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Let $S^{1} := \lbrace(\cos\alpha, \sin\alpha) \subset \mathbb{R}^{2} | \alpha \in \mathbb{R}\rbrace$ be the circumference of radius $1$ and $f: S^{1} \to \mathbb{R}$ a continuous function. Prove that there exist two points diametrically opposed at which $f$ assumes the same value.

My idea for solution: Define $\varphi: S^{1} \to \mathbb{R}$ as$$\varphi(\cos\alpha, \sin\alpha) = f(\cos\alpha, \sin\alpha) - f(-\cos\alpha, -\sin\alpha).$$ If $f(\cos\alpha, \sin\alpha) = f(-\cos\alpha, -\sin\alpha)$ for all $\alpha \in \mathbb{R}$, the result follows. Otherwise, there exist $\alpha_{1},\alpha_{2}$ such that $$f(\cos\alpha_{1}, \sin\alpha_{1}) - f(-\cos\alpha_{1}, -\sin\alpha_{1}) > 0$$ and $$f(\cos\alpha_{2}, \sin\alpha_{2}) - f(-\cos\alpha_{2}, -\sin\alpha_{2}) < 0.$$ Applying the intermediate value theorem, we prove the result.

Is this a correct idea? I appreciate your corrections. Thanks!

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The idea is good, but the intermediate value theorem applies to functions mapping a closed interval $I \subset \Bbb R$ to $\Bbb R$.

It would be possible to formulate a similar statement for functions $\phi: S^1 \to \Bbb R$, but it is simpler to consider $$ \phi: [0, \pi] \to \Bbb R, \quad \phi(\alpha) = f(\cos\alpha, \sin\alpha) - f(-\cos\alpha, -\sin\alpha) $$ instead, so that the IVT can be applied directly.

The remaining argument can also be simplified. It suffices to observe that $\phi(0) = - \phi(\pi)$, so that

  • either $\phi(0) = \phi(\pi) = 0$,
  • or $\phi(0)$ and $ \phi(\pi)$ have opposite sign, and the intermediate value theorem states that there is some $\alpha \in (0, \pi)$ with $\phi(\alpha) =0$.
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  • $\begingroup$ Perfect! Thanks for the help! $\endgroup$ – Lucas Corrêa Jan 31 '18 at 21:09
  • $\begingroup$ Or let $g(x)=f(\cos x,\sin x)$ and $h(x)=g(x)-g(x+\pi)$ for all $x\in \Bbb R. $ Then $g:\Bbb R\to \Bbb R$ is continuous so if $h(x)$ is never $0$ then $h(x)$ would be always positive or always negative. But $h(x)+h(x+\pi)=0.$ $\endgroup$ – DanielWainfleet Feb 1 '18 at 10:51

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