Let $f:[0,1]\to \mathbb R$ be a continuous function such that $f(0)=f(1)$.
Prove that $f(x)=f\left(x+\frac12\right)$ has a solution for $x\in [0,\frac12]$.
This question has to do with continuity and the intermediate value theorem.
I observed that $f(0)=f\left(\frac12\right)=f(1)$ but I don't see how to show that the function go through zero (i.e has a solution) for all we know it can be a straight line parallel to the $x$ axis in $[0,1]$.
Answer
Hint: put $g(x)=f(x)-f(x+\dfrac{1}{2})$ which is also continous .
Then $g(0)=f(0)-f(\dfrac{1}{2})$ anf $g(\dfrac{1}{2})=f(\dfrac{1}{2})-f(1)=-(f(0)-f(\dfrac{1}{2}))$ so that $g(0)g(\dfrac{1}{2}) <0$
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