Problem
Let $f(x)$ be twice-differentiable over $[-2,2]$, $|f(x)|\leq 1$ and $f^2(0)+[f'(0)]^2=4.$ Prove that $\exists \xi \in (-2,2):f(\xi)+f''(\xi)=0.$
My Proof
First，by Lagrange' Mean Value Theorem，we have $$\exists \xi_1 \in (-2,0),\xi_2 \in (0,2):f'(\xi_1)=\frac{f(0)-f(-2)}{2},f'(\xi_2)=\frac{f(2)-f(0)}{2}.$$ Thus $$|f'(\xi)|\leq \frac{|f(0)|+|f(-2)|}{2}\leq \frac{1+1}{2}=1,$$ and likewise $$|f'(\xi_2)|\leq 1.$$ Denote$$F(x):=f^2(x)+[f'(x)]^2.$$ It's easy to verify that $F(x)$ is continuous and differentiable over $[\xi_1,\xi_2]$. Therefore, $F(x)$ can reach its maximum value over the interval above. But $$F(\xi_1)\leq 1+f'^2(\xi_1)\leq 2<F(0),$$ and similarily $$F(\xi_2)\leq 2<F(0),$$ the maximum value can not come out at the end points of the interval.Hence, it must be taken over $(\xi_1,\xi_2)$. Let the point be $x=\xi\in(\xi_1,\xi_2)\subset (-2,2)$. Then $$F'(\xi)=0,$$ which implies that $$2f'(\xi)[f(\xi)+f''(\xi)]=0.$$
Now, Notice that $$4\leq F(\xi)=f^2(\xi)+f'^2(\xi)\leq 1+f'^2(\xi),$$ which implies $f'(\xi)\geq 3$，and hence$f'(\xi) \neq 0$. It follows that $$f(\xi)+f''(\xi)=0,$$ which is desired.
Please correct if I'm wrong. Hope to see other solutions. THX.