$$S^1=\{z\in \mathbb C : |z|=1\}$$ be the unit circle. Then which of the following is false $?$
Any continuous function from $S^1$ to $\mathbb R$ is
A. bounded
B. uniformly continuous.
C. has image containing a non empty open subset of $\mathbb R.$
D. has a point $z\in S^1$ such that $f(z)=f(-z)$
Since $S^1$ is compact any continuous function would be bounded or uniformly continuous so $A$ and $B$ are correct.
For $C$, the constant function does not have any open interval in its image. Thus, $C$ is the false statement.
That leaves $D$ to be correct. How can I prove the existence of a point $z$ having properties like said in $D$?
Best Answer
You prove D by the intermediate value theorem, used on the function $$ g(z)=f(z)-f(-z) $$ Pick a $z_0\in S^1$, and evaluate $g$ there. If you get $0$, then you're done. If not, then $g(z_0)=-g(-z_0)$, so $g$ changes sign. That means there must be a zero somewhere, and that somewhere is the $z$ you're looking for.