We define the following symbols:
-
$D = D(\mathbb{R}^n) = C_c^\infty$, the space of test functions from $\mathbb{R^n}\to\mathbb{C}$, equipped with the standard $C_c^\infty$ topology.
-
$D(U) = $ the space of test functions whose support is contained in $U$ as a subset,
-
$D'(U) = $ the space of distributions (continuous linear functionals) on $D(U)$, equipped with the weak-* topology,
-
$S = $ Schwartz Space, equipped with the semi-norms $\lVert{\cdot}\rVert_{(N,\alpha)}$, where
$$
\lVert{ f}\rVert_{(N,\alpha)} = \sup_{x\in\mathbb{R}^n} (1+\vert x\vert)^N\vert\partial^\alpha f(x)\vert
$$
where $\vert\cdot\vert$ is the standard Euclidean norm.
- $S' = $ the space of Tempered Distributions on $\mathbb{R}^n$, equipped with the weak-* topology.
The text I am following is Folland's Real Analysis. Errata here
I am trying to prove that the convolution between $F\in S'$ and $\psi\in S$ produces a $C^\infty$ function. Where
$$
(F\ast \psi)(x) = \langle F, \tau_x \widetilde{\psi}\rangle,\quad\text{where}\quad (\tau_x \widetilde{\psi})(y) = \psi(x-y)
$$
Excerpt of the Proposition below:
Proposition 9.9 concerns Density of $C_c^\infty$ in $S$, and the relevant parts of 9.3 are below
If we attempt the same technique as in 9.3, we get
$$\lim_{t\to 0}t^{-1}\biggl[(F\ast\psi)(x+te_j) – (F\ast\psi)(x)\biggr]=\lim_{t\to 0}\biggl\langle F, \: t^{-1}[\tau_{x+te_j}\widetilde{\psi} – \tau_{x}\widetilde{\psi}]\biggr\rangle$$
I do not see how we can push the limit inside the bracket, because we would require the difference quotient to converge in $S$ to the partial of $\phi$,
$$
t^{-1}\biggl[\tau_{x+te_j}\widetilde{\psi} – \tau_{x}\widetilde{\psi}\biggr](y)\overbrace{\longrightarrow}^{\text{in }S}\tau_x\widetilde{\partial/\partial x^j \psi}(y)\quad \text{as }t\to 0
$$
I have proven:
- If $f\in S$, then $f$ vanishes at infinity, which implies $f$ is uniformly continuous,
- If $f\in S$, so are all its derivatives $\partial^\alpha f\in S$ for every multi-index $\alpha$,
How do we show
$$
\sup_{y\in\mathbb{R}}(1+\vert y\vert)^N\Biggl[\dfrac{\partial^\alpha}{\partial y}\biggl(\dfrac{\psi(x+t e_j – y) – \psi(x – y)}{t} – \dfrac{\partial \psi(y)}{\partial x^j}\biggr)\Biggr]\to 0
$$
I also attempted an alternative line of argumentation, by borrowing properties of the convolution on $C_c^\infty$, but we require the uniform convergence of $\partial^\alpha(F\ast \psi_j)$. Let $\{\psi_j\}_{j\geq 1}\subseteq D(\mathbb{R}^n)$ be a sequence converging to $\psi$ in $S$. The restriction of the tempered distribution onto $D$ is a distribution, therefore
$$
(\partial^\alpha(F\ast \psi_j))(x) = (F\ast (\partial^\alpha \psi_j))(x) = \biggl\langle F, \: \tau_x\widetilde{\partial^\alpha\psi_j}\biggr\rangle
$$
The sequence $\{\tau_x\widetilde{\partial^\alpha \psi_j}\}\subseteq D(\mathbb{R}^n)$ converges in $S$ to $\tau_x\widetilde{\partial^\alpha \psi}$ which is almost what we want, but we still cannot claim
$$
\lim \partial^\alpha (F\ast \psi_j)(x) = \partial^\alpha \lim (F\ast\psi_j)(x)
$$
Since $F$ is endowed with the weak-* topology, I don't see where we can get this 'uniformity' in convergence.
