$f : S^1 \to\mathbb R$ is continuous then $f(x)=f(-x)$ for some $x\in S^1$ Question is to prove :
$f : S^1 \to \mathbb R$ is continuous then $f(x)=f(-x)$ for some $x\in S^1$
I guess it would be helpful to use intermediate value theorem
Assuming $f(x)\neq f(-x)$ then given any $p\in (f(-x),f(x))$ (assuming $f(-x)<f(x)$) there exists $y\in (-x, x)$ such that $f(y)=p$
I am not very sure of how to use this.. 
It would be helpful if someone can give some hint which would help me to solve this..
Thank you.
 A: You can parameterize $S^1$ in the usual way, so that $f$ is a function from $[0,2\pi]$ to $\mathbb{R}$ with $f(0)=f(2\pi)$.
Then define $g(x):[0,\pi]\to\mathbb{R}$, $g(x) = f(x+\pi)-f(x)$.


*

*How does $g(0)$ relate to $g(\pi)$?

*What can you conclude from the intermediate value theorem?

A: Hint: Let $\alpha:[0,1]\to S^1$ parametrize the unit circle, and take $g(z)=f(z)-f(-z)$. Apply the Intermediate Value Theorem to $g\circ\alpha$.
Second hint: If $\alpha(a)=1$ and $\alpha(b)=-1$, analyze $g\circ\alpha(a)$ and $g\circ\alpha(b)$.
A: Define $g:[0,1]\to \mathbb{R}$ as $g(t):=f(e^{2\pi it})-f(-e^{2\pi it})$. 
Clearly $g$ is also continuous. 
We claim that $g$ is zero at some point. 
If that is true, say $g(t_0)=0$, then it means $f(e^{2\pi it_0})=f(-e^{2\pi it_0})$, hence we get what we wanted.
Suppose $g(t)\neq 0\; \forall\; t\in S^1$, in particular $g(0)\neq 0$. 
Observe that $g(0)=f(1)-f(-1)$ and $g(\frac{1}{2})=f(-1)-f(1)=-g(0)$.
Hence if $g(0)>0$, then $g(\frac{1}{2})<0$, and similarly, if $g(0)<0$, then $g(\frac{1}{2})>0$. 
Thus by the intermediate value theorem, there must be a point say $t_0\in (0,\frac{1}{2})$ such that $g(t_0)=0$, which gives a contradiction.
Observe that in the same way you can obtain not just a $t_0\in (0,\frac{1}{2})$ such that $g(t_0)=0$, but another point $t_1\in (\frac{1}{2},1)$ such that $g(t_1)=0$, because $g(1)=g(0)$.
Note: Instead of defining $g$, you may define, $h:S^1\to \mathbb{R}$, as $h(x)=f(x)-f(-x)$, and follow the same steps as above. The intermediate value theorem can be applied to the function $h$ also, because $S^1$ is a path-connected space.
A: I just wanted to expand on what Quickbeam2k1 wrote at the top -  while he's not 100% right (it isn't clear from this particular example that there exist two antipodal points on earth with the same temperature),  (he is indeed correct, as he pointed out) he's leading you to the right conclusion - that this phenomena you described in your question happens to generalize to higher dimensions. If you're interested, you should look at the wikipedia page for the Borsuk-Ulam theorem, which I think is exactly what Quickbeam2k1 is leading you towards. 
