1
$\begingroup$

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!

$\endgroup$
1
  • $\begingroup$ This differential equation isn't too difficult via the method of the characteristic equation. $\endgroup$
    – Mark
    Apr 12 '17 at 23:31
2
$\begingroup$

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 $|f(x_0)|$.

$\endgroup$
3
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.