# Existence of the function

$$Given \ f: \mathbb{R} \mapsto [-1,1]$$ a function differentiable up to the second order and satisfying the following relation: $$(f(0))^2+(f'(0))^2=4 \tag{1}$$ Is it true that if f satisfies equation (1) then there must be a c∈$$\mathbb{R}$$ which satisfies equation (2)? $$f(c)+f''(c)=0 \tag{2}$$

I tried to solve this problem by trying to solve the equation 2, but I had problems to limit the domain.

I'd like to solve this problem without using the theory of differential equations, but I have no idea of how to do that.

• Try $y=\sin(ax)$ and try to find a value of $a$ that works. – John Wayland Bales Apr 7 at 0:04
• In the problem, I'd like to show that this is always true or find a counterexample. With this approach, I don't show what I want, do I? – Matheus Domingos Apr 7 at 0:08
• By "always true" do you mean for any value of $c$? – John Wayland Bales Apr 7 at 0:15
• No, I mean: Is always there a function f that satisfies that problem? In this problem the main issue is about the existence of the function, I guess I hadn't made a good translation, sorry! – Matheus Domingos Apr 7 at 0:17
• Did you try my suggestion? What value of $a$ works for equation (1)? Using that value of $a$ is there a $c$ that works for equation (2)? How many? – John Wayland Bales Apr 7 at 0:18

## 1 Answer

Argue by contradiction. Replace $$f$$ by $$-f$$ if necessary, we assume $$f+f''>0$$ everywhere. Replace $$f$$ by $$f(-x)$$ if necessary, we can assume $$f'(0)>0$$. Actually $$f(0)^2+f'(0)^2=4$$ and $$f$$ ranges in $$[-1, 1]$$ implies $$|f'(0)|\geq \sqrt 3$$; in particular $$f'(0)\geq \sqrt 3$$. Now observe $$[f^2+(f')^2]'=2f(x)f'(x)+2f'(x)f''(x)=2f'(x)[f(x)+f''(x)].$$ Let $$T$$ be the biggest number so that $$f'\geq 1$$ on $$[0, T]$$, we see $$f^2+(f')^2$$ is increasing on $$[0, T]$$, so $$f^2+(f')^2>4$$ at $$T$$, which then implies $$f'(T)>\sqrt 3$$, thus $$f'>1$$ on some $$[0, T+\delta]$$. This implies actually $$T=\infty$$. Now $$f'>1$$ on $$[0, \infty)$$ and this makes $$f$$ not bounded from above, contradicts the fact $$f$$ ranges in $$[-1, 1]$$.

This argument should work if you replace $$4$$ by any number $$>1$$.