Tempered distributions

A good first reference for distributions is [2], [4] gives a more exhaustive treatment.

The complete metric topology on $\mathcal{S}(\mathbb{R}^n)$ is described above. Next I want to try to convice you that elements of its dual space $\mathcal{S}'(\mathbb{R}^n),$ have enough of the properties of functions that we can work with them as `generalized functions'.

First let me develop some notation. A differentiable function $\varphi : \mathbb{R}^n \to \mathbb{C}$ has partial derivatives which we have denoted $\partial \varphi /\partial x_j : \mathbb{R}^n \to \mathbb{C}$. For reasons that will become clear later, we put a $\sqrt{-1}$ into the definition and write

$$D_j \varphi = \frac{1}{i} \frac{\partial \varphi}{\partial x_j} .$$ (67)

We say $\varphi$ is once continuously differentiable if each of these $D_j \varphi$ is continuous. Then we defined $k$ times continuous differentiability inductively by saying that $\varphi$ and the $D_j \varphi$ are $(k-1)$-times continuously differentiable. For $k=2$ this means that

$$D_j D_k \varphi \hbox{ are continuous for } j,k=1, \cdots , n .$$

Now, recall that, if continuous, these second derivatives are symmetric:

$$D_j D_k \varphi = D_k D_j \varphi .$$ (68)

This means we can use a compact notation for higher derivatives. Put $\mathbb{N}_0 = \left\{ 0,1, \ldots \right\}$; we call an element $\alpha \in \mathbb{N}^n_0$ a `multi-index' and if $\varphi$ is at least $k$ times continuously differentiable, we set¹²

$${D^{\alpha}} \varphi = \frac{1}{i^{\left\vert \alpha \right\vert}... ...alpha_1}}{\partial x_1} \cdots \frac{\partial^{\alpha_n}}{\partial x_n} \varphi \left\vert \alpha \right\vert = \alpha_1 + \alpha_2 + \cdots + \alpha_n \leq k.$$ (69)

Now we have defined the spaces.

$$\mathcal{C}^k_0 (\mathbb{R}^n) = \left\{ \varphi : \mathbb{R}^n \... ..._0 (\mathbb{R}^n) \ \forall\ \left\vert \alpha \right\vert \leq k \right\} \, .$$ (70)

Notice the convention is that ${D^{\alpha}} \varphi$ is asserted to exist if it is required to be continuous! Using $\langle x \rangle = (1 + \left\vert x \right\vert^2)$ we defined

$$\langle x \rangle^{-k} \mathcal{C}^k_0 (\mathbb{R}^n) = \left\{ \... ...\, \langle x \rangle^k \varphi \in \mathcal{C}^k_0 (\mathbb{R}^n) \right\} \, ,$$ (71)

and then our space of test functions is

$$\mathcal{S} (\mathbb{R}^n) = \bigcap_k \langle x \rangle^{-k} \mathcal{C}^k_0 (\mathbb{R}^n) \, .$$

Thus,

$$\varphi \in \mathcal{S} (\mathbb{R}^n) \Leftrightarrow {D^{\alpha... ...b{R}^n) \ \forall\ \left\vert \alpha \right\vert \leq k \hbox{ and all } k \, .$$ (72)

Proof. We first check that

$$\varphi \in \mathcal{C}^0_0 (\mathbb{R}^n) \, , \ D_j (\langle x \rangle \varphi) \in \mathcal{C}^0_0 (\mathbb{R}^n) \, , \, j=1 , \cdots , n$$

$$\Leftrightarrow \varphi \in \mathcal{C}^0_0 (\mathbb{R}^n) \, , \... ...le D_j \varphi \in \mathcal{C}^0_0 (\mathbb{R}^n) \, , \, j=1 , \cdots , n \, .$$

Since

$$D_j \langle x \rangle \varphi = \langle x \rangle D_j \varphi + (D_j \langle x \rangle) \varphi$$

and $D_j \langle x \rangle = \frac{1}{i} x_j \langle x \rangle^{-1}$ is a bounded continuous function, this is clear. Then consider the same thing for a larger $k$:

$${D^{\alpha}} \langle x \rangle^p \varphi \in \mathcal{C}^0_0 (\mathbb{R}^n) \ \forall\ \left\vert \alpha \right\vert =p \, , \ 0 \leq p \leq k$$ (73)

$$\Leftrightarrow \langle x \rangle^p {D^{\alpha}} \varphi \in \mat... ...b{R}^n) \ \forall\ \left\vert \alpha \right\vert =p \, , \ 0 \leq p \leq k \, .$$

$\qedsymbol$

I leave you to check this as Problem 6.1.

