A continuous function $f:S^1\to R$ then $\exists z\in S^1$ such that $f(z)=f(-z)$
My attempt:
On contrary suppose $f(z)-f(-z)\neq 0$ so either it is $>0$ or $<0$
WLOG $f(z)-f(-z)>0$
But I could not proceed further .
Please give me hint.
2026-04-07 11:18:36.1775560716
$f:S^1\to R$ then $\exists z\in S^1$ such that $f(z)=f(-z)$
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You proved correctly that, if the statement was false, then you must always have $f(z)-f(-z)>0$ or always have $f(z)-f(-z)<0$. Suppose the the first case holds. But then it will also be true that you always have $f(-z)-f\bigl(-(-z)\bigr)>0$, which means that $f(-z)-f(z)>0$, which is equivalent to $f(z)-f(-z)<0$. So, a contradiction is reached.