# Norms and Hilbert spaces

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In summary, the problem asks to show that for vectors u and v in a Hilbert space V, the statement ||u-4v|| = 2||u-v|| is equivalent to ||u|| = 2||v||. The proof involves using the parallelogram identity and squaring both sides of the equation to manipulate the norms. The backward direction can be shown by writing ||x-y||^2 as <x-y, x-y> and calculating ||u-4v||^2-4||u-v||^2.

## Homework Statement

Given a Hilbert space ##V## and vectors ##u,v\in V##, show $$\|u-4v\| = 2\|u-v\| \iff \| u \| = 2 \| v\|.$$

## Homework Equations

The parallelogram identity $$2\| x \|^2+2\| y \|^2 = \| x-y \|^2 + \| x+y \|^2$$

## The Attempt at a Solution

Forward:
$$\|u-4v\| = 2\|u-v\| \implies\\ \|u-4v\|^2 = 4\|u-v\|^2\implies\\ \int(u-4v)^2 = 4\int(u-v)^2 \implies\\ \int(u^2-4v^2) = 0\implies\\ \int u^2 = 4\int v^2\implies\\ \| u \| = \| 2 v\|.$$

Backwards is tough. All I can think of is the parallelogram identity, but it is not immediately obvious to me if this helps. I can square the first equation to have a similar form as the parallelogram identity, but cannot see how it applies. I really just don't know how to break apart the summed terms inside the norm aside from inequalities, which won't help since I must show equality. I just read the text and nothing pops up.

EDIT: I must be slow today, backward is clear now (same thing I posted but opposite). Sorry for not seeing this earlier, but thanks for stopping by to help :)

Last edited by a moderator:
Write ##||x-y||^2 = \langle x-y,x-y\rangle## and calculate ##||u-4v||^2-4||u-v||^2##.