Consider $x:\mathbb{R}_+ \to \mathbb{R}_+$, where $x\not\in\mathcal{L}^1$, i.e., $$\int_0^\infty x(s)ds = \infty.$$ Consider the convolution $$h(t) = \int_0^t e^{-(t-s)}x(s)ds.$$ I am wondering if it is possible that $h\in\mathcal{L}^1$, i.e., $$\int_0^\infty h(s)ds = C < \infty.$$ I guess that yes, but I cannot find an example.
I know that for $x\in\mathcal{L}^1$ it holds $h \in \mathcal{L}^1$ but is the opposite true?
On
you could rewrite: $$h(t)=e^{-t}\int_0^t e^sx(s)ds$$ Now consider the conditions which are required for: $$t\to\infty,h\to0$$
it looks like the integral, and so the integrand, would have to diverge slower than $e^t$, in other words: $$O\left(e^tx(t)\right)<O(e^t)$$
Basically what this means is that if you want the integral of $h$ to converge, the integral of $x$ must also.
Since the functions under consideration are non-negative, you can do the following computation in $[0,+\infty]$ : \begin{align*} \int_0^\infty h(t) dt &= \int_0^\infty \left(\int_0^t e^{-(t-s)} x(s) d s \right) d t \\ &= \int_0^\infty \left(\int_s^\infty e^{-(t-s)} d t \right) x(s) d s \\ &= \int_0^\infty x(s) d s \end{align*}
So $x$ is integrable if, and only if, $h$ is.