$f$ continuous on a circle, then there is a diameter at which ends $f$ has the same value

Let $f$ be a continuous function on a circle.
Show that there is a diameter for which the value of $f$ are equal at the diameter ends.

I don't know how to approah this problem.
Is a diameter any chord of the circle?
I tried to consider $f(\theta) = e^{i\theta}$ which satisfies the condition for the chord $y=f(\pi/4)$.
I don't see how to extend this to a general continuous function.

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@julien Actually, for $\mathbb R^2$ valued functions the statement is false: Simply take the embedding $S^1\to\mathbb R^2$, which is of course injective. – Hagen von Eitzen Feb 8 '13 at 21:59
@HagenvonEitzen Oh boy, I'm having a bad day. Thanks! – 1015 Feb 8 '13 at 22:02

(By function I assume you mean a function from $S^1\rightarrow\mathbb{R}$.

A diameter is a chord of maximal length. For example, on the unit circle in $\mathbb{C}$, diameters all have endpoints of the form $\{e^{2\pi i x}, -e^{2\pi i x}\}$.

The idea is that any such function can be thought of as a periodic function from $\mathbb{R}\rightarrow\mathbb{R}$, and for any periodic function (say with period $p$) (think about $\sin$ and $\cos$), you can always find a point such that their value at $x$ is the same as their value at $x+p/2$.

If you want to think about how to translate this into a proof, stop reading. The actual solution might proceed as follows:

Suppose you have a continuous function $f : S^1\rightarrow\mathbb{R}$. Suppose we view $S^1$ as the unit circle sitting in $\mathbb{C}$ (this doesn't change anything since we can always do things up to homeomorphism).

Consider the function $g : \mathbb{R}\rightarrow\mathbb{R}$ defined by $g(x) = f(e^{2\pi ix}) - f(-e^{2\pi ix})$. (This is just the difference of the values of $f$ at the endpoints of a diameter). We want to show that $g$ must have a zero. Now, $g$ is continuous and periodic with period 1. Thus, it's clear from the definition of $g$ that

$$g(x+1/2) = f(e^{2\pi i(x+1/2)})- f(-e^{2\pi i(x+1/2)}) = f(-e^{2\pi ix}) - f(e^{2\pi ix}) = - g(x)$$

Now fix some value for $x$, say 1. Then, if $g(1) = 0$, then and we're done. Otherwise, since $g(x+1/2) = -g(x)$, we have $g(1.5) = -g(1)$, so by the intermediate value theorem, $g(x)$ must have a zero between 1 and 1.5.

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oh man, when I started writing this, there were no answers. Oh well. – oxeimon Feb 8 '13 at 22:01

A diameter is a chord that passes through the centre of the circle.

HINT: For $0\le\theta\le\pi$ let $$g(\theta)=f\big(e^{i\theta}\big)-f\big(e^{i(\theta+\pi)}\big)\;,$$ the difference in the values of $f$ at the opposite ends of a diameter. Use the intermediate value theorem.

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@julien: Since nothing in the questions indicates otherwise, I assume that it’s about real-valued functions. – Brian M. Scott Feb 8 '13 at 21:52