i know that if we assume $T:[a,b] \to [a,b] $ and if $|T'(x)| ≤ α \space \forall \space a≤x≤b$ then T is a contraction .
but unsure of how to apply that to this question
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Find $T'(x)$ to begin with. How large can its modulus be? – Apr 15 '15 at 01:14
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You may run into trouble if $ T'x \geq 1 $ – gary Apr 15 '15 at 01:14
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I think that you should add $\alpha < 1$ in your statement. So, you just have to show that there exists at least one $x\in[0, 2\pi]$ such that this inqequality doesn't hold. Do you have any ideas? Just take the derivative of the $\sin(2014x)$.
Igor Deruga
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@Anon: Well, you're almost right. Supremum is a tricky thing: the function might never reach it (not in your case, though). So, I suggest just taking the biggest value at $x = 0$: $|2014\cos(2014\cdot 0)| = |2014*\cos(0)| = |2014\cdot 1| = 2014 > 1$. One point is enough to show that it is not a contraction. – Igor Deruga Apr 16 '15 at 00:13
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I just did |T'(X)| = |2014 cos(2014x)| < sup (2014 cos(2014x)) and it was not smaller than 1 so T is not a contraction is that right?
Anon
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Note: I don't think that you are supposed to answer your question. I mean, you can, but not with another question :) You can use the comments instead. – Igor Deruga Apr 16 '15 at 00:15