4 Distributions

Laurent Schwartz introduced the notion of distributions two talk about generalized solutions to differential equations. For his work he got the fields medal.

4.1 Space of Distributions

Definition 4.1

$D (Ω) = C_{c}^{\infty} (Ω)$ is the set of test functions together with a topology determined by its converging sequences : $ϕ_{n} \to ϕ$ in $D (Ω)$ if

there exists a compact subset $K \subset Ω$ such that $Supp (ϕ_{n}) \subset K$ .
For all multiindices $α$ we have $\partial_{α} ϕ_{n} \to \partial_{α} ϕ$ in uniformly.

Convention: if $x \in R^{d} ∖ Ω$ , then $ϕ (x) := 0$ .
$D^{'} (Ω)$ is the topological dual space, i.e. the space of continuous linear functionals $D (Ω) \to C$ with the weak-*-convergence, i.e. pointwise convergence.

Remark 4.2

A notion of converging sequence on a set $X$ is

The constant function on $x$ converges to $x$
A sequence converges to $x$ iff any subsequence has a subsequence converging to $x$ .

Note, that any subsequence of a converging sequence converges to the same limit. It induces a closure operator on $X$ (for $\overset{―}{A \cup B} \subset \bar{A} \cup \bar{B}$ you use, that if a sequence in $A \cup B$ converges to $x$ , then it has a subsequence (converging to $x$ ) lying in $A$ or in $B$ ).

Example 4.3

Every locally integrable function $f \in L_{l o c}^{1} (Ω)$ gives us a distribution $Λ f \in D^{'} (Ω)$ defined by

Λ f (ϕ) = \int_{Ω} f (x) ϕ (x) d x

which is well-defined because $ϕ$ has compact support.

Example 4.4

Let $μ$ be a Radon measure on $Ω$ (or more generally a signed Borel measure which is finite on compact subsets of $Ω$ ). Then it defines a distribution:

T_{μ} (ϕ) = \int_{Ω} ϕ d μ

Proof ▶

As Borel sets are $μ$ -measurable, every continuous function is $μ$ -measurable. Let $K := Supp ϕ \subset Ω$ . Because $μ (K) < \infty$ and $max ϕ (K) < \infty$ it follows that $ϕ$ is $μ$ -integrable. Linearity follows from linearity of the integral. If $ϕ_{j} \to ϕ_{*}$ unifomrly, then $Supp ϕ_{j} \subset K$ for all $j$ and $T_{μ} ϕ_{j} \to Λ_{μ} ϕ_{*}$ .

We have the following important special case:

Example 4.5 Dirac-

δ

We have $δ \in D^{'} (Ω)$ given by

δ (ϕ) := ϕ (0)

4.1.1 Convolution

Notation 1

For $ϕ \in D$ define $ϕ^{R} \in D$ as $ϕ^{R} (x) = ϕ (- x)$ and for $x \in R^{d}$ we have the shift $τ_{x} (ϕ) \in D$ given by $τ_{x} (ϕ) (y) = ϕ (y - x)$ .

Example 4.6

For $f \in L_{l o c}^{1} (Ω), g \in D (Ω)$ we have

(f * g) (x) = Λ f (τ_{x} (ψ^{R}))

Proposition 4.7

Let $F \in D^{'} (Ω), ψ \in D$ . The following two distributions coincide:

The distribution determined by the smooth function $x \mapsto F (τ_{x} (ψ^{R}))$ .
The distribution $ϕ \mapsto F (ψ^{R} * ϕ)$ .

We write this distribution as $F * ψ$ .

Proof ▶

$ζ := ψ^{R}$ The function $x \mapsto F (τ_{x} (ψ^{R}))$ is smooth :

It is continuous: If $x_{n} \to x$ , then $τ_{x} (ζ) - τ_{x_{n}} (ζ) \to 0$ uniformly and the same holds for partial derivatives. Now $F$ is continuous hence $F (τ_{∙} (ψ^{R}))$ is continuous and $F$ preserves the difference quotients, hence it will be also smooth.
The two distributions coincide:

