Product of tempered distribution is again a tempered distribution?

Question

I have the next doubts:

If $f$ is bounded and continuous function is a tempered distribution?

The product of two tempered distribution is again a tempered distribution?

Answer 1

I'll just work in $\mathbb{R},$ but it doesn't matter.

To answer your first question, note that if $\|f\|_\infty=M,$ then for any Schwartz function $\varphi$, \begin{align*}|\langle f,\varphi\rangle|&\leq\int\limits_{-\infty}^\infty |f(x)\varphi(x)|\, dx\leq M\int\limits_{-\infty}^\infty |\varphi(x)|\, dx\\ &=M\int\limits_{-\infty}^\infty |\varphi(x)|\langle x\rangle^2 \langle x\rangle^{-2}\, dx\\ &\leq M\sup_x|\langle x\rangle^2\varphi(x)| \int\limits_{-\infty}^\infty \langle x\rangle^{-2}\, dx\\ &=M\pi \sup_x|\langle x\rangle^2\varphi(x)|\leq M\pi\|\varphi\|_2,\end{align*} where $ \langle x\rangle=(1+|x|^2)^{1/2}$, and $$\|\varphi\|_k=\sum\limits_{|\alpha|\leq k}\sup_x| \langle x\rangle^k \partial^\alpha \varphi|$$ denotes the $k$th Schwartz semi-norm. So, the answer is yes, as this proves continuity (the only "difficult" thing to show). This is what @reuns is getting at with their comment. Also, this generalizes to $\mathbb{R}^n$ by putting an appropriate power of $\langle x\rangle$ (to make it integrable).

For multiplication, you'd need to be more clear as to what you mean when you talk about multiplying distributions.