5

Question: $f\in C^2(-\infty,+\infty)$, $|f(x)|\leq 1$, $[f(0)]^2+[f'(0)]^2=4$. Prove that there exists $\xi$ s.t. $$ f(\xi)+f''(\xi)=0. $$

I find this question in the chapter about mean value theorem, and I set up function $G=f^2+f'^2$ as usual. WLOG set $f(0)\geq 0$ and $f'(0)\geq 0$. Use Taylor mean value theorem I get $$ f(1)=f(0)+ f'(0)+\frac{1}{2}f''(\eta). $$ Set $x=f(0)\geq 0$ I get $$ f''(\eta)\leq 1-x-\sqrt{4-x^2}\leq -1. $$ So I get $$ f(\eta)+f''(\eta)\leq 0 $$ I want to prove by contradition， then I assume $\forall x\in(-\infty,+\infty)$, $f(x)+f''(x)<0$. Then I have no idea.

My question is: Is this idea correct? How to use the function $f^2+f'^2$(I haven't use it yet)?

md2perpe
  • 26,770
yahoo
  • 769

1 Answers1

6

Let $G=f^2+f'^2\in C^1(-\infty,+\infty)$. If $f'(x)\geq 1$ for all $x\geq 0$ then $$f(x)=f(0)+\int_0^x f'(t) dt\geq f(0)+x\to +\infty$$ which contradicts the fact that $|f(x)|\leq 1$. Similarly, it can not be that $f'(x)\leq -1$ for all $x\geq 0$. Hence there exists $x_1>0$ such that $|f'(x_1)|\leq 1$, which implies that $G(x_1)\leq 1+1=2$. In the same way, we prove that there is $x_2<0$ such that $|f'(x_2)|\leq 1$ and $G(x_2)\leq 2$. Since $G(0)=4$, by continuity, there exist $a,b\in \mathbb{R}$ such that $$a:=\max \{x<0: G(x)=2\}\quad,\quad b:=\min \{x>0: G(x)=2\}.$$ Note that $a<0<b$, $G(a)=G(b)=2$ and $G(x)>2$ for $x\in (a,b)$.

By the MVT, there is $c\in (a,b)$ such that $$G'(c)=2f'(c)(f(c)+f''(c))=0.$$ If $f'(c)=0$ then $G(c)\leq 1$ which implies that there is $d\in (a,b)$ such that $G(d)=2$ which is impossible. Hence $f(c)+f''(c)=0$.

Robert Z
  • 145,942