a linear map (because
we can easily translate) means that the inverse image of a ball
around the origin in B contains a ball around the origin in
S(Rn ). Thus, there must exist > 0 such that d(φ, 0) < implies ku(φ)kB < 1 – assuming continuity of u that is. Now, if
N is chosen so large that 2−N < /2 then kφkN < /2N implies
– splitting the sum in two as before - that
X
kφkM
(19)
d(φ, 0) =
2−M
< N kφN + /2 < .
1 + kφkM
M ≥0
Thus for this N, kphikN < /2N implies ku(φ)kB < 1 which by
scaling gives (16) .
Conversely (16) implies (sequential) continuity (at 0 suffices),
since if d(φn , 0) → 0 then kphin kN → 0 and then (16) implies
that ku(φn )kB → 0.
(10) The arguments for the other cases are similar, for instance for
(17) continuity means that the inverse image of {ψ; d(ψ, 0) <
} must, for every > 0 contain some {φ; d(φ, 0) < δ}, δ >
0
0. Choosing < 2−N −1 the -ball contains the set {ψkN 0 <
/2N 0 } so its inverse image must contain some set {kφkN < δ}
for N sufficiently large and δ sufficiently small. This gives (17).
Conversely, given that we already know that convergence in
S(Rn ) is just convergence with respect to each norm, (17) shows
that if φn → 0 then T φn → 0 showing continuity a the origin
and hence everywhere (for a linear map).
(11) I leave (18) to you, it is no harder.
(12) Now, we discuss some examples of elements of S 0 (Rn ) and continuous maps. One important example is integration,
Z
(20)
: S(Rn ) −→ C
Rn
where this is an improper (i.e. limit of) Riemann integral. Continuity follows from an estimate such as
Z
Z
(21) |
φ| ≤ (1 + |x|2 )n+1 |φ|(1 + |x|2 )−n−1
Rn
Z
≤ (1 + |x|2 )−mn−1 sup(1 + |x|2 )n+1 |φ| ≤ Ckφ|2n+2 .
Rn
(13) Another example is the Dirac delta ‘function’ at any point p ∈
Rn :
(22)
δp : S(Rn ) 3 φ(p) ∈ C, δp ∈ S 0 (Rn ).
LECTURE 2, 18.155, 13 SEPTEMBER 2011
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Clearly, |δp (φ)| ≤ kφk0 gives continuity.
(14) The maps
(23)
× xβ : S(Rn ) −→ S(Rn ), ∂ α : S(Rn ) −→ S(Rn )
are continuous – we will come back to these!
(15) Finally the Fourier transform is
Z
(24)
φ̂(ξ) = e−ix·ξ φ(x)
and I will show on Thursday that it gives a topological isomorphism of S(Rn ).