Below is what i have proved:

  1. If $V$ is a normed vector space over $\mathbb{R}$ satisfies parallelogram equality, then there exists an inner product $\langle \bullet,\bullet\rangle$ such that $\langle \bullet,\bullet \rangle = \lVert \bullet \rVert^2$

  2. If $V$ is an inner product space over $\mathbb{R}$ and $\lVert \bullet \rVert \triangleq \sqrt{\langle \bullet \rangle}$, then $\lVert \bullet \rVert$ is a norm and $\langle x,y\rangle = \frac{1}{4}(\lVert x+y\rVert^2 - \lVert x-y\rVert^2)$.

I thought defining a norm as square root of inner product $\langle x,x\rangle$ is just one possinle way to define norm on an inner product space.

However, the below is the article in wikipedia about impossibility of defining $p-norm$ on an inner product space:

"p ≠ 2 is a normed space but not an inner product space, because this norm does not satisfy the parallelogram equality required of a norm to have an inner product associated with it"

I don't think what i proved above give this conclusion in wikipeia.. What should i prove to conclude this?

Plus, it's on wikipedia that $2$-notm is completely natural to define as so, but it's not that natural to take square root to inner product to define norm on an inner product space to me. Why do we specifically use 2-norm?

Moreover, what's the point of relating two concepts inner product and norm, literally?


1 Answer 1


Just think of the ordinary scalar product in $2$ or $3$ dimension. This is where these concepts (inner product, norm) originally started out..

If we have an inner product on a space, then geometrically it's more or less nothing else but the ability to talk about angles and lengths (length=norm). By the Cauchy-Schwarz inequality, we have $$\langle x,y\rangle^2\le \langle x,x\rangle\cdot \langle y,y\rangle\,$$ that is, $\left|\displaystyle\frac{\langle x,y\rangle}{\|x\|\cdot\|y\|} \right|\le 1$, supposed that we define the norm as the 2d and 3d cases geometrically motivates (by the Pythagorean theorem). So there is an angle $\theta$ such that $\cos\theta=\left|\displaystyle\frac{\langle x,y\rangle}{\|x\|\cdot\|y\|} \right|$. And, most importantly, we have the notion of orthogonality (which happens to coincide with the case $\theta=90^\circ$).

If we have only a norm $\|\_\|$ on a real vector space, as you stated you proved, the sufficient and necessary condition to the existence of an inner product $\langle -,-\rangle$ such that $\|x\|^2=\langle x,x\rangle$, is that the norm satisfies the parallelogram law.

As wikipedia says, this is not the case for the $p$-norms $p\ne 2$.

  • 1
    $\begingroup$ What's the reason for defining $\lVert - \rVert$ as $\sqrt{\langle -,-\rangle}$, not another function $f(\langle -,- \rangle)$? I'm confused of the meaning of "a norm induced by inner product".. $\endgroup$
    – Jj-
    May 10, 2013 at 17:50
  • $\begingroup$ Well, how to say, this is a good question. The primarily answer is the pure analogy to 2d-3d geometry, applying the Pythagorean theorem. However, it could perhaps make sense to consider any other function of the (quadratic form of the) inner product, that will make a norm. I don't state, only $f=\sqrt{\cdot}$ can do it, good question. $\endgroup$
    – Berci
    May 11, 2013 at 10:12

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .