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Let $P_n^1$ be the set of all polynomials of degree $n$ with leading coefficient $1$ and define $\lVert p \lVert_{\infty}=\sup_{x\in [-1,1]}\lvert p(x)\lvert$. Compute $\inf_{p\in P_n^1} \lVert p\lVert_{\infty}$.

I know that $\inf_{p\in P_n^1} \lVert p\lVert_{\infty}=\frac{1}{2^{n-1}}\lVert T_n\lVert_{\infty}=2^{1-n}$ where $T_n$ is the Chebyshev polynomial of the first kind (proof by the alternation theorem).

Is there a alternate direct proof?

From functional analysis I know that for a normed space $X$ and a bounded linear functional $f$ in it's dual space $X'$ the following holds: $\lvert f(x)-c\lvert=\lVert f \lVert$ dist$(x,f^{-1}(\{c\}))$. So with a functional $f$ on the space of polynomials satisfying $f^{-1}(\{c\})=P_{n-1}$ I could compute $\inf_{p\in P_n^1} \lVert p\lVert_{\infty}=\inf_{p\in P_{n-1}} \lVert x^n-p \lVert_{\infty}=$ dist$(x^n,f^{-1}(\{c\}))=\frac{\lvert f(x^n)-c\lvert}{\lVert f\lVert}$

Can you give me advice for choosing $f$? I thought about $f(p)=\lVert p^{(n)}\lVert_{\infty}$ where $P_{n-1}=f^{-1}(\{0\})$, but I didn't see how I could compute $\lVert f\lVert$. I guess $f$ wouldn't even be bounded. Perhaps it would be better to restrict $f$ to the space of polynomials of degree $\leq n$ and choose $f(p)=a_n$, i.e. the coefficient of $x^n$ where $P_{n-1}=$ker $f$. But also not a bounded operator I guess...

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$f$ is continuous (it's more easy to check it with the norm $\max_{1\leq j\leq n}|a_j|$, which is equivalent to the initial norm, since we are in a finite dimensional space. But I'm not sure the computation of the norm of $f$ is easier than the initial problem. – Davide Giraudo Jun 4 '12 at 9:55
Probably... What would be a good approach to the problem which is not dependend on the alternation theorem? – Julian Jun 4 '12 at 10:02
What do you exactly mean by alternation theorem (maybe there are several versions)? If the $\inf$ is reached at the Chebychev polynomial and not at any other, I don't see an other fundamentally different approach (which of course doesn't imply this one doesn't exist). – Davide Giraudo Jun 4 '12 at 12:31

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