Compact integral operator on $H^1(\mathbb{R})$ Consider the operator
$$
{\mathcal{L}}v=e^{-x}\int_{0}^x v(y)\, dy.
$$
Is the operator ${\mathcal{L}}$ compact as an operator from $H^1({\mathbb{R}^+})$ to itself?
To give some context to the problem above, let me explain why I am interested in it. I have an operator of the form
$$
L\equiv{\mathcal{L}}_0+Q,
$$
where ${\mathcal{L}}_0$ is a differential operator I can deal with, i.e. I can compute the spectrum of it, while $Q$ is an integral operator. I would like to prove $Q$ is compact (or not). I was able to prove most of it is compact. The part above given by the integral operator $\mathcal{L}$ is the only term I could not deal with in $Q$. The difficulty seems to be that it is on $H^1(\mathbb{R^+})$.
 A: Since $H^1(\mathbb{R}^+)$ (I will omit $\mathbb{R}^+$ later on) is Hilbert, proving compactness is equivalent to prove that $\mathcal{L}$ is weak-strong continuous, i.e. for every sequence $u_n$ converging weakly to some $u\in H^1$ we have $\mathcal{L}u_n\to\mathcal{L}u$ strongly in $H^1$.
By linearity let us consider a sequence $u_n\to 0$ weakly in $H^1$: we would like to prove that
$$||\mathcal{L}u_n||_{H^1}^2=2\int_{\mathbb{R}^+}e^{-2x}\biggl|\int_0^xu_n(y)dy\biggr|^2dx-2\int_{\mathbb{R}^+}e^{-2x}u_n(x)\biggl[\int_0^xu_n(y)dy\biggr] dx+\int_{\mathbb{R}^+}e^{-2x}u_n(x)^2dx\to 0.$$
Let us call $F_n(x):=\int_0^xu_n(y)dy$: it is apparent that this term converges pointwise to $0$ thanks to the assumptions so that the first term on the r.h.s. goes to $0$ by Dominated Convergence. The third term can be handled by noting that $u_n$ converges locally uniformly to $0$ (thanks to the boundedness of the $H^1$-norm + Ascoli-Arzelà), therefore we have
$$\limsup_{n\to+\infty}\biggl|\int_{\mathbb{R}^+}e^{-2x}u_n(x)^2dx\biggr|\leq\limsup_{n\to+\infty}\int_0^ke^{-2x}u_n(x)^2dx+\limsup_{n\to+\infty}e^{-2k}\int_k^{+\infty}u_n(x)^2dx\leq Ce^{-2k},$$
for every $k\in\mathbb{N}$, therefore it goes to $0$.
For the second term it is again just a matter of computations: we have
$$L:=\limsup_{n\to+\infty}\biggl|\int_{\mathbb{R}^+}e^{-2x}u_n(x)\biggl[\int_0^xu_n(y)dy\biggr] dx\biggr|\leq \int_k^{+\infty}e^{-2x}|u_n(x)|\biggl[\int_0^x|u_n(y)|dy\biggr]dx,$$
where I got rid of the other integral thanks to local uniform convergence. Apply Cauchy-Schwarz inequality  two times to get
$$L\leq||u_n||_{L^2}^2\biggl(\int_k^{+\infty}e^{-4x}x dx\biggr)^{1/2}\leq C\biggl(\int_k^{+\infty}e^{-4x}x dx\biggr)^{1/2}$$
and finally note that the above tends to $0$ as $k\to+\infty$ thanks to integrability properties of the integrand.
EDIT: I apologize for changing my answer but I made a terrible computation mistake yesterday night, as it has been pointed out in the comments by the author of the post.
