Transformation of a continious function

Question

Suppose that $f:[0,2\pi$] $\rightarrow \mathbb{R}$ is continuous and $f(0)=f(2\pi)$.

Show that there exists an $x\in[0,\pi$] such that $f(x)=f(x+\pi)$.

I simply have no idea where to start, any help would be appreciated. I have run the transformation through the definition of continuity but I couldn't see that it helped at all.

This may give you some ideas. – David Mitra Jan 25 '14 at 22:16 — David Mitra, Jan 25 '14 at 22:16

score 3 · Accepted Answer · answered Jan 25 '14 at 22:16

3

Define $g(x) = f(x)-f(x+\pi)$. Then $$g(0) = f(0)-f(\pi)$$

and

$$g(\pi) = f(\pi)-f(2\pi) = f(\pi)-f(0) = -(f(0)-f(\pi)).$$

Do you see what this tells you?

answered Jan 25 '14 at 22:16

Cameron Williams

29,432

It shows that $g(0)=-g(\pi)$ and so if $g(0)$ doesn't equal $0$, then using Intermediate value theorem there must be a point between $g(0)$ and $g(\pi)$ that equals $0$, but I am not sure if that is what you meant . – Wonkman123 Jan 25 '14 at 22:25
That's exactly what I mean. – Cameron Williams Jan 25 '14 at 22:50
I understand this may be a stupid question but how exactly does this help? – Wonkman123 Jan 25 '14 at 22:56
It says that there is a point $x$ where $g(x) = 0$ or in other words.. $f(x)-f(x+\pi) = 0$. – Cameron Williams Jan 25 '14 at 22:58
I see it now.You've been a great help, thank you. – Wonkman123 Jan 25 '14 at 23:00

Transformation of a continious function

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