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In my course I have come across the following problem:

The Chebyshev polynomial of degree $n$, $T_n(x)$, is defined on $[-1,1]$ by $T_n(x)=\cos n\theta$.

Let $q_{n+1}(x)$ be any monic polynomial of degree $n+1$. Consider $2^n q_{n+1} - T_{n+1}$ to prove that $\|q_{n+1}\|\ge 2^{-n}$.

I have attempted to prove this in the following way:

$0\le \|2^n q_{n+1} -T_{n+1}\| \le \|2^n q_{n+1}\| - \|T_{n+1}\|$

$\implies \|T_{n+1}\|\le |2^n|\|q_{n+1}\|$

And using the fact that $\|T_{n+1}\| = 1 \space(*)$, we have:

$2^{-n}\le\|q_{n+1}\|$

Which was to be shown. However, I am not convinced by the step at $(*)$. I believe this is true for the infinity norm, but is it true for a norm in general? Are there any other steps in the proof that require more thought?

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    Some comments: First, the inequality $\|2^nq_{n+1} - T_{n+1}\|\leq \|2^nq_{n+1}\|-\|T_{n+1}\|$ seems highly suspect to me. Second, when you say "I believe this is true for the infinity norm, but is it true for a norm in general?", it makes it not so clear what norm you are using. So you should specify in the problem: what is $\|\cdot\|$?2012-12-17

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