# Find $\sup_{a a^* + b b^* = 1} | a^2x + aby |$ for fixed $x,y \in \mathbb{C}$, $|x| = 1$, $y \ne 0$.

While trying to come up with an answer to this question without using the theorem used in the existing answer, whose proof is non-trivial, I tried to find $$\tag{1} \sup_{a a^* + b b^* = 1} | a^2x + aby |$$ for some fixed complex numbers $$x, y \in \mathbb{C}$$ In the question the conditions $$|x| = 1$$ and $$y \ne 0$$ are added but I am interested if there's also a more general solution without those constrains. The task of the question is to prove that $$(1)> |x|.$$

My progress Aiming to use Lagrange multipliers, I expanded the term $$a^2x + aby$$ in terms of $$n_1 := \Re(n)$$ and $$n_2 := \Im(n)$$ for $$n \in \{a,b,x,y\}$$ to then use that $$\tag{2} |z| = \sqrt{z_1^2 + z_2^2}$$ for all $$z \in \mathbb{C}$$. I ended up with $$\Re(a^2x + aby) = a_1 b_1 y_1 + a_1 b_2 y_2 + a_2 b_1 y_2 - a_2 b_1 y_2 + a_2^1 x_1 + a_2^2 x_1$$ and $$\Im(a^2 x + aby) = a_1 b_1 y_2 - a_1 b_2 y_1 + a_2 b_1 y_1 + a_2 b_2 y_2 + a_1^2 x_2 + a_2^2 x_2.$$ Plugging this into (2) yields the square root of more than 40 terms. Is there a nice way to factor this or to approach this?

If this were only a real problem, we'd have $$\sup_{a^2 + b^2 = 1} ax^2 + a b y = \sup_{|a| \le 1} a + a \sqrt{1 - a^2} y$$ Setting the derivative w.r.t to $$a$$ equal to zero we obtain $$0 = \sqrt{1 - a^2} y + 1 - \frac{a^2 y}{\sqrt{1 - a^2}},$$ giving $$a^2 = \frac{\pm \sqrt{8 y^2 + 1} - 1}{8y^2} + \frac{1}{2}.$$ Plugging the positive root in gives $$a + a \sqrt{1 - a^2} y = \left(\sqrt{\frac{\pm \sqrt{8 y^2 + 1} - 1}{8y^2} + \frac{1}{2}}\right)\left(1 + \sqrt{\frac{1}{2} - \frac{\pm \sqrt{8 y^2 + 1} - 1}{8y^2}} \cdot y\right),$$ which looks approximately linear in $$y$$ when plotted with WolframAlpha. Unfortunately, $$\Re\left(\sqrt{\frac{1}{2} - \frac{\pm \sqrt{8 y^2 + 1} - 1}{8y^2}}\right) = 0$$ for all $$y \in \mathbb{R}$$.

Write $$a=e^{i\alpha}\cos(\psi),\qquad b=e^{i\beta}\sin(\psi),\qquad 0\leq\psi\leq{\pi\over2}\ .$$ Then $$a\bar a+b\bar b=1$$. We have to find the sup of the quantity $$Q:=\left|e^{2i\alpha}\cos^2(\psi) \>x+e^{i(\alpha+\beta)}\cos(\psi)\sin(\psi)\>y\right|\ ,$$ whereby $$\alpha$$, $$\beta$$, and $$\psi$$ are variables. As $$|x|=1$$ and $$y\ne0$$ we can write $$x=e^{i\xi},\qquad y=|y|e^{i\eta}\ ,$$ so that $$Q =\left|\cos^2(\psi) \>e^{i\xi}+\cos(\psi)\sin(\psi)\>|y|\>e^{i(\beta-\alpha+\eta)}\right|\ .$$ Since $$\xi$$ and $$\eta$$ are given, and $$0\leq\psi\leq{\pi\over2}$$, this $$Q$$ is maximal when $$\beta-\alpha=\xi-\eta$$ mod $$2\pi$$. It follows that $$\max_{\alpha, \beta,\psi}Q =\max_{0\leq\psi\leq\pi/2}\left(\cos^2(\psi)+\cos(\psi)\sin(\psi)\>|y|\right)\ .$$ In this way we are left with a single variable ($$\psi$$) extremal problem: We have $$\frac{\textrm{d}}{\textrm{d} \phi} (\cos^2(\psi)+\cos(\psi)\sin(\psi) \cdot |y|) = | y | \cos(2\phi) - \sin(2\phi) \overset{!}{=} 0 \implies x = \frac{1}{2} \tan^{-1}(|y|)$$ (and $$\frac{\textrm{d}^2}{\textrm{d} \phi^2} \big(\cos^2(\psi)+\cos(\psi)\sin(\psi)|y|\big) \big|_{x = \frac{1}{2} \tan^{-1}(|y|)} < 0$$). Plugging in gives $$\max_{\alpha, \beta, \phi} Q = \frac{\sqrt{|y|^2 + 1} + 1}{2},$$ which is always greater as 1 and only equal to one if $$y = 0$$, which we have excluded. Plotted as a function of its real $$(=x)$$ and imaginary part $$(=y)$$, it looks like this:

• Are those all solutions to $a\bar a+b\bar b=1$? $a=e^{i\alpha}\cos\psi,b=e^{i\beta}\sin\psi$ is sufficient. Is it necessary? – marty cohen Oct 8 '19 at 18:58
• @martycohen: From $|a|^2+|b|^2=1$ it follows that we can write $|a|=\cos\psi$, $|b|=\sin\psi$ with some $\psi\in[0,\pi/2]$. – Christian Blatter Oct 8 '19 at 19:04
• It's interesting that calculus is not needed to determine the max. – marty cohen Oct 8 '19 at 19:28
• Why is $\mathcal{Q}$ maximal when $\beta - \alpha = \xi - \eta \mod 2\pi$? – Viktor Glombik Oct 8 '19 at 20:00
• Because the two summands then show in the same direction. – Please restrict edits of other people's answers to the correction of typos, and similar. – Christian Blatter Oct 9 '19 at 7:03

Here is the solution to Christian Blatter's maximization problem, writing $$p$$ for $$\psi$$.

$$f(p) =\cos^2(p)+\cos(p)\sin(p)y =\frac12(\cos(2p)+1)+\frac12\sin(2p)y$$ so maximizing this is the same as $$g(q) =\cos(q)+\sin(q)y$$ with $$p = q/2$$.

Let $$\tan(r) = y$$, so $$\sin(r) =\frac{y}{\sqrt{1+y^2}}$$ and $$\cos(r) =\frac{1}{\sqrt{1+y^2}}$$.

Then

$$\begin{array}\\ g(q) &=\cos(q)+\sin(q)y\\ &=\sqrt{1+y^2}(\frac1{\sqrt{1+y^2}}\cos(q)+\frac{y}{\sqrt{1+y^2}}\sin(q))\\ &=\sqrt{1+y^2}(\cos(r)\cos(q)+\sin(r)\sin(q))\\ &=\sqrt{1+y^2}\cos(r-q)\\ \end{array}$$

This is maximized when $$r-q=0$$ so that $$q = r =\arctan(y)$$ or $$2p = r$$ or $$p = \frac12\arctan(y)$$.

• Aren't you missing some absolute values around the $y \in \mathbb{C}$? – Viktor Glombik Oct 8 '19 at 19:32