Convolution converging uniformly on real line

Question

I'm working on this question and stuck with the following part:

Suppose $f\in L^\infty(\mathbb{R})$ and $K,K_1,K_2,\ldots\in L^1(\mathbb{R})$ with $K_n\rightarrow K$ in $L^1$. Why is it true that $f\ast K_n$ converges uniformly on the real line to $f\ast K$?

Answer 1

For a fixed $x$, $$|f\star K_n(x)-f\star K(x)|=\left|\int_\mathbb R (K_n(x-t)-K(x-t))f(t)\mathrm dt\right|\leqslant \lVert f\rVert_{\infty}\int_{\mathbb R}|K_n(x-t)-K(x-t)|\mathrm dt.$$ The last integral is, by the substitution $y=x-t$, the $\mathbb L^1$ norm of $K-K_n$. We thus get a bound uniform with respect to $x$ and which goes to $0$ as $n$ goes to infinity.