Suppose that the function $f$ is continuous on $[0,1]$ and $f(0)=f(1)$, prove that there exists a number $c$ such that $$f(c-\frac{1}{8})=f(c+\frac{1}{8})$$
I am not sure how to prove its existence, but what I did was to find the possible range that $c$ is inside:
$0\le c-\frac{1}{8} \le1$ and $0\le c+\frac{1}{8} \le1$, and solving both inequalities gives: $$\frac{1}{8}\le c\le\frac{7}{8}$$
So far I have learnt the Extreme Value Theorem, Rolle's Theorem, Mean Value Theorem, Fermat's Theorem.
Appreciate it if you can help me on what should I do!
Best Answer
Let $g:[\frac{1}{8},\frac{7}{8}]\to \mathbb{R}$ be defined by $\displaystyle{g(x) = f(x+{\small{\frac{1}{8}}}) - f(x-{\small{\frac{1}{8}}})}$.
The goal is to show that $g(c) = 0$, for some $c$.
If $g(\frac{k}{8})=0$, for some $k \in \{1,3,5,7\}$, then let $c=\frac{k}{8}$, and we're done.
Suppose then that $g(\frac{1}{8}),g(\frac{3}{8}),g(\frac{5}{8}),g(\frac{7}{8})$ are all nonzero. \begin{align*} \text{Then}\;& g({\small{\frac{1}{8}}})+ g({\small{\frac{3}{8}}})+ g({\small{\frac{5}{8}}})+ g({\small{\frac{7}{8}}}) \\[4pt] =\;\,& \left(f({\small{\frac{2}{8}}}) - f(0)\right) + \left(f({\small{\frac{4}{8}}}) - f({\small{\frac{2}{8}}})\right) + \left(f({\small{\frac{6}{8}}}) - f({\small{\frac{4}{8}}})\right) + \left(f(1) - f({\small{\frac{6}{8}}})\right) \\[4pt] =\;\,&f(1) - f(0)\\[4pt] =\;\,&0\\[4pt] \end{align*} Since $g(\frac{1}{8}),g(\frac{3}{8}),g(\frac{5}{8}),g(\frac{7}{8})$ sum to zero, and since they're all nonzero, it follows that at least one of them is positive, and least one is negative.
Since $f$ is continuous on $[\frac{1}{8},\frac{7}{8}]$, so is $g$.
Hence, by the Intermediate Value Theorem, we must have $g(c) = 0$, for some $c$.
This completes the proof.