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Bernstein's Theorem (Polynomial norms inequality)
04-15-2010, 02:33 PM
Post: #1
Bernstein's Theorem (Polynomial norms inequality)
Theorem Let \(deg P = n\), then \(\left \| P'\right \| \le n \left \| P\right \|\) where \(\left \| f\right \| = \underset{|z|\le 1}{max} |f(z)|\)
Lemma Let \(deg P = n\) and \(z_1,\cdots,z_n\) are roots of \(z^n+1=0\),then
\[tP'(t)=\frac{n}{2} P(t)+\frac{1}{n}\sum_{k=1}^n P(tz_k) \frac{2z_k}{(z_k-1)^2} \]
Proof of lemma. Let \(g_t(z)=\displaystyle{\frac{P(tz)-P(z)}{z-1}}\), then \(deg(g_t) \le n-1\) and \(g_t(1)=tP'(t)\). By Lagrange's interpolation formula we get \[g_t(z)=\sum_{k=1}^n g_t(z_k) \frac{z^n+1}{(z-z_k) nz_k^{n-1}} = \frac{1}{n} \sum_{k=1}^n g_t(z_k) \frac{z^n+1}{z_k-z} z_k \](since \(z_k^{n-1}=-1/z_k\)) Putting \(z=1\) we get \[ tP'(t)=\frac{1}{n}\sum_{k=1}^n g_t(z_k) \frac{2z_k}{(z_k-1)}=\frac{2}{n} \sum_{k=1}^n \frac{P(tz_k)-P(t)}{(z_k-1)^2}z_k=\frac{2}{n}\sum_{k=1}^n \frac{z_kP(tz_k)}{(z_k-1)^2}-\frac{P(t)}{n}\sum_{k=1}^n \frac{2z_k}{(z_k-1)^2}\]If \(P(t)=t^n\), then \(P(tz_k)=-t^n\) and the above gives
\(nt^n=\displaystyle{\frac{-2t^n}{n}\sum_{k=1}^n\frac{2z_k}{(z_k-1)^2}}\) or \(\displaystyle{\sum_{k=1}^n\frac{2z_k}{(z_k-1)^2} = -\frac{n^2}{2}} \) and the lemma follows
Suppose \(|t|\le 1\), from the lemma we have \[|P'(t) | \le \left( \frac{n}{2}+\frac{1}{n}\sum_{k=1}^n \left|\frac{2z_k}{(z_k-1)^2}\right|\right )\left \| P\right \|\] Since \(z_k=e^{iw}\ne 1\), \(\displaystyle{\frac{2z_k}{(z_k-1)^2}=\frac{1}{\cos w -1}} < 0\) and so from the proof of the lemma, we get \(\displaystyle{\frac{1}{n}\sum_{k=1}^n \left|\frac{2z_k}{(z_k-1)^2}\right| = \frac{n}{2}}\) and this proves the theorem.
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08-11-2010, 10:53 AM
Post: #2
RE: Bernstein's Theorem (Polynomial norms inequality)
What's the version of the inequality in real field?
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