The Parallelogram Identity for the Norm Induced by an Inner Product

The Parallelogram Identity for the Norm Induced by an Inner Product

Recall from The Normed Space Induced by an Inner Product page that if $H$ is an inner product then the norm induced by the inner product on $H$, $\| \cdot \| : H \to \mathbb{C}$, is defined for all $x \in H$ by:

(1)
\begin{align} \quad \| x \| = \langle x, x \rangle^{1/2} \end{align}

We will now prove that this norm satisfies a very special property known as the parallelogram identity.

Theorem 1 (The Parallelogram Identity): Let $H$ be an inner product space and let $\| \cdot \|$ be the norm induced by the inner product. Then for all $x, y \in H$, $\| x + y \|^2 + \| x - y \|^2 = 2 \| x \|^2 + 2 \| y \|^2$.
  • Proof: We have that:
(2)
\begin{align} \quad \| x + y \|^2 + \| x - y \|^2 &= \langle x + y, x + y \rangle + \langle x - y, x - y \rangle \\ &= \langle x, x \rangle + 2 \mathrm{Re} \langle x, y \rangle + \langle y, y \rangle + \langle x, x \rangle + \langle x, -y \rangle + \langle -y, x \rangle + \langle -y, -y \rangle \\ &= \langle x, x \rangle + 2 \mathrm{Re} \langle x, y \rangle + \langle y, y \rangle + \langle x, x \rangle - \langle x, y \rangle - \langle y, x \rangle + \langle y, y \rangle \\ &= \| x \|^2 + 2 \mathrm{Re} \langle x, y \rangle + \| y \|^2 + \| x \|^2 - 2 \mathrm{Re} \langle x, y \rangle + \| y \|^2 \\ &= 2 \| x \|^2 + 2 \| y \|^2 \end{align}
Theorem 2: Let $X$ be a normed linear space. Then $X$ is an inner product space if and only if the parallelogram identity holds.

If $X$ is a normed linear space in which the parallelogram identity holds for all $x, y \in X$ then it can be shown that the function defined an inner product on $X$ that induces the norm on $X$:

(3)
\begin{align} \quad \langle x, y \rangle = \frac{1}{4} \| x + y \|^2 - \frac{1}{4} \| x - y \|^2 \end{align}
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