if $|az^2+bz+c|\le 1$, find the maximum of $|a|+|b|$

Question

Let $a,b,c$ be complex numbers, and $f(z)=az^2+bz+c$, such that if the complex number $|z|\le 1$, then we have $|f(z)|\le 1$. Find the maximum of $|a|+|b|$.

if we $a,b,c$ be real and $z$ be real, I can find the maximum is $2$,because $f(z)=2z^2-1$ such it. and $$a=\dfrac{1}{2}[f(1)+f(-1)]-f(0),b=\dfrac{1}{2}[f(1)-f(-1)]$$ so $$|a|+|b|=\dfrac{1}{2}|f(1)+f(-1)-2f(0)|+\dfrac{1}{2}|f(1)-f(-1)|\le 2$$But for complex, maybe this answer is also $2?$,I'm not sure.

Answer 1

By replacing $f(z)$ with $$ \tilde f(z) = e^{i\phi} f(e^{i\psi} z) = e^{i(\phi+2\psi)}a z^2 + e^{i(\phi+\psi)}b z + c = \tilde a z^2 + \tilde b z + c $$ with suitably chosen $\phi, \psi \in \Bbb R$ we can assume that both $a$ and $b$ are real and $\ge 0$.

Now let $\omega = e^{i\pi/3} = \frac 12 + \frac i2 \sqrt 3$ be the primitive $6^{\text{th}}$ root of unity. Then $\omega^2 = \omega - 1$ and $ \omega^{10} = \omega^5 - 1$. Therefore $$ f(\omega) = (a+b)\omega + c - a \\ f(\omega^5) = (a+b)\omega^5 + c - a $$ which implies $$ a + b = \frac{f(\omega)- f(\omega^5)}{\omega - \omega^5} = \frac{f(\omega)- f(\omega^5)}{i \sqrt 3} \, . $$ Now use that $|f(z)| \le 1$ on the unit circle, this gives the estimate $$ |a| + |b| = a + b \le \frac{2}{\sqrt 3} \, . $$

And this is the actual maximum. Credit for the following example goes to achille hui: Let $p(z) = 2z^2+4z - 1$, then $$ \begin{align}|p(e^{it})|^2 &= |2e^{it} - e^{-it} + 4|^2 = |\cos(t) + 4 + 3i\sin(t)|^2\\ &=(\cos(t)+4)^2 + 9\sin(t)^2 = 25 +8\cos(t)(1-\cos(t))\\ &\le 25 + \frac{8}{4} = 27 \, ,\end{align} $$ so that $$ f(z) = \frac{2z^2 + 4z - 1}{3\sqrt{3}} $$ satisfies $|f(z)| \le 1$ on the unit circle and therefore – due to the maximum principle – for all $z$ in the unit disk.

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Remark: This is how I came up with above approach: In order to compute $a+b$ from two equations $$ f(z_1) = a(z_1^2-z_1) + (a+b)z_1 + c \\ f(z_2) = a(z_2^2-z_2) + (a+b)z_2 + c $$ we need different $z_1, z_2$ with $z_1^2-z_1 = z_2^2-z_2$, or $z_1 + z_2 - 1 = 0$. Also $z_1, z_2$ should be of absolute value $\le 1$, and their difference as large as possible.

This eventually led to the choice $z_1 = \frac 12 + \frac i2 \sqrt 3$ and $z_2 = \frac 12 - \frac i2 \sqrt 3$.

Answer 2

The minimization problem can be stated as

$$ \max_{z,a,b} \vert a\vert + \vert b\vert $$

subject to

$$ \vert z \vert \le 1\\ \vert a z^2+b z + c \vert \le 1 $$

We will relax the problem to a more amenable form as

$$ \max_{z,a,b} \vert a\vert^2 + \vert b\vert^2 $$

subject to

$$ \vert z \vert = 1\\ \vert a z^2+b z + c \vert = 1 $$

according to this set of restrictions the maximum is located at the feasible region boundary

Forming the Lagrangian

$$ L = \vert a\vert^2 + \vert b\vert^2 + \lambda (\vert a z^2+b z + c \vert -1) $$

After some cumbersome algebraic operations we get at

$$ \max_{z,a,b} \vert a\vert^2 + \vert b\vert^2 = \frac{1}{2}(1+\vert c\vert^2+2\vert c\vert) $$

