LECTURE 10, 18.155, 13 OCTOBER 2011
I tried to cover a bit too much I think, although none of it was
conceptually hard (maybe that is the problem!)
(1) The support of a smooth (or just a continuous) function defined on an open set – maybe Rn – can be defined either as the
closure of the set where it is non-zero, or as the complement of
the largest open set on which it is identically zero. For distributions the first approach does not make sense since the ‘value’
at a point is undefined but the second works if we do a little
preparation.
(2)
Lemma 1. If Ω ⊂ Rn is open and K b Ω is a compact subset
then there exists χ ∈ Cc∞ (Ω) such that χ ≡ 1 on a neighbourhood
of K.
Proof. For each x ∈ K there exists x > 0 such that B(x, x ) ⊂
Ω. By the compactness of K there is a finite subset of K such
that the corresponding B(xj , j /4) cover K. Take our model
non-negative function of compact support µ which is 1 in the
unit ball and vanish outside the ball of radius 2 and scale it to
χj (x) = µ(4(x
P − xj )/j ) so that has support in B(xj , j /2). The
sum φ1 =
χj is then strictly positive in a neighbourhood of
j
K and has compact support in Ω.
Now, apply the same argument to the compact set K2 =
{|x| ≤ R; x ∈
/ B(xj , j /2)} ⊂ Ω2 = (B(0, 2R) \ K) where R > 0
is chosen large enough that B(xj , j ) ⊂ B(0, R) for all j. This
gives a second function φ2 ∈ Cc∞ (Rn ) which is strictly positive on
K2 but with support not meeting K. It follows that φ1 + φ2 > 0
on B(0, R) and we may define φ = φ1 /(φ1 + φ2 ) in B(0, R)
and zero otherwise. It is smooth since the first definition make
it vanish near the boundary of B(0, R) and it satisfies all the
requirements.
(3) Using this we can decompose a function with compact support:1
2
LECTURE 10, 18.155, 13 OCTOBER 2011
n
Lemma
P 2. If Ωa , a ∈ A are open∞ subsets of R and φ ∈
∞
Cc (
Ωa ) then there exist φj ∈ Cc (Ωaj , j = 1, . . . , N such
a∈A P
that φ = φj .
j
Proof. The support of φ is covered by a finite number, Ωaj of the
Ωa so we can simply assume that A is finite. Then we proceed
succesively, since if we have established the result for two open
sets the general case follows by it applying it repeatedly to one
set and the union of the others. Thus suppose that supp(φ) ⊂
Ω1 ∪ Ω2 . Set K = supp(φ) \ Ω2 b Ω1 and apply the preceding
lemma to construct χ ∈ Cc∞ (Ω1 ) with χ = 1 in a neighbourhood
of supp(φ) \ Ω2 . Clearly then φ1 = χφ ∈ Cc∞ (Ω1 ) and φ2 =
(1 − χ)φ ∈ Cc∞ (Ω1 ∪ Ω2 ) must vanish where either χ = 1 or
φ = 0, including in a neighbourhood of each point of Ω1 \ Ω2 .
Thus φ2 ∈ Cc∞ (Ω2 ).
(4) Now we say that u ∈ S 0 (Rn ) vanishes in a given open set Ω if
u(φ) = 0 for all φ ∈ Cc∞ (Rn ) with supp(φ) ⊂ Ω. The decomposition above shows that
[
(1)
u = 0 in Ωa , a ∈ A =⇒ u = 0 in
Ωa .
a
S
Indeed, given the first condition and φ ∈ C ∞ ( a Ωa ) it follows
N
P
that φ =
φj where each φj ∈ Cc∞ (Ωaj ) for some aj ∈ A. Thus
Pj=1
S
u(φ) = u(φj ) = 0 and u = 0 in a Ωa .
j
In particular there is a largest open set on which u = 0 defined
as the union of all the open sets on which u = 0.
(5) So we can finally define the support of a (tempered) distribution
by
(2) supp(u) = Rn \ Ω, Ω ⊂ Rn the largest open set on which u = 0.
You should check that a distribution ‘has compact support’ in
the sense discussed earlier iff its support in this sense is compact.
(6) I did not do this in lectures, since it should wait until after you
do the homework for the coming week.
Proposition 1. An element u ∈ S 0 (Rn ) has compact support
contained in Ω iff it extends by continuity (using the density
of the range of the restriction may from S(Rn ) to C ∞ (Ω)) to
C ∞ (Ω).
LECTURE 10, 18.155, 13 OCTOBER 2011
3
(7) If the support of a distribution is empty it follows that the
distribution vanishes. The simplest non-trivial case corresponds
to support just being one point.
(3)
Lemma 3. If u ∈ S 0 (Rn ) and supp(u) ⊂ {p} then
X
u=
cα Dα δp , δp (φ) = φ(p).
|α|≤N
Proof. If u ∈ §0 (Rn ) has support only at p then, by definition,
u(φ) = 0 for any φ ∈ Cc∞ (Rn ) such that p ∈
/ supp(φ) and it
follows by density that u(φ) = 0 for any φ ∈ S(Rn ) with p ∈
/
supp(φ). Take some χ ∈ Cc∞ (Rn ) with χ(x) = 1 in |p − x| < 1/2
and χ(x) = 0 in |p − x| ≥ 1, it follows that χu = u. The
continuity estimate for u implies that
|u(φ)| ≤ CkφkN =⇒
X
(4) |u(φ)| = |u(χφ)| ≤ C 0 kχφ|N ≤ C
sup |Dα φ| ∀ φ ∈ S(Rn ).
|α|≤N
|x−p|≤1
Thus |u(φ)| is bounded by a constant multiple of the C N norm
of φ on {|x − p| ≤ 1}.
