Showing a convolution of a $f\in C_c(\mathbb{R}^n)$ with $g\in L^p_{loc}(\mathbb{R}^n)$s continuous

Question

I am trying to use Lebesgue's dominated convergence theorem to show that if $f\in C_c(\mathbb{R}^k)$ and $g\in L_{loc}^{1}(\mathbb{R}^k)$ then we will have that $f* g\in C(\mathbb{R}^k)$. Now my idea was to take a sequence $x_n\rightarrow x$ and show that $(f*g)(x_n)\rightarrow (f*g)(x)$.

We have that $(f*g)(x_n)=\int_{\mathbb{R}^n}f(x_n-y)g(y)\,dy$ and $(f*g)(x)=\int_{\mathbb{R}^n}f(x-y)g(y)\,dy$, and so we want to see that $\lim_{n\rightarrow \infty}\int_{\mathbb{R}^n}f(x_n-y)g(y)\,dy=\int_{\mathbb{R}^n}f(x-y)g(y)\,dy$, and so the idea would be to apply the dominated convergence theorem to $h_n(y):=f(x_n-y)g(y)$ and $h(y):=f(x-y)g(y)$

Now Using the fact $f$ has compact support and is continuous, I was able to prove that $f(x_n-y)\rightarrow f(x-y)$ uniformly.

First we check that we have that $f(x_n-y)\rightarrow f(x-y)$ uniformly. Since the function $f$ is continuous we have that for every fixed $y$, $f(x_n-y)\rightarrow f(x-y)$, and around each point we can find a neighborhood $V_y$ such that if $y'\in V_y$, $|f(x_n-y')-f(x_n-y)|<\epsilon_y/3$ and $|f(x-y')-f(x-y)|<\epsilon_y/3$. Now since $f(x_n-y)\rightarrow f(x-y)$ there exists $N_y\in \mathbb{N}$ such that for every $n>N_y$ we will have that $|f(x_n-y)-f(x-y)|<\epsilon_y/3$. Using the triangle inequality we see that for every $y'\in V_y$ we have that $|f(x_n-y)-f(x-y)|<\epsilon_y$ for every $n>N_y$. Now since the support of $f$ is compact , we can take a finite subcovering of this covering by open sets $\{V_y\}$ and choose $N:=\max N_y$ and $\epsilon:=\min \epsilon_y$, so that for every $z\in K:=supp f$ and $n>N$ we have that $|f(x_n-z)-f(x-z)|<\epsilon$. Since the choice of $\epsilon$ was arbitrary we get that $f(x_n-y)\rightarrow f(x-y)$ uniformly.

Now I am not sure this can give me that $h_n\rightarrow h$ almost everywhere.And also we would need the fact that there exists an $h'\in L^1(\mathbb{R}^n)$ such that $|h_n(x)|\leq |h'(x)|$ almost everywhere in $\mathbb{R}^n$ for every $n$, but this seems to be a bit hard since I only know that $g\in L_\text{loc}^1(\mathbb{R}^n)$. Does anyone have any suggestions ? Thanks in advance.

Attempt at solving this difficulties :

Since $x_n\rightarrow x$ we can find an $N$ such that $x_n\in K$ for every $n\geq N$, where $K$ is a compact set and so since $f$ has compact support we will have that $h_n$ will have it's support inside a compact set $K'$ for every $n$ greater than $N$. And so know we can use the hypothesis that $g\in L_{loc}^1(\mathbb{R}^n)$ to get that there exists $M>0$ such that $|g(y)|\leq M$ for every $y\in K'$, and so we can get that $h_n\rightarrow h$ since $|f(x_n-y)g(y)-f(x-y)g(y)|\leq |f(x_n-y)-f(x-y)||M|$ for every $x\in \mathbb{R}^n$, and that $|h_n|\leq M'$ for every $n$ since $f$ has compact support.

What do you guys think ? I would appreciate some input just to see if I made any mistake or not.

Answer 1

Once you proved all this, it's easy to finish:

1. Since $f(x_n-y) \to f(x-y)$ pointwise (even uniformly), then also $f(x_n-y)g(y) \to f(x-y)g(y)$ pointwise. Possibly not uniformly, but this is not necessary here.
2. Since you already observed that $h_n$ are supported in some compact set $K'$ (for large enough $n$), the distinction between $L^1_{loc}$ and $L^1$ disappears. To be precise, $|f|$ is bounded by some constant $M$, so $|h_n(y)| \le M|g(y)|$ everywhere. Knowing that $|h_n(y)| = 0$ outside $K'$, we have $$ |h_N(y)| \le M |g(y)| \cdot \chi_{K'} =: h'(y), $$ and $h'$ is the integrable function we seek. ($\chi_{K'}$ denotes the characteristic function of $K'$)