The Question
Let $f,\varphi\in\mathcal S(\mathbb R^d)$, $\varepsilon>0$ and denote $\varphi_\varepsilon(x)=\frac{1}{\varepsilon^d}\varphi\left(\frac x\varepsilon\right)$. Prove that $\varphi_\varepsilon\ast f\to f\int\varphi$ as $\varepsilon\to 0$ in $\mathcal S(\mathbb R^d)$.
Some Ideas!
I've been working around this question in a course in Fourier Analysis. Some of the ideas that I've worked around are as such.
First we've got to see what $\varphi_\varepsilon\ast f$, so we expand it in terms of the integral:
$$(\varphi_\varepsilon\ast f)(x)=\int\limits_{\mathbb R^d}\frac{1}{\varepsilon^d}\varphi\left(\frac y\varepsilon\right)f(x-y)\ dy=\int\limits_{\mathbb R^d}\varphi\left(y\right)f(x-\varepsilon y)\ dy,$$
where the last equality is true after a change of variables $y\mapsto \varepsilon y$. So, if we were searching for pointwise convergence we would be done. Applying Lebesgue's DCT we obtain that the last term converges to $f\int\varphi$ as $\varepsilon\to 0$.
However, reality sets in and then I realized that this was not the question I was supposed to answer. So I go back to Grafakos' book to check the definitions and now I've got to prove that $$\rho_{\alpha,\beta}\left(\varphi_\varepsilon\ast f- f\int\varphi\right),$$ where $\rho_{\alpha,\beta}$ is a seminorm in $\mathcal S(\mathbb R^d)$, is a really small quantity as $\varepsilon\to 0$. Writing this out without a supremum we obtain that its absolute value is bounded above by
$$C\|x\|^{|\alpha|}\left|\partial^\beta\int\limits_{\mathbb R^d}\varphi(y)(f(x-\varepsilon y)-f(x))\ dy\right|.$$
Now, this quantity converges to zero pointwise when $x$ is fixed. It doesn't converge uniformly because of the norm outside the integral and I require that because of the supremum in the seminorm.
One idea could be to use the Fourier transform in $\mathcal S$, since I know that if $f_n\to f$ in $\mathcal S$, then $\hat f_n\to \hat f$ in $\mathcal S$ as well (the same result holds for the inverse transform). However, I'm not fully in control of that because I still don't know how to handle the Fourier transform very well.
There's also the fact that applying the Fourier transform four times in a row is the identity operator. But I can't see how to use that.
I also thought for a moment that I could bound the norm using a compact support, but then I recalled that $\mathcal S\subseteq \mathcal C_0$ and not $\mathcal C_c$. So my argument went down.
Any hints, suggestion or comments are welcome. Kind thanks for reading the whole thing.
Let $C=\int_{\Bbb{R}^d} \varphi(x)dx$.
$f \ast \varphi_\epsilon \to f$ in the $\sup$ norm, this follows from $$(\varphi_\varepsilon\ast f)(x)-C f(x)=\int_{\mathbb R^d}\varphi\left(y\right)(f(x-\varepsilon y)-f(x))\ dy$$ that $\varphi\in L^1$ and $f$ is bounded uniformly continuous so that $\sup_{x,y}|f(x-\varepsilon y)-f(x)|\to 0$ as $\epsilon\to 0$.
$(1+|x|^k) (f \ast \varphi_\epsilon) \to (1+|x|^k) f$ in the $\sup$ norm: $$(1+|x|^k)((\varphi_\varepsilon\ast f)(x)-C f(x))=\int_{\mathbb R^d}\varphi\left(y\right)(1+|x|^k)(f(x-\varepsilon y)-f(x))\ dy$$ $$\sup_{x,y}\frac{1+|x|^k}{1+|y|^k}|f(x-\varepsilon y)-f(x)|\to 0, \qquad (1+|y|^k) \varphi(y) \in L^1$$ ($f$ is rapidly decreasing)
Finally use that $$\partial^\alpha(f \ast \varphi_\epsilon)=(\partial^\alpha f) \ast \varphi_\epsilon \to C \partial^\alpha f$$