Convolution a Schwartz function and a rapidly decreasing function

Question

Let $\phi$ be a Schwartz function and $g$ a rapidly deceasing function in $\Bbb{R}^d$,that is $g(x)=f(x)$ where $|f(x)||x|^k \in L^{\infty}(\Bbb{R}^d), \forall k=0,1,2...$.

Prove that $f \ast g$ is infinitely differentiable.

Here is my proof:

Let $x \in \Bbb{R}^d$.We will prove that $D^a(g \ast \phi)(x)=(g \ast D^a \phi)(x)$ for every multi-index $a \in \Bbb{N_0}^d$.

It suffices to prove that $$\frac{\partial(g \ast \phi)(x)}{\partial x_j}=(g \ast \frac{\partial \phi}{\partial x_j}) (x)$$ for an arbitrary $j \in \Bbb{N}.$ Then we can proceed by induction of a general multi-index.

Now let $h_n \in \Bbb{R}, h_n \to 0, j \in \Bbb{N}$. Then $$\frac{(g \ast \phi)(x+h_ne_j)-(g\ast \phi)(x)}{h_n}$$ $$=\int_{\Bbb{R}^d}g(y) \frac{\phi(x-y+h_ne_j)-\phi(x-y)}{h_n}dy$$ $$=^{M.V.T}\int_{\Bbb{R}^d}g(y)\frac{\partial \phi(x-y+c(h_n)e_j)}{\partial x_j}dy$$ where $|c(h_n)| \leq |h_n|$. We have that $$\lim_n \frac{\partial \phi(x-y+c(h_n)e_j)}{\partial x_j} = \frac{\partial \phi(x-y)}{\partial x_j}$$ since $\frac{\partial \phi}{\partial x_j}$ is continuous.

Also $|g(y)\frac{\partial \phi(x-y+c(h_n)e_j)}{\partial x_j}| \leq M|g(y)|$, for some $M>0$, $\forall x \in \Bbb{R}^d,\forall n \in \Bbb{N}$, since $\phi$ is a Schwartz function.

Note that $$\int_{\Bbb{R}^d}|g(y)|dy=\int_{\{|y| \leq 1\}}|g(y)|dy+\int_{\{|y|>1\}}|g(y)|dy$$ $$\leq ||g||_{\infty}|B(0,1)| +C_g \int_{\{|y|>1\}}\frac{1}{|y|^{[d]+1}}dy <+\infty$$

So we may have the conclusion by dominated convergence.

Is this proof correct? (One thing that concerns me is the step of the proof in pink)If it's not,then what can i do correct it?

Thank you in advance.

Answer 1

What maybe really isn't that obvious is that in $$\left|g(y) \frac{\partial \varphi(x -y + c(h_n)e_j)}{\partial x_j}\right| \leq M |g(y)|$$ the constant $M$ is really independent of $h_n$, which you would need to use dominated convergence! If you use the Lipschitz property you get it easier just by $$ \left|\frac{\varphi(x - y -h_n e_j) - \varphi(x -y)}{h_n} \right| \leq \left| \text{Lip}(\varphi)\frac{x-y - h_n e_j - x -y}{h_n}\right| = |\text{Lip}(\varphi)|, $$ where $\text{Lip}(\varphi)$ is the Lipschitz-constant of $\varphi$.

(Also notice that in your last estimate you missing a constant in the second integral (which is not at all important).

Answer 2

The conclusion holds for any $g\in L^1.$ As you noticed, the difference quotients are dominated in absolute value by $|g(x)|$ times $\sup |D_k\phi(x)|,$ where the $\sup$ is taken over all $k\in \{1,2,\dots,d\}$ and $x\in \mathbb R^d.$ Because $\phi \in \mathcal S,$ that supremum is finite. The DCT then shows

$$D_k(\phi*g)(x) = (D_k\phi)*g(x).$$

Since each $ D_k\phi$ is in the Schwartz class, this argument can be repeated to give $\phi*g \in C^\infty.$