Does it matter if functional analysis was introduced from a normed space versus a metric space formulation? Are all major theorems from functional analysis (such as Banach contraction mapping, Hahn Banach theorem...) directly transferable if we exchanged every instance of norm with metric?

I am asking because strictly speaking a metric is different than a norm. But do these differences such as the existence of discrete metric effect functional analysis in a way that a person has to relearn functional analysis if things were introduced as norms instead metric?

Ultimately, which perspective would be more beneficial from a didactic point of view?

  • $\begingroup$ Henning, if you don't mind, I suggest answering through the "answer" box and not in the comments. $\endgroup$ – Martin Argerami Aug 15 '15 at 21:48
  • $\begingroup$ @Martin: As you wish. Though I thought someone who actually knows functional analysis should have a chance to answer first. :-) $\endgroup$ – hmakholm left over Monica Aug 15 '15 at 22:24

Yes, insisting upon a normed vector space rather than a metric vector space introduces important additional properties.

For instance, $\mathbb{R}^n$ is always complete when equipped with a norm, because every norm on a finite-dimensional real (complex) vector space is equivalent, and $\mathbb{R}^n$ has at least one norm under which it is complete, namely the Euclidean norm.

On the other hand, there are metrics under which $\mathbb{R}^n$ is not complete. One example is covered in Why $\mathbb R$ is not complete with the metric $d(x,y)=|\phi(x)-\phi(y)|$ where $\phi(x)=\frac{x}{1+|x|}$?.

There are also complete metrics on $\mathbb{R}^n$ that are not norms; the discrete metric is one easy example.

Therefore if you want to study $\mathbb{R}^n$ as a metric space only, then you must give up on results that depend critically on a norm structure. If you allow yourself incomplete metrics, then you cannot have a norm; and even if you give yourself completeness, a norm is not guaranteed.

Here is also an example of a theorem for finite-dimensional real vector spaces that works for norms but not quite for metrics: Are all the finite dimensional vector spaces with a metric isometric to $\mathbb R^n$. And it is quite easy to see that not all metrics on $\mathbb{R}^n$ are equivalent; the discrete metric cannot be metric-equivalent to any norm. Since a norm scales and the discrete metric does not, no constant $C$ can satisfy $ n(x,y) \leq C d(x,y)$ for all $x,y$, where $d$ is the discrete metric and $n$ is the norm. (Take $x\neq y$ so that the RHS is $C$, then look at $\alpha x,\alpha y$ for $\alpha\to +\infty$.)


A metric space doesn't come with a vector space structure, without which there can't be much analysis going on. It woulds just be a corner of topology, then. You can imagine equipping a vector space with a separate metric, but by the time you require that the metric be compatible with the linear structure -- $d(v+w,u+w)=d(v,u)$ and $d(λv,λu)=|λ|d(v,u)$ are both natural expectations -- you'll find that you've actually just introduced a norm under a different name ...

  • $\begingroup$ The second expectation is not necessary. Consider any nonnormable Frechet space. Anyway: (+1) $\endgroup$ – C-Star-W-Star Aug 16 '15 at 4:53

At the very, veery heart functional analysis is about topological vector spaces (TVS).
So basically the answer is almost 'no'. Browse through your notes and look for theorems that concern more than continuity. There are however concepts, i.e. techniques, that require seminorms resp. norms like integration resp. differention.

Now, there are TVS which are not first countable thus not metrizable.
As an example consider any function space with pointwise convergence.

Next, there are metrizable TVS which have no bounded neighborhoods thus not normable.
As an example consider any function space with local convergence.

In fact, most interesting examples don't fall into the category of Banach spaces:
The holomorphic functions, the analytic functions, the smooth functions, the test functions, etc.

For further reading: TVS vs. HS, TVS vs. UVS


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