Here's the set-up: Let $f\in L^{\infty}(\mathbb{R})$ and uniformly continuous, and let $g\in L^1(\mathbb{R})$ Show that the convolution $$f*g(t)= \int_{-\infty}^{\infty}f(t-s)g(s)ds$$ is uniformly continuous on $\mathbb{R}$
Here's my reasoning so far:
Let $\epsilon>0$ be given. Since $f$ is uniformly continuous choose $\delta$ corresponding to $\lambda=\dfrac{\epsilon}{\|{g}\|_1}$, i.e. $|t_1-t_2|< \delta \implies |f(t_1)-f(t_2)|<\lambda$
$$ |f*g(t_1)-f*g(t_2)| = \bigg| \int_{-\infty}^{\infty}f(t_1-s)g(s)-f(t_2-s)g(s) ds\text{ }\bigg|$$ $$\leq \int_{-\infty}^{\infty}|f(t_1-s)g(s)-f(t_2-s)g(s)| ds \leq \int_{-\infty}^{\infty}|g(s)||f(t_1-s)-f(t_2-s)|ds$$ Now note that $|t_1-t_2|=|(t_1-s)-(t_2-s)|<\delta \implies |f(t_1-s)-f(t_2-s)|<\lambda$ so we can therefore substitute making the inequality become $$ \int_{-\infty}^{\infty}|g(s)||f(t_1-s)-f(t_2-s)|ds < \int_{-\infty}^{\infty}|g(s)|\lambda ds = \|g\|_1\lambda = \epsilon$$ Doesn't this therefore show that the convolution is uniformly continuous? The $\delta$ depends only on $\epsilon$ since $f$ was uniformally continuous. I never used the fact that $f\in L^{\infty}(\mathbb{R})$, and this seems suspiciously too simple to be correct, where is the flaw?
I think this is the general idea, but it might be missing some subtleties. On the very last line I think you need to use Hölder's inequality for $p=1$ and $q=\infty$. You can also consider it "sup'ing out" $|f(t_1-s)-f(t_2-s)|$. Since $f$ is $L^\infty$ and uniformly continuous, you can find $\delta>0$ small enough such that $\|f(t_1-s)-f(t_2-s)\|_\infty<\lambda$. Specifically,
$$\int_{-\infty}^\infty |g(s)||f(t_1-s)-f(t_2-s)| \leq \int_{-\infty}^\infty |g(s)|\|f(t_1-s)-f(t_2-s)\|_\infty \\ =\|f(t_1-s)-f(t_2-s)\|_\infty \|g\|_1.$$
I hope this is helpful!