Convolution $f*g(x)$ is continuous if $f \in L^1(\mathbb{R}^n)$ and $g \in L^\infty(\mathbb{R}^n)$

Question

I am stuck in what I believe to be the last part of this problem.

If $f \in L^1(\mathbb{R}^n)$ and $g \in L^\infty(\mathbb{R}^n)$, then $f*g(x):=\int_{\mathbb{R}^n} f(x-y)g(y) \mathrm{d}y$

(Throughout this problem, we integrate with respect to $\mathbb{R}^n$ and the Lebesgue measure)

Since $g \in L^\infty(\mathbb{R}^n)$, then there exists a real number $M$, such that $g \leq M \ \text{a.e.}$

Therefore, for $x, x_0 \in \mathbb{R}^n$

$$|f*g(x) - f*g(x_0)| = \left| \int f(x-y)g(y) \mathrm{d}y - \int f(x_0-y)g(y)\mathrm{d}y \right|$$ $$\leq M \left| \int \left( f(x-y)-f(x_0-y) \right) \mathrm{d}y \right|$$

and this is finite since $f \in L^1(\mathbb{R}^n)$

But, since we are not making any assumptions about the continuity of $f$, I am stuck.

Answer 1

Let $x_0 \in \Bbb{R}^n$. Then $$|(f\ast g)(x)-(f\ast g)(x_0)| \leq ||g||_{\infty}\int|f(x-y)-f(x_0-y)|dy$$

Make the change of variable: $t=x_0-y$ thus $y=x_0-t$

So $$\int|f(x-y)-f(x_0-y)|dy=\int|f((x-x_0)+t)-f(t)|dt=I(x)$$

By continuity of translations on $L^1$ we have that $I(x) \to ^{x \to x_0}0$

This part need approximation by continuous functions with compact support. But if you have seen this property and its proof then you don't have to do the proof again.