Convolution and density

We have defined an inclusion map

$$\mathcal{S}(\mathbb{R}^n) \ni \varphi \longmapsto u_{\varphi} \in... ...R}^n} \varphi (x) \psi (x) dx \forall \psi \in \mathcal{S} (\mathbb{R}^n).$$ (91)

This allows us to `think of' $\mathcal{S}(\mathbb{R}^n)$ as a subspace of $\mathcal{S}'(\mathbb{R}^n)$; that is we habitually identify $u_{\varphi}$ with $\varphi.$ We can do this because we know (7.1) to be injective. We can extend the map (7.1) to include bigger spaces

$$\begin{aligned}\mathcal{C}^0_0 (\mathbb{R}^n) &\ni \varphi \l... ...} (\psi) = \int_{\mathbb{R}^n} \psi d \mu , \end{aligned}$$

but we need to know that these maps are injective before we can forget about them.

We can see this using convolution. This is a sort of `product' of functions. To begin with, suppose $v \in \mathcal{C}^0_0 (\mathbb{R}^n)$ and $\psi \in\mathcal{S}(\mathbb{R}^n)$. We define a new function by `averaging $v$ with respect to $\psi$:'

$$v * \psi (x) = \int_{\mathbb{R}^n} v (x-y) \psi (y) dy .$$ (93)

The integral converges by dominated convergence, namely $\psi(y)$ is integrable and $v$ is bounded,

$$\left\vert v (x-y) \psi (y) \right\vert \leq \Vert v \Vert _{\mathcal{C}^0_0} \, \left\vert \psi (y) \right\vert \, .$$

We can use the same sort of estimates to show that $v * \psi$ is continuous. Fix $x \in \mathbb{R}^n$,

$$\begin{multline} v* \psi (x+x') - v * \psi (x) \\ = \int (v (x+ x' -y) - v (x-y)) \psi (y) dy . \end{multline}$$

To see that this is small for $x'$ small, we split the integral into two pieces. Since $\psi$ is very small near infinity, given $\epsilon >0$ we can choose $R$ so large that

$$\Vert v \Vert _{\infty} \cdot \int_{\left\vert y ]\right\vert \geq R} \left\vert \psi (y) \right\vert dy \leq \epsilon /4 .$$ (94)

The set $\left\vert y \right\vert \leq R$ is compact and if $\left\vert x \right\vert \leq R'$, $\left\vert x' \right\vert \leq 1$ then $\left\vert x+x' -y \right\vert \leq R+R'+1$. A continuous function is uniformly continuous on any compact set, so we can chose $\delta > 0$ such that

$$\sup_{\overset{\left\vert x' \right\vert < \delta}{\left\vert y \... ...\right\vert \leq R} \left\vert \psi (y) \right\vert dy < \epsilon /2 .$$ (95)

Combining (7.5) and (7.6) we conclude that $v * \psi$ is continuous. Finally, we conclude that

$$v \in \mathcal{C}^0_0 (\mathbb{R}^n) \Rightarrow v* \psi \in \mathcal{C}^0_0 (\mathbb{R}^n) \, .$$ (96)

For this we need to show that $v * \psi$ is small at infinity, which follows from the fact that $v$ is small at infinity. Namely given $\epsilon >0$ there exists $R>0$ such that $\vert v(y)\vert\le\epsilon$ if $\vert y\vert\ge R.$ Divide the integral defining the convolution into two

$$\begin{multline*} \vert v*\psi (x)\vert\le\int_{\vert y\vert>R} u(y)\psi (x-y)dy... ... \Vert _\infty+\Vert u\Vert _\infty \sup_{B(x,R)}\vert\psi\vert. \end{multline*}$$

Since $\psi \in\mathcal{S}(\bbR^n)$ the last constant tends to 0 as $\vert x\vert\to\infty.$

We can do much better than this! Assuming $\left\vert x' \right\vert \leq 1$ we can use Taylor's formula with remainder to write

$$\psi (z+x')- \psi (z) = \int'_0 \frac{d}{dt} \psi (z+tx') dt= \sum^n_{j=1} x_j \cdot \tilde{\psi}_j (z,x') .$$ (97)

As Problem 23 I ask you to check carefully that

$$\psi_j (z;x') \in \mathcal{S} (\mathbb{R}^n) \hbox{ depends continuously on } x' \left\vert x' \right\vert \leq 1 .$$ (98)

Going back to (7.3))we can use the translation and reflection-invariance of Lebesgue measure to rewrite the integral (by changing variable) as

$$v* \psi (x) = \int_{\mathbb{R}^n} v(y) \psi (x-y) dy .$$ (99)

This reverses the role of $v$ and $\psi$ and shows that if both $v$ and $\psi$ are in $\mathcal{S}(\mathbb{R}^n)$ then $v * \psi = \psi * v$.

