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 Bernstein's Theorem (Polynomial norms inequality)
04-15-2010, 02:33 PM (This post was last modified: 04-15-2010 09:36 PM by elim.)
Post: #1
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
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.
08-11-2010, 10:53 AM
Post: #2
 elim Moderator Posts: 544 Joined: Feb 2010 Reputation: 0
RE: Bernstein's Theorem (Polynomial norms inequality)
What's the version of the inequality in real field?
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