Best Answer
DEFINITION 1
For $\psi\in C^{\infty}(\mathbb{R}^n,\mathbb{C}) = C^{\infty}$,
$$ \lVert\psi\rVert_{(N,\alpha)} = \sup_{x\in\mathbb{R}^n}(1+\vert{x}\vert)^{N}\vert\partial^{\alpha}\psi(x)\vert, $$ and $$ \lVert\psi\rVert_u = \sup_{x\in\mathbb{R}^n}\vert\psi(x)\vert. $$ Define the Schwartz functions
$$ \mathcal{S} = \{\psi\in C^{\infty}(\mathbb{R}^n,\mathbb{C}),\ \lVert\psi\rVert_{(N,\alpha)}<+\infty,\ \text{for all }N,\alpha\}, $$
is a Frechet space with $\psi_{n}\to \psi$ in $\mathcal{S}$ if for all $(N,\alpha)$, the difference $\lVert\psi_{n}-\psi\rVert_{(N,\alpha)}$ shrinks to $0$. Its continuous dual is the space of tempered distributions, $\mathcal{S}'$ and has the weak-* topology. For any distribution $F\in \mathcal{S}'$, the scalar product is written $F(\psi) = \langle F(y),\ \psi(y) \rangle$ whenever $\psi$ is a function of $y$, and $\tau_{x}\widetilde{\psi}(y) = \psi(x-y)$. The convolution between $F, \psi$ at $x$ is computed using
$$ G(x) = (F\ast \psi)(x) = \langle F(y),\ \tau_{x}\widetilde{\psi}(y)\rangle. $$
END OF DEFINITION 1
It turns out that there is no ball trick, and the missing piece is the following estimate, which holds for all $y,z\in\mathbb{R}^n$, $N\geq 1$ (Folland Equation 8.12). We need the following equation.
Equation (2)
$$ (1+\vert y \vert)^{N}\leq(1+\vert z\vert)^{N}(1+\vert y-z \vert)^{N}. $$
Proposition
Let $\psi\in \mathcal{S}$ be a function of $y$, and $x_0, x_1\in\mathbb{R}^n$, $\delta = x_1 - x_0$.
(CLAIM 1) $\tau_{x_{1}}\widetilde{\psi}(y)\overset{\mathcal{S}}{\longrightarrow}\tau_{x_{0}}\widetilde{\psi}(y)$ as $\delta\to 0$.
(CLAIM 2) $(\psi(x + t e_{j} - y) - \psi(x-y))t^{-1}\overset{\mathcal{S}}{\longrightarrow}\partial_{j}\psi(x-y)$ for $j\leq n$, as $t\to 0$.
BEGIN OF PROOF
PROOF OF FIRST CLAIM
Fix a norm $(N,\alpha)$, write $\psi_{\beta} = \partial^{\beta}\psi$ and $\delta_j = \langle \delta, \ e_j\rangle$. $$ \sup_y(1+\vert y \vert)^N\left\vert\partial^{\alpha}\psi(x_1 - y) - \partial^{\alpha}\psi(x_0 - y) \right\vert = \sup_y (\quad)^N \left\vert\psi_\alpha(x_1 - y) - \psi_\alpha(x_0 -y)\right\vert. $$ The difference can be replaced with a line integral, parameterized by \begin{align*} z(t) &= (1-t)(x_0 - y) + (t)(x_1 - y) \\ &= [(1-t)x_0 + (t)x_1] - y. \end{align*} Coordinatewise: $$ \left\lVert\tau_{x_1}\widetilde{\psi} - \tau_{x_0}\widetilde{\psi}\right\rVert_{(N,\alpha)} = \sup_y(\quad)^N\left\vert \sum_1^n\delta_j\int_0^1\psi_{\alpha+e_j}(z(t))dt\right\vert. $$ Using (Equation 2), we can push $(1+\vert y \vert)^N$ into the integral, and use the decay properties of $\psi_{\alpha+e_j}$ $$ \sum_1^n\delta_j\int_0^1(1+\vert z(t)\vert)^N\biggl\vert\psi_{\alpha + e_j}(z(t))\biggr\vert (1+\vert y-z(t)\vert)^N dt. $$ The integrand has $\infty$-bound independent of $\delta$, and $1$-bound which shrinks as $\delta\to 0$. The last displayed expression is bounded above by, for some $\omega(\delta)$ bounded for all sufficiently small $\delta$. $$ \sum_1^n\delta_j \left[\int_0^1(1+\vert z\vert)^N\vert\partial^{\alpha + e_j}(z)\vert dt\right]\omega(\delta)\lesssim{\psi,\alpha,j}\vert \delta\vert\omega(\delta) $$ where $\lesssim_{x}$ indicates a dependence on $x$ (in this case it is the sum of the required norms).
