# Parallelogram law valid in banach spaces?

It is known that the parallelogram law

$\|x-y\|^2+\|x+y\|^2 = 2(\|x\|^2 + \|y\|^2)$

holds in any space with an inner product (the norm being induced by this inner product). Is this formula valid in general Banach spaces? Or are there counterexamples?

Thank you! :)

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A Banach space is a Hilbert space if and only if the Parallelogram law holds. See here. Also, this post should prove helpful. – David Mitra Feb 4 '13 at 14:46

Counterexample: Consider the space of continuous functions in $[0,1]$ $\mathscr{C}[0,1]$ with the supremum norm $$\left\|f\right\|_{\infty}=\max_{x\in [0,1]}\left|f(x)\right|$$ Prove that $\left\|\cdot\right\|_{\infty}$ is a norm and $(\mathscr{C}[0,1],\left\|\cdot\right\|_{\infty})$ is a Banach space. Then consider the functions $f(x)=1-x$ and $g(x)=x$. They are obviously in $\mathscr{C}[0,1]$ but $$\left\|f-g\right\|^2_{\infty}+\left\|f+g\right\|^2_{\infty}=\left\|1-2x\right\|^2_{\infty}+\left\|1\right\|^2_{\infty}=1+1=2$$ but $$2(\left\|f\right\|^2_{\infty}+\left\|g\right\|^2_{\infty})=2\left\|1-x\right\|^2_{\infty}+2\left\|x\right\|^2_{\infty}=2+2=4$$
The parallelogram law holds if and only if the norm is induced by an inner product. Namely, if $\Vert \cdot \Vert$ satisfies the parallelogram law then $$\langle x, y \rangle := \frac{1}{4} \Vert x+y \Vert^2 - \frac{1}{4} \Vert x-y \Vert^2$$ defines an inner product that induces $\Vert \cdot \Vert$.
There are Banach spaces that do not satisfy the parallelogram law, e.g. $\mathbb R^n$ with the $\ell^p$-norm $$\Vert (x_1,\ldots,x_n)\Vert_p = (|x_1|^p + \ldots + |x_n|^p)^{1/p}$$ for $n \geq 2$ and $p \neq 2$ (take $x = (1,0,0,\ldots,0)$ and $y = (0,1,0,\ldots,0)$ to see that the parallelogram law fails).