Prove that if $f:[-1,1] \to \mathbb{R}$ is continuous and satisfies $f(-1)=f(1)$, then $f(B)=f(B-1)$. Suppose $f:[-1,1] \to \mathbb{R}$ is continuous and satisfies $f(-1)=f(1)$. Prove that $\exists B \in [0,1]$ such that $f(B)=f(B-1)$.
I tried by looking at a new function $g(x)=f(B)-f(B-1), x\in [0,1]$. In order for $f(B)=f(B-1)$ to be true, then $g(x)=0, x\in [0,1]$. Since $f(-1)=f(1)$, I'll try to assume that it's an event function or even periodic, though nothing suggests that, only at a point where it's symmetric. Then, for any $B \in [0,1]$, we have a function where at it's extremities, is equal to each other, and since it's continuous, maybe I can use some $\epsilon$ and $\delta$ to prove that $f(B)=f(B-1)$?
 A: Define the function:
$$ g(x) = f(x) - f(x-1) \quad x \in [-1,1]$$
This function is continuous on$[-1,1]$ since $f$ is continuous. Also using the fact that $f(1)=f(-1)$ we see that: $g(0)g(1) = -(f(1) -f(0))^2 \leq 0$.
So, we have the following two cases:


*

*$g(0)g(1) < 0$ then applying Bolzano's theorem (or the Intermediate value property) we gat that there exist $\displaystyle{ B \in (0,1) \subset (-1,1): g(B)=0 \Longleftrightarrow f(B)=f(B-1)}$.


2.
$g(0)g(1)=0 \Longleftrightarrow g(0)=0 \quad \text{or} \quad g(1)=0$, and in this case $B=0$ or $B=1$.
Therefore, in any case there exist such $B$.
A: HINT: Use the fact that $f(x)$ is continuous.
First, if $f(x)$ is constant, then the result is obviously true. 
Now if $f(x)$ is nonconstant, then because $f(x)$ is continuous it achieves a maximum value on $(-1,1)$. Call this value $M$ at let it occur at $x=x_0$. Then you essentially have two intervals to look at: $(-1,x_0),(x_0,1)$. Think about what $f(x)$ being continuous and all the values it has to achieve on the way to $M$ on $(-1,x_0)$ and what happens then on $(x_0,1)$ means for your desired result. 
Also, note that I ignored $[-1,1]$ because if the maximum occurs at $x=-1$ or $x=1$, then $f(1)=f(-1)=M$ it is easy to show that the result holds because then just repeat the idea above with the minimum value $m$ that must occur on $(-1,1)$. Why can't $m$ occur at $x=-1$ and $x=1$? Because them $f(-1)=f(1)=M=m$ and $f$ would then have to be constant, contradicting the fact that you assumed $f$ was nonconstant. 
A: We have that  $g(0)=f(0)-f(-1),g(1)=f(1)-f(0)$ then $g(0)\cdot g(1)<0$ and from Bolzano's Theorem we have that $g$ has a root in $(0,1)$.
A: generalization
$f$ is continuous on $[0,n], f(0)=f(n)$, then there are $(x_i,y_i),i=1,2,\dotsc,n$ such that
$$x_i-y_i\in\Bbb Z ,  f(x_i)=f(y_i)$$
A: You mean $g(x)=f(x)-f(x-1)$.  Now what can you say about $g(0)$ and $g(1)$?
Btw, this is one formulation of Borsuk-Ulam in dimension $1$ in slight disguise.
