Suppose $f:\mathbb R \rightarrow \mathbb R$ is differentiable with $f(0)=f(1)=0$ and $\{x: f^{'}(x)=0\} \subset \{x: f(x)=0\}$.

Question

Suppose $f:\mathbb R \rightarrow \mathbb R$ is differentiable with $f(0)=f(1)=0$ and $\{x: f^{'}(x)=0\} \subset \{x: f(x)=0\}$. Show that $f(x)=0$ for all $x \in [0,1]$.

Since $f$ is differentiable on $[0,1]$, by Rolle's theorem, there exist a $c \in(0,1)$ such that $f^{'}(c)=0$. Let $A= \{x: f^{'}(x)=0\}$ and $B=\{x: f(x)=0\}$. Now, $c\in A \subset B$. This implies $f(c)=0$.

How I claim that $f(x)=0$ for all $x \in [0,1]$?

Answer 1

You are almost there! Note that since $[0,1]$ is compact, the function $f$ takes its highest and lowest values on the interval.

By proof of contradiction and WLOG, suppose that the maximal value that $f$ takes is $r>0$. So, there exists some $z\in[0,1]$ such that $f(z)=r$. Note that $z\in(0,1)$ since $f(0)=f(1)=0$.

Now suppose that $f^{'}(z)\neq0$. WLOG suppose it is greater than $0$. That implies that there is some $\epsilon>0$ such that $f(z+\epsilon)>f(z)$, a contradiction. So, $f^{'}(z)=0$, which is a contradiction.

So, the highest and lowest values the function takes on the interval are both $0$, proving the desired result.

Answer 2

Fix an $n$, and consider the partition $\{0,1/2^{n},2/2^{n},...,2^{n}/2^{n}=1\}$ of $[0,1]$. For any subinterval $[(i-1)/2^{n},i/2^{n}]$, there is a $c_{i,n}\in[(i-1)/2^{n},i/2^{n}]$ such that $f(c_{i,n})=0$. Indeed, if $f'((i-1)/2^{n})=0$ or $f'(i/2^{n})=0$ will do by the assumption. Or else, say, $f'((i-1)/2^{n})>0$ and $f'(i/2^{n})<0$, then by Intermediate Value Theorem for derivative (Darboux's Theorem), there is some $c_{i,n}\in[(i-1)/2^{n},i/2^{n}]$ such that $f'(c_{i,n})=0$, and again by the assumption we have found that point.

As the union of all such points $\{0,1/2^{n},2/2^{n},...,2^{n}/2^{n}=1\}$ with $n=1,2,...$ are dense in $[0,1]$, so are those $\{c_{i,n}\}$. By the continuity of $f$, we must have $f=0$.