Suppose $f:[0,1] \rightarrow \mathbb{R}$ is twice differentiable and $f = f''$, $f(0) = f(1) = 0$, then $f$ is constant.

Question

Suppose $f:[0,1] \rightarrow \mathbb{R}$ is twice differentiable and $f = f''$, $f(0) = f(1) = 0$.
Show that $f$ is constant.

The main approach we are taught to use at these kind of problems is using Rolle's theorem/MVT. I am not sure how this this idea can be applied here and moreover I struggle to relate it to the condition $f = f''$.
I hope to get an idea about it, thanks

Answer 1

As $f$ is continuous, $|f|$ attains a maximum somewhere. If $f$ is not constant, it must happen at some $c\in(0,1)$.

WLOG, let's say $f(c)>0$ (otherwise consider $-f$). Then, $c$ is a maximum of $f$. But $f''(c)=f(c)>0$. As $f$ is continuous, $f''$ is too and thus near $c$ $f''$ is positive, that is, $f'$ is strictly increasing.

As $f'(c)=0$ being a maximum, $f'(x)>0$ for $x$ to the right of $c$ (sufficiently close to $c$).

Then, $f$ increases and $c$ cannot be a maximum.

Answer 2

Perhaps using differential equation is quite straight forward (but quite boring). Or we use analytical method. First, if $(c,f(c))$ is a critical point, then by second derivative test, if $f(c)>0$, then $f$ is a local minimum, vice versa.

We first claim that there is a zero $s\in(0,1)$. Then we can keep apply the same argument on the $(0,s)$, $(s,1)$, and so on. This means $f\equiv0$.

By Rolle's theorem on $[0,1]$, there exists $c\in(0,1)$ where $f'(c)=0$. If $f(c)=0$, then we are done!

• If $f(c)\ne0$, then WLOG $f(c)>0$, this means $f$ has a local minimum at $x=c$. But $f(0)<f(c)$, so there must be a local maximum attained by $f$ at $d$ in $(0,c)$, then $f"(d)\le0$. Hence, $f(d)\le0$. If $f(d)=0$, then we are done!
• If $f(d)<0$, then by intermediate value theorem on $(d,c)$, we have a zero in it, let say $e$. Then we are done!