LECTURE 11, 18.155, 18 OCTOBER 2011
(1)
Lemma 1. For any s ∈ R, m ∈ N, u ∈ H s (Rn ) and Dα u ∈
H s (Rn ) for all α ∈ N with |α| = m is equivalent to u ∈
H s+m (Rn ). In fact it suffices to have u ∈ H s and Djm u ∈ H s
for j = 1, . . . , n.
Proof. hξis û ∈ L2 and ξjm hξis û ∈ L2 j = 1, . . . , n together imply
that hξis+m û ∈ L2 .
k
Lemma 2. The elements of C∞
(Rn ) – the functions with all
derivatives up to order k ∈ N bounded and continuous – are
multipliers on H s (Rn ) provided |s| ≤ k.
Proof. Certainly true for s = k, since the product exists in
S 0 (Rn ) and expanding out a derivative of it, up to order k, by
Leibniz’ formula gives a sum where each term is in L2 .
Next consider the case 0 < s < k = 1. Then we may use the
characterization of H s you did in the homework last week. This
shows that
Z
|u(x) − u(y)|2
2
kukL2 +
dxdy
n+2s
|x−y|≤1 |x − y|
is the square of an equivalent norm on H s (Rn ). Here I have
dropped the other part of the ‘Hölder’ type integral since it is
controlled by the L2 norm anyway:
Z
Z
|u(x) − u(y)|2
1
|u(x)|2
dxdy ≤ 4
dxdy ≤ Ckuk2L2 .
n+2s
n+2s
|x − y|
|x−y|≥1 |x − y|
|x−y|≥1
1
Now observe that if u ∈ H s and f ∈ C∞
(bbRn ), then using the
estimate |f (x) − f (y)| ≤ sup |Df ||x − y|
Z
|f (x)u(x) − f (y)u(y)|2
dxdy
|x − y|n+2s
|x−y|<1
Z
|f (x)u(x) − f (x)u(y) + f (x)u(y) − f (y)u(y)|2
=
dxdy
|x − y|n+2s
|x−y|<1
Z
Z
|u(x) − u(y)|2
|u(y)|2
≤ 2 sup |f |
dxdy
+
2
sup
|Df
|
dxdy
n+2s
n+2s−2
|x−y|<1 |x − y|
|x−y|≤1 |x − y|
1
2
LECTURE 11, 18.155, 18 OCTOBER 2011
so this is bounded in terms of H s and we can use density of
S(Rn ) in H s .
Now, the case of general s > 1 follows by considering the
largest integer l ≤ s and combining these two arguments to
show that Dα (f u) ∈ H s−l for all |α| = l, so f u ∈ H s (Rn ).
The case of negative s follows by duality since u ∈ H s with
s < 0 iff if
|u(φ)| ≤ CkφkH −s
and then
|f u(φ)| = |u(f φ)| ≤ Ckf φk)H −s ≤ C 0 kf kC∞
k kφkH −s
by the discussion for s > 0.
(2) Now, back to homogeneity. We are in the middle of discussing
the space of distributions, on the line, which are homogeneous
of degree z ∈ C – meaning
u(φt ) = tz+1 u(φ), ∀ φ ∈ Cc∞ (R), φt (x) = φ(x/t), t > 0
(where the +1 comes from the homogeneity of dx.)
The convergence of the difference quotient
φt (x) − φ(x)
→ −xφ0 (x) as t → 1
t−1
is uniform with all derivatives and uniform support – certainly
in S(R) but actually in Cc∞ (R). Anyway, we are assuming u ∈
S 0 (R) (although this is unnecessary) so from (2) it follows that
d
u(φt ) − u(φ)
→ u(−xφ0 (x) = ( (xu))(φ) = (z + 1)u(φ)
t−1
dx
so u satisfies the Euler’s differential equation
du
x
= zu on R.
dx
So this is a little exercise in differential equations. Consider
what happens in x > 0 – in fact we can see that if χ ∈ Cc∞ (R)
has support in x > 0 then
1
dχu
= zχu − χ0 u.
dx
x
We need to check that χu ∈ C 1 and then equation holds in the
classical sense and since we can take χ = 1 on any interval we
want,
du
x
= zu =⇒ u(x) = cxz+ on x > 0
dx
LECTURE 11, 18.155, 18 OCTOBER 2011
3
by integration.
So, it is only the regularity we need. Fix χ ∈ Cc∞ (R) with
support contained in [a, b] where b > a > 0. We can choose
another χ1 ∈ Cc∞ (Rn ) with χ0 = 1 on [a − , b + ] for some a >
> 0 and then successive χk ’s, k ≥ 1, with ‘nested supports’
χk (x) = 0 if x < a − /k or x > b + /k and χk = 1 if x ≥
a − /(k + 1) or x ≤ b + /(k + 1). Thus χk χk+1 = χk+1 .
