mathematical analysis problem about intermediate value theorem

Question

Suppose $f(x)\in C^2(-\infty ,\infty )$, $|f(x)|\le 1$, and $(f(0))^2+(f'(0))^2=4$. Prove that $\exists \xi $ such that $f(\xi )+f''(\xi )=0$.

I think the function $(f(x))^2+(f'(x))^2$ may help but I don't know how to use this function.

Answer 1

Let $g(x)=f(x)^2 + f'(x)^2$. We know $g(0)=4$ and $g'(x)=2f'(x)(f(x)+ f''(x))$.

We can prove the proposition by analyzing several cases.

First case: $g'(0)=0$. Then either $f'(0)=0$ or $f(0)+f''(0)=0$. But $f'(0)=0\implies f(0)=2$ which can't be. Hence $f(0)+f''(0)=0$.

From now on, assume by contradiction $f(x)+f''(x)\ne 0$ for all $x$. Since $f+f''$ is a continuous function, it must have only one sign over $(-\infty, \infty)$ (this, again, can be proved with the intermediate value theorem).

Second case: $g'(0)>0$. Since $|f(x)|\le 1$ for all $x$, we can show that there must be some $x>0$ s.t. $g'(x)<0$. By intermediate value theorem, there is some $y\in (0,x)$ s.t. $g'(y)=0$. Assume this is the "first $y>0$" s.t. $g'(y)=0$(or, the minimum of such values, if you will). Then $g'>0$ on $[0,y)$. Hence $g(y)>g(0)=4$. But since we assumed $f(y) + f''(y)\ne 0$, then $f'(y)=0$, hence $4 < g(y)=f(y)^2$ which can't be.

The case $g'(0)<0$ is analogous. By the same argument of the previous case, there must be some $y<0$ s.t. $g'(y)=0$. Assume this is "the last" $y<0$ s.t. $g'(y)=0$ (or, the maximum of such values, if you will). Then $g'<0$ on $(y,0]$. Hence $g(y)>g(0)=4$. But since we assumed $f(y) + f''(y)\ne 0$, then $f'(y)=0$, hence $4 < g(y)=f(y)^2$ which can't be.