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.