I meet with this problem in my homework.
Let
$\varphi \in S(\mathbb{R}^n)$ be such that $\int \varphi = 1$, and set $\varphi_\epsilon(x) := \epsilon^{-n} \varphi(\epsilon^{−1}x)$. Then
$f∗\varphi_\epsilon \to f$ in $S(\mathbb{R}^n)$ for all $f\in S(\mathbb{R}^n)$ as $\epsilon \to 0$.
I try to imitate the proof for the case when functions are smooth and compactly supported. But I failed to handle the unbounded part $(1+|x|)^N$, which need a finer approximation.
I would appreciate if anyone could give a proof or a reference.
Best Answer
We split up the integral in Question.
Let $R>0$. We will estimate $\lvert\lvert \phi_\epsilon *f - f\rvert\rvert_\infty$ (multiplying by a monomial/taking a derivative will not change the result; we will see in a second why). Note that continuous functions, that vanish at infinity, are uniformly continuous. Hence $f$ is and thus we can find a $\delta >0$, s.t. $\lvert f(x-y) - f(x)\rvert < R$ will hold for any $x\in\mathbb{R}^n$ and $y\in\mathcal{B}_\delta (0)$. Now since $\phi$ is also a Schwartz-Function (therefore the following integral finite) we obtain a constant $c_1>0$ and the following estimate:
\begin{equation} \int_{\lvert y\rvert \leq \delta} \lvert\phi_\epsilon(y) \rvert \lvert f(x-y) - f(x)\rvert dy < c_1 R. \end{equation}
Of Course this estimate is uniform in $x$ and $y$. Now we need to handle the part outside the ball somehow. For this observe via the Transformation Formula
\begin{equation} \int_{\lvert y\rvert > \delta} \lvert\phi_\epsilon(y) \rvert = \int_{\lvert y\rvert > \frac{\delta}{\epsilon}} \lvert\phi(y)\rvert. \end{equation}
This integral vanishes as $\epsilon$ gets smaller, since $\phi\in L^1(\mathbb{R}^n)$. Therefore we find an $\epsilon$, s.t.
\begin{equation} \int_{\lvert y\rvert > \delta} \lvert\phi_\epsilon(y) \rvert \leq R. \end{equation}
Additionally we find constant $c_2>0$, which is an upper Bound for $f$, since $f$ is Schwartz. Combining everything with the fact that $\int\phi =1$ and the triangle ineqhality you get:
\begin{align} \lvert\phi_\epsilon *f(x) - f(x)\rvert &\leq \int_{\mathbb{R}^n} \lvert\phi_\epsilon(y) \rvert \lvert f(x-y) - f(x)\rvert dy\\ &= \int_{\lvert y\rvert \leq \delta} \lvert\phi_\epsilon(y) \rvert \lvert f(x-y) - f(x)\rvert dy + \int_{\lvert y\rvert > \delta} \lvert\phi_\epsilon(y) \rvert \lvert f(x-y) - f(x)\rvert dy\\ &< c_1R + c_2\int_{\lvert y\rvert > \delta} \lvert\phi_\epsilon(y) \rvert\\ &< (c_1 + c_2)R. \end{align}
This is uniform in $x\in\mathbb{R}^n$. So for any $R>0$ we can find an $\epsilon >0$ small enough s.t. $\lvert\lvert \phi_\epsilon *f - f\rvert\rvert_\infty < R$. This gives the convergence in the Supremum norm. Again note that replacing $f(x)$ by $x^\alpha \partial^\beta f(x)$ in the proof doesn't Change anything, because it doesn't destroy the property 'uniform continuity' and 'boundedness', which we use.