Norm satisfying Parallelogram Law induced by Inner Product/Real Case

== Theorem ==
Let $V$ be a vector space over $\R$.
Let $\norm \cdot : V \to \R$ be a norm on $V$ such that: 
:$\norm {x + y}^2 + \norm {x - y}^2 = 2 \paren {\norm x^2 + \norm y^2}$
for each $x, y \in V$.
Then the function $\innerprod \cdot \cdot : V \times V \to \R$ defined by: 
:$\ds \innerprod x y = \frac {\norm {x + y}^2 - \norm {x - y}^2} 4$
for each $x, y \in V$, is an inner product on $V$.
Further, $\norm \cdot$ is the inner product norm of $\struct {V, \innerprod \cdot \cdot}$.
=== Corollary ===
== Proof ==
We first verify symmetry. 
Let $x, y \in V$. 
Then we have: 
We now show non-negative definiteness and positiveness.
We have for each $x \in V$: 
Since norms are positive definite, we then have: 
:$\innerprod x x \ge 0$
for each $x \in V$, with: 
:$\innerprod x x = 0$
 $x = 0$.
We finally show linearity in the first argument. 
We first show that for each $x, y, z \in V$ we have: 
:$\innerprod {x + y} z = \innerprod x z + \innerprod y z$
We have: 
Note that , we have: 
:$\norm {f + g}^2 + \norm {f - g}^2 = 2 \paren {\norm f^2 + \norm g^2}$
for each $f, g \in V$.
So , setting $f = x + z$ and $g = y$, we have: 
:$\norm {x + y + z}^2 + \norm {x - y + z}^2 = 2 \paren {\norm {x + z}^2 + \norm y^2}$
similarly, setting $f = x - z$ and $g = y$:
:$\norm {x + y - z}^2 + \norm {x - y - z}^2 = 2 \paren {\norm {x - z}^2 - \norm y^2}$
Setting $f = x$ and $g = y + z$ we also obtain:
:$\norm {x + y + z}^2 + \norm {x - y - z}^2 = 2 \paren {\norm x^2 + \norm {y + z}^2}$
Finally setting $f = x$ and $g = y - z$ we get:
:$\norm {x + y - z}^2 + \norm {x - y + z}^2 = 2 \paren {\norm x^2 + \norm {y - z}^2}$
Putting this all together, we have:
We now show that: 
:$\innerprod {\alpha x} y = \alpha \innerprod x y$
for any $\alpha \in \R$ and $x, y \in V$. 
We then have the :
=== Lemma ===
We then have, for $q \in \N$: 
:$\ds q \innerprod {\frac x q} y = \innerprod x y$
so:
:$\ds \innerprod {\frac x q} y = \frac 1 q \innerprod x y$
So, for any integer $p \ge 0$, we have: 
:$\ds \innerprod {\frac p q x} y = \frac p q \innerprod x y$
That is: 
:$\innerprod {\alpha x} y = \alpha \innerprod x y$
for any rational number $\alpha \ge 0$.
We now note that:
We can therefore obtain, for any rational number $\alpha \ge 0$:
That is: 
:$\innerprod {\alpha x} y = \alpha \innerprod x y$
for all $\alpha \in \Q$.
We now conclude with a density argument using Inner Product is Continuous.
Let $\alpha \in \R$ and $x, y \in V$. 
From Rationals are Everywhere Dense in Reals: Normed Vector Space:
:for each $n \in \N$, we can pick $\alpha_n \in \Q$ such that $\size {\alpha - \alpha_n} < \dfrac 1 n$ 
Then we have: 
:$\alpha_n \to \alpha$
So from Convergence of Product of Convergent Scalar Sequence and Convergent Vector Sequence in Normed Vector Space, we have: 
:$\alpha_n x \to \alpha x$ 
Then we have, from Inner Product is Continuous: 
:$\ds \lim_{n \mathop \to \infty} \innerprod {\alpha_n x} y = \innerprod {\alpha x} y$
while:
:$\innerprod {\alpha_n x} y = \alpha_n \innerprod x y$
So, from the Multiple Rule for Real Sequences, we have: 
:$\ds \innerprod {\alpha x} y = \paren {\lim_{n \mathop \to \infty} \alpha_n} \innerprod x y = \alpha \innerprod x y$
for all $x, y \in V$ and $\alpha \in \R$. 
So we have that $\innerprod \cdot \cdot$ is an inner product. 
Finally, note that since: 
:$\innerprod x x = \norm x^2$
We immediately obtain: 
:$\norm x = \sqrt {\innerprod x x}$
Since $\innerprod \cdot \cdot$ is an inner product, we have that $\norm \cdot$ is the inner product norm for $\struct {V, \innerprod \cdot \cdot}$.
Category:Norm satisfying Parallelogram Law induced by Inner Product
Category:Normed Vector Spaces
Category:Inner Product Spaces