PROOF OF SECOND CLAIM
Fix a norm $(N,\alpha)$, and let $j = 1,\ \ldots, n$. The technique is the same as above, first we replace $\psi$ with $\psi_\alpha$, the difference with the integral, and use (Equation 1) in order to estimate each of the coordinates by factoring $$ (1+\vert y \vert)^N \leq (1+\vert y-z \vert)^N(1+\vert z\vert)^N. $$ For $t\neq 0$, \begin{multline*} \left\lVert\frac{\psi(x+te_j - y) - \psi(x-y)}{t} - \partial^{e_j}\psi(x-y)\right\rVert_{(N,\alpha)}\\ =\left\lVert(-1)^{\vert \alpha\vert}\left[\frac{\psi_{\alpha}(x+te_j - y) - \psi_{\alpha}(x-y)}{t} - \psi_{\alpha+e_j}(x-y)\right]\right\rVert_{(N,0)}. \end{multline*} For each $y\in\mathbb{R}^n$, we find $\vert t_y\vert\leq \vert t\vert$ so that the previous expression is equal to \begin{align*} \left\lVert\psi_{\alpha+e_j}(x+t_ye_j - y) - \psi_{\alpha+e_j}(x-y)\right\rVert_{(N,0)} &= \left\lVert(1+\vert y \vert)^N t_y\int_0^1\psi_{\alpha+2e_j}(x-y + st_ye_j)\mathrm{d} s\right\rVert_u\\ &\leq \vert t\vert\left\lVert\int_0^1(1+\vert z_y(s)\vert)^N\psi_\beta(z_y(s))(1+\vert y-z_y(s)\vert)^N\mathrm{d} s\right\rVert_u, \end{align*} where $\beta = \alpha + 2e_j$ and $z_y(s) = z(y,s) = x-y + st_ye_j$. The last integral (the one inside the norm) is bounded above by the product: \begin{align*} \int_0^1 (1+\vert z_y\vert)^N\vert \psi_{\beta\vert(z_y)}(1+\vert y-z_y\vert)^N\mathrm{d} s&\leq \left\lVert\psi\right\rVert_{(N,\beta)}\int_0^1 (1+\vert y-z_y\vert)\mathrm{d} s\\ &\lesssim_\psi \omega(\vert t\vert) \end{align*} Therefore $$ \left\lVert\frac{\psi(x+te_j - y) - \psi(x-y)}{t} - \partial^{e_j}\psi(x-y)\right\rVert_{(N,\alpha)}\lesssim_{\psi,\alpha,j}\vert t\vert\omega(\vert t\vert). $$
End of Proof
With this, the following is immediate.
Corollary
Let $F,\psi,G$ be as in DEFINITION 1, then $G(x)\in C^\infty(\mathbb{R}^n,\ \mathbb{C})$ and $\partial^{\alpha}G(x) = (F\ast \partial^{\alpha}\psi)(x)$.