Apologies for the embarrassingly bad pun in the title.

Distributions in general

First, it’s necessary to talk about distributions on an arbitrary open set ${\Omega \subset \mathbb{R}^n}$, which are not necessarily tempered. In particular, they may “grow arbitrarily” as one approaches the boundary. So, instead of requiring a functional on a Schwarz space, we consider functionals on ${C_0^{\infty}(\Omega),}$ the space of smooth functions compactly supported in ${\Omega}$. However, we need some notion of continuity, which would require a topology on ${C_0^{\infty}(\Omega)}$. There is now the tricky question of how we would require completeness of the topological vector space ${C_0^{\infty}(\Omega)}$, which we of course desire. We can get such a topology by talking about “strict inductive limits” and whatnot, but since I don’t really find that particularly fun, I’ll sidestep it (but not really—most of the ideas will still remain).

Anyway, the idea here will be to consider auxiliary spaces ${C^{\infty}(K)}$ for ${K \subset \Omega}$ compact. This is the space of smooth functions ${f: \Omega \rightarrow \mathbb{R}}$ which are supported in ${K}$. We give the space a Frechet topology by the family of seminorms

$\displaystyle ||f||_a := \sup_K |D^a f|.$

Now, any functional ${\phi: C_0^{\infty}(\Omega) \rightarrow \mathbb{R}}$ induces functionals on the spaces ${C^{\infty}(K)}$. We say that ${\phi}$ is a distribution if all these restrictions are continuous, whenever ${K \subset \Omega}$ is compact. The space of distributions is denoted ${\mathcal{D}'(\Omega)}$. For instance, any tempered distribution on ${\mathbb{R}^n}$ is a distribution. (Note incidentally that if two tempered distributions induce the same (untempered) distribution, they are also equal as tempered distributions. This is because ${C_0^{\infty}(\mathbb{R}^n)}$ is dense in ${\mathcal{S}}$ in the Schwarz topology.) More interestingly, any locally integrable function or measure is a distribution.

As before, we can multiply a distribution by a function, and we can differentiate a distribution. We can also talk about convolutions of a distribution with a compactly supported distribution or with afunction, and these satisfy all the usual identities. (I’m not inclined to write out all the details, though.) So it is still interesting to talk about fundamental solutions to PDE which may not be tempered.

Malgrange-Ehrenpreis

So, let’s fix a constant-coefficient differential operator

$\displaystyle P(D) = \sum_{a: |a| \leq k} C_a D^a.$

The big theorem here is:

Theorem 1 (Malgrange-Ehrenpreis)

There is a distribution ${\phi \in \mathcal{D}(\mathbb{R}^n)}$ with$\displaystyle P(D) \phi = \delta,$
i.e. a fundamental solution.

The precise statement is proved in the book by Hormander (Linear Partial Differential Operators), where he proves in fact that the fundamental solution belongs locally to a generalization of the Sobolev spaces that we shall meet presently. However, I’m not inclined to write out all the details right now, so I’ll prove the following weaker statement:

Theorem 2 Given ${\Omega \subset \mathbb{R}^n}$ bounded, there is a ${\phi \in \mathcal{D}(\mathbb{R}^n)}$ with ${P(D) \phi = \delta}$.

Note that since convolution still works with arbitrary distributions and compactly supported functions, we get the previous theorem about local solvability of ${P(D) f = g}$ for ${f,g}$ smooth. And more.

This is the key lemma:

Lemma 3

Let ${K \subset \mathbb{R}^n}$. Then there exist ${N,M}$ such that$\displaystyle |f(0)| \leq M \sum_{a:|a| \leq N} ||P(D)f||_a$
for ${f \in C^{\infty}(K)}$ and the seminorms defined as before.

What does this mean? The map ${P(D) f \rightarrow f(0)}$ thus defines a map continuous on the linear subspace of ${C^{\infty}(K)}$ spanned by ${P(D)f}$. By the Hahn-Banach theorem, we can extend it continuously to all of ${C^{\infty}(K)}$. Now, if ${\Omega \subset \mathbb{R}^n}$ is bounded, then we get such a linear form on ${C^{\infty}(\overline{\Omega})}$, which induces a distribution on ${\Omega}$ that solves the constant-coefficient equation. So we just have to prove the lemma.

Polynomials

To get the estimates, we will use

$\displaystyle f(0) = \int_{\mathbb{R}^n} \hat{f}(x) dx ,$

together with ${P \cdot\hat{f} = \widehat{P(D)f}}$ to get bounds with ${P}$ as in the “symbol” of the differential operator. So we need losely to get estimates of the form “if something times a polynomial has small integrals, so does that something.” This is what the next few lemmas from complex analysis do.

Lemma 4

Let ${u}$ be analytic in a neighborhood of the closed unit disk, and ${P}$ be a one-variable polynomial with leading coefficient ${A}$. Then$\displaystyle | A u(0)| \leq \frac{1}{2\pi} \int_0^{2 \pi} | P(e^{it}) u(e^{it})| dt .$

This looks a lot like Cauchy’s formula, and indeed it reduces to that if we replace the polynomial ${P}$ (of degree ${m}$, say) with ${Q(z) := z^{-m} \bar{P}(z)}$ where ${\bar{P}}$ denotes that all the coefficients have been complex conjugated. Since ${|Q(z)| = |P(z)|}$ for ${|z| = 1}$ and ${Q(0)=1}$, the lemma now follows from the Cauchy formula.

However, we have the problem that polynomials in general have roots, which will cause some problems in our estimates. All the same, if ${P}$ is a polynomial in many variables, then

$\displaystyle \tilde{P} := \sum_a |P^{(a)}|$

is bounded below and polynomially at ${\infty}$.

Lemma 5

Let ${P}$ be a one-variable polynomial of degree ${m}$, ${u}$ as before. Then there is a constant ${C_m}$ such that for any ${r}$, we have$\displaystyle | u(0) P^{(r)}(0)| \leq C_m \int_0^{2 \pi} | P(e^{it}) u(e^{it})| dt.$

We can of course assume that we have an expression ${P(z) = \prod (z-z_i)}$. Then ${P^{(r)}(0)}$ is a linear combination of products ${z_{i_1} \dots z_{i_r}}$ with coefficients depending only on ${m}$. In particular, if we can show

$\displaystyle | u(0) z_{i_1} \dots z_{i_r} | \leq \frac{1}{2\pi} \int_0^{2 \pi} | P(e^{it}) u(e^{it})| dt$

we will be done. This follows from observing that we may consider the pair

$\displaystyle v(z) = u(z) (z - z_{i_1}) \dots (z-z_{i_r}), \ \ Q(z) = \prod_{i \neq i_1, \dots, i_r} (z-z_i).$

If we now apply the previous lemma to this pair, we find the claim.

Next time, we’re going to use these lemmas to get the appropriate bounds necessary to prove Theorem 2.