1
$\begingroup$

This question comes from the paper: B. Simon, Schrodinger Semigroups, Bull. A.M.S., (1982) Vol. 7 (3).

On the Theorem C.3.4(subsolution estimate) of the paper, it says that: Let $Hu=Eu$ and $u\in L^2$, where $H=-\Delta+V$ for some bounded continuous function $V$, if $E$ is in discrete spectrum of $H$. Then for some $C,~\delta>0$ $$|u(x)|\leq Ce^{-\delta|x|}.$$

Q

  • How to get constant $C$, does it depend $\delta$?

  • Is there any estimate about $C$, especially without the condition on the compactness of $supp(E-V)_+$?

PS: I can follow how to get $\delta$. I can understand that $e^{\delta |x|}u\in L^\infty$. I do not know how to get the constant $C$, as the equation is linear, if we multiply any positive number $k$ with $u$, it is also a solution for the equation, where I can not follow. And I think $C$ depends on the $L^2$-norm of $u$.

$\endgroup$
  • 1
    $\begingroup$ Assuming $u\in \mathbb{R}^n$, this comes from an argument using the Agmon metric: since this bound is easier to show in the 1-dimensional case, you extend the result to higher dimensions by approximating your eigenfunction using the WKB method. A good place to read about this is Hislops "Introduction to Spectral Theory." Theorem 3.10 is the result you want, and the geometric hypothesis (the condition on $\text{supp}(E-V(x))_+$) of this theorem requires is implied by your potential $V$ being bounded. The constant $C$ does depend on $\delta$. $\endgroup$ – Hadrian Quan Mar 4 at 15:35
  • $\begingroup$ @HadrianQuan Thank you for the comment and reference. But I still have one question, if we multiply any positive number $k$ with the function $u$, it is still a solution of that equation.So, how to get the constant ($\delta$ comes from the spectrum of $H$). $\endgroup$ – DLIN Mar 5 at 2:28
  • 1
    $\begingroup$ @DLIN $u$ is assumed to be normalized. This also gives you a hint on the dependence on $\delta$ $\endgroup$ – lcv Mar 5 at 14:47
  • $\begingroup$ @lcv Well, from that paper, I have not seen any hypothesis on the norm of $u$. By the way, do you mean the theorem needs the assumption that $\|u\|_{L^2}=1$. $\endgroup$ – DLIN Mar 6 at 0:44
  • $\begingroup$ The theorem as stated does not need it. But if you want to see how $C$ depends on $\delta$ you can simply normalize both sides. $\endgroup$ – lcv Mar 6 at 4:41

Your Answer

By clicking "Post Your Answer", you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.