# Showing inner product comes from a norm defined using Polarization Identity [duplicate]

Possible Duplicate:
Norms Induced by Inner Products and the Parallelogram Law

Trying to prove if an norm satisfies the Parallelogram Law then a norm arises from an inner product.

Let $||\cdot||$ be any norm on $\mathcal{X}$, namely any function that satisfies:

1.$||x|| >0 \ \forall x \neq 0, \\$

2.$||\alpha x|| = |\alpha|*||x|| \ \ \ \forall \alpha \in \mathbb{C}, x\in \mathcal{X},$

3.$||x_1 + x_2|| \leq ||x_1|| + ||x_2||\ \forall x_1,x_2 \in \mathcal{X},$

4.$||x_1 + x_2||^2 + ||x_1 - x_2||^2 = 2||x_1||^2 + 2||x_2||^2 \ \forall x_1, x_2 \in \mathcal{X}.$

$\mathbf{If}$ an inner product existed such that it would arise from a norm, we define this inner product uniquely in terms of our norm by Polarization Identity namely:

$\langle x_1,x_2 \rangle = \frac{1}{4}(||x_1 + x_2||^2 - ||x_1 - x_2||^2 + i ||x_1 - ix_2||^2 - i||x_1 + ix_2||^2)$.
Show this inner product satisfies $||x|| = \sqrt{ \langle x,x \rangle}$ and also satisfies the four properties of an inner product:

i. $\langle x_1,x_2+x_3 \rangle = \langle x_1,x_2 \rangle + \langle x_1,x_3 \rangle \ \forall x_1,x_2,x_3 \in \mathcal{X}$

ii.$\langle x_1, \alpha x_2 \rangle = \alpha \langle x_1,x_2 \rangle \ \forall x_1,x_2\in \mathcal{X}, \ \forall \alpha \in \mathbb{C}$

iii.$\langle x_1, x_2 \rangle = \langle x_2,x_1 \rangle^* \ \forall x_1,x_2\in \mathcal{X}$

iv.$\langle x, x \rangle > 0 \ \forall x \neq 0.$

Hint:\ Use assumptions 1 to 4 to prove i to iv. Use(4) to prove (i). To prove (ii), fix any $x_1, x_2$ and consider the set $\lbrace \alpha \in \mathbb{C}: \langle x_1, \alpha x_2 \rangle = \alpha \langle x_1,x_2 \rangle \rbrace.$ Show that this set contains $\alpha = 0,\alpha = \pm 1$ and $\alpha = i$. Moreover, show that this set is closed under addition(using(i)),subtraction, multiplication and division. Finally, show this set is closed under limits. Use these facts to show that this set contains all integers, and thus all rational numbers, and thus all real numbers, and thus all complex numbers.

I have been able to prove iii and iv but have been stuck trying to prove i and ii. I am having trouble showing how to bring the alpha from inside the inner product to the outside using the Polarization identity. Thinking some way to factor it out would work but not sure if making a substitution would work. I know I need 4 to prove i but having troulbe finding how to simplify when I have $x_1, x_2$ and $x_3$ and I need to separate from three to two variables. Any help is greatly appreciated.

## marked as duplicate by kahen, Noah Snyder, rschwieb, Hagen von Eitzen, EmilyOct 19 '12 at 14:58

• Use the symmetry $(y,x) = \overline{(x,y)}$ together with additivity in the first variable. – commenter Oct 19 '12 at 5:33