$\begin{aligned} F (ψ^{R} * ϕ) & = F (\int (τ_{x} ψ^{R} (∙) ϕ (x) d x)) \\ \overset{!}{=} \int F (τ_{x} ψ^{R} (∙) ϕ (x)) d x \\ = \int F (τ_{x} ψ^{R}) ϕ (x) d x \\ = Λ (F (τ_{∙} ψ^{R})) (ϕ) \end{aligned}$

Where we are allowed to pull out the integral by 4.8

Lemma 4.8

Let $ϕ \in C_{c}^{\infty} (Ω \times Ω)$ . Then for any $F \in D^{'} (Ω)$ we have

F (\int_{Ω} ϕ (x,_) d x) = \int_{Ω} F (ϕ (x,_)) d x

Proof ▶

Consider $S_{ε} \in D$ defined by

S_{ε}^{ϕ} (y) = ε^{d} \sum_{n \in Z^{d}} ϕ (n ε, y)

which is a finite sum as $ϕ$ has compact support. Then for all $ϕ \in C_{c}^{\infty} (Ω \times Ω)$ one has a limit in $D$

\int_{Ω} ϕ (x,_) d x = lim_{ε \to 0} S_{ε}^{ϕ}

Hence by continuity of $F$

\begin{aligned} F (\int_{Ω} ϕ (x,_) d x) & = F (lim_{ε \to 0} S_{ε}^{ϕ}) \\ = lim_{ε \to 0} F (S_{ε}^{ϕ}) \\ = lim_{ε \to 0} ε^{d} \sum_{n \in Z^{d}} F (ϕ (n ε,_)) \\ = lim_{ε \to 0} S_{ε}^{F \circ ϕ} \\ = \int_{Ω} F (ϕ (x,_)) d x \end{aligned}

Using the first definition we learn the following two things

Example 4.9

From the description 4.6: Writing $Λ f * g$ is unambiguous.

Example 4.10

We have $δ_{0} * f = Λ f$ for all $f \in D$ .

Lemma 4.11

Convolution $F * ψ$ is continuous in both variables.

Proof ▶

Continuity in the distribution variable is clear by pointwise convergence.
For the continuity in the test function variable one uses that convolution with a fixed test function is a continuous function $D \to D$ and distributions are continuous.

Proposition 4.12

There exists a sequence $ψ_{n} \in C_{c}^{\infty} (Ω)$ such that $Λ ψ_{n} \to δ_{0}$ in $D^{'} (Ω)$ .

Proof ▶

Fix some $ψ \in D$ with $\int ψ (x) d x = 1$ . Define $ψ_{n} (x) := n^{d} ψ (n x)$ . Then

(Λ ψ_{n} - δ_{0}) (ϕ) = \int n^{d} ψ (n x) ϕ (x) d x - ϕ (0) = \int ψ (x) \cdot (ϕ (x / n) - ϕ (0)) d x \to 0

where we used that $ϕ (x / n) - ϕ (0) \to 0$ uniformly.

The following is how we view $L_{l o c}^{1} (Ω) \subset D^{'} (Ω)$ .

Corollary 4.13

Let $f, g \in L_{l o c}^{1} (Ω)$ such that $Λ_{f} = Λ_{g}$ . Then $f = g$ almost everywhere.

Proof ▶

We have $0 = Λ (f - g) (τ_{∙} ψ_{n}^{R}) = ψ_{n} * (f - g) \to δ_{0} * (f - g) = f - g$ in $L_{l o c}^{1}$ , hence $f = g$ almost everywhere.

Example 4.14

$δ$ is not a function! I.e. its not of the form $Λ f$ for some $f \in L_{l o c}^{1}$

Proof ▶

We have $Δ |_{Ω ∖ {0}} = Λ 0$ so if it would be a function, then it would be zero almost everywhere, hence 0. But this is a contradiction.

Corollary 4.15

Then $C^{\infty} (R^{d})$ is dense in $D^{'} (R^{d})$ .

Proof ▶

We know by 4.12 that there exists $Λ ψ_{n} \to δ_{0}$ in $D^{'}$ . Let $F$ be a distribution on $R^{d}$ . Now setting $F_{n} := F * ψ_{n}^{R}$ , yields pointwise

F_{n} (ϕ) = F (ψ_{n} * ϕ) \to F (δ_{0} * ϕ) = F (ϕ)

by 4.11 hence $F_{n} \to F$ in $D^{'}$ .

