supremum of two sets

Question

Let $$l_1=\sup\{2|a|^2+|b|^2;\;\;(a,b)\in \mathbb{C}^2,\;|a|^2+|b|^2=1\},$$ and $$l_2=\sup\{|a|^2+|a\bar{b}+b\bar{a}|^2;\;\;(a,b)\in \mathbb{C}^2,\;|a|^2+|b|^2=1\}.$$

It is possible that $l_1=l_2$?

Thank you.

Answer 1

Hint : $$l_1=\sup\{|a|^2+1; \ |a|^2\leq 1\} = 2$$

For $l_2$, let $a=r_1e^{i\theta_1}$ and $b=r_2e^{i\theta_2}$ then $$ |a|^2+ |b|^2=1 \quad \Rightarrow \quad r^2_1+r_2^2 =1$$ and $$|a|^2+|a\bar{b}+b\bar{a}|^2= r^2_1+r^2_1r_2^2|e^{i(\theta_1-\theta_2)}+e^{i(\theta_2-\theta_1)} | \leq r^2_1+2r^2_1r_2^2 $$ so \begin{align*} l_2\leq l_3 & = \sup\{r^2_1+2r^2_1r_2^2; \ r_1,r_2\geq 0; \ r^2_1+r_2^2 =1\} \\ & = \sup\{r^2_1+2r^2_1(1-r_1^2); \ 1\geq r_1\geq 0\} \end{align*}

So you have to study the real valued function $$ x\mapsto x^2+2x^2(1-x^2)-2 $$ over $[0,1]$. The derivative is $6x-8x^3$ and you will find the maximums at $$ x=\pm\frac{\sqrt{3}}{2} $$ with value $-7/8$.

Finally (since the converse $l_3\leq l_2$ is clear) $$ l_2=l_3= 2-\frac{7}{8}<l_1$$