Proof. Any reasonable proof of (6.2) shows that the norms

$$\Vert \langle x \rangle^k \varphi \Vert _{\mathcal{C}^k} \hbox{ a... ...ight\vert \leq k} \Vert \langle x \rangle^{k} D^{\beta} \varphi \Vert _{\infty}$$

are equivalent. Since there are positive constants such that

$$C_1 \left( 1+ \sum_{\left\vert \alpha \right\vert \leq k} \left\v... ...\left\vert \alpha \right\vert \leq k} \left\vert x^{\alpha} \right\vert \right)$$

the equivalent of the norms follows.

$\qedsymbol$

Proof. This is just the equivalence of the norms, since we showed that $u \in \mathcal{S}' (\mathbb{R}^n)$ if and only if

$$\left\vert u (\varphi) \right\vert \leq C \Vert \langle x \rangle^k \varphi \Vert _{\mathcal{C}^k}$$

for some $k$.

$\qedsymbol$

Proof. This is Problem 6.2. $\qedsymbol$

All this messing about with norms shows that

$$x_j : \mathcal{S} (\mathbb{R}^n) \to \mathcal{S} (\mathbb{R}^n) \hbox{ and } D_j : \mathcal{S} (\mathbb{R}^n) \to \mathcal{S} (\mathbb{R}^n)$$

are continuous.

So now we have some idea of what $u \in \mathcal{S}' (\mathbb{R}^n)$ means. Let's notice that $u \in \mathcal{S}' (\mathbb{R}^n)$ implies

$$x_j u \in \mathcal{S}' (\mathbb{R}^n) \forall j=1 , \cdots , n$$ (74)

$$D_j u \in \mathcal{S}' (\mathbb{R}^n) \forall j=1 , \cdots , n$$ (75)

$$\varphi u \in \mathcal{S}' (\mathbb{R}^n) \forall \varphi \in \mathcal{S} (\mathbb{R}^n)$$ (76)

where we have to define these things in a reasonable way. Remember that $u \in \mathcal{S}' (\mathbb{R}^n)$ is ``supposed'' to be like an integral against a ``generalized function''

$$u(\psi) = \int_{\mathbb{R}^n} u(x) \psi (x) dx \forall \psi \in \mathcal{S} (\mathbb{R}^n).$$ (77)

Since it would be true if $u$ were a function we define

$$x_ju(\psi) = u (x_j\psi ) \forall \psi \in \mathcal{S} (\mathbb{R}^n).$$ (78)

Then we check that $x_ju \in \mathcal{S}'(\mathbb{R}^n)$:

$$\left\vert x_ju(\psi) \right\vert = \left\vert u(x_j\psi ) \right\vert$$

$$\leq C\sum_{\vert\alpha \vert \leq k, \vert\beta\vert\leq k} \sup_{\mathbb{R}^n} \left\vert x^{\alpha} D^{\beta} (x_j\psi) \right\vert$$

$$\leq C'\sum_{\vert\alpha \vert \leq k+1, \vert\beta\vert\leq k} \sup_{\mathbb{R}^n} \left\vert x^{\alpha} D^{\beta} \psi \right\vert .$$

Similarly we can define the partial derivatives by using the standard integration by parts formula

$$\int_{\mathbb{R}^n} (D_j u) (x) \varphi (x) dx = - \int_{\mathbb{R}^n} u(x) (D_j \varphi (x)) dx$$ (79)

if $u\in\mathcal{C}^1_0(\mathbb{R}^n).$ Thus if $u \in \mathcal{S}' (\mathbb{R}^n)$ again we define

$$D_j u (\psi ) = - u(D_j \psi ) \forall \psi \in \mathcal{S} (\mathbb{R}^n).$$

Then it is clear that $D_ju \in \mathcal{S}' (\mathbb{R}^n).$

Iterating these definition we find that ${D^{\alpha}}$, for any multi-index $\alpha$, defines a linear map

$${D^{\alpha}} : \mathcal{S}' (\mathbb{R}^n) \to \mathcal{S}' (\mathbb{R}^n) .$$ (80)

In general a linear differential operator with constant coefficients is a sum of such ``monomials''. For example Laplace's operator is

$$\Delta =- \frac{\partial^2}{\partial x^2_1} - \frac{\partial^2}{\... ...cdots - \frac{\partial^2}{\partial x^2_n} = D^2_1 + D^2_2 + \cdots + D^2_n .$$

We will be interested in trying to solve differential equations such as

$$\Delta u = f \in \mathcal{S}' (\mathbb{R}^n) .$$

We can also multiply $u \in \mathcal{S}' (\mathbb{R}^n)$ by $\varphi \in \mathcal{S}(\mathbb{R}^n)$, simply defining

$$\varphi u(\psi)=u(\varphi \psi)\ \forall\ \psi\in\mathcal{S}(\mathbb{R}^n).$$ (81)

For this to make sense it suffices to check that

$$\sum_{\overset{\left\vert \alpha \right\vert \leq k ,}{\left\vert... ... \leq k}} \sup_{\mathbb{R}^n} \left\vert x^{\alpha} D^{\beta} \psi \right\vert.$$ (82)

This follows easily from Leibniz' formula.

