Define convolution $f$ of two functions $f_1, f_2: \mathbb R \rightarrow \mathbb R$: $$f(x) = \int_{-\infty}^{\infty} f_1(t) f_2(x-t)dt$$
Show that if $f_1$ is bounded, then $f$ is continuous.
I'm totally out of ideas to be honest. I only suppose that the previous problem in my assignment might have to do something with this (I'm allowed to link to this problem if needed):
Let $f_1, f_2 \in L_1(\mathbb R)$. Show that $f_1(t)f_2(x-t)$ is measurable for all $t\in\mathbb R$ and $f(x) \in L_1(\mathbb R)$
Use the fact that for every $f\in L_1(\mathbb{R})$ and $t\in \mathbb{R}$, if $f_t(x)=f(x-t)$ then $$\tag{1}||f-f_t||_1=\int_{-\infty}^{\infty}|f(x)-f(x-t)|dx\to 0$$ as $t\to 0$. (1) can be proved by using the fact that continuous functions with compact support are dense in $L_1(\mathbb{R})$. For these functions its easier to prove (1).
Now fix $\delta\in \mathbb{R}$ and $x$ and write \begin{align} |f(x)-f(x-\delta)|=&\biggl|\int_{-\infty}^{\infty}f_1(t)[f_2(x-t)-f_2(x-\delta-t)]dt\biggr|\\ &\leq M\int_{-\infty}^{\infty}|f_2(x-t)-f_2(x-\delta-t)|dt \end{align}
where $M=\sup\{|f_1(t)|:t\in \mathbb{R}\}$. Now use (1) and conclude that $f$ is indeed continuous.