0
$\begingroup$

Why $$\left\|\sum_{i=1}^nx_i\right\|^2\leq n\sum_{i=1}^n\|x_i\|^2$$ for arbitrary norm on the inner product space over the real field? My attempt

$$\left\|\sum_{i=1}^nx_i\right\|^2 = \sum_{i=1}^n\sum_{j=1}^n\langle x_i,x_j\rangle = \sum_{i=1}^n\|x_i\|^2 + \sum_{i=1}^n\sum_{j\ne i}\langle x_i,x_j\rangle \\ \leq \sum_{i=1}^n\|x_i\|^2 + \sum_{i=1}^n\sum_{j>i}\left(\|x_i\|^2 + \|x_j\|^2\right)$$

but then I fail to asses the second term.

$\endgroup$
4
  • 1
    $\begingroup$ That's not an arbitrary norm, though. How are you defining norm? $\endgroup$ Dec 18, 2014 at 15:10
  • 2
    $\begingroup$ Assuming you're willing to use some form of the Cauchy Schwartz inequality, just look at $\sum 1 \cdot x_i$ $\endgroup$
    – jxnh
    Dec 18, 2014 at 15:11
  • $\begingroup$ @Thomas The norm is the one associated with the inner product $\|x_i\| = \sqrt{\langle x_i,x_i\rangle }$ $\endgroup$
    – Aad
    Dec 18, 2014 at 15:38
  • $\begingroup$ @JHance This is very nice. Is it ok, if $x_i$ are, for example, $L^2$ functions? I'm confused by the fact that we define the euclidean norm for vector, whose points are not in $\mathbb{R}$, but functions. $\endgroup$
    – Aad
    Dec 18, 2014 at 15:49

1 Answer 1

2
$\begingroup$

Use this identity

$$\left|\sum_{i=1}^{n}a_{i}b_{i}\right|^2+\sum_{i<j}^{n}\left|a_{i}\overline{b_{j}}-a_{j}\overline{b_{i}}\right|^2=\left(\sum_{i=1}^{n}|a_{i}|^2\right)\left(\sum_{i=1}^{n}|b_{i}|^2\right)$$ let $b_{i}=1$

see:http://en.wikipedia.org/wiki/Lagrange%27s_identity

$\endgroup$

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .