# Boundedness of solutions to second order ODE

Let $$q(x)$$ be a probability density function over $$[0,1]$$. Let $$\lambda > 0$$ and $$f: [0,1] \to \mathbb{R}$$ be any solution to the following ODE: $$\lambda f''(x) + q(x) f(x) = 0, \text{for all }x \in [0,1]; \\ f(0) = 0; \\ \int_{0}^{1} x f(x) q(x) dx = \lambda f(1); \\ \int_{0}^{1} f^{2}(x) q(x) dx = 1.$$ Does there exist an absolute constant $$C$$ such that any solution pair $$(\lambda, f)$$ satisfies $$\sup_{x \in [0,1]} |f(x)| \leq C?$$

Special case: This conclusion holds true for $$q(x)=1$$, i.e., the uniform distribution. Under such circumstances, the solution pair is given by $$(\mu, \sqrt{2}\sin(x/\sqrt{\mu}))$$, where $$1/\sqrt{\mu}=\frac{2k+1}{2}\pi$$ for $$k\geq 0$$. As one can see, all the solutions are bounded in magnitude by $$\sqrt{2}$$.

• Could you explain why you want to know that / why you think it is true, what you tried so far to establish it, etc? Commented Aug 6, 2021 at 17:44
• You're imposing three conditions on the solution of a second order ODE, so typically one would not expect that there is such a solution. Commented Aug 6, 2021 at 17:48
• @username Thanks for your interest. This ODE originates from the first-order Sobolev function space, namely the functions from $[0,1]$ to $\mathbb{R}$ that are absolutely continuous and $\int_{0}^{1} [f'(t)]^{2} dt$ is finite. Given any distribution over [0,1], one can find eigenbasis associated with the function space under the distribution; guaranteed by Mercer's theorem. The ODEs above correspond to the eigenequations. See en.wikipedia.org/wiki/Mercer%27s_theorem Commented Aug 8, 2021 at 21:20
• @ChristianRemling Thanks for your reply. As I explained above, the ODEs orginate from Mercer's theorem. The existence is guaranteed. The uniform distribution (i.e., $q(x)=1$) is often used in textbooks to show that all the eigenfunctions $f(x)$ is bounded. I'm wondering if this boundedness property still holds regardless of the distribution being used. Commented Aug 8, 2021 at 21:22
• For small $\lambda$ (large $k$), the method suggested here should help. Commented Aug 13, 2021 at 21:18