$E = L^2(-\infty,\infty)$, if $f \in E,g(t)=\int_{-\infty}^{t}e^{x-t}f(x)dx$, could I get $g\in E$? 
$E = L^2(-\infty,\infty)$, if $f \in E,g(t)=\int_{-\infty}^{t}e^{x-t}f(x)dx$, could I got $g\in E$?

I do something like:
$$\int_{-\infty}^{t}|f(x)|^2dx\le \int_{-\infty}^{\infty}|f(x)|^2dx=M<\infty,$$
then
$$|g(t)|\le\left(\int_{-\infty}^{t}|e^{x-t}|^2dx\right)^{\frac{1}{2}}\left(\int_{-\infty}^{t}|f(x)|^2dx\right)^{\frac{1}{2}}\le \sqrt{M}\left(\int_{-\infty}^{t}e^{2(x-t)}dx\right)^{\frac{1}{2}}=\sqrt{\frac{M}{2}}<\infty.$$
but I can not get $g\in E$. Is that true, and how to prove that please?
 A: Let $y=t-x$, then
$$g(t)=\int_{-\infty}^t e^{x-t}f(x)\,dx=\int_0^\infty e^{-y}f(t-y)\,dy=\int_{\mathbb R}\chi_{[0,\infty)}(y)e^{-y}f(t-y)\,dy.$$
Let $F(t,y)=\chi_{[0,\infty)}(y)e^{-y}f(t-y)$, then by Minkowski's integral inequality, we have
\begin{align*}
\|g\|_{L^2}&=\left(\int_{\mathbb R}\left|\int_{\mathbb R}F(t,y)\,dy\right|^2\,dt\right)^{\frac12}\\
&\leq\int_{\mathbb R}\left( \int_{\mathbb R}|F(t,y)|^2\,dt\right)^{\frac12}\,dy\\
&=\int_0^\infty e^{-y}\,dy\|f\|_{L^2}=\|f\|_{L^2}<\infty.
\end{align*}
Therefore, $g\in E=L^2(\mathbb R)$.
Simpler proof (but essentially the same as before): Let $h(t)=\chi_{[0,\infty)}e^{-t}$, then
$$g(t)=h*f(t):=\int_{\mathbb R}h(x)f(t-x)\,dx=\int_{\mathbb R}h(t-x)f(x)\,dx.$$By Young's inequality, we have $\|h*f\|_{L^2}\leq \|h\|_{L^1}\|f\|_{L^2}=\|f\|_{L^2}$, giving that $g\in L^2$.
Remark. The inequality $\|h*f\|_{L^2}\leq \|h\|_{L^1}\|f\|_{L^2}$ can be proved not only using Minkowski's integral inequality, but also using Fourier transform:
$$\|h*f\|_{L^2}=\|\widehat{h*f}\|_{L^2}=\|\hat h\hat f\|_{L^2}\leq\|\hat h\|_{L^\infty}\|\hat f\|_{L^2}\leq \|h\|_{L^1}\|f\|_{L^2}.$$
Here our Fourier trnasform is defined by
$$\hat \varphi(\xi)=\int_{\mathbb R}\varphi(x)e^{-2\pi ix\xi}\,dx,\qquad \xi\in\mathbb R, \varphi\in L^1(\mathbb R),$$
and standard extension to $\varphi\in L^2(\mathbb R)$. We also used the Plancherel theorem: $\|\varphi\|_{L^2}=\|\hat\varphi\|_{L^2}$ and the equality $\widehat{\phi*\varphi}=\hat\phi\cdot\hat\varphi$.
