Assume $f$ is continuous with period $T$. Prove for every $n \in \Bbb N~ \exists x \in [0, T]$ such that $f(x) = f(x + \frac Tn)$

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I tried using another function $g(x) = f(x + T/n) - f(x)$, and found that $g(0) + g(T/n) + ... g(T(n-1)/n) = 0$, but I don't know where to go from there. Any ideas on how prove this?

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If there is no such $x$ then $f(x+T/n)-f(x)$ is always positive or always negative. Suppose it is always positive. Then $$0=f(T)-f(0)=[f(\frac {nT} n)-f(\frac {(n-1) T)} n]$$ $$+[f(\frac {(n-1)T} n)-f(\frac {(n-2)T} n)]+...+[f(\frac T n)-f(0)]>0,$$ a contradiction. The other case is similar.