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

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 $f\ast g$.
