Is a convolution of an $L^2[0,\infty)$ function $L^2[0,\infty)$?

Question

Suppose $f\in L^2[0,\infty)$. Is $\int_0^x f(y)e^{-(x-y)}dy\in L^2[0,\infty)$? What if we replace $e^{-x}$ by a general $L^2[0,\infty)$ function?

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Had $L^2[0,\infty)$ been replaced by $L^1[0,\infty)$, the answer is affirmative. Let $\mathcal L[f]$ be the Laplace transform of $f$, we have $\mathcal L[f*g]=\mathcal L[f]$$\mathcal L[g]$. But I do not see how this helps in answering the question.

Answer 1

Is $F(x)=\int_{0}^{x}f(y)e^{-(x-y)}dy$ in $L^2$? The Laplace transform $\mathscr{L}\{F\}$ is $\mathscr{L}\{f\}\mathscr L\{e^{-x}\}$, and $$ \mathscr{L}\{e^{-x}\}=\int_{0}^{\infty}e^{-sx}e^{-x}dx=\frac{1}{s+1} $$ $f\in L^2[0,\infty)$ iff $\mathscr{L}\{f\}$ is in the Hardy class $H^2(R)$ on the right half-plane $R$. Multiplying by $\frac{1}{s+1}$ gives another function that is in $H^2$, which means that $F \in L^2$.

The Laplace transform of the convolution $f\star g$, where $f,g\in L^2[0,\infty)$, is the product of two functions in $H^2$. Such a product is not necessarily in $H^2$, which means $f\star g$ is not necessarily in $L^2[0,\infty)$. For example, $$ F(s) = \frac{1}{s^{2/5}(1+s)} \in H^2, $$

but $F(s)F(s)\notin H^2$.