Now, χ0 u has compact support so is in some H s where s may
be negative as far as we know. The differential equation (2) for
χ = χk shows that
dχk u
1
1
= zχk u − χ0k u = zχk (χk−1 u) − χ0k (χk−1 u).
dx
x
x
Proceeding inductively this shows that χk u ∈ H s+k for all k.
Namely, it is true for k = 0 and if true for k − 1 then χk u =
χk χk−1 u ∈ H s+k−1 and (2) shows that dχk u/dx ∈ H s+k−1 as
well, hence χk u ∈ H s+k .
So, by the Sobolev embedding theorem, taking k > |s| + 1,
this actually shows that u ∈ C 1 (a, b) on any interval (a, b) with
b > a > 0 and the argument leading to (2) is justified. The
same thing works in x < 0 of course, so we conclude that if u
is homogeneous of degree z then
(
c+ xz+ if x > 0
u=
c− cz− if x < 0.
Now, let’s assume that Re z > −1 since then we showed last
time that xz± are well-defined tempered distributions and what
(2) shows is that
(1)
v = u − c+ xz+ − c− xz− ∈ S 0 (R)
has support contained in {0}.
So, in fact we know that v can only be a finite sum of multiples
of derivatives of the delta function. However, we can check
immediately that
δ(φt ) = φt (0) = φ(0) = δ(φ)
which means that δ is homogeneous of degree −1. (Not 0 becuase of the z + 1 in the definition of distributional homogeneity.) The derivative dk δ/dxk clearly has homogeneity −1 − k by
the same argument, so no combination of these can be homogeneous of degree z with Re z > −1 (except the zero combination).
4
LECTURE 11, 18.155, 18 OCTOBER 2011
So, finally we see that v = 0 in (1) and
(2)
The space of homogeneous distributions on R of degree z, Re z > −1
is of dimension two spanned by xz± .
(3) Now, to the case Re z ≤ 1. The Fourier transform of test functions gives, by changing variable,
Z
b
φt = e−ixξ φ(x/t)dx = tφ̂(ξt).
So, if u is homogeneous of any degree z,
(3)
û(φt ) = u(φbt ) = tu(φ̂(·t)) = t(1/t)z+1 u(φ̂) = t−z û(φ),
shows that û is homogeneous of degree Z = −z − 1 (so that
t−z = tZ+1 .
Thus, if Re z < 0 then û is homogeneous of degree Z with
Re Z = − Re z −1 > −1. There is some overlap! So by applying
the Fourier transform we conclude that the spac of tempered
distributions homogeneous of degree z is always of dimension
two – although the spanning part of (2) is not always true.
(4) What really happens? If φ ∈ S(R) then
Z ∞
z
F (z) = x+ (φ) =
xz φ(x)dx for Re z > −1
0
is actually a holomorphic function there and extends to a meromorphic function of z ∈ C with poles at z ∈ −R. Holomorphy
in Re z > 0 follows by differntiating the integral – which is
easily seen to be once continuously differentiable in u and v if
z = u + iv and to satisfy ∂¯z F (z) = 0 which is one definition of
holomorphy. Moreover the ‘functional equation’
xz+ (φ0 ) = −zxz−1 (φ), Re z > 0
follows by integration by parts. This however can be written
1
xz+ (φ) = −
xz+1 (φ0 ) Re z > 1
z+1 +
and then iterated to get
(4)
xz+ (φ)
= (−1)
k
k
Y
1 z+k dk φ
x (
), Re z > k
z + j + dxk
j=1
for any positive integer k. However, the right side is then holomorphic, as a function of z, in Re z > −k − 1 so, by the uniqueness of holomorphic extension, the left side must extend to a
LECTURE 11, 18.155, 18 OCTOBER 2011
5
holomorphic function as stated – with the only poles arising
from the vanishing of the z + j.
Proposition 1. The distribution xz+ is well-defined by analytic
continuation, i.e. (4) for all z ∈ C \ (−N) and the residue at
k−1
z = −k is a multiple of ddk−1 xδ .
Proof. The continuity of φ 7−→ xz+ (φ) also follows from (4). The
residue at z = −k is
dk φ
1
dk−1 φ
1
(−1)k−1 dk−1 δ
−
x0+ ( k ) =
δ( k−1 ) =
(φ).
(k − 1)!
dx
(k − 1)! dx
(k − 1)! dxk−1
Essentially the same thing is true for xz− with sign changes.
From this you can check that a certain combination xz− + ak xz−
has no pole at z = −k and therefore the value there defines
a homogeneous distribution of degree −k. The upshot is that
xz± span the homogeneous distributions for z ∈
/ −N but for the
dk−1 δ
negative integers this combination, plus dxk−1 span the homogeneous distributions. Just to make it doubly clear
The function x−k defined on x > 0 is not the restriction of homogeneous distribution (which would have
to be of degree −k) which vanishes in x < 0.
However, it is the restriction to x > 0 of an element of S 0 (R)
which vanishes in x < 0 it just cannot be homogeneous.
Exercise 1. Can you find a distribution with this property?
Hint: Subtract (z+k) times the residue and evaluate at z = −k.