Calculate a supremum over the unit sphere of $\mathbb{C}^2$

Question

I want to calculate $$K=\sup\left\{\left|y\overline{x}+|y|^2\right|^2+1;\;\;(x,y)\in \mathbb{C}^2,\;|x|^2+|y|^2=1\right\}.$$

I try to solve the problem as follows:

let $x=r_1e^{i\theta_1}$ and $y=r_2e^{i\theta_2}$ then $$ |x|^2+ |y|^2=1 \quad \Rightarrow \quad r^2_1+r_2^2 =1.$$ and $$y\overline{x}+|y|^2= r_2^2+r_1r_2e^{i(\theta_2-\theta_1)}.$$ so $$|r_2^2+r_1r_2e^{i(\theta_2-\theta_1)}|^2\leq r_2^4+r_1^2r_2^2.$$ Hence

$$K\leq\sup\{r_2^4+r_1^2r_2^2;\;\;r_1,r_2\geq0,\;r^2_1+r_2^2 =1\}.$$

Answer 1

You can simplify the problem a little first: Note that $|y \bar{x} +|y|^2| = |y (\bar{x}+\bar{y})| = |y| |x+y|$. We can choose the phase of $x,y$ arbitrarily, and $|x+y| \le |x|+|y|$ with equality when they are colinear. so we can choose $x,y$ to be real.

In particular, the problem reduces to $\max \{ y (x+y) | x,y \in \mathbb{R}, x^2+y^2 = 1 \}$.

You can solve the equivalent problem $ \max_t \sin t (\cos t + \sin t) = \max_t {1 \over 2} (1+\sin (2t) - \cos (2t)) = \max_t {1 \over 2}(1+\sqrt{2}\sin (2x - {\pi \over 4}))$.

Answer 2

We have

$$ f(x,y) = (y\bar x+y\bar y)(\bar y x+\bar y y)+1 = |y|^2\left(|x|^2+|y|^2+y\bar x+\bar y x\right)+1 = |y|^2\left(y\bar x+\bar y x+1\right)+1 $$

now calling

$$ x = \rho_x e^{i\phi}\\ y = \rho_y e^{i\psi} $$

$$ f(\rho_x e^{i\phi},\rho_y e^{i\psi}) = \rho_y^2\left(2\rho_2\rho_y\cos(\phi-\psi)+1\right)+1 $$

so the problem now reads

$$ \sup f(\rho_x e^{i\phi},\rho_y e^{i\psi}) \;\;\;\mbox{subjected to}\;\;\; \rho_x^2+\rho_y^2 = 1 $$

Here $\rho_x \ge 0, \rho_y \ge 0$ so the problem can be simplified to

$$ \sup \rho_y^2\left(2\rho_x\rho_y+1\right)+1 \;\;\;\mbox{subjected to}\;\;\; \rho_x^2+\rho_y^2 = 1 $$

This problem can be solved easily with the Lagrange multipliers technique giving the results

$$ \rho_x = \frac{84-28 \sqrt{2}-7 \sqrt{7} \left(1+2 \sqrt{2}\right)^{3/2}+3 \sqrt{7 \left(1+2 \sqrt{2}\right)}+6 \sqrt{14 \left(1+2 \sqrt{2}\right)}}{28 \left(\sqrt{2}-\sqrt{7 \left(1+2 \sqrt{2}-3\right)}\right)}\\ \rho_y = \frac{1}{14} \left(6-2 \sqrt{2}+7 \sqrt{\frac{4}{7}+\frac{8 \sqrt{2}}{7}}\right) $$

and the maximum is

$$ \frac{3283+392 \sqrt{2}+10 \sqrt{7 \left(835+872 \sqrt{2}\right)}}{2401}\approx 2.09937 $$

Answer 3

You had $$y\overline{x}+|y|^2= r_2^2+r_1r_2e^{i(\theta_2-\theta_1)}\ .$$ Here the right hand side has maximal absolute value when $e^{i(\theta_2-\theta_1)}=1$. It follows that $$\sup\nolimits_{\theta_1,\>\theta_2}\,\bigl|y\overline{x}+|y|^2\bigr|=r_2(r_1+r_2)\ .$$ We now have to maximize the RHS under the constraint $r_1^2+r_2^2=1$. To this end we write $$r_1=\sin u,\quad r_2=\cos u, \qquad{\rm with}\quad 0\leq u\leq{\pi \over2}\ ,$$ and then have $$r_2(r_1+r_2)={1\over2}\bigl(\sin(2u)+\cos(2u)+1\bigr)={1\over2}\bigl(\sqrt{2}\cos({\textstyle{\pi\over4}}-2u)+1\bigr)\ .$$ Here the right hand side is maximal when $u={\pi\over8}$, so that $$\max\bigl(r_2(r_1+r_2)\bigr)={\sqrt{2}+1\over2}\ .$$ It follows that $$K=\left({\sqrt{2}+1\over2}\right)^2+1={7+2\sqrt{2}\over4}\ .$$