TITLE: Schrödinger eigenfunctions are bounded
QUESTION [6 upvotes]: 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 $n\geq 0$, 
$$|| \phi_i ||_\infty <M $$

REPLY [2 votes]: 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 $j$.