f is continuous on $\mathbb{R}$ and we have $\lim_{x\rightarrow \infty}f(x)=0$. $g$ is continuous on $[0,\infty)$ and $\int_0^\infty |g(x)|dx$ is finite. $h(x)=\int_0^\infty f(y-x)g(y)dy$, Is $h(x)$ continuous and uniformly continuous on $[0,\infty)?$
From the condition, I can find $f(x)$ is uniformly continuous on some intervals. Also from the limit I can get:
For every $\epsilon>0$, there exists $N_1>0$, if $x \geq N_1$ then $|f(x)|<\epsilon$
From the finite integral I can get: there exists $N_2>0$, if $x \geq N_2$ then $\int_{N_2}^\infty |g(x)|dx < \epsilon$
If I take $N=\max\{N_1,N_2\}$, let $M,K$ be upper bound of $f$ and $\int_0^\infty |g(x)|dx$ respectively, for any $c \in [0,\infty)$, I want to have $|h(x)-h(c)|<\epsilon \cdot \text{number}$ as $(x-c) \rightarrow 0$, but I can only get $$|h(x)-h(c)|<\int_0^\infty g(y)(f(y-x)-f(y-c))dx=\int_0^N g(y)(f(y-x)-f(y-c))dx+\int_N^\infty g(y)(f(y-x)-f(y-c))dx$$
Even if I substitute the $\epsilon, K$ in the equation, I still cannot find how to get the expected result. If I can show it is continuous and the limit at infinity is $0$, then it should be uniformly continuous.
Can someone help me with how to identify the continuity? Thanks.
Since $f$ is continuous in $[0,+\infty)$ with $\lim_{x\rightarrow \infty}f(x)=0$ then $|f|$ is bounded by some constant $M$ in $[0,+\infty)$. Hence \begin{align}\left|\int_N^\infty g(y)(f(y-x)-f(y-c))dy\right|&\leq \int_N^\infty |g(y)|(|f(y-x)|+|f(y-c)|)dy\\&\leq 2M\int_N^\infty |g(y)|dy<2M\epsilon. \end{align} Moreover, $f$ is uniformly continuous in $[0,N]$ and $|f(y-x)-f(y-c)|<\epsilon$ when $|(y-x)-(y-c)|=|x-c|<\delta$. Therefore \begin{align} \left|\int_0^N g(y)(f(y-x)-f(y-c))dy\right|&<\int_0^N |g(y)||f(y-x)-f(y-c)|dy\\& <\epsilon \int_0^N |g(y)|dy. \end{align} Therefore $h(x)$ is uniformly continuous on $[0,\infty)$.