If every inner product space can be converted into a norm space, then why is there a distinction between the two?

Question

If every inner product space can be converted into a norm space, then why is there a distinction between the two? $$\|x\| = \sqrt{\langle x,x\rangle }$$

Answer 1

Even in the realm of finite spaces, one can define infinitely many norms. For example for each integer $n$, the norm

$$ \|x\|= \left( \sum_{k=1}^n \left| x_k \right| ^n \right)^{\frac 1 n}$$ is a norm. The inner product norm is the case of $n=2$.

Answer 2

Because there are normed spaces whose norm does not arise from an inner product. An example of this is $\ell^1$, the space of absolutely convergent sequences.

One easy way to check if a norm arises from an inner product is the Parallelogram Law.

Answer 3

Yes, there are normed spaces $(X,\|\cdot\|)$ whose norms are not generated by inner products. For a complex inner product space, the norm is generated by an inner-product iff the norm satisfies the the parallelogram law: $$ \|x+y\|^{2}+\|x-y\|^{2}=2\|x\|^{2}+\|y\|^{2} \mbox{ for all } x,y \in X. $$ This condition can be checked for a norm without any mention of an inner product.

One of the things that's very different about a Hilbert space is that, if $M$ is a closed subspace of $X$, then, for every $x \in X$, there is a unique $y \in M$ which is closest to $x$. That is, there exists a unique $m \in M$ such that $$ \|x-m\| \le \|x-m'\| \mbox{ for all } m' \in M \mbox{ with equality iff } m'=m. $$ Furthermore, if $d=\inf_{m\in M}\|x-m\|$, and if you choose $\{ m_{n}\}_{n=1}^{\infty} \subseteq M$ such that $\|x-m_{n}\|\rightarrow d$, then $\{ m_{n} \}_{n=1}^{\infty}$ automatically converges in norm to $m$. That's a strong statement about arbitrary dimensional Hilbert Space.

The closest point projection may not exist or may not be unique for general normed spaces. Even for a two-dimensional space $\mathbb{C}^{2}$ endowed with the norm $|(w_{1},w_{2})|=|w_{1}|+|w_{2}|$, there are infinitely many closest point projections of $(1,1)$ onto the subspace $M=\{ (-z,z) : z \in \mathbb{C} \}$. For example, if $z$ is real and satisfies $-1 \le z \le 1$, then $$ \|(-z,z)-(1,1)\|_{1} = |1-z|+|1+z|=1-z+1+z=2. $$ And 2 is the closest distance, even though there are many points in the subspace $M$ which are at a distance of 2 from $(1,1)$. The closed Euclidean ball centered at $(1,1)$ which is generated from the Euclidean inner-product of radius $r=\sqrt{2}$ touches $M$ at exactly one point--namely, the origin.

The geometry of the unit ball drastically affects closest-point projection, distance, convergence; and nothing is much nicer than for an inner-product space.