Let $f$ be twice twice continuously differentiable on the interval $[0,1]$ and suppose that $\exists c>0$ such that:

$$ \frac{d^2}{dx^2}f(x) = c f(x) $$

I would like to show that $|f(x)| \le |f(0)|$ or $|f(x)| \le |f(1)|$, or more succinctly that

$$|f(x)| \le \max\{|f(0)|, |f(1)|\}$$

I have tried attacking this using the fundamental theorem of calculus but no success. I even solved the differential equation involving the 2nd derivative but that seems like an awfully long route. I am wondering whether the is a succinct approach to this. I would appreciate any hints, or references to useful theorems, but please, do not provide a fully-fledged solution. Much obliged!

  • $\begingroup$ This differential equation isn't too difficult via the method of the characteristic equation. $\endgroup$
    – Mark
    Apr 12 '17 at 23:31

If $f$ is constant, the claim holds. Assume then that $f$ is not constant.

It's enough to show that $$\max_{x\in [0,1]} |f(x)| \in \{|f(0)|,|f(1)|\}$$.

Suppose $\max_{x\in [0,1]} |f (x)|$ is attained at $x_0\in (0,1)$. Equivalently, $\max_{x\in [0,1]} f^2(x)$ attained at $x_0$ (allowing us to differentiate). This implies $\frac{d}{dx} f^2 (x_0)= 2f'(x_0)f(x_0) =0$. Yet,since $f$ is not constant, $f(x_0)\ne 0$, and so $f'(x_0)=0$. The differential equation then implies $f''(x_0)>0$ if $f(x_0)>0$, and $f''(x_0)<0$ if $f(x_0)<0$. By the second derivative test, the former case corresponds to $x_0$ being a local minimum (with function strictly larger than $f(x_0)$ in some punctured neighborhood of $x_0$), and the latter $x_0$ being a local maximum (with function strictly smaller than $f(x_0)$ in some punctured neighborhood of $x_0$). Each of the two alternatives is in violation to the maximality of $|f(x_0)|$.

  • $\begingroup$ Why is it necessary that $f(x_0)\neq 0$? Isn't it sufficient that at a maximum of $f$ we necessarily have $f'(x_0)=0$? Also what happens in the case that $f''(x_0)=0$? $\endgroup$
    – Matt
    Apr 13 '17 at 3:17
  • $\begingroup$ Remember: $|f(x_0)|$ is the maximum of $|f(x)|$ over all $x\in [0,1]$. That is $|f(x)|\le |f(x_0)|$ for all $x\in [0,1]$. So if $f(x_0)=0$, then $f(x)=0$ for all $x$, a constant function. Since $f(x_0)\ne 0$, the differential equation guarantees that $f''(x_0)\ne 0$. $\endgroup$
    – Fnacool
    Apr 13 '17 at 3:23
  • $\begingroup$ Ahhh very simple! Thanks! $\endgroup$
    – Matt
    Apr 13 '17 at 3:24

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