Let $\bar D$ denote the closed unit disc in the complex plane. Consider the function $f:\bar D\longrightarrow \mathbb{C}$, defined as $f(z)=z$ for all $z\in \bar D$.

Let $n\in \mathbb{N}$. For $1\leq i\leq n$, let $p_i:\bar D\longrightarrow \mathbb{C}$ be monic complex polynomials on $\bar{D}$, such that $p_i$’s have no roots in $\bar{D}$ and $c_i\in\mathbb{C}$ such that $$f=\sum_{i=1}^{n}c_ip_i.$$ Can we say that for any such representation of $f$, the value $$K=|c_1|\|p_1\|_\infty+|c_2|\|p_2\|_\infty+\cdot+|c_n|\|p_n\|_\infty\geq 2?$$

$\|.\|_\infty$ denotes the uniform norm, defined as $\|p\|_\infty=sup\{|p(t)|: t\in \bar D\}$ for any complex polynomial $p$ on $\bar{D}$.


The question basically boils down to how large the coefficient at $z$ of a polynomial $p(z)$ that has no zeroes in the circle can be compared to the uniform norm of the polynomial. Indeed, one direction is clear. For the other direction, notice that if we have a polynomial $P$ of degree $n$ with the coefficient $1$ at $z$ and without zeroes, then for every $N>n$ we can write $$ z=\frac 1N\sum_{\zeta:\zeta^N=1}\zeta^{-1}P(\zeta z) $$ (the restriction that $p_j$ are monic is totally pointless because only the products $c_jp_j$ matter in the whole problem setup).

Now, let's try to figure it out in this new formulation. Nothing is special about polynomials here, we can always just take a long enough Taylor series of a function analytic in a slightly larger disk and we can always reduce the radius a bit and expand afterwards. Also, any function $F$ without zeroes in the disk is just $e^g$. So the question becomes something like that: given a function $g=az+\dots$ in the unit disk, what is the least left half-plane we can squeeze its image into. The answer (by the Schwarz lemma) is given by any conformal mapping of the disk to a left half-plane preserving the origin and having the first Taylor coefficient of some fixed absolute value $A$, which is, say, $g(z)=\frac {Az}{z+1}$ corresponding to the left half plane $\Re z\le\frac A2$. Thus, the best constant is $\min_A\frac{e^{\frac A2}}A=\frac e2\approx 1.36$, which is still noticeably above $1$, but definitely not $2$.

  • $\begingroup$ @WilsonGeorge Yes, that's what I'm claiming. Initially I made an error thinking that you can normalize $|a|=1$, but, obviously, once you have normalized $F(0)=1$, you have no more freedom except rotations, so you have to consider all possible values of $|a|$ and compare them to the corresponding minima of the $H^\infty$ norms. Anyway, if something is unclear, you are always welcome to ask questions :-) $\endgroup$ – fedja Jun 3 '17 at 21:31
  • $\begingroup$ @WilsonGeorge "How large the coefficient at $z$ of ... can be compared to the uniform norm of ..." = "what is the supremum of the ratio $|a_1|/\|p\|_\infty$ where $p(z)=a_0+a_1z+\dots$ has no zeroes in the unit disk". Sorry for my poor English :-) $\endgroup$ – fedja Jun 4 '17 at 12:43
  • $\begingroup$ @WilsonGeorge Have you read the next two sentences? $\endgroup$ – fedja Jun 4 '17 at 17:14
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    $\begingroup$ @WilsonGeorge Yes, exactly. The expression just comes from the fact that $\frac 1N\sum_{\zeta:\zeta^N=1}\zeta^m=1$ if $N|m$ and $0$ otherwise. So, what would it be for the reference: Fourier or Vieta? :-) $\endgroup$ – fedja Jun 5 '17 at 12:34
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    $\begingroup$ @WilsonGeorge $P$ (or, rather, $F$) at that moment is $e^g$. We can always normalize $F(0)=1$, so $g(0)=0$ and $g(z)=az+\dots$. Now $|F'(0)|=|a|$ while $\|F\|_\infty=e^{\sup\Re g}$. Thus, if we fix $|a|$, we would like to minimize $\sup\Re g$, i.e., to "squeeze the image...". $\endgroup$ – fedja Jun 6 '17 at 15:14

I am not sure whether a counterexample can be found but you can find one if you relax the polynomial condition and allow general continuous invertibles.

There exists a variation on a partition of unity of the circle, $\mathbb T$, call it $u_1, u_2, u_3 \in C(\mathbb T)$ such that $\|u_j\|_\infty = 1/2$, $u_1+u_2+u_3 = z$ and $u_j = 0$ on the arc $[2\pi(j-1)/3, 2\pi(j)/3]$. Basically, each $u_j$ can be just thought of as $\frac{1}{2}z$ on the other two thirds of the circle.

Define functions $g_j \in C(\overline{\mathbb D})$ by $g_j(\lambda e^{i\theta}) = \lambda u_j(e^{i\theta})$. Thus, $$g_1(z) + g_2(z) + g_3(z) = f(z) = z$$ and we still have that $\|g_j\|_\infty = 1/2$.

Finally, define $$ f_j(z) = g_j(z) - \frac{1}{9}e^{i(2\pi j/3 - \pi/3)}$$ which is invertible in $C(\overline{\mathbb D})$ because of the definition of $u_j$. Therefore, $$\|f_j\|_\infty \leq \frac{1}{2} + \frac{1}{9}$$ which gives that $$\|f_1\|_\infty + \|f_2\|_\infty + \|f_3\|_\infty = \frac{3}{2} + \frac{1}{3} < 2$$ and $$ f_1(z) + f_2(z) + f_3(z) = f(z) = z$$.

By chopping the circle into $n$ pieces and defining $n$ functions in the above manner you can probably bring the sum of the norms asymptotically to 1.

  • $\begingroup$ You cannot uniformly approximate on the unit circle by holomorphic polynomials; rather, you need polynomials in $z,z^{-1}$. $\endgroup$ – Christian Remling Apr 2 '17 at 20:53
  • $\begingroup$ @ChristianRemling Of course, I was rushing before the baby woke up. It is corrected now. $\endgroup$ – Chris Ramsey Apr 3 '17 at 0:10

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