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Suppose $P(z)$ is a polynomial satisfying $|P(z)|\leq 1$ for all $|z|=1$, prove that all its coefficients are bounded by $1$.

So far, I've been thinking something like this:

proof: Suppose $g(z)=\frac{P(z)}{z}$. We observe that $$ |g(z)|=\left|\frac{f(z)}{z}\right|\leq 1 \quad \forall z\in \partial D_{1}(0). $$ First, we see $g$ has a singularity at $z=0$, however it is removable since it is bounded there. Thus, $g$ is analytic in $D_1(0)$. Second (I'm not sure if I can apply the MMP), since $g$ attains its maximum there, by MMP it must be a constat.

Any hint or observation would be very welcome.

Thanks

Charles Seife
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  • Look at Cauchy's estimates to bound each coefficient. – copper.hat Jan 18 '20 at 22:11
  • (As a previous commenter noted, if $P$ is not constant then in fact $P(z)=z^n$ for some $n$.) – copper.hat Jan 18 '20 at 22:15
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    @copper.hat: I had linked to https://math.stackexchange.com/q/326832/42969, but then noticed that that question assumes that the main coefficient is 1, so this here is slightly different. – Martin R Jan 18 '20 at 22:26
  • Thank you so much. It seems that $Q(0)=a_n$ if I follow the same notation and reasoning. I looks like that $P(z)=a_nz^n$. – Charles Seife Jan 19 '20 at 00:46
  • Take any nonzero polynomial $Q$ and then if $M$ is the maximum on the unit circle, $P=\frac{Q}{M}$ will satisfy the hypothesis, so lots of such $P$ exist – Conrad Jan 19 '20 at 02:58

2 Answers2

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Note that if $Q$ is an arbitrary polynomial which is not the constant zero, and $M$ the maximum of $Q$ on the unit circle which is strictly positive because $Q$ is not the zero polynomial, then $P=\frac{Q}{M}$ satisfies the hypothesis so there are tons of such polynomials.

$P(z)=\frac{1+z}{2}$ shows that it is simply not true that $P(0)$ must be zero as incorrectly proved in the OP post.

To prove that the coefficients are bounded by $1$ under the hypothesis here, the simple identity:

If $P(z)=a_0+a_1z+..a_nz^n$, then $2\pi ia_k=\int_{|z|=1}P(z)z^{-k-1}dz$ and the triangle inequality for integrals suffices.

(much more it is true but less trivially to prove, for example $|a_0|+|a_n| \le 1$ if $P$ has degree $n$ and satisfies $|P(z)| \le 1$ if $|z|=1$ which obviously implies that if $|a_n|=1$ then $P(z)=a_nz^n$ by a simple induction; similar inequalities $|a_j|+|a_k| \le 1$ are true for other pairs of coefficients $a_j, a_k, j\ne k$ but under some arithmetic conditions regarding $|j-k|$)

Conrad
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  • Thank you so much. I don't know why I didn't think about it. Now, everything is clear. – Charles Seife Jan 20 '20 at 16:21
  • Sorry for misusing your statement "much more is true" as a motivation to ask for a reference for a compilation of such bounds, or for advanced methods. What one typically finds in textbooks are Cauchy's estimates in basic form, i.e. $|a_n| \le \frac{\max_{|z| = r} |P(z)|}{r^n}$. I'd like to study, following your examples, "mean" estimates when using $r=1$, like (proposition 1): $\max_{|z| = 1} |P(z)| \ge \frac{1}{n}\sum_{k=0}^n |a_k|$ or (proposition 2): $\max_{|z| = 1} |(1-z)P(z)| \ge \frac{2}{n+1}\sum_{k=0}^n |a_k|$. Would you have a reference? Thanks! – Andreas Nov 24 '21 at 22:24
  • @Andreas The most comprehensive reference I know is the book Analytic Theory of Polynomials by Rahman and Schmeisser; https://global.oup.com/academic/product/analytic-theory-of-polynomials-9780198534938?cc=us&lang=en& – Conrad Nov 24 '21 at 22:43
  • Thanks for the reference! Would you have particular ideas on my two propositions? – Andreas Nov 24 '21 at 23:01
  • @Andreas - I suggest making them (separate) questions here on MSE so one can discuss them; note that from Schwarz we have $\sum_{k=0}^n |a_k| \le \sqrt {n+1}||P||2\le \sqrt {n+1}||P||{\infty}$ which should give you the estimate required (though it is far from optimal as $1/\sqrt {n+1}$ is actually optimal, but that is much harder to prove) since $n \ge \sqrt {n+1}, n \ge 2$ while for $n=1$ it is trivial that $||P||_{\infty}=|a_0|+|a_1|$ by choosing $|z|=1$ so $a_0, a_1z$ are in the same direction – Conrad Nov 24 '21 at 23:35
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One can use Parseval's identity:

Let $f(z) = \sum_{n=0}^\infty a_n z^n$ be holomorphic in $B(0,R)$. Then $$ \sum_{n=0}^\infty |a_n|^2 r^{2n} = \frac{1}{2 \pi }\int_0^{2\pi}|f(re^{it})|^2dt. $$ for $0 \le r < R$.

For an elementary proof see for example Proof of Parseval's identity.

It follows that if $f$ is holomorphic in a neighborhood of the closed unit disk and satisfies $|f(z)| \le 1$ for $|z| = 1$ then $$ \sum_{n=0}^\infty |a_n|^2 \le 1 $$ which in particular implies that $|a_n| \le 1$ for all $n$.

Martin R
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