# Schrödinger eigenfunctions are bounded

Let $V:\mathbb{R}\rightarrow \mathbb{R}^{+ *}$ a real positive function such that $\displaystyle \lim_{ x \to \pm\infty} V(x)= +\infty$.

Then the Schrödinger operator $H=-\frac{d^2}{dx^2}+V(x)$ has compact resolvant, in particular it has a pure discrete spectrum $(\lambda_i)_{i\geq 0}$ such that $\displaystyle \lim_{i\to +\infty} \lambda_i = + \infty$. Associated to each eigenvalue, there is an eigenfunction $\phi_i\in L^2(\mathbb{R})$, ie $$-\phi_i''(x)+V(x)\phi_i(x)=\lambda_i \phi_i(x), \quad \forall x\in\mathbb{R}$$

satisfying $||\phi_i||_{L^2(\mathbb{R})}=1$,

My question is: Are eigenfunctions uniformly bounded? i.e. Does exist $M>0$ such that for all $n\geq 0$,

$$|| \phi_i ||_\infty <M$$

• the eigenfunctions could have a square-integrable singularity, couldn't they? – Carlo Beenakker May 14 '17 at 14:20
• Relevant for free Schrödinger problem on manifolds of dimension at least two: projecteuclid.org/euclid.dmj/1087575454 – Neal May 14 '17 at 14:20
• @CarloBeenakker: No, being solutions to an ODE, they have a second derivative that is locally integrable (so in particular are continuous). – Christian Remling May 18 '17 at 23:06

In general, there won't be a uniform bound on all eigenfunctions simultaneously. If $[a,b]$ is a short interval with Dirichlet boundary conditions $y(a)=y(b)=0$ and constant potential $V=c$, then the ground state (normalized) eigenfunction $$\phi(x)=\sqrt{\frac{2}{b-a}}\, \sin\pi\frac{x-a}{b-a}$$ is quite large pointwise.
A very high steep potential wall has approximately the same effect as a Dirichlet boundary condition, so we can easily cook up a $V$ where we are close to this situation infinitely many times, with arbitrarily small interval lengths.
• The bound M of the question is a bound $M = M(V)$. For your example,there is such an M: $M = \sqrt{2/(b-a)}$, also valid for the higher eigenvalues which are this M multiplied by $sin(k\pi (x-a)/(b-a))$, $k=2,3,...$ so your counterexample' is in fact an example of this bounded phenomenon. – Richard Montgomery May 15 '17 at 2:30
• @RichardMontgomery: You completely misunderstood my example. I construct a whole line potential $V$ that contains parts that look like $V=c_n$ on $I_n=(a_n,b_n)$, $V\gg c_n$ on two intervals of length $L_n\gg 1$ to the left and right of $I_n$. This will have approximately the eigenfunctions I stated, but for arbitrarily small $b_n-a_n$. – Christian Remling May 15 '17 at 16:31
• Yes. I did not read your last sentence Christian. Using your trick I see you can do as you say, having a potential which is non-negative, piecewise constant, and zero on a countable collection of intervals $I_j$ whose size tends to zero. There will be eigenfunctions supported on each interval and their max will go like $\sqrt{2/|I_j|}$. – Richard Montgomery May 16 '17 at 3:05
• @RichardMontgomery: I don't really have a precise answer, but I can't imagine that having any consequences on this issue. In fact, maybe even a totally explicit (and real analytic) version of my example such as $V(x)=(2+\sin x^2) e^{x^2}$ would work. – Christian Remling May 16 '17 at 16:17
This is Christian Remling's answer. I am just adding this to make it more explicit and underline what he wrote, so vote him up' not me. Take the potential $V(x)$ to be zero on a countable collection of disjoint intervals $I_j$, $j =1, 2, \ldots$ with $\lim_j |I_j| = 0$, and take $V$ postive and tending to infinity on the complement of these intervals. For example, if we assume that $\Sigma_j |I_j| < 1$ we could place all the intervals within the interval $[1,2]$ and take $V(x) = x^2 +1$ on the complement of the $I_j$. The `ground state for $I_j$'', which I will call $\psi_j$, and for which Christian writes down a formula when $I_j = (a,b)$, is zero outside of $I_j$, vanishes on the endpoints of $I_j$, forms the positive arc of a sine wave inside $I_j$, has maximum $\sqrt{2/| I_j|}$ occuring at the midpoint of $I_j$, and has $L_2$ norm equal to $1$. Since the $|I_j| \to 0$ the maxima of these $\psi_j$ tend to $\infty$ with $j$.