Real analysis proof verification 2

I am taking a real analysis course and I'm supposed to solve a problem and present in class, but I'm not sure if this proof is good. This is the problem:

Let a function $f$ be continuous on $[0,1]$ and differentiable on $(0,1)$. Assume that $f(0)=f(1)$. Prove that for any $n\in N$ there exist points $0<c_1<...<c_n<1$ such that $\sum_{k=1}^nf'(c_k)=0$. Hint: apply Rolle's theorem, induction and Darboux's theorem.

This is what I have done:
Let $g(x):=f(x)-f(1)$, then $g$ is continuous on $[0,1]$ and differentiable on $(0,1)$. $g(1)=g(0)=0$ so Rolle's theorem gives that there exists a point $c_1\in (0,1)$ s.t. $g'(c_1) =f'(c_1) =0$. So the statement is true for $n=1$.

If $f(x)=C, C\in R, \forall x\in [0,1]$ then the statement is true for any points in $(0,1)$.
Assume that $f$ is not constant.

Assume that there exists $n$ points, $0<c_1<...<c_n<1$ s.t. $\sum_{k=1}^nf'(c_k)=0$.
Since $f(0)=f(1)$ the function is not increasing or decreasing on the whole interval $[0,1]$. If there's no point in this sum where the derivative is zero, we can add that point to the sum and thus the statement is true for $n+1$ points. Assume there is a point $c_i$ s.t. $f'(c_i)=0$ in the sum. Since $f$ is increasing on some interval of $[0,1]$ and decreasing on some interval of $[0,1]$ there exists points $d_1, d_2 \in(0,1)$ s.t. $f'(d_1)<0$ and $f'(d_2)>0$.
$f'(d_1)<0=f'(c_i)<f'(d_2)$. If $|f'(d_1)|=f'(d_2)$ then $f'(c_i)=f'(d_1)+f'(d_2)$ and so we can exchange the point $c_i$ with $d_1$ and $d_2$ and so the statement is true for $n+1$ points.
If $|f'(d_1)|>f'(d_2)$ Darboux's theorem gives that there exists a point $d_3$ bewtween $d_1$ and $d_2$ s.t. $f'(d_1)<f'(d_3)<0<f'(d_2)$ and $|f'(d_3)|=f'(d_2)$. Then we can exchange the point $c_i$ with $d_2$ and $d_3$ and the statement is true for $n+1$ points.
If $|f'(d_1)|<f'(d_2)$ we can do the same thing with Darboux's theorem.

• From your argument I suppose you are not allowed to use mean value theorem? – user99914 Oct 15 '15 at 4:57
• I have made some changes now to the proof. – efk Oct 15 '15 at 14:37

For a simpler alternative using MVT, there exist points $(k-1)/n \leqslant c_k \leqslant k/n$ such that
$$\begin{split} 0 &= f(1) - f(0) \\ &= \sum_{k=1}^n [f(k/n) - f((k-1)/n)]\\ &= \sum_{k=1}^n f'(c_k)[k/n - (k-1)/n] \\ &= \frac1{n}\sum_{k=1}^n f'(c_k) \\ \implies &\sum_{k=1}^n f'(c_k)= 0 \end{split}$$