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$$ Given \ f: \mathbb{R} \mapsto [-1,1] $$ 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.

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  • $\begingroup$ Try $y=\sin(ax)$ and try to find a value of $a$ that works. $\endgroup$ – John Wayland Bales Apr 7 at 0:04
  • $\begingroup$ 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? $\endgroup$ – Matheus Domingos Apr 7 at 0:08
  • $\begingroup$ By "always true" do you mean for any value of $c$? $\endgroup$ – John Wayland Bales Apr 7 at 0:15
  • $\begingroup$ 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! $\endgroup$ – Matheus Domingos Apr 7 at 0:17
  • $\begingroup$ 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? $\endgroup$ – John Wayland Bales Apr 7 at 0:18
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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 $>1$.

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