Sum of the norm of polynomials

Question

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 $\bar{D}$.

Answer 1

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 $2$.

Answer 2

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.