Suppose that $g\in \mathcal{S}(\mathbb{R^n})$ (Schwartz space) and $f\in L^p(\mathbb{R^n}).$ The idea is to prove that the differential operator $D^\alpha$ does not follow the Leibniz rule when applied to the convolution of $g$ and $f,$ in other words to show that: ($*$ stands for convolution here) $$ D^\alpha (g*f) = (D^\alpha g)*f \tag{1} $$ - a) Does one first have to show that $g*f \in L^p(\mathbb{R^n})\cap C^\infty (\mathbb{R^n})?$
What I have tried so far:
We know that for a function $r$ in $L^1 (\mathbb{R^n}),$ $r$ convolved with $f$ is in $L^p (\mathbb{R^n}.)$ Thus for $g \in \mathcal{S}(\mathbb{R^n}) \subset L^1 (\mathbb{R^n})$ we know at least that $g*f\in L^p (\mathbb{R^n}),$ but remains to show how $C^\infty$ comes into play (see a)).
On the other hand, to prove (1), I've started from:
$$ \mathcal F \left(D^\alpha (g*f)\right) = (ik)^\alpha \mathcal F (g*f) $$ and since $g$ is a Schwartz function, we can rewrite the above as:
$$ \mathcal F \left((D^\alpha g)*f\right)= (\mathcal F(D^\alpha g))\mathcal F(f) = (ik)^\alpha \hat{g}\hat{f} $$ Where I'm assmuing $\mathcal F(g*f)=\hat{g}\hat{f}$ holds here since $g \in \mathcal{S}(\mathbb{R^n}) \subset L^1 (\mathbb{R^n}).$
- Am I on the right path to showing (1)? I'm sort of stuck, not knowing how to make progress.
You have the right idea for proving $D^\alpha (g*f) = (D^\alpha g)*f $ "formally" using the Fourier transform. You're also right that you need to show that both sides of the equation are well-defined. The right side is clearly well-defined since $g\in \mathcal{S}$. To show that $g*f\in C^\infty$, use the Fourier-side decay condition: if $\widehat{g*f}$ decays faster than any power of $k$, then $g*f\in C^\infty$. This should be easy to show using $\widehat{g*f} = \hat{g}\hat{f}$ and the fact that $g\in \mathcal{S}$.