Euclidean Sphere Consider the Euclidean sphere $S^n = \{x\in \mathbb{R}^{n+1}: ||x||_2=1\}$ of dimension $n\ge1$. Show that for every continuous function $f:S^n\longrightarrow \mathbb{R}$, there exists $x\in S^n$ such that $f(x)=f(-x)$.
I think I need to show that there exists $x_0\in  S^n$ with $g(x_0)=0$ with $g$ defined by $x\mapsto f(x)-f(-x)$. I have tried using the continuity of $g$ together with the connectedness of $S^n$ but i don't find a good result.
 A: By symmetry, we have
$$
\int_{S^n}\big(f(x)-f(-x)\big)\,\mathrm{d}x=0\tag{1}
$$
Assume that $g(x)=f(x)-f(-x)\ne0$ for any $x\in S^n$.
If $g(x)\gt0$ for all $x\in S^n$, then the integral in $(1)$ would be positive.
If $g(x)\lt0$ for all $x\in S^n$, then the integral in $(1)$ would be negative.
Therefore, there must be some $x_1\in S^n$ so that $g(x_1)\gt0$ and some $x_2\in S^n$ so that $g(x_2)\lt0$. Connect $x_1$ and $x_2$ by any path $\gamma$ in $S^n$. The image of $\gamma$ under $g$ is connected since $\gamma$ is connected and $g$ is continuous. Since the image of $\gamma$ includes $g(x_1)\gt0$ and $g(x_2)\lt0$, and any connected subset of $\mathbb{R}$ that contains a positive and negative number also contains $0$, the image of $\gamma$ contains $0$. Thus, there must be some $x_0\in\gamma\subset S^n$ so that $g(x_0)=0$. Contradiction.
Therefore, there must be some $x\in S^n$ so that $g(x)=0$; that is,
$$
f(x)-f(-x)=0\tag{2}
$$
A: Restrict $f$ to any great circle on $S^n$. If on this circle you can find antipodal points with equal images, then they'll work on $S^n$ -- because it was a great circle.
So, all you need to prove is case $n=1$.
Case $n=1$:
An $ f:S^1\longrightarrow \mathbb{R} $ is just an $ f:[0,2\pi]\longrightarrow \mathbb{R}$ with $f(0)=f(2\pi)$.
Antipodal points on $S^1$ are those corresponding to values in $[0,2\pi]$ that differ by $\pi.$
So, question is reduced to:
For any $ f:[0,2\pi] \longrightarrow \mathbb{R} $ with $f(0)=f(2\pi)$, there exists $ t \in [0,\pi] $ such that $f(t)=f(t+\pi)$.
To show this latter assertion, work with $g(t)=f(t)-f(t+\pi) $ defined over $[0,\pi] $. Notice that $ g(o)$ and $g(\pi)$ can't be of same sign -- draw a graph of $ f:[0,2\pi]\longrightarrow \mathbb{R}$ to see this. By intermediate value theorem then, $g$ must have a $0$.
Wow! Nice solution.
