$f \in C_0$, i.e. $f(x):=e^{x} \int_{\infty}^{x} e^{-t} g(t) dt -e^{-x} \int_{-\infty}^{x} e^{t} g(t) dt \rightarrow 0$ Let $f(x):=e^{x} \int_{\infty}^{x} e^{-t}  g(t) dt-e^{-x} \int_{-\infty}^{x} e^{t} g(t) dt$ then I want to show that $$\lim_{x \rightarrow \pm \infty} f(x)=0$$ if $g \in C_0(\mathbb{R}).$
Since somebody gave a wrong reply, notice that $C_0$ consists of all continuous functions $g$ such that $\lim_{|x|\rightarrow  \infty} g(x)=0.$
Does anybody know how to show this?
 A: You can write $f$ as
\begin{align}
            f(x) &= e^x\int_{\infty}^{x}e^{-t}g(t)dt-e^{-x}\int_{-\infty}^{x}e^{t}g(t)dt \\
                 &= -\int_{x}^{\infty}e^{-(t-x)}g(t)dt-\int_{-\infty}^{x}e^{-(x-t)}g(t)dt
\end{align}
In the first integral on the right, set $u=t-x$ or $t=u+x$. Then $dt=du$ and the new integral ranges over $[0,\infty)$. In the second integral on the right, let $u=x-t$ or $t=x-u$. Then $dt=-du$ and the new integral now ranges over $(\infty,0]$. The above becomes
$$
           f(x) = -\int_{0}^{\infty}e^{-u}g(u+x)du+\int_{\infty}^{0}e^{-u}g(x-u)du \\
             = -\int_{0}^{\infty}e^{-u}\{g(x+u)+g(x-u)\}du.
$$
To show that $\lim_{x\rightarrow\pm\infty}$, let $\epsilon > 0$ be given. Let $M$ be a uniform bound for $g$, which exists because $g$ vanishes at $\infty$ and is continuous. Choose $r > 0$ to be large enough that $2Me^{-r} < \epsilon/2$. Because $g$ vanishes at $\infty$, there exists $R > 0$ such that $|g(v)| < \epsilon/4$ whenever $|v| \ge R-r$. Then, for $|x| > R$,
\begin{align}
      |f(x)| &\le \int_{0}^{r}e^{-u}|g(x+u)|+|g(x-u)|du+\int_{r}^{\infty}e^{-u}2Mdu \\
      & \le \int_{0}^{r}e^{-u}\frac{\epsilon}{2}du+2Me^{-r}\int_{0}^{\infty}e^{-u}du \\
      & < \int_{0}^{\infty}e^{-u}du\frac{\epsilon}{2}+\frac{\epsilon}{2}=\epsilon\;\;\;\; \mbox{whenever } |x| > R.
\end{align}
Because $\epsilon > 0$ was arbitrary, it follows that
$$
                   \lim_{x\rightarrow\pm\infty}f(x)=0.
$$
