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:

\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}