4.1.2 Derivatives

For $ϕ, ψ \in D$ we have

\int_{Ω} \partial^{α} ϕ ψ d x = (- 1)^{| α |} \int_{Ω} ϕ \partial^{α} ψ d x

This motivates the definition

Definition 4.16

For a multiindex $α$ and a distribution $F$ define the distribution

\partial^{α} F (ϕ) = (- 1)^{| α |} (F \partial^{α} ϕ)

Remark 4.17

let $f \in L_{l o c}^{1} (Ω)$ . If there exists some $f^{'} \in L_{l o c}^{1} (Ω)$ , such that $Λ f^{'} = \partial^{α} Λ f$ as distributions, then we call $f^{'}$ the weak derivative of $f$ with respect to $α$ .

Proposition 4.18

For $F \in D^{'}, ϕ \in D$ , We have

\partial^{α} (F * ϕ) = (\partial^{α} F) * ϕ = F * \partial^{α} ϕ

Proof ▶

First note, that holds in the case where $F$ is a test function, so that we have ordinary convolution. Then check pointwise. After erasing the sign $(- 1)^{| α |}$

F (ϕ^{R} * \partial^{α} ψ) = F (\partial^{α} (ϕ^{R} * ψ)) = F ((\partial^{α} ϕ) * ψ)

4.1.3 Support

The Support of a distribution $F$ is the complement of the largest open subset $U$ , such that $F (ϕ) = 0 \forall ϕ \in D, Supp (ϕ) \subset U$ . This definition is unambiguous: If $F$ vanishes on each $U_{i}$ for some index set $I$ and $ϕ \in D$ such that the compact set $Supp ϕ \subset U = ⋃ U_{i}$ , we may assume that $I$ is finite and choose a partition of unity $Supp η_{i} \subset U_{i}$ and $\sum η_{i} = 1$ . Then $F (ϕ) = \sum F (ϕ η_{i}) = 0$ .

4.1.4 Tempered Distributions

We no enlarge our test space to the schwartz space $S = S (R^{d})$ consisting of smooth functions that are rapidly decreasing at $\infty$ with all derivatives.

Definition 4.19

Consider the increasing sequence of norms on $C^{\infty} (Ω)$ defined by

‖ ϕ ‖_{N} = sup {| x^{β} (\partial_{x}^{α} ϕ (x)) | ∣ x \in R^{d}, | α |, ‖ β ‖ \leq N}

S = {ϕ \in C^{\infty} (R^{d}) ∣ ‖ ϕ ‖_{N} < \infty \forall N}

with the obvious notion of convergence.

Lemma 4.20

We have a continuous inclusions $D \subset S$ , hence $S^{'} \subset D^{'} (R^{d})$ .
Moreover, this inclusion is dense. Hence being tempered is a property of distributions.

Lemma 4.21

Let $f \in L_{l o c}^{1} (R^{d})$ such that there exists $N \geq 0$ with

\int_{| x | < R} | f (x) | d x = O (R^{N}), as R \to \infty

The distribution $Λ_{f}$ is tempered.

Example 4.22

This condition holds for functions in $L^{p} (R^{d})$ for $p \in [1, \infty]$

Lemma 4.23

If $F \in D^{'}$ has compact support then it is tempered: Choose $η \in D$ such that $η |_{U} = 1$ on some neighborhood $U \supset Supp (F)$ . Then $F (η ϕ) = F (ϕ)$ for all $ϕ \in D$ and $ϕ \mapsto F (η ϕ)$ defines a continuous functional on $S$ . It does not depend on the choice of $η$ , because given another such $η^{'}$ and any $ϕ \in S$ we have $Supp ((η - η^{'}) \cdot ϕ) \cap Supp (F) = \emptyset$ .

Let $F$ denote a tempered distribution.

Lemma 4.24

All $\partial^{α} F$ are tempered.
Let $ψ \in C^{\infty}$ be slowly increasing, i.e. for each $α$ exists $N_{α}$ such that $\partial_{x}^{α} ψ (x) = O (| x |^{N_{α}})$ . Then $ψ F$ , defined by $(ψ F) (ϕ) (F (ψ ϕ)$ is tempered.

