Prove that this function define a norm on $\mathbb{R}^2$.

Question

Let $\| \;\|:\mathbb{R}^2\rightarrow\mathbb{R}$ be a function such that $ \| (x,y)\|=\sqrt{|x|^2+|y|^2}$ for all $(x,y)\in\mathbb{R}^2$.

I need to show that $\|\;\|$ defines a norm. I would like a hint in order to prove that $\parallel \; \parallel$ satisfies the triangle inequality:

$ \sqrt{|x_1+x_2|^2+|y_1+y_2|^2}\leq \sqrt{|x_1|^2+|y_1|^2}+\sqrt{|x_2|^2+|y_2|^2}$

Thanks!

Answer 1

Let $x, y\in \mathbb{R}^2$. Then

\begin{align} \|x+y\|^2&=\langle x+y, x+y\rangle\\&=\langle x, x\rangle + 2\langle x, y\rangle+\langle y, y\rangle \\ &\le \{\text{Cauchy-Schwarz}\} \\&\le \|x\|^2+2\|x\|\|y\|+\|y\|^2 \\ &=(\|x\|+\|y\|)^2, \end{align} which is your result.

Answer 2

You can do it directly, for $\mathbb{R}^2$; it becomes really complicated in higher dimension, where the Cauchy-Schwarz inequality is the best approach.

Square your inequality: you can because both terms are non negative. You can also remove the absolute values, because $(-a)^2=a^2$. Thus your inequality is equivalent to $$ (x_1+x_2)^2+(y_1+y_2)^2\leq x_1^2+y_1^2+x_2^2+y_2^2 +2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)}. $$ Expand the terms and you get, after rearranging terms, $$ 2(x_1x_2+y_1y_2)\le+2\sqrt{(x_1^2+y_1^2)(x_2^2+y_2^2)} $$ that's obviously satisfied when $x_1x_2+y_1y_2<0$. So we proceed under the additional assumption that $x_1x_2+y_1y_2\ge0$, so the inequality is equivalent to the one we obtain by squaring (first remove the common factor $2$): $$ (x_1x_2+y_1y_2)^2\le(x_1^2+y_1^2)(x_2^2+y_2^2) $$ Expand the terms $$ x_1^2x_2^2+2x_1x_2y_1y_2+y_1^2y_2^2\le x_1^2x_2^2+x_1^2y_2^2+x_2^2y_1^2+y_1^2y_2^2 $$ and transport all terms to the right hand side $$ 0\le x_1^2y_2^2-2x_1x_2y_1y_2+x_2^2y_1^2 $$ which can be written $$ 0\le (x_1y_2-x_2y_1)^2 $$ that's certainly true.