Using this formula on (7.4) we find

$$\begin{multline} v* \psi (x+x') - v* \psi (x) = \int v(y) (\psi (x+x'-y)- \psi ... ...}_j (x-y,x') dy = \sum^n_{j=1} x_j (v*\psi_j(\cdot;x')(x) . \end{multline}$$

From (7.9) and what we have already shown, $v*\psi(\cdot;x')$ is continuous in both variables, and is in $\mathcal{C}^0_0 (\mathbb{R}^n)$ in the first. Thus

$$v \in \mathcal{C}^0_0 (\mathbb{R}^n) \, , \, \psi \in \mathcal{S} (\mathbb{R}^n) \Rightarrow v* \psi \in \mathcal{C}^1_0 (\mathbb{R}^n) \, .$$ (100)

In fact we also see that

$$\frac{\partial}{\partial x_j} v* \psi = v* \frac{\partial \psi}{\partial x_j} .$$ (101)

Thus $v * \psi$ inherits its regularity from $\psi$.

Proof. This follows from (7.12), (7.13) and induction. $\qedsymbol$

Now, let us make a more special choice of $\psi.$ We have shown the existence of

$$\varphi \in \mathcal{C}^\infty_c (\mathbb{R}^n) , \varphi \... ...e{supp} (\varphi) \subset \left\{ \left\vert x \right\vert \leq 1 \right\} .$$ (102)

We can also assume $\int_{\mathbb{R}^n} \varphi dx =1$, by multiplying by a positive constant. Now consider

$$\varphi_t (x) = t^{-n} \varphi \left( \frac{x}{t} \right) 1 \geq t>0 .$$ (103)

This has all the same properties, except that

$$\operatorname{supp} \varphi_t \subset \left\{ \left\vert x \right\vert \leq t \right\} , \int \varphi_t dx =1 .$$ (104)

Proof. using (7.17) we can write the difference as

$$\begin{multline} \vert v_t(x)-v(x)\vert= \vert\int_{\mathbb{R}^n} (v(x-y)-v(x)) ... ...e \sup_{\vert y\vert\le t}\left\vert v(x-y)-v(x)\right\vert\to0. \end{multline}$$

Here we have used the fact that $\varphi_t\ge0$ has support in $\vert y\vert\le t$ and has integral $1.$ Thus $v_t \to v$ uniformly on any set on which $v$ is uniformly continuous, namel $\mathbb{R}^n$! $\qedsymbol$

Proof. Take $k=0$ first. The subspace $\mathcal{C}^0_c (\mathbb{R}^n)$ is dense in $\mathcal{C}^0_0 (\mathbb{R}^n)$, by cutting off outside a large ball. If $v \in\mathcal{C}^0_c(\mathbb{R}^n)$ has support in $\left\{ \left\vert x \right\vert \leq R\right\}$ then

$$v* \varphi_t \in \mathcal{C}^\infty_c (\mathbb{R}^n) \subset \mathcal{S}(\mathbb{R}^n)$$

has support in $\left\{ \left\vert x \right\vert \leq R+1 \right\}$. Since $v* \varphi_t \to v$ the result follows for $k=0$.

For $k \geq 1$ the same argument works, since ${D^{\alpha}} (v* \varphi_t ) = ({D^{\alpha}} V) * \varphi_t$. $\qedsymbol$

Now, we want the same result for $L^2 (\mathbb{R}^n)$ (and maybe for $L^p (\mathbb{R}^n)$, $1 \leq p< \infty$). I leave the measure-theoretic part of the argument to you.

This is Problem 71.

