Proving ||u-4v|| = 2||u-v|| iff ||u|| = 2||v|| in Hilbert spaces

[member 428835]

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 :)

 

[fresh_42]

Write $||x-y||^2 = \langle x-y,x-y\rangle$ and calculate $||u-4v||^2-4||u-v||^2$.