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Show that $d$ is a metric on $\mathbb{C^n}$

On $\mathbb{C^n}$, define $||z||=(\sum_{j=1}^{n}|z_{j}|^{2})^{1/2}$ and for $z,w\in\mathbb{C^n}$ define $d(z,w)=||z-w||$. Show that $d$ is a metric on $\mathbb{C^n}$.

My attempt:

(1) (nonnegativity) It is clear that for any $z,w\in\mathbb{C^n}$, $d(z,w)=||z-w|| = (\sum_{j=1}^{n}|z_{j}-w_{j}|^{2})^{1/2}\geq 0$ since $|z_{j}-w_{j}|^{2}\geq0$. Also, $||z-w||=0$ iff $(\sum_{j=1}^{n}|z_{j}-w_{j}|^{2})^{1/2}= 0$ iff $z=w$.

(2) (symmetry) $d(z,w)=||z-w|| = (\sum_{j=1}^{n}|z_{j}-w_{j}|^{2})^{1/2}=(\sum_{j=1}^{n}|w_{j}-z_{j}|^{2})^{1/2}=d(w,z)$ by properties of modulus in $\mathbb{C^n}$.

(3) (triangle inequality) $\forall w,z,v\in\mathbb{C^n}$, $$\begin{align*}d(z,w)&=||z-w||\\ &= \left(\sum_{j=1}^{n}|z_{j}-w_{j}|^{2}\right)^{1/2}\\ &=\left(\sum_{j=1}^{n}|z_{j}+v_{j}-v_{j}-w_{j}|^{2}\right)^{1/2}\\ &=\left(\sum_{j=1}^{n}|(z_{j}-v_{j})+(v_{j}-w_{j})|^{2}\right)^{1/2}\leq... \end{align*}$$

I'm not sure how to split up the sum here.

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marked as duplicate by Nate Eldredge, Srivatsan, Zev Chonoles Jan 28 '12 at 4:00

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You wrote an equality in (3) instead of an inequality. Anyway, $\mathbb{C}^n\cong\mathbb{R}^{2n}$ as additive groups, so you can just inherit the triangle inequality from the general Euclidean case (or at least transplant the proof of it). –  anon Jan 27 '12 at 22:09
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well...that's embarrassing :P –  Emir Jan 27 '12 at 22:42

2 Answers 2

If we consider a complex number as a pair of real numbers, then your metric is equivalent to the Euclidean distance in $\mathbb{R}^{2n}$

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Part (3) follows from Cauchy-Schwarz inequality. $\vert \langle z, w\rangle\vert^2\leq \langle z,z\rangle \langle w,w\rangle $

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