Using this we conclude that

$$\mathcal{S} (\mathbb{R}^n) \hookrightarrow L^2 (\mathbb{R}^n) \hbox{ is dense}$$ (105)

as before. First observe that the space of $L^2$ functions of compact support is dense in $L^2 (\mathbb{R}^n)$, since

$$\lim_{R \to \infty} \int_{\left\vert x \right\vert \geq R} \left\vert u (x) \right\vert^2 dx = 0 \forall u \in L^2 (\mathbb{R}^n) .$$

Then look back at the discussion of $v* \varphi$, now $v$ is replaced by $u \in L^2_c (\mathbb{R}^n).$ The compactness of the support means that $u\in L^1(\bbR^n)$ so in

$$u*\varphi(x)=\int_{\bbR^n}u(x-y)\varphi(y)dy$$ (106)

the integral is absolutely convergent. Moreover

$$\begin{multline*} \left\vert u* \varphi (x+x') - u * \varphi (x) \right\vert \\... ...R} \left\vert \varphi (x+x'-y)- \varphi (x-y) \right\vert \to 0 \end{multline*}$$

when $\left\{ \left\vert x \right\vert \leq R\right\}$ large enough. Thus $u* \varphi$ is continuous and the same argument as before shows that

$$u * \varphi_t \in \mathcal{S} (\mathbb{R}^n) .$$

Now to see that $u * \varphi_t \to u$, assuming $u$ has compact support (or not) we estimate the integral

$$\left\vert u* \varphi_t (x) - u (x)\right\vert = \left\vert \int (u(x-y)-u(x)) \varphi_t (y) dy \right\vert$$

$$\leq \int \left\vert u(x-y)-u(x) \right\vert \varphi_t (y) dy .$$

Using the same argument twice

$$\begin{multline*} \int \left\vert u* \varphi_t (x)-u(x) \right\vert^2 dx \ ... ...rt \leq t} \int \left\vert u (x-y)-u(x) \right\vert^2 dx . \end{multline*}$$

Note that at the second step here I have used Schwarz's inequality with the integrand written as the product

$$\left\vert u(x-y)-u(x) \right\vert \varphi^{1/2}_t (y) \varphi^{1... ...eft\vert u(x-y')-u(x) \right\vert \varphi^{1/2}_t (y) \varphi^{1/2}_t (y') .$$

Thus we now know that

$$L^2 (\mathbb{R}^n) \hookrightarrow \mathcal{S}' (\mathbb{R}^n) \hbox{ is injective.}$$

This means that all our usual spaces of functions `sit inside' $\mathcal{S}' (\mathbb{R}^n).$

Finally we can use convolution with $\varphi_t$ to show the existence of smooth partitions of unity. If $K \Subset U \subset \mathbb{R}^n$ is a compact set in an open set then we have shown the existence of $\xi \in \mathcal{C}^0_c (\mathbb{R}^n)$, with $\xi =1$ in some neighborhood of $K$ and $\xi =1$ in some neighborhood of $K$ and $\operatorname{supp} (\xi) \Subset U$.

Then consider $\xi * \varphi_t$ for $t$ small. In fact

$$\operatorname{supp} (\xi * \varphi_t) \subset \left\{ p \in \mathbb{R}^n ; \operatorname{dist} (p, \operatorname{supp} \xi ) \leq 2t \right\}$$

and similarly, $0 \leq \xi * \varphi_t \leq 1$ and

$$\xi * \varphi_t =1 \hbox{ at } p \hbox{ if } \xi =1 \hbox{ on } B (p, 2t) .$$

Using this we get:

Proof. By the compactness of $K$ we may choose a finite open subcover. Using Lemma 1.8 we may choose a continuous partition, $\phi'_i,$ of unity subordinate to this cover. Using the convolution argument above we can replace $\phi'_i$ by $\phi'_i*\varphi_t$ for $t>0.$ If $t$ is sufficiently small then this is again a partition of unity subordinate to the cover, but now smooth. $\qedsymbol$

