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

My attempt:

I need to show that:

(1) $\forall x,z\in\mathbb C^n$, $d(x,z)\geq0$ and $d(x,z)=0$ iff $x=z$.

$d(x,z)=||x-z||=(\sum_{j=1}^{n}|x_j-z_j|^2)^\frac{1}{2}\geq0.$ For $x=z$, $(\sum_{j=1}^{n}|x_j-z_j|^2)^\frac{1}{2}=\sum_{j=1}^{n}0 = 0$.

(2) $\forall x,z\in\mathbb C^n$, $d(x,z)=d(z,x)$.

$d(x,z)=||x-z||=(\sum_{j=1}^{n}|x_j-z_j|^2)^\frac{1}{2} = (\sum_{j=1}^{n}|z_j-x_j|^2)^\frac{1}{2} = d(z,x)$ by properties of modulus in $\mathbb{C}$.

(3) $\forall x,y,z\in\mathbb C^n$, $d(x,z)\leq d(x,y)+d(y,z)$.

This is always the tough one. Do I need Cauchy-Schwarz or Hoelder's Inequality? Attempt: $d(x,z)=||x-y+y-z||=(\sum_{j=1}^{n}|x_j-y_j+y_j-z_j|^2)^\frac{1}{2} =(\sum_{j=1}^{n}|x_j-y_j|+|y_j-z_j|^2)^\frac{1}{2}$,...

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Did you prove Cauchy-Schwarz Inequality? That should be an ideal starting point! – user21436 Jan 24 '12 at 16:57

I think you can also try to prove that $\parallel\cdot\parallel$ is a norm over $\mathbb{C}^n$. This is less uncomfortable regarding calculations. Then you simply note that your $d$ is the metric induced by the norm $\parallel\cdot \parallel$ and use triangular inequality without problems.

In other words, prove that for all $x\in \mathbb{C}^n$

$\displaystyle \parallel x + y \parallel \leq \parallel x\parallel + \parallel y\parallel$

A good trick could be also to prove the inequality holds for squares...

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Let $s_j=x_j-y_j$ and $t_j=y_j-z_j$.

Then: $$\eqalign{ d^2(x,z)&= \sum_{j=1}^n |s_j+t_j|^2\cr &\le\sum_{j=1}^n |s_j|^2+2\sum_{j=1}^n |s_j t_j| +\sum_{j=1}^n |t_j|^2 \cr &\le \sum_{j=1}^n |s_j|^2+2\Bigl(\,\sum_{j=1}^n |s_j|^2\,\Bigr)^{1/2}\Bigl(\,\sum |t_j|^2\,\Bigr)^{1/2}+\sum_{j=1}^n|t_j|^2\cr &=\biggl[\,\Bigl(\,\sum_{j=1}^n |s_j|^2\,\Bigr)^{1/2}+\Bigl(\,\sum_{j=1}^n |t_j|^2\,\Bigr)^{1/2}\,\biggr]^2\cr &=\bigl[\, d(x,y)+d(y,z)\,\bigr]^2; } $$ and the result follows after taking square roots.

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