# A continuous function on $S^1$- unit circle .

$$S^1=\{z\in \mathbb C : |z|=1\}$$ 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 $D$?

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.
$S^1$ is compact hence every continuous function on it is bounded and uniformly continuous. Certainly, the image of a constant function consists of one point only. As for the last one, have you heard about the intermediate value theorem?
$D$ is true. Proof:
Define the function $g:S^1\rightarrow \mathbb{R} \mid g(x)=f(x)-f(-x)$. Note that $g(S^1)\subset \mathbb{R}$ is convex, then $\forall x\in S^1,(\frac{1}{2}g(x)+\frac{1}{2}g(-x))\in g(S^1).$ Now $(\frac{1}{2}g(x)+\frac{1}{2}g(-x))=0\in S^1$, then exist $y\in S^1 \mid g(y)= 0$.