Schrödinger eigenfunctions are bounded

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.

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 $j$.