Next we can make a simple `cut off argument' to show

Proof. Choose $\varphi \in \mathcal{C}^\infty_c (\mathbb{R}^n)$ with $\varphi (x) =1$ in $\left\vert x \right\vert \leq 1$. Then given $\psi \in\mathcal{S}(\mathbb{R}^n)$ consider the sequence

$$\psi_n (x) = \varphi (x/n) \psi (x) .$$

Clearly $\psi_n = \psi$ on $\left\vert x \right\vert \leq n$, so if it converges in $\mathcal{S}(\mathbb{R}^n)$ it must converge to $\psi$. Suppose $m \geq n$ then by Leibniz's formula¹³

$$\begin{multline*} D^\alpha_x (\psi_n (x) - \psi_m (x) ) \\ = \sum_{\beta \leq... ... (\frac{x}{m}) \right) \cdot D^{\alpha - \beta}_x \psi (x) . \end{multline*}$$

All derivatives of $\varphi (x/n)$ are bounded, independent of $n$ and $\psi_n = \psi_m$ in $\left\vert x \right\vert \leq n$ so for any $p$

$$\left\vert D^\alpha_x (\psi_n (x) - \psi_m (x)) \right\vert \leq ... ...ngle x \rangle^{-2p} & \left\vert x \right\vert \geq n \end{array} \right. .$$

Hence $\psi_n$ is Cauchy in $\mathcal{S}(\mathbb{R}^n)$. $\qedsymbol$

Thus every element of $\mathcal{S}'(\mathbb{R}^n)$ is determined by its restriction to $\mathcal{C}^\infty_c (\mathbb{R}^n)$. The support of a tempered distribution was defined above to be

$$\operatorname{supp} (u) = \left\{ x \in \mathbb{R}^n ; \exists ... ...{R}^n) , \varphi (x) \neq 0 , \varphi u=0 \right\}^\complement .$$ (107)

Using the preceding lemma and the construction of smooth partitions of unity we find

Proof. From (7.23), if $\psi \in \mathcal{S} (\mathbb{R}^n),$ $\operatorname{supp} (\psi u) \subset \operatorname{supp} (u)$. If $x \ni \operatorname{supp} (u)$ then, by definition, $\varphi u =0$ for some $\varphi \in \mathcal{S}(\mathbb{R}^n)$ with $\varphi (x) \neq 0$. Thus $\varphi \neq 0$ on $B (x, \epsilon )$ for $\epsilon >0$ sufficiently small. If $\psi \in \mathcal{C}^\infty_c (\mathbb{R}^n)$ has support in $B (x, \epsilon )$ then $\psi u = \tilde{\psi} \varphi u = 0$, where $\tilde{\psi} \in \mathcal{C}^\infty_c (\mathbb{R}^n)$:

$$\tilde{\psi} = \left\{ \begin{array}{cl} \psi / \varphi & \hbox{in }B(x, \epsilon) \ 0 & \hbox{elsewhere} . \end{array} \right.$$

Thus, given $K \Subset \mathbb{R}^n$ we can find $\varphi_j \in \mathcal{C}^\infty_c (\mathbb{R}^n)$, supported in such balls, so that $\sum_j \varphi_j \equiv 1$ on $K$ but $\varphi_j u =0$. For given $\mu \in \mathcal{C}^\infty_c (\mathbb{R}^n)$ apply this to $\operatorname{supp} (\mu).$ Then

$$\mu = \sum_j \varphi_j \mu \Rightarrow u (\mu) = \sum_j (\phi_ju)(\mu)=0 .$$

Thus $u=0$ on $\mathcal{C}^\infty_c (\mathbb{R}^n)$, so $u=0$. $\qedsymbol$

The linear space of distributions of compact support will be denoted $\mathcal{C}^{-\infty}_c (\mathbb{R}^n)$; it is often written $\mathcal{E}' (\mathbb{R}^n)$.

Now let us give a characterization of the `delta function'

$$\delta ( \varphi ) = \varphi (0) \forall \varphi \in \mathcal{S} (\mathbb{R}^n) ,$$

or at least the one-dimensional subspace of $\mathcal{S}'(\mathbb{R}^n)$ it spans. This is based on the simple observation that $(x_j \varphi ) (0) =0$ if $\varphi \in \mathcal{S}(\mathbb{R}^n)$!

