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 $\struct {V, \innerprod \cdot \cdot}$.
Corollary
Let $\struct {V, \norm \cdot}$ be a normed vector space over $\R$.
Then there exists an inner product $\innerprod \cdot \cdot : V \times V \to \R$ such that:
- $\norm x = \sqrt {\innerprod x x}$
- $\norm {x + y}^2 + \norm {x - y}^2 = 2 \paren {\norm x^2 + \norm y^2}$
for each $x, y \in V$.
That is, a norm is induced by an inner product if and only if it satisfies the Parallelogram Law for Inner Product Spaces.
Proof
We first verify symmetry.
Let $x, y \in V$.
Then we have:
\(\ds \innerprod y x\) | \(=\) | \(\ds \frac {\norm {y + x}^2 - \norm {y - x}^2} 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\norm {x + y}^2 - \norm {-\paren {x - y} }^2} 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\norm {x + y}^2 - \norm {x - y}^2} 4\) | from the positive homogeneity of the norm $\norm \cdot$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \innerprod x y\) |
We now show non-negative definiteness and positiveness.
We have for each $x \in V$:
\(\ds \innerprod x x\) | \(=\) | \(\ds \frac {\norm {x + x}^2 - \norm {x - x}^2} 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {\norm {2 x}^2 - \norm 0^2} 4\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac {4 \norm x^2} 4\) | from the positive homogeneity of the norm $\norm \cdot$ | |||||||||||
\(\ds \) | \(=\) | \(\ds \norm x^2\) |
Since norms are positive definite, we then have:
- $\innerprod x x \ge 0$
for each $x \in V$, with:
- $\innerprod x x = 0$
if and only if $x = 0$.
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:
\(\ds \innerprod x y + \innerprod y z\) | \(=\) | \(\ds \frac 1 4 \paren {\norm {x + y}^2 - \norm {x - y}^2 + \norm {y + z}^2 - \norm {y - z}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 4 \paren {\norm {x + z}^2 + \norm y^2 - \norm {x - z}^2 - \norm y^2 + \norm x^2 + \norm {y + z}^2 - \norm x^2 - \norm {y - z}^2}\) | adding $\dfrac 1 4 \paren {\norm y^2 - \norm y^2 + \norm x^2 - \norm x^2} = 0$ |
Note that by hypothesis, we have:
- $\norm {f + g}^2 + \norm {f - g}^2 = 2 \paren {\norm f^2 + \norm g^2}$
for each $f, g \in V$.
So by 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:
\(\ds \frac 1 4 \paren {\norm {x + z}^2 + \norm y^2 - \norm {x - z}^2 - \norm y^2 + \norm x^2 + \norm {y + z}^2 - \norm x^2 - \norm {y - z}^2}\) | \(=\) | \(\ds \frac 1 8 \paren {\norm {x + y + z}^2 + \norm {x - y + z}^2 - \norm {x + y - z}^2 - \norm {x - y - z}^2 + \norm {x + y + z}^2 + \norm {x - y - z}^2 - \norm {x + y - z}^2 - \norm {x - y + z}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 2 8 \paren {\norm {x + y + z}^2 - \norm {x + y - z}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 4 \paren {\norm {x + y + z}^2 - \norm {x + y - z}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \innerprod {x + y} z\) |
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
- $\innerprod {n x} y = n \innerprod x y$
for each integer $n \ge 0$ and $x, y \in V$.
$\Box$
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:
\(\ds \innerprod {-x} y\) | \(=\) | \(\ds \frac 1 4 \paren {\norm {-x + y}^2 + \norm {-x - y}^2}\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds \frac 1 4 \paren {\norm {x - y}^2 - \norm {x + y}^2}\) | from the positive homogeneity of the norm $\norm \cdot$ | |||||||||||
\(\ds \) | \(=\) | \(\ds -\innerprod x y\) |
We can therefore obtain, for any rational number $\alpha \ge 0$:
\(\ds \innerprod {-\alpha x} y\) | \(=\) | \(\ds -\innerprod {\alpha x} y\) | ||||||||||||
\(\ds \) | \(=\) | \(\ds -\alpha \innerprod x y\) |
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 $\struct {V, \innerprod \cdot \cdot}$.
$\blacksquare$