Convolution between a Bounded Tempered Distribution and a $L^1$ function is a distribution. In this question we can see the definition of a convolution between a bounded tempered distribution and a $L^1$ function. In the comments, there are two possible ways to show that this is in fact a distribution.

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*One idea is to use some version of Banach-Steinhauss (Uniform Boundedness Principle), but I was not able to found a reference that applies in this case.


*The second ideia: note that $\sup\limits_{y \in \mathbb{R}^n}|D^{\alpha}(\tau_x \phi(y))|= \sup\limits_{y \in \mathbb{R}^n}|D^{\alpha}(\phi(y))|$. But the Schwartz semi-norms are given by $\| \phi \|_{(m, \alpha)} = \sup\limits_{y \in \mathbb{R}^n}(1+|y|)^m|D^{\alpha}(\phi(y))|$ and I can't conclude the result. Moreover, i think that the following is true:
$$\| \tau_x \phi \|_{(m, \alpha)} = \sup\limits_{y \in \mathbb{R}^n}(1+|y|)^m|D^{\alpha}(\tau_x \phi(y))| \leq (1 + |x|)^{m} \|\phi\|_{(m, \alpha)}$$
because $(1 + |y|) \leq(1 + |x|)(1 + |y - x|)$. Unfourtunately, the right side depends of $x$.
Can someone help me in one (or both) of this directions?
 A: I wanted to provide an answer since I ran into the same issue when encountering the previous posts mentioned. Here is one way to think about it with Banach-Steinhaus.
Consider the family of continuous linear functionals on $\mathcal{S}$, $\Lambda_x(\phi) = f * \phi(x)$ parametrized by $x \in \mathbb{R}^n$. Each of these is indeed a continuous linear functional, since if $\phi_n \rightarrow 0$ in $\mathcal{S}$ then for $x$ fixed, $f*\phi_n(x) \rightarrow 0$ in $\mathbb{C}$. Moreover, the assumption that $f$ is a bounded tempered distribution implies that for $\phi \in \mathcal{S}$ fixed, we have bounded orbits:
$$\sup_{x \in \mathbb{R}^n} |\Lambda_x(\phi)| = \sup_{x \in \mathbb{R}^n} |f*\phi(x)| = ||f * \phi||_\infty < \infty.$$
Hence the Banach-Steinhaus Theorem applies to give that the family $\{ \Lambda_x\}$ is equicontinuous on $\mathcal{S}$. In particular, for the unit ball $U \subset \mathbb{C}$, there is a neighborhood $W \subset \mathcal{S}$ of the origin so that if $\phi \in W$, then $\sup_{x \in \mathbb{R}^n}|\Lambda_x (\phi)| = ||f * \phi||_\infty \le 1$. Unwinding definitions of the topology on $\mathcal{S}$ gives that
$$||f* \phi||_\infty \le C p_k(\phi)$$
for some $C \ge 0$ and seminorm $p_k$ defining the topology on $\mathcal{S}$. This is the key estimate we need.
Finally, suppose $f$ is a bounded tempered distribution, and $h \in L^1(\mathbb{R}^n)$. Recall that $f*h$ is defined as a tempered distribution (point-wise) by
\begin{align*}
\langle f*h, \phi \rangle & \equiv \langle f*\tilde{\phi}, \tilde{h} \rangle \\
& = \int_{\mathbb{R}^n} (f * \tilde{\phi})(x) h(x)  dx
\end{align*}
where $\tilde{g}(x)= g(-x)$. This definition makes sense point-wise for $\phi \in \mathcal{S}$ since $f * \tilde{\phi} \in L^\infty$. This is indeed a tempered distribution (i.e., continuity holds) since we can estimate
\begin{align*}
\left| \langle f * h, \phi \rangle \right | & \le ||f*\tilde{\phi}||_\infty \int_{\mathbb{R}^n} |h(x)| dx \\
& \le C p_k(\tilde{\phi}) ||h||_1
\end{align*}
and $p_k(\tilde{\phi}) \rightarrow 0$ as $p_k(\phi) \rightarrow 0$, completing the proof.
