Consider a differential operator


$L=(-1)^ma(x)\frac{d^{2m}}{dx^{2m}}$, with boundary value conditions 


$u^{(j)}(0)=u^{(j)}(1)=0, j=m,m+1,\ldots,2m-1$,


where $m\ge1$ is an integer and $a(x)>0$ is continuous on $x\in[0,1]$.


Question: Are the normalized eigenfunctions $u\_1,u\_2,\ldots$ (with $\int\_0^1 u\_j(x)^2/a(x)dx=1$) of $L$ uniformly bounded in the sense that


$\sup\_{i\ge1}\sup\_{0\le x\le 1}|u\_i(x)|<\infty$?


Related questions have been asked a few times. For instance, [Uniform boundedness of an $L^2[0,1]$-ONB in $C[0,1]$](https://mathoverflow.net/questions/55235/uniform-boundedness-of-an-l20-1-onb-in-c0-1), [Uniform bound on the eigenfunctions of the Laplacian](https://mathoverflow.net/questions/118692/uniform-bound-on-the-eigenfunctions-of-the-laplacian) and [Boundness of Laplacian eigenfunctions](https://mathoverflow.net/questions/61031/boundness-of-laplacian-eigenfunctions). However, they are in more general situations, say, operator is general, or in $\mathbb{R}^n$ or manifolds where the operator becomes Laplacian. It is indeed true that in such general situations uniform boundedness may fail. But my question is less ambitious, i.e., the form of operator and boundary condition is simple and concrete, and the domain is simply [0,1]. So I was wondering if uniform boundedness of the eigenfunctions can still hold.


When $a(x)\equiv1$, uniform boundedness has been proven by Utreras, F. (1988): Boundary effects on convergence rates for Tikhonov regularization, Journal of Approximation Theory, Vol 54, 235--249. Lemma 6 therein with $j=0$ implies uniform boundedness. 


For general continuous coefficient $a(x)>0$, I have not found any results. This might be a question that deserves further investigation? 




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An update: finding eigenvalues of $L$ is equal to finding solutions to the ODE below:


$(-1)^m a(x) D^{2m}u=\lambda u$, $u^{(j)}(0)=u^{(j)}(1)=0, j=m,m+1,\ldots,2m-1$,


If we denote $v(x)=u^{(m)}(x)$, then differentiating $m$ times on both sides of the above equation we get that


$(-1)^m D^m(a(x) D^mv(x))=\lambda v(x)$, $v^{(j)}(0)=v^{(j)}(1)=0, j=0,1,\ldots,m-1$.----(1)


The latter is a Sturm-Liouville equation of order $m$. So it is sufficient to investigate the behavior of eigenfunctions of the last equation (1).


Sturm-Liouville equation is pretty magic. Though in (1) the coefficient $a(x)$ is inside $D^m(\cdot)$, it is not required that $a(x)$ is differentiable. Actually, at least when $m=1$, continuity or even integrability of $a(x)$ is enough to guarantee good properties of the solution; see Weyl (1911)'s famous work.


However, most historic efforts have been put in $m=1$ for S-L problem. Is there a reference systematically introducing S-L problem for general $m$, especially in terms of eigenvalue problems? Thanks. 



Uniform boundedness of eigenfunctions of an elementary differential operator
Here are a few remarks on more specialized situations; this is not a complete answer.


If $m=1$ and $a$ has additional smoothness (say $a\in C^2$, but I can really get away with $a'\in BV$ and probably somewhat less than that), I can run a [Kummer-Liouville transformation](https://encyclopediaofmath.org/wiki/Liouville_normal_form) to bring your equation $-(ay')'=\lambda y$ to Schrödinger form. More specifically, if we set
$$
t=\int\_0^x a^{-1/2}(s)\, ds, \quad\quad u = a^{1/4} y ,
$$
then $u$ solves $-d^2u/dt^2 + Vu = \lambda u$ where $V=a''/4 - a'^2/(16a)$. Now the uniform boundedness of the eigenfunctions is obvious because we have asymptotic expansions as $|\lambda|\to\infty$ and the asymptotics of the solutions are just those of the free case ($V=0$).


For general $m>1$, you could try similar transformations; my advisor has worked on this, though I'm not sure if this immediately applies to your problem. See [here.](https://mathscinet.ams.org/mathscinet-getitem?mr=2238021 "Behnke, Horst; Hinton, Don. Transformation theory of symmetric differential expressions. Adv. Differ. Equ. 11, No. 6, 601-626 (2006). zbMATH review at https://zbmath.org/1160.34076")


If $a$ is not smooth, the argument in this form doesn't go through without changes: sufficiently singular $V$'s (for example, $\delta'$ contributions) can change the asymptotics of the solutions.


