Convolution of Schwartz Function and Distribution of Compact Support

Question

From Stein-Shakarchi Functional Analysis Chapter 3 Exercise 12 and Exercise 13.

I'm having trouble proving that: If $F_1$ is a distribution with compact support and $\varphi\in \mathcal{S}$ is a Schwartz function, then $F_1*\varphi\in\mathcal{S}$.

They have a hint: For each $N$, there exists constant $c_N$ such that $\|\psi_y^{\sim}\|_N \leq c_N(1+|y|)^N\|\psi\|_N.$ Here, $\psi_y^\sim (x) = \psi(y-x)$ and I'm pretty sure that $\psi$ is compactly supported on the same set as $F$. Here, $\|\psi_y^\sim\|_N = \sup\{|x^\beta(\partial_x^\alpha \psi_y^\sim)(x)|:x\in\mathbb{R}^d,|\alpha|,|\beta|\leq N\}.$ I think here I can write $\|\psi_y^\sim\|_N = \sup\{|(y-x)^\beta(-1)^{|\alpha|}(\partial_x^\alpha \psi)(x)|:x\in\mathbb{R}^d,|\alpha|,|\beta|\leq N\},$ but I can't think of a clever way to pull out $(1+|y|)^N$. Any help proving the hint would be appreciated!

Now, after I have the hint, I just need to show that $F*\varphi$ is continuous in $\mathcal S$. I think the equivalent statement is simply $\|(F*\varphi)(\psi)\|_N <\infty$ for all $N$ and $\psi\in\mathcal S$ (not sure 100% what it means for a distribution to be in $\mathcal S$). I know that $(F*\varphi)(\psi) = F(\varphi^{\sim}*\psi)$ but I don't really see how this follows from the hint.

Answer 1

Since you are using Stein and Shakarchi's Functional Analysis, I will use the following results already proved in the book.

Proposition 1.5 Suppose $F$ is a tempered distribution and $\psi \in \mathcal{S}$. Then $F * \psi$ is a slowly increasing $C^\infty$ function, which when considered as a tempered distribution satisfies $(F * \psi )^\wedge = \psi^\wedge F^\wedge$.

Proposition 1.6 If $F$ is a distribution of compact support then its Fourier transform $F^\wedge$ is a slowly increasing $C^\infty$ function.

Since $F_1$ is a distribution of compact support, $F_1^\wedge$ is slowly increasing. Noticing $\phi^\wedge \in \mathcal{S}$, we have $(F * \psi )^\wedge = F^\wedge \phi^\wedge \in \mathcal{S}$. (It is easy to show that the product of a slowing increasing function and a function in $\mathcal{S}$ is in $\mathcal{S}$.) Therefore $F * \psi \in \mathcal{S}$.