# If $f$ is a bounded tempered distribution and $g \in L^1$ is then $\int_{\Bbb R^n}(f\ast\tilde\varphi)(x)\tilde g(x)\,dx$ a tempered distribution?

Let $f$ be a bounded tempered distribution, that is, $f\ast\varphi \in L^\infty(\mathbb R^n)$ for every Schwartz function $\varphi$. If $g \in L^1(\mathbb R^n)$, does the following definition define a tempered distribution:

$$\langle f \ast g,\varphi \rangle = \int\limits_{{\mathbb R^n}} {(f \ast \tilde \varphi } )(x)\tilde g(x) \, dx$$

where $\varphi$ is a Schwartz function and $\tilde \varphi(x)=\varphi(-x)$?

• I changed $<\langle f * g,\varphi >$ to $\langle f * g,\varphi \rangle$. That is standard usage. – Michael Hardy Oct 3 '12 at 0:32

Yes. Take a look on this book in the part of tempered distributions:

"Michael Eugene Taylor - Partial Differential Equations Volume I Basic Theory"

Ok lets try. Let $S$ be the space of Schwartz functions.

Note first that this number is well defined:

\begin{eqnarray} |\langle f\ast g,\phi\rangle| &\leq& \int_{\mathbb{R}^{n}}|f\ast \tilde{\phi}||\tilde{g}| \nonumber \\ &\leq& \|f\ast \tilde{\phi}\|_{\infty}\|\tilde{g}\|_{1} \nonumber \end{eqnarray}

On the other hand, as you can see in that book, $f\ast \tilde{\phi}$ is a tempered distribution, so its is continuous i.e. $$(\forall\phi)\Bigl(\phi\in S\Rightarrow |f\ast \tilde{\phi}|\leq Cp_{k}(\tilde{\phi})\Bigr)$$ where $p_{k}(\phi)$ is defined as there "in the book".

Now

\begin{eqnarray} |\langle f\ast g,\phi\rangle| &\leq& \int_{\mathbb{R}^{n}}|f\ast \tilde{\phi}||\tilde{g}| \nonumber \\ &\leq& Cp_{k}(\tilde{\phi})\|\tilde{g}\|_{1} \nonumber \end{eqnarray}

With the last inequality you can conclude.

• I had a look in the book as far as it was available online. But I couldn't see an explicit statement for the continuity in $\varphi \in \mathcal{S}$, as - for me - this is not obvious. – Vobo Oct 3 '12 at 12:04
• Even though we know $S \rightarrow S'$ which is given by $\phi\rightarrow f*\phi$ is continuous, it does not say $\phi\rightarrow f*\phi$ is continuous as $S \rightarrow L^\infty$. – Hezudao Oct 3 '12 at 18:19
• Let me try explain. A priori $f\ast \tilde{\phi}$ isnt a function, so what you have to do? You have to take its representative. What is the representative? $(f\ast \tilde{\phi})(x)$ is defined by $$(f\ast \tilde{\phi})(x)=\langle f,\tau_{x}\phi\rangle$$ where $\tau_{x}\phi=\phi(x+y)$. Now use the continuity of $f$ as a distribution. – Tomás Oct 3 '12 at 23:07
• I know $f*\tilde\phi(x)$ is a function (actually it is smooth), and this function is bounded because of the assumption about $f$. But it is not clear if $\phi\rightarrow f*\phi$ is continuous, that is, $||f*\tilde\phi||_\infty \leq C p_k(\phi)$ where $p_k$ is some seminorm in $S$. – Hezudao Oct 4 '12 at 5:06
• @Hezudao, the last comment of Tomas has the key argument: You know that $|<f,\tau_x\phi>| <= Cp_k(\tau_x\phi)$ for all $x$, and hence it is bounded uniformly. – Vobo Oct 4 '12 at 6:08