A question about a positive continuous function 
Suppose $f \in C[0, 1]$ satisfies following properties:
1)$f(0) = f(1) = 0$
2)$\forall x \in (0, 1)$ $f(x) > 0$
Do there always exist such $c$ and $d$ in $(0, 1)$ that $f(c) = f(d) = d - c$?

I  have tried to consider the function $g(x) = \max\{t \in [0, 1]|f(t) = f(x)\} - x - f(x)$ which is strictly positive in $0$ and negative in the point of largest point of maximum. Thus if that function were always continuous, our problem would have been solved. Unfortunately, it isn’t.
 A: This is a particular case of Toeplitz's conjecture, which if the argument below is correct, is way simpler than the cases in which the conjecture is known to be true. the conjecture asks for the existence of a non-degenerate square inscribed in any simple closed curve of the plane.
In our case the simple closed curve is formed by the graph of the function $f$ and the segment $[0,1]\times \{0\}$.

Let's extend $f(x)$ to a function on $\mathbb{R}$ by defining it to be $0$ outside of $[0,1]$.
Define $g(x)=f(x+f(x))-f(x)$.
The function $g$ is continuous, since it is a composition of continuous functions.
It satisfies that $g(0)=f(f(0))-f(0)=0$ and $g(1)=f(1+f(1))-f(1)=f(1)-f(1)=0$.
Observe that also, if $x\in (0,1)$ and $x+f(x)\notin[0,1]$, then $g(x)\neq0$.
Therefore, if for some $x\in(0,1)$ we have $g(x)=0$, then $x+f(x)\in(0,1)$ and the points $c=x, d=x+f(x)$ satisfy that $f(c)=f(d)=d-c$.

To get a contradiction, assume that $g(x)>0$ on $(0,1)$.
Let $x_0\in(0,1)$ be the supremum of the points in which $f$ attains its maximum. Note that the set where $f$ attains its maximum is compact and contained in $(0,1)$.
Then $g(x_0)>0$ implies that $f(x_0+f(x_0))>f(x_0)>0$. Therefore, $x_0+f(x_0)\in(0,1)$ and $f$ attains on it a larger value than that on $x_0$. This is a contradiction. Therefore we cannot have $g(x)>0$ for $x\in(0,1)$.

Assume now that $g(x)<0$ for $x\in (0,1)$. 
Pick any point $x_0\in (0,1)$. And without loss of generality we can assume that $x_0$ is the infimum of $\{x\in[0,1]:\ f(x)=f(x_0)\}\subset (0,1)$. By assumption $g(x_0)<0$ and therefore $f(x_0+f(x_0))<f(x_0)$.
Consider the set $K_0=\{x\in[0,1]:\ x+f(x)\geq x_0\}$. This set is non-empty (contains $x_0$) compact, contained in $(0,1)$. Observe that it must contain some points $<x_0$, otherwise it will force $f(x_0)=0$, which is not allowed.
Let $x_1=\inf K_0<x_0$. Then $x_1+f(x_1)=x_0$. Since $g(x_1)<0$ we have that $f(x_0)=f(x_1+f(x_1))<f(x_1)$. But then, by continuity, there must be a point in $(0,x_1)$ where $f$ takes the value $f(x_0)$. This is a contradiction with the choice of $x_0$.

Since assuming $g(x)>0$ on $(0,1)$ and assuming $g(x)<0$ on $(0,1)$ are both impossible, we must have $g(x)=0$ for some point in $c\in(0,1)$. Then define $d=c+f(c)$ and we will have $f(c)=f(d)=d-c$.
