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
2026-04-11 16:51:19.1775926279
Prove or disprove that T:[0,2π] -> [0,2π] given by Tx = sin(2014x) is a contraction
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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 $\sin(2014x)$.