Is all normed space also inner product space?

Question

1) I know that all inner product space is also a normed space with the norm induce by the scalar product, but is the reciprocal true ? I mean, is all normed space also a inner product space ?

2) I know that all normed space is a metric space with the metric induced by the norm. Is the reciprocal true ? I mean, is all metric space also a normed space ?

Answer 1

The answer is no for both questions.

For the first one, any norm induced by an inner product must satisfy the identity $$2\lVert u\rVert^2 + 2\lVert v\rVert^2 = \lVert u + v\rVert^2 + \lVert u - v\rVert^2;$$ but some norms do not satisfy this, and hence do not come from an inner product. For example, the supremum norm on $\Bbb R^2$ does not satisfy the identity : $$2\|(1,0)\|_\infty^2+2\|(0,1)\|^2_\infty=4\neq \|(1,1)\|_\infty^2+\|(1,-1)\|_\infty^2=2.$$ However, any norm satisfying the identity does come from an inner product; see this question.

For the second one, note that metric space are not even required to be vector spaces, so for example the set $\{1,\ldots,10\}$ equipped with the metric $d(x,y)=|x-y|$ is not a normed space. You can also consider any set with its discrete metric.

Answer 2

#1: No. Norms which are induced by inner products are exactly those satisfying the "parallelogram law": $2 \| x \|^2 + 2 \| y \|^2 = \| x+y \|^2 + \| x-y \|^2$. In this case you have an inner product defined by the "polarization identity" $\langle x,y \rangle = \frac{1}{4} \left ( \| x+y \|^2 - \| x - y \|^2 \right )$ (with some small change in the complex case). Two norms on $\mathbb{R}^n$ for $n>1$ that do not have this property are the $1$ norm, $\| x \|_1=\sum_{i=1}^n |x_i|$, and the $\infty$ norm, $\| x \|_\infty = \max_i |x_i|$.

#2: No, there are many metric spaces which are not normed spaces. Many of these are not even vector spaces, but we can even equip vector spaces with metrics which are not compatible with norms (in the sense that there is no norm such that $\| x \|=d(x,0)$). For example, if we equip any vector space with positive dimension with the discrete metric, then the homogeneity property of the norm is violated: if $x \neq 0,\alpha \neq 0,|\alpha| \neq 1$, then $d(\alpha x,0)=d(x,0)=1 \neq |\alpha|$.

Answer 3

A norm is induced b yan inner product iff it satisfies the paralellogram law

$$2||u||^2 + 2||v||^2 = ||u -v||^2 + ||u+v||^2$$

And e.g. the supremum norm on $\mathbb{R}^2$ already fails this.

Even if we have a metric topological vector space with a translation invariant metric $(V,d)$ which is moreover complete (a so-called Fréchet space), so we have compatibility with the linear operations, $d$ need not be induced by a norm.

A normed space has the property that every open neighbourhood $U$ of $0$ is bounded (in the sense that $\forall x, \exists t \in \mathbb{R}: tx \in U$, while this fails for many Fréchet spaces like $\mathbb{R}^\mathbb{N}$. The $\ell^p$ spaces for $0 < p < 1$ fail normability as well, for other reasons (not locally convex), even though they have a nice metric structure.