NOTE

$$ \vert z \vert = 1 \Rightarrow z = e^{i\phi} $$

so the problem changes again to

$$ \max_{\phi,a,b} \vert a\vert^2 + \vert b\vert^2 $$

subject to

$$ \vert a e^{2i\phi}+b e^{i\phi} + c \vert = 1 $$

Answer 3

A comment on the uniqueness of the solution of achile hui: and @Martin R: and a generalization

Take $f(z) = az^2 + bz + c$ with $a$, $b\ge 0$, $|f(z)|\le 1$ for $|z|\le 1$ ( enough for $|z|=1$) and such that $a+b=\frac{2}{\sqrt{3}}$, the maximum,

With $\omega=e^{i \pi/3}$ we have $$a+b = \frac{f(\omega)-f(\omega^{-1})}{i \sqrt{3}}$$ so its absolute value does not exceed $\frac{2}{\sqrt{3}}$. Since we have equality we conclude $f(\omega)=i $ and $f(\omega^{-1})=-i$.

So we must have $$f(z) = \frac{2z^2 +4 z -1}{3\sqrt{3}} + \eta\cdot ( z^2-z+1) = f_0(z) + \eta\cdot ( z^2-z+1)$$ for some $\eta$ real.

The function $|f_0(e^{i \theta})|$ achieves a maximum at $\theta = \pi/3$ so the tangent to the curve $f_0(e^{i\theta)}$ is horizontal. Since $f$ is also an extreme, the tangent to $f(e^{i\theta})$ at $\theta = \pi/3$ has the same direction. Recall that $\frac{d}{d\theta} f(e^{i \theta})= i e^{i \theta} f'(e^{i \theta})$. Therefore, we must have $$\frac{ \eta (2 \omega -1) }{4 \omega + 4} \in \mathbb{R}$$ and that implies $\eta = 0$. Therefore $f=f_0$.

The plot of $\theta \mapsto \frac{2 e^{2 i \theta} + 4 e^{i \theta} -1}{3 \sqrt{3}}$ is a cardioid ( see pic )The fixed circle has center $(-\frac{1}{3\sqrt{3}},0)$ and radius $\frac{2}{3\sqrt{3}}$. The parametrization will match if we start rolling the circle from the right.

ADDED: After some calculations I found the maximum of $s|a| + |b|$, where $s\ge 0$ is a constant. There are two cases:

1. $0\le s \le \sqrt{2}$, so $s = 2 \cos \theta$, where $\theta \in [\pi/4, \pi/2]$. Then $\max (2 \cos\theta |a| + |b|) =\frac{1}{\sin \theta}$.

2. If $s\ge \sqrt{2}$ then $\max (s|a|+|b|)=s$.

The inequality $2 \cos \theta |a| + |b| \le \frac{1}{\sin \theta}$ is proved in the same way by evaluating the polynomial $f$ at $\omega, \bar \omega = \cos \theta \pm i \sin \theta$. If moreover $\theta \in [\pi/4, \pi/2]$, one checks that the value $\frac{1}{\sin \theta}$ is achieved for the polynomial

$$f_{\theta}(z) = \frac{\cos \theta\cdot z^2 - 2 \cos 2 \theta \cdot z + \cos \theta \cos 2 \theta }{2\sin^3 \theta}$$

$f_{\theta}$ was found so that $f(e^{\pm i \theta}) = \pm i$ and the curve $t \mapsto f(e^{\pm i t})$ has a horizontal tangent at $t = \phi$.

One checks that we have the equality $$ 1- |f_{\theta}(e^{it})|^2 = \frac{- \cos (2 \theta) \cos^2 \theta\, (\cos t - \cos \theta)^2 }{\sin^6 \theta}$$ (so the need for $\theta \ge \pi/4$ ). The uniqueness of the optimal polynomial (up to some phase changes) also holds.

Note that for $\theta = \pi/4$ we get $\max \sqrt{2} |a| + |b| = \sqrt{2}$ achieved for the (essentially) unique polynomial $z^2$. If $s\ge \sqrt{2}$ we have $$s|a| + |b| = \frac{s}{\sqrt{2}} \cdot ( \sqrt{2}|a| + \frac{\sqrt{2}}{s}|b|)\le \frac{s}{\sqrt{2}} (\sqrt{2}|a|+|b|)\le \frac{s}{\sqrt{2}} \cdot \sqrt{2} = s$$ Therefore, the maximum of $s|a| + |b|$ is $s$ ( achieved for $z^2$).