Proof. The main work is in characterizing the null space of $\delta$ as a linear functional, namely in showing that

$$\mathcal{H} = \left\{ \varphi \in \mathcal{S} (\mathbb{R}^n) ; \varphi (0) =0 \right\}$$ (108)

can also be written as

$$\mathcal{H} = \left\{ \varphi \in \mathcal{S} (\mathbb{R}^n) ; ... ..._{j=1} x_j \psi_j , \varphi_j \in \mathcal{S} (\mathbb{R}^n) \right\} .$$ (109)

Clearly the right side of (7.25) is contained in the left. To see the converse, suppose first that

$$\varphi \in \mathcal{S} (\mathbb{R}^n) , \varphi =0 \hbox{ in } \left\vert x \right\vert <1 .$$ (110)

Then define

$$\psi = \left\{ \begin{array}{ll} 0 & \left\vert x \right\vert <1 ... ...vert x \right\vert^2 & \left\vert x \right\vert \geq 1 . \end{array} \right.$$

All the derivatives of $1 / \left\vert x \right\vert^2$ are bounded in $\left\vert x \right\vert \geq 1$, so from Leibniz's formula it follows that $\psi \in\mathcal{S}(\mathbb{R}^n)$. Since

$$\varphi = \sum_j x_j (x_j \psi )$$

this shows that $\varphi$ of the form (7.26) is in the right side of (7.25). In general suppose $\varphi \in \mathcal{S}(\mathbb{R}^n)$. Then

$${\begin{gathered}\varphi (x) - \varphi (0) = \int^t_0 \frac{d}{... ...c{\partial \varphi}{\partial x_j} (tx) dt . \end{gathered}}$$ (111)

Certainly these integrals are $\mathcal{C}^\infty$, but they may not decay rapidly at infinity. However, choose $\mu \in \mathcal{C}^\infty_c (\mathbb{R}^n)$ with $\mu = 1$ in $\left\vert x \right\vert \leq 1$. Then (7.27) becomes, if $\varphi (0) =0$,

$$\varphi = \mu \varphi + (1- \mu ) \varphi$$

$$= \sum^n_{j=1} x_j \psi_j + (1- \mu ) \varphi , \psi_j = \m... ...{\partial \varphi}{\partial x_j} (tx) dt \in \mathcal{S} (\mathbb{R}^n) .$$

Since $(1- \mu) \varphi$ is of the form (7.26), this proves (7.25).

Our assumption on $u$ is that $x_j u=0,$ thus

$$u( \varphi ) =0 \forall \varphi \in \mathcal{H}$$

by (7.25). Choosing $\mu$ as above, a general $\varphi \in \mathcal{S}(\mathbb{R}^n)$ can be written

$$\varphi = \varphi(0) \cdot \mu + \varphi' , \varphi' \in \mathcal{H} .$$

Then

$$u( \varphi) = \varphi(0) u( \mu) \Rightarrow u= c \delta , c= u( \mu ) .$$

$\qedsymbol$

This result is quite powerful, as we shall soon see. The Fourier transform of an element $\varphi \in \mathcal{S}(\mathbb{R}^n)$ is¹⁴

$$\hat{\varphi} (\xi) = \int e^{-ix \cdot \xi} \varphi (x) dx , \xi \in \mathbb{R}^n .$$ (112)

The integral certainly converges, since $\left\vert \varphi \right\vert \leq C \langle x \rangle^{-n-1}$. In fact it follows easily that $\hat{\varphi}$ is continuous, since

$$\left\vert \hat{\varphi} (\xi) - \hat{\varphi} (\xi') \right\vert... ... e^{ix-\xi} -e^{-x \cdot \xi'} \right\vert \left\vert \varphi \right\vert dx$$

$$\to 0 \hbox{ as } \xi' \to \xi .$$

In fact

Proof. Differentiating under the integral¹⁵ sign shows that

$$\partial_{\xi_j} \hat{\varphi} (\xi) = -i \int e^{-ix \cdot \xi} x_j \varphi (x) dx .$$

Since the integral on the right is absolutely convergent that shows that (remember the $i$'s)

$$D_{\xi_j} \hat{\varphi} = -\widehat{x_j \varphi} , \forall \varphi \in \mathcal{S} (\mathbb{R}^n) .$$ (113)