Example 4.25

If $ψ \in S$ , then $F (τ_{∙} (ψ^{R}))$ is slowly increasing. The other formulation is still valid because $S$ is stable under $*$ .

What is the point of the Schwartz space?

Definition 4.26

The fourier transformation is a continuous bijection

\begin{aligned} S & \to S \\ ϕ & \mapsto \hat{ϕ} = (ξ \mapsto \int_{R^{d}} ϕ (x) e^{- 2 π i x ξ} d x) \end{aligned}

Lemma 4.27

We have

Λ_{\hat{ψ}} (ϕ) = \int_{R^{d}} \hat{ψ} (x) ϕ (x) d x = \int_{R^{d}} ψ (x) \hat{ϕ} (x) d x = Λ_{ψ} (\hat{ϕ})

So its easy to define a compatible generalization to the tempered distributions:

Definition 4.28

Define

\hat{F} (ϕ) = F (\hat{ϕ})

and similarly for the inverse transform $f \mapsto \overset{ˇ}{f}$ .

We automatically have the inversion theorem for distributions, because it holds for test functions.

Example 4.29

If $1 \in S$ denotes the constant function at $1$ , then

\hat{δ} = Λ_{1}

because

\hat{δ} (ϕ) = \hat{ϕ} (0) = \int 1 ϕ (x) d x = Λ_{1} (ϕ)

4.2 Fundamental solutions

In this chapter we may omit the $Λ$ . Fix a partial differential operator $L$

L = \sum_{| α | \leq m} a_{α} \partial^{α} on R^{d}

with $a_{α} \in C$ .

Definition 4.30

A fundamental solution of $L$ is a distribution $F$ such that $L (F) = δ$

The reason why this is interesting:

Lemma 4.31

The operator

f \mapsto T (f) := F * f

defines an inverse to $L$

Proof ▶

because

\partial^{α} (F * f) = (\partial^{α} F) * f = F * (\partial^{α} f)

Summing over $α$ gives

L (F * f) = δ * f = F * L f

Now recall $δ * f = f$ .

Definition 4.32

The characteristic polynomial of $L$ is

P (ξ) = \sum_{| α | \leq m} a_{α} (2 π i ξ)^{α}

It is defined it such a way, that $\hat{L f} = P \cdot \hat{f}$ . So our hope is

F := \widecheck 1 / (P (ξ)) = \int \frac{1}{P (ξ)} e^{2 π i x ξ} d ξ

The problem is, that the zeros of $P$ result in difficulties to define $F$ , even as a distribution.

4.2.1 Laplacian

If $L = Δ = \sum_{i = 1}^{d} \frac{\partial^{2}}{\partial^{2} x_{i}}$ So $1 / P (ξ) = 1 / (- 4 π^{2} | ξ |^{2})$ which lies in $L_{l o c}^{1}$ for $d \geq 3$ . To calculate our distribution the following is helpful

Theorem 4.33

For $λ > - d$ , Let $H_{λ}$ be the tempered distribution associated to $| x |^{λ} \in L_{l o c}^{1}$ .
If $- d < λ < 0$ then

\hat{H_{λ}} = c_{λ} H_{- d - λ}

with

c_{λ} = \frac{Γ ((d + λ) / 2)}{Γ (λ / 2)} π^{- d / 2 - λ}

Set $λ := - d + 2$ . Hence we can find an appropriate constant $C_{d}$ such that $F (x) = C_{d} | x |^{- d + 2}$ is a fundamental solution, because $\hat{F} (ξ) = 1 / (- 4 π^{2} | ξ |^{2})$ . So writing out $\hat{Δ F} = 1$ .

Proposition 4.34

If $d = 2$ , the function $F := 1 / (2 π) \log | x | \in L_{l o c}^{1}$ is a fundamental solution of $Δ$

Proof ▶

Sketch. One can actually compute $\hat{F} = - 1 / (4 π^{2}) [\frac{1}{| x |^{2}}] - c^{'} δ$ for some constant $c^{'}$ , where $[\frac{1}{| x |^{2}}]$ is a distribution, that replaces the (non locally integrable) function $1 / | x |^{2}$ in an appropriate way:

[\frac{1}{| x |^{2}}] = \int_{| x | \leq 1} \frac{ϕ (x) - ϕ (0)}{| x |^{2}} d x + \int_{| x | > 1} \frac{ϕ (x)}{| x |^{2}} d x

Notice, that on the complement of zero, this distribution coincides with $1 / | x |^{2}$ .

