[proofplan] We prove two directions. The forward direction (parallelogram law in inner product spaces) is a direct algebraic expansion. The converse (Jordan--von Neumann theorem) defines a candidate inner product via the polarisation identity $\frac{1}{4}(\|x+y\|^2 - \|x-y\|^2)$ and verifies it satisfies symmetry, positive definiteness, and bilinearity. The main difficulty is additivity, which requires the parallelogram law to establish $2$-homogeneity and then inductively extend to full linearity. [/proofplan] [step:Prove the parallelogram law holds in any inner product space] For any $x, y \in H$, expand using bilinearity and symmetry of the inner product: \begin{align*} \|x + y\|^2 + \|x - y\|^2 &= (x+y, x+y) + (x-y, x-y) \\ &= \|x\|^2 + (x,y) + (y,x) + \|y\|^2 + \|x\|^2 - (x,y) - (y,x) + \|y\|^2 \\ &= 2\|x\|^2 + 2\|y\|^2. \end{align*} The cross terms $(x,y) + (y,x)$ cancel between the two expansions. [/step] [step:Define the candidate inner product via the polarisation identity] In a real [Banach space](/page/Banach%20Space) $(X, \|\cdot\|)$ satisfying the parallelogram law, define \begin{align*} (x, y) := \frac{1}{4}\left(\|x+y\|^2 - \|x-y\|^2\right). \end{align*} [/step] [step:Verify symmetry and positive definiteness] **Symmetry.** $(x,y) = \frac{1}{4}(\|x+y\|^2 - \|x-y\|^2) = \frac{1}{4}(\|y+x\|^2 - \|-(y-x)\|^2) = \frac{1}{4}(\|y+x\|^2 - \|y-x\|^2) = (y,x)$. **Positive definiteness.** $(x, x) = \frac{1}{4}(\|2x\|^2 - 0) = \|x\|^2 > 0$ for $x \ne 0$. [/step] [step:Establish bilinearity using the parallelogram law] [claim:Bilinearity of the polarisation form] If the parallelogram law holds, then the form $(x,y)$ defined above is bilinear. [/claim] [proof] The key difficulty is **additivity**: $(x + z, y) = (x, y) + (z, y)$. **$2$-homogeneity.** Apply the parallelogram law with $a = x+y$ and $b = x-y$ to establish $(2x, y) = 2(x, y)$. By induction, $(nx, y) = n(x, y)$ for all $n \in \mathbb{N}$, then for $n \in \mathbb{Z}$ (using $(0, y) = 0$ and $(-x, y) = -(x,y)$), then for $n \in \mathbb{Q}$ (writing $n = p/q$ and using $q(x/q, y) = (x, y)$). Continuity of the norm extends this to $(\lambda x, y) = \lambda(x, y)$ for all $\lambda \in \mathbb{R}$. **Additivity.** The parallelogram law applied to suitable substitutions yields \begin{align*} \|x + z + y\|^2 + \|x + z - y\|^2 &= 2\|x + z\|^2 + 2\|y\|^2, \\ \|x + y\|^2 + \|z + y\|^2 &= 2\left\|\frac{x+z}{2} + y\right\|^2 + 2\left\|\frac{x-z}{2}\right\|^2. \end{align*} Combining these identities with the $2$-homogeneity just established gives $(x+z, y) = (x, y) + (z, y)$, completing the proof of bilinearity. [/proof] [/step] [step:Conclude that $X$ is a Hilbert space] Since $(x, y)$ is bilinear, symmetric, and positive definite, it is an inner product on $X$. The induced norm satisfies $(x,x)^{1/2} = \|x\|$, recovering the original norm. Since $(X, \|\cdot\|)$ is a Banach space (complete), it is a [Hilbert space](/page/Hilbert%20Space) with this inner product. [/step]