Similarly, if we multiply by $\xi_j$ and observe that $\xi_j e^{-ix \cdot \xi} = i \frac{\partial}{\partial x_j} e^{-ix \cdot \xi}$ then integration by parts shows

$$\xi_j \hat{\varphi} = i \int (\frac{\partial}{\partial x_j} e^{-ix \cdot \xi}) \varphi (x) dx$$ (114)

$$= -i \int e^{-ix \cdot \xi} \frac{\partial \varphi}{\partial x_j} dx$$

$$\widehat{D_j \varphi} = \xi_j \hat{\varphi} , \forall \varphi \in \mathcal{S} (\mathbb{R}^n) .$$

Since $x_j \varphi$, $D_j \varphi \in \mathcal{S} (\mathbb{R}^n)$ these results can be iterated, showing that

$$\xi^{\alpha} D^{\beta}_{\xi} \hat{\varphi} = \mathcal{F} \left( (-1)^{\left\vert \beta \right\vert} {D^{\alpha}}_x x^{\beta} \varphi \right) .$$ (115)

Thus $\left\vert \xi^{\alpha} D^{\beta}_{\xi} \hat{\varphi} \right\vert \leq C_{\a... ...beta \right\vert} \varphi \Vert _{\mathcal{C}^{\left\vert \alpha \right\vert}},$ which shows that $\mathcal{F}$ is continuous as a map (7.32).

$\qedsymbol$

Suppose $\varphi \in \mathcal{S}(\mathbb{R}^n)$. Since $\hat{\varphi} \in \mathcal{S} (\mathbb{R}^n)$ we can consider the distribution $u \in \mathcal{S}' (\mathbb{R}^n)$

$$u ( \varphi ) = \int_{\mathbb{R}^n} \hat{\varphi} (\xi) d \xi .$$ (116)

The continuity of $u$ follows from the fact that integration is continuous and (7.29). Now observe that

$$u(x_j \varphi ) = \int_{\mathbb{R}^n} \widehat{x_j \varphi} (\xi) d \xi$$

$$=- \int_{\mathbb{R}^n} D_{\xi_j} \hat{\varphi} d \xi =0$$

where we use (7.30). Applying Proposition 7.10 we conclude that $u=c \delta$ for some (universal) constant $c$. By definition this means

$$\int_{\mathbb{R}^n} \hat{\varphi} (\xi) d \xi = c \varphi (0) .$$ (117)

So what is the constant? To find it we need to work out an example. The simplest one is

$$\varphi = \exp (- \left\vert x \right\vert^2 /2) .$$

Proof. There are two obvious methods -- one uses complex analysis (Cauchy's theorem) the other, which I shall follow, uses the uniqueness of solutions to ordinary differential equations.

First observe that $\exp (- \left\vert x \right\vert^2/2) = \prod_j \exp (-x_j^2/2)$. Thus¹⁶

$$\hat{\varphi} (\xi) = \prod^n_{j=1} \hat{\psi}(\xi_j) , \psi (x) = e^{-x^2 /2} ,$$

being a function of one variable. Now $\psi$ satisfies the differential equation

$$\left( \partial_x +x \right) \psi = 0 ,$$

and is the only solution of this equation up to a constant multiple. By (7.30) and (7.31) its Fourier transform satisfies

$$\widehat{\partial_x \psi} + \widehat{x \psi} = i \xi \hat{\psi} + i \frac{d}{d \xi} \hat{\varphi} =0 .$$

This is the same equation, but in the $\xi$ variable. Thus $\hat{\psi} = c e^{- \left\vert \xi \right\vert^2 /2}.$ Again we need to find the constant. However,

$$\hat{\psi}(0) = c = \int e^{-x^2/2} dx = (2 \pi)^{1/2}$$

by the standard use of polar coordinates:

$$c^2 = \int_{\mathbb{R}^n} e^{-(x^2 +y^2)/2} dx dy = \int^{\infty}_0 \int^{2 \pi}_0 e^{- r^2 /2} r dr d\theta = 2\pi .$$

This proves the lemma.

$\qedsymbol$

Thus we have shown that for any $\varphi \in \mathcal{S}(\mathbb{R}^n)$

$$\int_{\mathbb{R}^n} \hat{\varphi} (\xi) d \xi = (2 \pi )^n \varphi (0) .$$ (118)

Since this is true for $\varphi = \exp (- \left\vert x \right\vert^2/2)$. The identity allows us to invert the Fourier transform.