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This is from Putnam:

If a function $f: \mathbb{R} \to [-1, 1]$ is such that $f''(x)$ exists for all $x \in \mathbb{R}$ and $\{f(0)\}^{2} + \{f'(0)\}^{2} = 4$ then prove that there is a point $c$ such that $f(c) + f''(c) = 0$.

Now the condition concerning values of $f(0), f'(0)$ (and the fact that range of $f$ is subset of $[-1, 1]$) implies that $f'(0) \neq 0$ so that function is not a constant. If we take $g(x) = \{f(x)\}^{2} + \{f'(x)\}^{2}$ then we can see that $g' = 2f'(f + f'')$. One hope of solving the problem is to show that $g$ attains local extrema at some point different from an extrema of $f$. This would ensure that $g'$ vanishes without making $f'$ vanish and that will lead to $f + f'' = 0$. However the constraint on $f(0), f'(0)$ does not help in analyzing the extrema of $f$ or $g$.

Another line of thought which could be helpful here is to consider $F(x) = f'(x)\sin x - f(x)\cos x$ so that $F'(x) = \sin x\{f(x) + f''(x)\}$ but in this case I am not able to see how to use the condition on values $f(0), f'(0)$ to impose some constraint on $F(x)$.

Perhaps both the approaches which came to my mind are not in right direction (or may be they are, but I am unable to see). Please provide any hints.

Update: Based on the answer by Sandeep Thilakan, I have dropped the requirement for continuity of $f''(x)$.

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  • $\begingroup$ The solutions look like sine or cosine. $\endgroup$ – Lord Soth Mar 18 '14 at 17:59
  • $\begingroup$ @LordSoth: I have a hunch like that but its not exactly trying to solve the differential equation $y'' + y = 0$, we need to show that there is at least one point where $y'' + y$ vanishes. $\endgroup$ – Paramanand Singh Mar 18 '14 at 18:01
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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 $f(x) \leq 1$.

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  • $\begingroup$ Your proof shows that the problem remains valid even if we replace the constant $4$ by some $a > 0$ and the range $[-1, 1]$ by $[-b, b]$ where $b^{2} < a$. Thus effectively we need $f'(0) \neq 0$. $\endgroup$ – Paramanand Singh Mar 19 '14 at 3:39
  • $\begingroup$ Also the continuity of $f''(x)$ is not needed in any fundamental way. The intermediate value property (IVP) is satisfied by derivatives so we don't need the continuity of $f''(x)$ in order to let $H(x)$ have IVP. $\endgroup$ – Paramanand Singh Mar 19 '14 at 3:47

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