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Prove: If $f : [0, 1]\to [0, 1]$ is continuous and $f(0) = f(1)$, then there exists a point $0 \le x\le \frac{1}{2}$ such that $f(x) = f(x+\frac{1}{2})$

Hint: Consider the function $g(x) = f(x) - f(x+\frac{1}{2})$.

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    What's $g(0) + g(1/2)$? What can you conclude? – 2017-01-14
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    You seem to have a perfectly good hint there. What is your question? – 2017-01-14

1 Answers 1

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$$g(x)=f(x)-f\left(x+\frac12\right)$$ Put $x=0$, $$g(0)=f(0)-f\left(\frac12\right)$$ Put $x=\frac12$, $$g\left(\frac12\right)=f\left(\frac12\right)-f(1)$$ Since $f(0)=f(1)$, we have, $$g\left(\frac12\right)=f\left(\frac12\right)-f(0)=-g(0)$$

By the intermediate value theorem, which can be applied due to the continuity of $f$ and hence $g$, there exists an $x\in\left[0,\frac12\right]$ such that $g(x)=0$, which gives $$f(x)=f\left(x+\frac12\right)$$