Notation 2

If $ϕ \in C^{\infty} (Ω)$ slowly increasing (i.e. all derivatives are bounded by polynomials), define $ϕ F \in S^{'}$ as $ϕ F (ψ) = F (ϕ ψ)$ .

\begin{aligned} \hat{Δ F} & = - 4 π^{2} | x |^{2} \hat{F} \\ = | x |^{2} [\frac{1}{| x |^{2}}] - 4 π^{2} c^{'} \underset{= 0}{\underset{⏟}{| x |^{2} δ}} \\ = 1 \end{aligned}

4.2.2 Heat operator

$L = \frac{\partial}{\partial t} - Δ_{x}$ taken over $(x, t) \in R^{d + 1} = R^{d} \times R$ i.e. we want to solve the homogeneous initial value problem

{\begin{cases} L (u) = 0 & , t > 0 \\ u (x, 0) = f (x) & , t = 0 \end{cases}

for some initial value $f \in S$ .
We have

(\frac{\partial}{\partial t} \hat{H_{t}}) (ξ) = \hat{\frac{\partial}{\partial t} H_{t}} (ξ) = \hat{Δ_{x} H_{t}} (ξ) = - 4 π^{2} | ξ |^{2} \hat{H_{t}} (ξ)

and this is obviously solved by $H_{t} = e^{- 4 π^{2} | ξ |^{2} t}$ . We may call this the heat kernel

\hat{H_{t}} (ξ) = e^{- 4 π^{2} | ξ |^{2} t}

Note that for $t = 0$ , we have $\hat{H_{0}} = 1$ , hence $H_{0} = δ$ , so $u (x, t) = (H * f) (x)$ solves the equation $L (u) = 0$ and $u (x, t) \to f (x)$ in $S$ as $t \to 0$ .

Remark 4.35

$H_{t} \to δ$ in $S^{'}$ as $t \to 0$ and $\int_{R^{d}} H_{t} (x) d x = 1$ for all $t$

Now define

F (x, t) := {\begin{cases} H_{t} (x) & , i f t > 0, \\ 0 & , t \leq 0 \end{cases}

$F$ is locally integrable on $R^{d + 1}$ and it actually holds

\int_{| t | \leq R} \int_{R^{d}} F (x, t) d x d t \leq R

so $F$ defines a tempered distribution by 4.21.

Theorem 4.36

$F$ is a fundamental solution of $L = \frac{\partial}{\partial t} - Δ_{x}$ .

Proof ▶

Denote $L^{'} = - \frac{\partial}{\partial t} - Δ_{x}$ , then we have to see the last equation

L F (ϕ) = F (L^{'} (ϕ)) = lim_{ε \to 0} \int_{t \geq ε} \int_{R^{d}} F (x, t) (- \frac{\partial}{\partial t} - Δ_{x}) ϕ (x, t) d x d t \overset{!}{=} δ (ϕ)

Integration by parts

\begin{aligned} \int_{t \geq ε} \int_{R^{d}} F (x, t) (- \frac{\partial}{\partial t} - Δ_{x}) ϕ (x, t) d x d t \\ = - \int_{R^{d}} (\int_{t \geq ε} H_{t} \frac{\partial}{\partial t} + (Δ_{x} H_{t}) ϕ d t) d x \\ = - \int_{R^{d}} (\int_{t \geq ε} H_{t} \frac{\partial}{\partial t} + (\frac{\partial}{\partial t} H_{t}) ϕ d t) d x \\ = \int_{R^{d}} H_{ε} (x) ϕ (x, ε) d x & ∣ | ϕ (x, ε) - ϕ (x, 0) | \leq O (ε) uniformly in x \\ = \int_{R^{d}} H_{ε} (x) (ϕ (x, 0) + O (ε)) d x & ??? \\ \to ϕ (0, 0) \end{aligned}

where in the last line we let $ε \to 0$ □