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 $\sqrt{2}$.