Show that if $f\in L^1(\mathbb{R})$ and $g\in S(\mathbb{R})$. then $f*g\in C(\mathbb{R})$

Question

Schwartz functions :$S(\mathbb{R})$

$f \in S(\mathbb{R})$ if $f$ is smooth and sup$_{x\in \mathbb{R}}|x|^k|f^l(x)| <\infty, \ \forall k \geq0, l\geq0$.

Show that if $f\in L^1(\mathbb{R})$ and $g\in S(\mathbb{R})$. Then $f*g\in C(\mathbb{R})$

What I have attempted is $$|(f*g)(x+h)-(f*g)(x)|=| \int_{\mathbb{R}} f(y) \{g(x+h-y)-g(x-y)\} dy|$$

Now $g\in S(\mathbb{R})$ imply $g$ is uniform continuous. Hence given $\epsilon >0$ there exist $\delta>0$ such that $|g(x+h-y)-g(x-y)|< \epsilon,$ whenever$|h|<\delta$.

$$\int_{\mathbb{R}} f(y) \{g(x+h-y)-g(x-y)\} dy=\int_{|h|\leq \delta} f(y) \{g(x+h-y)-g(x-y)\} dy +\int_{|h|\geq \delta} f(y) \{g(x+h-y)-g(x-y)\} dy$$

Now first integral is bounded by $ \ \epsilon||f||_1$ But how to bound second integral such that it gives continuity of $f*g$

Any hint or some other better way to do this problem?

Answer 1

\begin{align*} |(f\ast g)(x+h)-(f\ast g)(x)|&=\left|\int_{{\bf{R}}}f(y)(g(x+h-y)-g(x-y))dy\right|\\ &\leq\int_{{\bf{R}}}|f(y)||g(x+h-y)-g(x-y)|dy\\ &\leq\|f\|_{L^{1}}\|g(\cdot+h)-g(\cdot)\|_{L^{\infty}}, \end{align*} where \begin{align*} |g(x+h)-g(x)|&=\left|\dfrac{g(x+h)-g(x)}{h}\cdot h\right|\\ &=|g'(\xi_{h})||h|\\ &\leq\|g'\|_{L^{\infty}}|h|. \end{align*}