$f$ is a continuous function from $[0,1]$ to $\mathbb{R}$ s.t. $f(0)=f(1)$, prove $\exists c$ in $[0,\frac{1}{2}]$ such that $f(c)=f(c+ \frac{1}{2})$ $f$ is a continuous function from $[0,1]$ to $\mathbb{R}$ and $f(0)=f(1)$, prove that exists $c$ in $[0,\frac{1}{2}]$ such that $f(c)=f(c+ \frac{1}{2})$
I have no idea about this question,anyone could help me?
 A: We wish to prove the function $g(x)=f(x)-f(x+\frac{1}{2})$ defined in $[0,\frac{1}{2}]$ is zero at some point. Notice this function is continuous. Assume the function is never zero. Then it must always be positive or always negative (since otherwise we can apply the intermediate value theorem and find a zero).
We do the case when it is always positive.
Then $f(0)>f(\frac{1}{2})$ and $f(\frac{1}{2})>f(1) \implies f(0)>f(1)$ a 
contradiction!
the other case is analogous.

As an extra, the problem can be solved similarly when $2$ is replaceed by $n$  a natural number and the interval is changed from $[0,\frac{1}{2}]$ to $[0,\frac{n-1}{n}]$. And it turns to be false for any $\alpha$ that is not a natural number(although the construction for a counter-example is rather erratic)
A: Hint: Consider the function $g(x) = f(x+1/2) - f(x)$. If $g(0) = 0$ you're done. If not, then ...
A: Consider the function $g:[0,\frac{1}{2}]\to\mathbb{R}$ defined as
$$g(x)=f(x)-f(x+\frac{1}{2}).$$
Then $g$ is continuous since $f$ is continuous, and $g(0)=f(0)-f(\frac{1}{2})$ and $g(\frac{1}{2})=f(\frac{1}{2})-f(1)=f(\frac{1}{2})-f(0)$ by assumption. 
Now if $f(0)=f(\frac{1}{2})$, then we can take $c=0$ and $f(c)=f(c+ \frac{1}{2})$. If $f(0)-f(\frac{1}{2})\neq 0$, then $g(0)$ and $g(\frac{1}{2})$ have opposite sign, by intermediate value theorem, there exists $c\in[0,\frac{1}{2}]$ such that $g(c)=0$, i.e. $f(c)=f(c+ \frac{1}{2})$
