Problem involving twice continously differentiable function

Question

Let f be twice continuously differentiable on [0,1]. Suppose that $$f''(x) +xf(x)=0 \tag1$$ $$f'(0)=0, \text{ and} \tag2$$ $$\int_0^1f(x)dx=0 \tag3$$ Prove that $$|f(1)|\leq \frac{\sqrt5}{2} \Vert f \Vert_{L^2([0,1])}$$

I have found different bounds for this function, many of which are tighter than the one given. Since this is a qualifying exam problem, I was wondering if someone could illuminate the idea behind getting this bound exactly.

A couple of remarks are in order. Condition (1) implies $|f(1)|=|f''(1)|$ and $f''(0)=0$. Condition (3) implies that there exists $x_o \in [0,1]$ s.t $f(x_0)=0$. Furthermore, since $f''$ is the product of two continuously differentiable functions, then $f'''(x)=f(x) + xf'(x)$ on $(0,1)$. How can we put all of this together to make this bound appear?

Also, on another note, is it plausible to believe that this function will be identically $0$? I have found very tight bounds for this function, but maybe I am just wrong.

Answer 1

This is a partial answer. The general solution of $f''+x f = 0, f'(0)=0$ can be explicitly written in terms of Airy functions, namely $$ f(x) = \frac{1}{3} \left(3 c_1 \text{Ai}\left(\sqrt[3]{-1} x\right)+\sqrt{3} c_1 \text{Bi}\left(\sqrt[3]{-1} x\right)\right)$$

Computing $\int_0^1 f(x) dx$, numerically, for different values of $c_1$ indicates that the integral is zero only when $c_1=0$, hence, $f(x)\equiv 0$.

Answer 2

In terms of regularized hypergeometric functions, we have $$f(x)=c_1\,\, _0\tilde{F}_1\left(;\frac{2}{3};-\frac{x^3}{9}\right)$$ which gives $$\int f(x)\,dx=c_1\frac{\sqrt{3}\, \Gamma \left(\frac{1}{3}\right) }{2 \pi }\,x\,\, _1F_2\left(\frac{1}{3};\frac{2}{3},\frac{4}{3};-\frac{x^3}{9} \right)$$ $$\int_0^1 f(x)\,dx=-c_1\, \frac{\Gamma \left(-\frac{2}{3}\right)}{\pi\sqrt{3} }\, _1F_2\left(\frac{1}{3};\frac{2}{3},\frac{4}{3};-\frac{1}{9} \right)$$ $$\int_0^1 f(x)\,dx\sim 0.70829821 \,c_1$$ So, for the strange second condition, a strange function.