How these two quantities compare in $\mathbb{C}^n$

Question

Let $z=(z_1,…,z_n),w=(w_1,…,w_n) \in \mathbb{B}$ (the open unit ball in $\mathbb{C}^n$. I want to compare the quantities $A=|1-\left\langle z,w\right\rangle|$ and $|z-w|$ where $|z|^2=|z_1|^2+ \ldots +|z_n|^2$ and $\langle z,w\rangle =z_1\overline{w_1}+…+z_n\overline{w_n}.$

I am stuck as I want to find an inequality between these quantities. Is an inequality of the following type achievable in any way inside the unit ball? $C_1|z-w|\leq |1-\langle z, w\rangle| \leq C_2|z-w|$? Than you for taking the time to look my question.

Answer 1

• If $w$ or $z$ are not on the sphere the inequalities are false.

• If $w$ and $z$ are on the sphere the right inequality will be true. Indeed,$$\left|1 - \left\langle z, \, w\right\rangle\right| = \left|\left\langle w, \, w\right\rangle - \left\langle z, \, w\right\rangle\right| = \left|\left\langle w - z, w \right\rangle\right| \le \left|z\right|\left|z-w\right| \le \left|z-w\right|$$ However the left one can not be true. Let $t = \left\langle z, w\right\rangle$ and $v = z - t w$ (the projection of $z$ onto $w^{\perp}$), $\left|v\right|^2 + t^2\left|w\right|^2 = \left|z\right|^2$ So $$\left|v\right|^2 = 1 - t^2 $$ Since $v\perp w$ then, $$\left|z - w\right|^2 = \left|v\right|^2 + \left(1 - t\right)^2 \left|w\right|^2 = 1 - t^2 + \left(1 - t\right)^2 = \left(1 - t\right)\left(2 + t\right)$$ so $$\left|\frac{1 - \left\langle z, w\right\rangle}{\left|z-w\right|}\right| = \frac{1-t}{\sqrt{(1-t)(2+t)}} = \sqrt{\frac{1-t}{2+t}} \underset{t\to 1}{\to} 0$$ This proves that $$\inf_{z\neq w,\, z\in \mathbb S,\, w\in\mathbb S} \left|\frac{1 - \left\langle z, w\right\rangle}{\left|z-w\right|}\right| = 0.$$

Answer 2

If the inequality were true, then taking $z=w$ would give $0\leq 1-|z|^2\leq 0$, i.e $1-|z|^2=0$ and hence $|z|=1$ which contradicts that $z$ is in the open unit ball.