If $\{f(0)\}^{2} + \{f'(0)\}^{2} = 4$ then there is a $c$ with $f(c) + f''(c) = 0$

The solution presented here is not mine and has been reproduced from Alexanderson et al. The solution is a gem as it uses several key concepts in analysis.

Let $G(x) = [f(x)]^2 + [f'(x)]^2$ and $H(x) = f(x) + f''(x)$. Since $H$ is continuous, it suffices to show that it changes sign. So, lets assume $H(x) > 0$ for all $x$ or $H(x) < 0$ for all $x$ and obtain a contradiction.

Since $|f(0)| \leq 1$ and $G(0) = 4$, either $f'(0) \geq \sqrt{3}$ or $f'(0) \leq -\sqrt{3}$. We will show for the case in which $H(x) > 0$ and $f'(0) \geq \sqrt{3}$; the other cases are similar.

Assume that the set $S$ of positive $x$ with $f'(x) < 1$ is nonempty and let $g$ be the greatest lower bound of $S$. Then $f'(0) \geq \sqrt{3}$ and continuity of $f'(x)$ imply $g > 0$. Now, $f'(x) \geq 0$ and $H(x) \geq 0$ for $0 \leq x \leq g$ lead to

$G(g) = 4 + 2 \int_{0}^{g} f'(x)[f(x) + f''(x)] dx \geq 4$

Since $|f(g)| \leq 1$, this implies $f'(g) \geq \sqrt{3}$. Then continuity of $f'(x)$ tells us that there is an $a > 0$ such that $f'(x) \geq 1$ for $0 \leq x < g+a$. This contradicts the definition of $g$ and hence $S$ is empty. Now $f'(x) \geq 1$ for all $x$ and this imples that $f(x)$ is unbounded, contradicting $f(x) \leq 1$.