Norm satisfying Parallelogram Law induced by Inner Product

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 $\struct {V, \innerprod \cdot \cdot}$.

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 by hypothesis, we have:


 * $\norm {f + g}^2 + \norm {f - g}^2 = 2 \paren {\norm f^2 + \norm g^2}$

for each $f, g \in V$.

By the hypothesis, 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:

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 Combination Theorem for Sequences: Real Multiple Rule, 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 $\struct {V, \innerprod \cdot \cdot}$.