Is there such thing as an unnormed vector space?

Question

I learned about Banach spaces a few weeks ago. A Banach space is a complete normed vector space. This of course made me wonder: are there unnormed vector spaces? If there are, can anyone please provide any examples?

Some thoughts:

A complete space is where all Cauchy sequences converge.

A normed vector space is a vector space (say, over $\mathbb{R})$ on some norm $N$ (which is a function that maps $N\to\mathbb{R}$), where the norm obeys the triangle inequality, the norm of a vector is non-negative, and if you have a scalar being multiplied by a vector, you can factor the scalar out, but it'll have absolute value braces.

I'm not really sure what is needed in order to have an unnormed vector space (perhaps the vector space necessarily needs to be infinite dimensional?). Perhaps something really weird like the zero space?

Thanks for any insight.

Answer 1

While your question could have multiple answers, perhaps the closest to what you are looking for is the notion of a non-metrizable vector space.

In the general setting of topological vector spaces, we consider (as one might guess from the name) vector spaces endowed with a topology so that we can discuss ideas like the continuity of linear operators. Normed vector spaces are examples of topological vector spaces where the topology is induced by a given norm.

A non-metrizable vector space is a topological vector space whose topology does not arise from any metric. These are rather common in functional analysis. For example, if $X$ is a Banach space, then the weak-* topology on $X^*$ is never metrizable unless $X$ is finite-dimensional. Another family of examples are locally convex spaces, a natural generalization of Banach spaces, which are not metrizable unless their topology is generated by a countable collection of seminorms that separate points.

Answer 2

If $V$ is finite dimensional, it is normable, in the sense that you can use an isomorphism into $\mathbb R^n$ to pull back the $\mathbb R^n$-norm . And then this norm is equivalent to any other norm topology-wise, i.e., in a finite- dimensional space, all norms are equivalent in this sense.

Answer 3

Every (real or complex) vector space admits a norm. Indeed, every vector space has a basis you can consider the corresponding «$\ell^1$» norm.

Answer 4

Of course there are. :-) First of all, there are linear spaces which are not endowed by a topology, and there are topological vector spaces which are not endowed by a norm. So a wise question is: when a topology of a topological vector space $X$ is generated by a norm? It seems it is iff $X$ is a locally convex Hausdorff space containing such a neighborhood $U$ of the zero that for each neighborhood $V$ of the zero there exists a scalar $\lambda>0$ such that $\lambda U\subset V$.

Answer 5

Vector spaces are, by default, unnormed. A norm is extra structure we add to a vector space, to define a normed vector space.

Answer 6

Outlining the construction by Mariano Suárez-Alvare...

Problem

Does every plain vector space admit a norm?

Construction

Given a plain vector space $V$.

Choose a Hamel basis $\mathcal{B}$.

Denote functions with finite support by $\mathbb{R}^\mathcal{B}_0$.

Decide for a norm there $\|(\lambda_b)_{b\in\mathcal{B}}\|$.

Now, regard the isomorphism: $$\Phi:V\to\mathbb{R}^\mathcal{B}_0:\sum_{b\in\mathcal{B}}\lambda_bb\mapsto\left(\lambda_b\right)_{b\in\mathcal{B}}$$

Then, it pulls back the norm: $\|\sum_{b\in\mathcal{B}}\lambda_bb\|:=\|(\lambda_b)_{b\in\mathcal{B}}\|$

Special Example

Take the norm of $\ell^2$.

The norm becomes: $\|\sum_{b\in\mathcal{B}}\lambda_bb\|_2^2=\sum_{b\in\mathcal{B}}|\lambda_b|^2$

Especially, it becomes an orthonormal basis: $\langle b,b'\rangle=\delta_{b,b'}$

Conclusion

Every plain vector space admits a norm - no matter of its dimension.

Outview

Does every algebra with unit admit a compatible norm: Existence of Norm for C*-Algebras

Answer 7

While you might be able to define a norm, there are real world circumstances where you might not want to. Let's take an example motivated more from science/engineering than mathematics. Consider the vector space defined by the tuple x = (a mm, b kg) where a and b are real numbers and mm and kg mean "millimeter" and "kilogram." To a scientist/engineer, the notion of a naive L1 metric might seem ludicrous, because you can't add 1mm + 1kg. Yes, the metric $\lVert x \rVert = |a| + |b|$, which scales the first element by 1/mm and the second by 1/kg, is a metric, but it's a completely arbitrary one: why is 1mm in some sense "equivalent" to 1kg? You might have a hard time arguing with me if I instead declare that you should scale the second component by 1/g rather than 1/kg, but of course that will give us very different notions of distance. (There are exceptions: e.g., in relativity theory, the speed of light makes 1s equivalent to 3e8m, but in other contexts one might not agree with that equivalence.)

From the standpoint of a mathematician, I suppose the question is what is the right topology for such a vector space. If you decide to allow open sets of the form $I_a \otimes I_b$, where $I_a$ and $I_b$ are open intervals with respect to the first and second tuple element and $\otimes$ indicates the Cartesian product, then the entire space is second-countable and satisfies the Urysohn metrization theorem. However, if instead you say that the topology is defined through open sets only of the form $I_a \otimes b$ and $a \otimes I_b$ (i.e., have zero "width" along one of the two coordinates), and if the numerical coefficients can be any real number, then I think the space is rigorously non-metrizable.