Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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! :)

share|improve this question
8  
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
add comment

2 Answers

up vote 4 down vote accepted

No. This formula holds if and only if an inner product induces the norm, that is your Banach space is actually a Hilbert space.

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 $$2(\left\|f\right\|^2_{\infty}+\left\|g\right\|^2_{\infty})=2\left\|1-x\right\|^2_{\infty}+2\left\|x\right\|^2_{\infty}=2+2=4$$

share|improve this answer
add comment

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).

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.