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Note: This is a simplified version of the following question. I did not get a full response and realized can make it simpler to have my main interrogation answered. I decided to write it as a separate question because, if I just edit my previous question, it will not get any attention.

Question: Consider $$ v(x)\equiv e^x\int_x^\infty \frac{f(x')}{e^{x'}}dx'. $$ where $f\in H^1(\mathbb{R}^+)\cap C^1(\mathbb{R}^+)$. I want to prove that $v\in L^2(\mathbb{R}^+)$.

Context: I am computing the resolvent set of a differential operator and I computed the solution of the resolvent equation. Using variation of parameters, I end up with expressions similar to what is above. Among other things, I need to determine if those expressions give me functions that are in $L^2(\mathbb{R}^+)$.

Reasoning: Considering the behavior at $x=\infty$, it "seems" fine to me also since the integral will go to zero exponentially fast and fast enough to cancel the behavior of the exponential outside and make $v$ be in $L^2(\mathbb{R})$. Now, obviously, this last sentence is a heuristic argument, which needs to be made formal, if true at all. I need help with that.

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  • $\begingroup$ In this context, which space is $H^1$? The Sobolev space of functions such that $f, f'$ are both in $L^2$? $\endgroup$
    – Willie Wong
    Commented Aug 4, 2022 at 12:22
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    $\begingroup$ You're considering the integral operator with kernel $K(x,y)=e^{x-y}\chi_{y>x}$. Schur's test shows that this is bounded $L^2\to L^2$: en.wikipedia.org/wiki/… $\endgroup$
    – Christian Remling
    Commented Aug 4, 2022 at 14:53
  • $\begingroup$ @WillieWong Yes, that's it. $\endgroup$
    – Gateau au fromage
    Commented Aug 4, 2022 at 19:11
  • $\begingroup$ @Christian Remling Let me see if I get this straight. I am looking at the first inequality of the section entitled "Common usage and Young's inequality". If the two "sups" on its RHS are finite, then it gives me an upper bound on the operator bound AND it shows that the operator is well-defined in all of $L^2$ and is continuous. Is that right? $\endgroup$
    – Gateau au fromage
    Commented Aug 4, 2022 at 19:16
  • $\begingroup$ @Christian Remling I looked at the books given in reference to that Wikipedia article and found the precise statements. Thank you very much for the information. I had never seen that result before. $\endgroup$
    – Gateau au fromage
    Commented Aug 4, 2022 at 19:49

1 Answer 1

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From Theorem 0.3.1 of [1], we have that if a kernel operator $$ T(f)\equiv \int_0^\infty K(x,x') f(x')dx',\;\;K(x,y)=e^{x-x'}\chi_{y>x>0} $$ is such that $$ \sup_{x'}\left(\int_0^\infty K(x,x') dx\right),\;\; \sup_{x}\left(\int_0^\infty K(x,x') dx'\right)\leq C $$ then $$ \|T(f)\|_{L^2(\mathbb{R}^+)}\leq C \|f\|_{L^2(\mathbb{R}^+)}. $$ Thus $T$ defines a continuous operator on $L^2(\mathbb{R}^+)$. In my specific case, $K(x,y)=e^{x-x'}\chi_{y>x>0}$, $C=1$ and thus that answers my question!

A big thank you to Christian Remling to point me toward that theorem.

[1] Sogge, Christopher D., Fourier integrals in classical analysis, Cambridge Tracts in Mathematics. 105. Cambridge: Cambridge University Press. x, 237 p. (1993). ZBL0783.35001.

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