Show that $h(x) = h(x + \frac{1}{2013})$ for some x in $[0,1]$ Edit: I very much appreciate alternate solutions to the problem, but I would also like to know if there are any problems or suggestions regarding the way I solved it. 
This is a problem I'm attempting from a set of practice Putnam questions (it's also very similar to a problem in Spivak's Calculus.  I was wondering if there were any problems with my solution.
Problem:
Let $h(t)$ be a continuous function on the interval $[0, 1]$ such that $h(0) = h(1) = 0$.  Show that there exists a real number $x \in  [0, \frac{2012}{2013}]$ such that $h(x) = h(x + \frac{1}{2013})$.
My attempt at a solution:
I'm going to generalize it with the case $h(x) = h(x + \frac{1}{n})$ where $n \in \mathbb{N}$ (assuming that won't cause any difficulties).
Let $$g(x) = h(x) - h(x + \frac{1}{n}) \text{ on } [0, \frac{n-1}{n}]$$
We now attempt to prove the problem by contradiction. Assume $g(x) \neq 0$ for all $x$ in $[0, \frac{n-1}{n}]$.Then $h(x) \neq h(x + \frac{1}{n})$ for all $x$ in $[0, \frac{n-1}{n}]$.Certainly $g$ is continuous on the interval, since $h$ is continuous on the interval, so by the intermediate value theorem, it cannot be both positive and negative on the interval (for if it were, then we would have $g(a) > 0$ and $g(b) < 0$ (or vice versa) on some interval, which would mean that there is some $c \in [a, b]$ such that $g(c) = 0$, which we are assuming is not true).  First we assume that $g(x) > 0$ on $[0, \frac{n-1}{n}]$.  If this were the case, then $h(x) > h(x + \frac{1}{n})$ on $[0, \frac{n-1}{n}]$. This implies that $h$ is always decreasing on $[0, \frac{n-1}{n}]$, which is not possible, since $h(0) = h(1)$.  If we assume the opposite case, that $g(x) < 0$ on $[0,\frac{n-1}{n}]$, then $h(x + \frac{1}{n}) > h(x)$ on $[0, \frac{n-1}{n}]$, which is also not possible, because it would mean $h$ is always increasing on $[0,\frac{n-1}{n}]$.  Since neither of these cases is possible, then $g(x)$ must equal $0$ at some point in $[0, \frac{n-1}{n}]$, so by extension, $h(x) = h(x + \frac{1}{n})$ for some $x \in [0, \frac{n-1}{n}]$. To solve the given problem, we simply let $n = 2013$.
 A: By contradiction, assume that:
$$
  \forall x \in \left[0, \frac{n-1}{n} \right] \; h(x) > h \left(x + \frac{1}{n}\right)
$$
Then
$$
  h(0) > h \left( \frac{1}{n} \right) > h \left( \frac{2}{n} \right) > \dots > h(1)
$$
But $h(0) > h(1)$ is obviously false since $h(0) = h(1) = 0$, hence our assumption must be false, that is:
$$
 \exists x \in \left[0, \frac{n-1}{n} \right] : h(x) \leq h \left(x + \frac{1}{n}\right)
$$
And with an analogous argument we can prove that:
$$
 \exists x \in \left[0, \frac{n-1}{n} \right] : h(x) \geq h \left(x + \frac{1}{n}\right)
$$
This proves that the continuous function
$$
 g(x) = h(x) - h \left( x + \frac{1}{n} \right)
$$
assumes both positive and negative values on the interval (or is constant), and hence it must have a zero.
A: $$g(x)=h(x)-h\left(x+\frac{1}{n}\right)$$
$$g(0)=-h\left(\frac{1}{n}\right)$$
$$g\left(1-\frac{1}{n}\right)=h\left(\frac{1}{n}\right)$$
Because $g(0)$ and $g\left(1-\frac{1}{n}\right)$ have opposite signs...
A: Let:
$$g:x\in[0,1-\dfrac{1}{n}]\longmapsto h\left(x+\dfrac{1}{n}\right)-h(x)$$
Evaluate $g$  in $n$ different points:
$$\left\lbrace\begin{array}{rcl} g(0)&=&{\color{green}{h\left(\dfrac{1}{n}\right)}}-0 \\
g\left(\dfrac{1}{n}\right)&=&{\color{blue}{h\left(\dfrac{2}{n}\right)}}-{\color{green}{h\left(\dfrac{1}{n}\right)}}\\
g\left(\dfrac{2}{n}\right)&=&h\left(\dfrac{3}{n}\right)-{\color{blue}{h\left(\dfrac{2}{n}\right)}}\\
\dots & = & \dots  \\
g\left(\dfrac{n-1}{n}\right)  & = & 0-h\left(1-\dfrac{1}{n}\right) \\ 
\end{array}\right.$$
Let's sum:
$$\sum_{k=0}^{n-1} g\left(\dfrac{k}{n}\right)=0$$
So $g$ is either the zero function, or has strictly positive and strictly negative values. Conclude with continuity of $g$ and the intermediate value theorem!