Now, to start thinking of $u \in \mathcal{S}' (\mathbb{R}^n)$ as a generalized function we first define its support. Recall that

$$\operatorname{supp}(\psi)=\operatorname{clos}\left\{x\in\mathbb{R}^n; \psi (x) \neq 0 \right\}.$$ (83)

We can write this in another `weak' way which is easier to generalize. Namely

$$p \notin \operatorname{supp} (u) \Leftrightarrow \exists \varph... ... \mathcal{S}(\mathbb{R}^n) , \varphi (p) \neq 0 , \varphi u =0 .$$ (84)

In fact this definition makes sense for any $u\in\mathcal{S}'(\mathbb{R}^n).$

Proof. The set defined by (6.19) is closed, since

$$\operatorname{supp}(u)^\complement = \left\{ p \in \mathbb{R}^n;\... ...varphi \in \mathcal{S} (\mathbb{R}^n), \varphi(p)\neq 0, \varphi u=0 \right\}$$ (85)

is clearly open -- the same $\varphi$ works for nearby points. If $\psi \in\mathcal{S}(\mathbb{R}^n)$ we define $u_{\psi} \in \mathcal{S}' (\mathbb{R}^n),$ which we will again identify with $\psi$, by

$$u_{\psi} (\varphi) = \int \varphi (x) \psi (x) dx .$$ (86)

Obviously $u_{\psi}=0 \Longrightarrow \psi =0$, simply set $\varphi = \overline{\psi}$ in (6.21). Thus the map

$$\mathcal{S} (\mathbb{R}^n) \ni \psi \longmapsto u_{\psi}\in \mathcal{S}' (\mathbb{R}^n)$$ (87)

is injective. We want to show that

$$\operatorname{supp} (u_{\psi}) = \operatorname{supp} (\psi)$$ (88)

on the left given by (6.19) and on the right by (6.18). We show first that

$$\operatorname{supp}(u_{\psi}) \subset \operatorname{supp} (\psi ).$$

Thus, we need to see that $p \notin \operatorname{supp} (\psi) \Rightarrow p \notin \operatorname{supp} (u_{\psi}).$ The first condition is that $\psi (x) =0$ in a neighbourhood, $U$ of $p$, hence there is a $\mathcal{C}^\infty$ function $\varphi$ with support in $U$ and $\varphi (p) \neq 0.$ Then $\varphi \psi \equiv 0$. Conversely suppose $p \notin \operatorname{supp} (u_{\psi}).$ Then there exists $\varphi \in \mathcal{S}(\mathbb{R}^n)$ with $\varphi (p) \neq 0$ and $\varphi u_{\psi} =0$, i.e., $\varphi u_{\psi} (\eta ) =0 \forall \eta \in \mathcal{S} (\mathbb{R}^n).$ By the injectivity of $\mathcal{S} (\mathbb{R}^n) \hookrightarrow \mathcal{S}' (\mathbb{R}^n)$ this means $\varphi \psi =0$, so $\psi \equiv 0$ in a neighborhood of $p$ and $p \notin \operatorname{supp} (\psi ).$ $\qedsymbol$

Consider the simplest examples of distribution which are not functions, namely those with support at a given point $p.$ The obvious one is the Dirac delta `function'

$$\delta_p (\varphi ) = \varphi (p) \forall \varphi \in \mathcal{S} (\mathbb{R}^n) .$$ (89)

We can make many more, because ${D^{\alpha}}$ is local

$$\operatorname{supp} ({D^{\alpha}} u) \subset \operatorname{supp} (u) \forall u \in \mathcal{S}' (\mathbb{R}^n) .$$ (90)

Indeed, $p \notin \operatorname{supp} (u) \Rightarrow \exists \varphi \in \mathcal{S} (\mathbb{R}^n),$ $\varphi u \equiv 0,$ $\varphi (p) \neq 0.$ Thus each of the distributions $D^\alpha \delta_p$ also has support contained in $\{p\}.$ In fact none of them vanish, and they are all linearly independent.