I'm having trouble understanding this proof for the reverse triangle inequality of the complex numbers.
Suppose for any $z, w \in \mathbb{C}$, we have $|z + w| \leq |z| + |w|$ (the triangle inequality).
Prove $|\,|z| – |w|\,| \leq |z – w|$.
To prove this, we know $|z + w| \leq |z| + |w|$. Then letting $v = z + w$ gives $z = v – w$, and so we get the inequality $|v| \leq |v – w| + |w|$, or equivalently, $|v| – |w| \leq |v – w|$.
Great, so now we have $|v| – |w| \leq |v – w|$. For some reason, this implies that $|w| – |v| \leq |w – v|$, and then this clearly implies $|\,|w| – |v|\,| \leq |w – v|$.
Two questions:
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We proved that given arbitrary $z, w$ in $\mathbb{C}$, if $v = z + w$, then $|v| – |w| \leq |v – w|$. Why does this imply we can say $|w| – |v| \leq |w – v|$?
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We only proved the inequality for $v$ of the form $z + w$. How does this magically extend to holding for all complex numbers?
I have a feeling someone will answer 2 with "any arbitrary vector in $\mathbb{C}$ can be expressed as $z + w$ since $z$, $w$ were arbitrary" or something like this. If this is true, please explain further.
Best Answer
You know that $|x| \le |x-y|+|y|$ and so $|x|-|y| \le |x-y|$. The same argument with $x,y$ switched gives $|y|-|x| \le |y-x| = |x-y|$. Hence $||x|-|y|| \le |x-y|$.
This is true for any norm, not just the modulus. The essential element here is the triangle inequality.