Let \(f: S^1 \to \mathbb{R}\) be a continuous map. Show there exists a point \(x\) of \(S^1\) such that \(f(x) = f(-x)\).
Solution
Let \(g\colon S^1\to\mathbb R\), \(g(x)=f(x)-f(-x)\).
1. \(g\) is continuous and \(g(-x)=-g(x)\).
The antipodal map \(x\mapsto-x\) is continuous, so \(g\) is a difference of continuous maps. The identity is immediate from the definition.
2. Q.E.D.
Fix \(x_0\in S^1\). If \(g(x_0)=0\), take \(x=x_0\). Otherwise \(g(x_0)\) and \(g(-x_0)=-g(x_0)\) have opposite signs. A semicircular arc \(A\subseteq S^1\) from \(x_0\) to \(-x_0\) is connected, so \(g(A)\) is an interval containing both values, hence \(0\). A point \(x\in A\) with \(g(x)=0\) satisfies \(f(x)=f(-x)\).