Lemma 4. If φ ∈ S(Rn ) then there is a sequence φj ∈ S(Rn )
with p ∈
/ supp(φj ) for all j and Dα (phi − φj ) → 0 uniformly on
|x − p| ≤ 1 if (and only if ) Dα φ(p) = 0 for all α with |α| ≤ N.
(5)
Proof. Taylor’s theorem with remainder means that the last
condition implies that
X
φ=
(x − p)β φβ , in |x − p| ≤ 2
|β|=N +1
with the φβ smooth in this ball. Thus if χ ∈ C ∞ (Rn ) is equal
to 1 in |x − p| ≤ 1 and has support in |x − p| ≤ 2 then
X
χφ =
(x − p)β ψβ , ψβ ∈ Cc∞ (Rn ).
|β|=N +1
(6)
Now, choose µ ∈ Cc∞ (Rn ) with µ(x) = 0 in |x| < 1/2 and with
support in |x| ≤ 1 and consider φj = χφ(1 − µ(j(x − p)). Each
φj vanishes in |x − p| < 1/2j and the difference can be written
X
χφ − φj =
(x − p)β µ(j(x − p))ψβ .
|β|=N +1
So consider the supremum norm of a derivative, of order at most
N of one of the terms on the right. Expanding out by Leibniz’
4
LECTURE 10, 18.155, 13 OCTOBER 2011
formula
(7) Dα ((x − p)β µ(j(x − p))ψβ ) =
X
dα,γ Dγ (xβ µ(j(x − p)))Dα−γ φβ .
γ≤α
The last terms are again just smooth functions of compact support so are bounded and it is enough to consider
X
(8) D(x) = |Dγ (xβ µ(j(x − p)))| ≤
cδ,γ xβ−δ Dδ µ(j(x − p))
δ≤γ
=
X
cδ,γ (x − p)β−δ j |δ| (Dδ µ)(j(x − p)).
δ≤γ
The support of µ(j(x − p) lies in |x − p| ≤ 1/j and |β − δ| > 1
so using the fact that |x − p| is bounded by 1/j
|D(x)| ≤ Cj |β−δ| j |δ| ≤ Cj −1 → 0.
Thus, all of the finitely many terms on the right of (6) tend to
zero in supremum norm and it follows that
|χφ − φj kC N → 0 in |x − p| ≤ 1.
Since we may add (1 − χ)φ to each φj without changing the
behaviour at x = p this shows that φ is the limit of (1−χ)φ+φj
with respect to the C N norm (even globally).
From this lemma it follows that for u with support at p there
is an N (any for which the continuity bound (4) holds) such
that
(9)
(10)
φ ∈ S(Rn ), Dα φ(p) = 0 ∀ |α| ≤ N =⇒ u(φ) = 0.
Now, we can again apply Taylor’s formula with remainder,
without assuming that the derivatives vanish, to see that any
φ ∈ S(Rn ) is of the form
X 1
φ=
(i)|α| (x − p)α Dα φ(p)χ + φ0
α!
|α|≤N
(11)
where u(φ0 ) = 0 because of (9) and χ ∈ Cc∞ (Rn ) is a fixed
function of compact support (with χ = 1 near p). Thus
X 1
X
u(φ) =
(i)|α| Dα φ(p)u((x − p)α χ) =
dα (Dα δp )(φ)
α!
|α|≤N
|α|≤N
LECTURE 10, 18.155, 13 OCTOBER 2011
5
where the constant dα include contributions from the u((x −
p)α χ) and we use the definition of the Dirac delta and derivations which shows that
(12)
(−1)|α| Dα φ(p) = δp ((−D)α φ) = Dα δp (φ).
(8) Next I want to discuss homogeneous distributions, to get more
some more examples. First for n = 1. We can allow complex
homogeneity, so a function on R is said to be (positively) homogeneous of degree z ∈ C if
(13)
(14)
f (tx) = tz f (x) ∀ t > 0 and x ∈ R, tz = exp(z log t)
using the standard (real) branch of the logarithm. The obvious
examples of such functions are the powers themselves
(
(
z
x
in
x
>
0
0
in x ≥ 0
xz+ =
xz− =
z
0
in x ≤ 0,
(−x)
in x < 0.
These functions are clearly smooth and of polynomial growth
away from the origin and are locally integrable there if and only
z
if Re z > −1 (notice that |xz± | = xRe
± ).
So this means that, for Re z > −1 they are well-defined elements of S 0 (R).
(9) A homogenous function has to be a constant multiple (possibily different on the two sides) of xz± for ±x > 0 and so, if
locally integrable, does define a tempered distibution and we
see that there is a weak version of homogeneity. Namely, if we
set φt (x) = φ(x/t) for t > 0 then
Z
Z
(15)
f (φt ) = f (x)φ(x/t)dx = f (tX)φ(X)tdX = tz+1 f (φ).
We adopt this as the definition of homogeneity for a distribution.
(10)
Proposition 2. For any z ∈ C the (tempered but that is actually automatic) distributions which are homogeneous of degree
z form a two-dimensional space.