Calculate $\sup_{(a,b)\in B_2(0,1)}\left\{\max(|a|,|a+b|)\right\}$

Question

Let $$B_2(0,1):=\{(a,b)\in \mathbb{C}^2;\;|a|^2+|b|^2<1\}.$$

I want to calculate $$M:=\sup_{(a,b)\in B_2(0,1)}\left\{\max(|a|,|a+b|)\right\}.$$

I hope to show that $M\neq 1$.

Answer 1

The supremum is $M = \sqrt{2}$.

We have

$$|a| \le |a+b| \le |a| + |b| \stackrel{CSB}\le \sqrt{2}\sqrt{|a|^2+|b|^2} < \sqrt{2}$$

so $M \le \sqrt{2}$.

On the other hand, for the points $(a,a)$ with $a \in \left[0, \frac{\sqrt{2}}2\right\rangle$ we have $(a,a) \in B_2(0,1)$ and

$$\max\{|a|, |a+a|\} = |a + a| = 2|a| = 2a \xrightarrow{a \to \frac{\sqrt{2}}2} \sqrt{2}$$

so $M \ge \sqrt{2}$.

Answer 2

Since for all $(a,b)\in B_2(0,1)$ we have $$\max\{|a|,|a+b|\}\leq|a|+|b|\leq\sqrt{2}\sqrt{|a|^2+|b|^2}<\sqrt{2}$$ it follows that $M\leq\sqrt{2}$.

On the other hand, if $0\leq a<1$ and $0\leq b<1$ then $\max\{|a|,|a+b|\}=a+b$. It follows that for all $r<1$ we have $$M\geq \max\bigl\{a+b\bigm| a^2+b^2=r^2\bigr\}=\sqrt{2} r\ .$$ Since this is true for all $r<1$ we conclude that $M\geq\sqrt{2}$, so that in fact $M=\sqrt{2}$.