I would like to ask a question with possibly a reference. If we have a Schrödinger operator $-\Delta+V$ on an interval $[0,L]$ with $V$ continous and Dirichlet conditions, can we state that the eigenfunctions of such operator are uniformly bounded, i.e. there exists $M>0$ such that the eigenfunctions $\{\phi_n\}_n$ satisfy

\begin{equation*} \sup_n \lvert\lvert \phi_n \rvert\rvert_\infty\leq M \end{equation*} ?


1 Answer 1


Yes, this follows because asymptotically, as $|z|\to\infty$, the solutions of $-y''+Vy=zy$ look like those of the free equation $V\equiv 0$, and the eigenfunctions of $-y''=zy$, $\phi_n=(2/(L\pi ) )^{1/2}\sin n\pi x/L$, are uniformly bounded.

In fact, they are uniformly bounded not just in $n$, but also in the potential $V$ as long as we impose a uniform bound on $\|V\|_1$.

(I'm assuming here that you normalize your eigenfunctions as usual, $\|\phi_n\|_2=1$, and then you are asking about $\|\phi_n\|_{\infty}$.)

  • $\begingroup$ Thank you for your answer. Your assumptions are correct. I understand what you mean and numerical simulations point that way but I could not manage to find precise proof of that. Could you suggest any reference? $\endgroup$ May 15, 2022 at 20:18
  • $\begingroup$ The book by Poschel and Trubowitz discusses asymptotics in some detail (though with an $L^2$ assumption on $V$, which isn't necessary; you can adapt the proofs if required) and it's what I usually quote when I use these things, but there are many other references, so keep searching if you don't like this (Marchenko's book is another option). $\endgroup$ May 15, 2022 at 20:25
  • $\begingroup$ @Christian Remling Christian, do you know examples where $\sup_n \|\phi_n\|_\infty$ really depends on $\|V\|_1$ ($V \geq 0$)? $\endgroup$ May 16, 2022 at 17:10
  • 1
    $\begingroup$ @GiorgioMetafune: I think one way of doing this is to take $V=0$ on $(0,a)$ and $V=N\to\infty$ on $(a,L)$. In the limit $N\to\infty$, this will simulate a Dirichlet boundary condition at $x=a$, and if $a$ is small, then $\|\phi_n\|_{\infty}/\|\phi_n\|_2$ will become large. $\endgroup$ May 16, 2022 at 17:22
  • $\begingroup$ @ChristianRemling Thank you, right! $\endgroup$ May 16, 2022 at 17:28

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .