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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 $H^1(\mathbb{R^+})$.

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  • $\begingroup$ Perhaps you've already thought through the point that if the kernel is not in $H^1(\mathbb R\times\mathbb R)$ then it's literally not Hilbert-Schmidt on $H^1(\mathbb R)$. Sure, there are compact operators that are not Hilbert-Schmidt... :) $\endgroup$ Commented Apr 7, 2022 at 1:06
  • $\begingroup$ It's too late in the day for me to see whether or not your operator is normal... But, if it were, one could imagine showing that its eigenvectors were actually in $H^1$, and show that (with normalization for $H^1$) the eigenvalues went to $0$, proving compactness... $\endgroup$ Commented Apr 7, 2022 at 1:38
  • $\begingroup$ @paul garrett First, are you saying that if the kernel is in $H^1(\mathbb{R})\cap L^1(\mathbb{R})$, then it is ``Hilbert-Schmidt'' on $H^1$ and thus compact on $H^1$? Not that it matters her since my kernels are discontinuous. But I thought Hilbert-Schmidt only applied to $L^2$. $\endgroup$ Commented Apr 7, 2022 at 20:08
  • $\begingroup$ "Hilbert-Schmidt" (suitably formulated) applies to any Hilbert space, not just concrete $L^2$'s. $\endgroup$ Commented Apr 7, 2022 at 20:15
  • $\begingroup$ If you could mention what $\mathcal L_0$ and $Q$ are explicitly, it might be helpful. $\endgroup$ Commented Apr 22, 2022 at 11:36

1 Answer 1

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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.

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  • $\begingroup$ First, thank you so much for your time. On the line where you take the $H^1$ norm, I am looking on the second term of the RHS. I think you should be taking $x$ derivatives there, not $y$. Is that right or perhaps there is something I do not understand? $\endgroup$ Commented Apr 23, 2022 at 23:23
  • $\begingroup$ I modified my answer since as you pointed out I made a terrible mistake in the calculations. You don't have to thank me, I hope it is right now and that you will find it useful. $\endgroup$
    – Gauge_name
    Commented Apr 24, 2022 at 11:38
  • $\begingroup$ I will thank you anyway. I am looking at the derivative term in the $H^1$ norm. It should be $|({\mathcal{L}}u_n)_x|^2$ that is integrated on $\mathbb{R}^+$. What seems to be there is $u_n({\mathcal{L}}u_n)_x$. Again, as I said before, I may be wrong. $\endgroup$ Commented Apr 24, 2022 at 13:24
  • $\begingroup$ Do we aree that $(\mathcal{L}u_n)_x=-e^{-x}\int_0^xu_n(y)dy+e^{-x}u_n(x)$? If so we have $(\mathcal{L}u_n)_x^2=e^{-2x}|\int_0^x u_n(y)dy|^2+e^{-2x}|u_n(x)|^2-2e^{-2x}u_n(x)\int_0^xu_n(y)dy$, right? $\endgroup$
    – Gauge_name
    Commented Apr 24, 2022 at 13:58
  • $\begingroup$ OK, I see, you grouped the terms together in one step. I read the first term too fast and did not see the "2" at the beginning of the $H^1$ norm. The only thing is that we should write the the last term in your comment as $-4{{\Re}\!\left({\overline{u_n}\int_0^x u_n dy}\right)}$, which corresponds to the next to last term in your $H^1$ norm. But, I don't think it changes anything in your argument. Let me read your argument carefully and I will come back to you later. $\endgroup$ Commented Apr 24, 2022 at 17:13

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