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I need to show that the l2 norm : $\|x\|_2 = \sqrt{ x_1^2 + x_2^2 + ... + x_n^2}$ has the following properties:

i) $\|x\| > 0$ if $x\neq 0$

ii) $\|\lambda x\| = |\lambda|\cdot\|x\|$ for $\lambda \in \mathbb{R}$

iii) $\|x+y\| \leq \|x\| + \|y\|$

I have solved the first two but I am having trouble with (iii). I tried doing out the triangle inequality but didn't get anywhere. Could someone possibly help explain this last step?

BlackAdder
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user87803
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1 Answers1

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You need to show: $\sqrt{(x_1+y_1)^2 +(x_2+y_2)^2 +...+(x_n+y_n)^2 } \le \sqrt{x_1^2 +x_2^2 +...+x_n^2} + \sqrt{y_1^2 +y_2^2 +...+y_n^2} $

Claim 1) It's equivalent to:

$(x_1y_1+x_2y_2+...+x_ny_n)^2 \le (x_1^2 +x_2^2 +...+x_n^2)(y_1^2 +y_2^2 +...+y_n^2)$

(hint: square both sides of the original inequality, break the brackets and simplify the terms $x_i^2, y_i^2$)

Claim 2) For any quadratic polynomial: $f(t) = a t^2 + b t + c$, with a>0
$f(t)\ge 0 $ for all t if and only if $b^2 - 4ac \ge 0$ (it can be seen geometrically, when does the bottom of the graph touch the x -coordinate?)

Claim 3) The inequality from Claim 1) is the famous Cauchy-Schwarz inequality, which can be proved using the quadratic polynomial $f(t) = (x_1 - ty_1)^2 + (x_2 - ty_2)^2 + ...+ (x_n - ty_n)^2 \ge 0 \forall t$ and Claim 2)

Long
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