Triangle inequality for the $l^2$ norm

Question

Consider the sequence space $l^2:=\{(x_n)_n\mid \sum^\infty_{n=0}x_n<\infty\}$ together with the norm $$ ||(x_n)_n||=(\sum^\infty_{n=0}|x_n|^2)^{1/2} $$ How can I show that the triangle inequality holds for $||\cdot||$?

Answer 1

Let $x,y\in\ell^2(\mathbb{N})$. Then we have $$\begin{align} \lVert x+y\rVert^2 &= \sum_{n=0}^\infty \lvert x_n+y_n\rvert^2 \leq \sum_{n=0}^\infty (\lvert x_n\rvert+\lvert y_n\rvert)^2 \tag{Triangle}\\ &= \lVert x\rVert^2+\lVert y\rVert^2 + 2\sum_{n=0}^\infty \lvert x_n\rvert\lvert y_n\rvert\\ &\leq \lVert x\rVert^2+\lVert y\rVert^2 + 2\sum_{n=0}^\infty \lvert x_n\rvert^2\sum_{n=0}^\infty\lvert y_n^2\rvert \tag{Cauchy-Schwarz}\\ &= \lVert x\rVert^2+\lVert y\rVert^2 + 2\lVert x\rVert\lVert y\rVert\\ &= \left(\lVert x\rVert+\lVert y\rVert\right)^2 \end{align}$$ (where the first inequality is the triangle inequality in $\mathbb{R}$); so that $$ \lVert x+y\rVert \leq \lVert x\rVert+\lVert y\rVert\,. $$

Answer 2

Hint Cauchy-Schwartz. You can either show that $l^2$ is an inner product space, or use the fact that for each $N$ you have by C-S in $\mathbb R^N$: $$(\sum^N_{n=0}|x_n+y_n|^2)^{1/2} \leq (\sum^N_{n=0}|x_n|^2)^{1/2}+(\sum^N_{n=0}|y_n|^2)^{1/2}\leq (\sum^\infty_{n=0}|x_n|^2)^{1/2}+(\sum^\infty_{n=0}|y_n|^2)^{1/2} $$