# Uniform boundedness of an $L^2[0,1]$-ONB in $C[0,1]$

Assume that we have an orthonormal basis of smooth functions in $L^2[0,1]$. Are there useful practical criteria to determine whether the sup-norm of the basis functions has a uniform bound? I am sure people investigated this but I do not manage to find a useful result. Can someone offer me references, where I can read about this or a connected problem?

I would like to investigate the properties of some fourth order differential operators (with the corresponding boundary conditions) in $C[0,1]$. Since these operators are nice selfadjoint operators with compact resolvent in $L^2[0,1]$, and the corresponding eigenfunctions are analytic, it would be straightforward to ask whether the properties I need can be carried over using the basis.

Edit on 02/18/2011: Though completely unrelated to my question, googling brought me the following interesting paper:

This gives me hope, however, that my question might have interested somebody...

-
It is clear (see also the answer by Piero) that you cannot expect boundedness in general. However, in many cases where you can explicitly calculate the eigenfunctions, they are bounded. So maybe there is a harmonic analysis guru who can help me... –  András Bátkai Feb 13 '11 at 17:48
Isn't there a theorem of Orlicz which states that there is no uniformly bounded unconditional basis for the standard Hilbert space on the unit interval? I think that this subject is covered in detail in the standard monograph of Lindenstrauss and Tzafriri but, unfortunately, cannot access it at present. –  jbc Dec 4 '12 at 8:37
The above comment is, of course, terribly wrong (the trigonometric basis). The theorem of Orlicz applies to $L^p$ with $p$ not equal to $2$. Sorry. –  jbc Dec 4 '12 at 9:33

I might be wrong, but there answer to you general question is 'no'. Take an arbitrary function with $L^2$ norm equal to one, then start an orthogonalization procedure (which leaves your first function fixed) and produce an ONB. You have no way to bound the sup norm of the first function.

But your actual needs seem much more natural, and standard. For an ONB which arises as the eigenfunction set of a given selfadjoint, positive, elliptic operator $L$ of order $m$, you can resort to an elementary Sobolev embedding. Let $f_j$ be the eigenfunction corresponding to the eigenvalue $c_j$. Then the sup norm of $f_j$ is less than the $H^{1/2+\epsilon}$ norm of $f_j$, which is equivalent to the $L^2$ norm of $L^{\frac 1{2m}+\epsilon}f_j=c_j^{\frac 1{2m}+\epsilon}f_j$ which is exactly $c_j^{\frac 1{2m}+\epsilon}$. In general I think you can not do better than that. When $L$ is the Laplace-Beltrami operator on some compact manifold, or powers thereof, there exist much more precise results.

-
Thanks. Yes, I wanted a method to decide, it is clear that in general the answer is no, it is easy to construct an artificial example. Do you have a reference on your last sentence? –  András Bátkai Feb 12 '11 at 23:03
You may start from <a href="arxiv.org/abs/0907.4827">; this </a> paper which has some sharp results and an history of the subject. There are other papers by the same author on arXiv with related results –  Piero D'Ancona Feb 13 '11 at 0:17
PS I do not know how to fix the link - hopefully you can work it out –  Piero D'Ancona Feb 13 '11 at 0:18
Thanks, I managed it. –  András Bátkai Feb 13 '11 at 17:44