Suppose $f$ is integrable on $\mathbb{R}$, and $g$ is locally integrable and bounded. Then $f*g$ is uniformly continuous and bounded?

Question

Suppose $f$ is integrable on $\mathbb{R}$, and $g$ is locally integrable and bounded. Then $f*g$ is uniformly continuous and bounded?

I don't even know where to start proving or disproving, but I feel it is true. Can somebody give me a hint?

Answer 1

Assuming $g$ bounded $$ |f\ast g(x)|\le\int_{\mathbb{R}}|g(y)|\,|f(x-y)|\,dy\le\|g\|_\infty\|f\|_1. $$ Next, for $h\in\mathbb{R}$ $$ |f\ast g(x+h)-f\ast g(x)|\le\|g\|_\infty\int_{\mathbb{R}}|f(x+h-y)-f(x-y)|\,dy=\int_{\mathbb{R}}|f(z+h)-f(z)|\,dz. $$ Sincc $f$ is integrable, $$ \lim_{h\to0}\int_{\mathbb{R}}|f(z+h)-f(z)|\,dz=0, $$ proving the uniform continuity of $f\ast g$.