The Unapologetic Mathematician

Mathematics for the interested outsider

Topological Vector Spaces, Normed Vector Spaces, and Banach Spaces

Before we move on, we want to define some structures that blend algebraic and topological notions. These are all based on vector spaces. And, particularly, we care about infinite-dimensional vector spaces. Finite-dimensional vector spaces are actually pretty simple, topologically. For pretty much all purposes you have a topology on your base field \mathbb{F}, and the vector space (which is isomorphic to \mathbb{F}^n for some n) will get the product topology.

But for infinite-dimensional spaces the product topology is often not going to be particularly useful. For example, the space of functions f:X\to\mathbb{R} is a product; we write f\in\mathbb{R}^X to mean the product of one copy of \mathbb{R} for each point in X. Limits in this topology are “pointwise” limits of functions, but this isn’t always the most useful way to think about limits of functions. The sequence

\displaystyle f_n=n\chi_{\left[0,\frac{1}{n}\right]}

converges pointwise to a function f(x)=0 for n\neq0 and f(0)=\infty. But we will find it useful to be able to ignore this behavior at the one isolated point and say that f_n\to0. It’s this connection with spaces of functions that brings such infinite-dimensional topological vector spaces into the realm of “functional analysis”.

Okay, so to get a topological vector space, we take a vector space and put a (surprise!) topology on it. But not just any topology will do: Remember that every point in a vector space looks pretty much like every other one. The transformation u\mapsto u+v has an inverse u\mapsto u-v, and it only makes sense that these be homeomorphisms. And to capture this, we put a uniform structure on our space. That is, we specify what the neighborhoods are of 0, and just translate them around to all the other points.

Now, a common way to come up with such a uniform structure is to define a norm on our vector space. That is, to define a function v\mapsto\lVert v\rVert satisfying the three axioms

  • For all vectors v and scalars c, we have \lVert cv\rVert=\lvert c\rvert\lVert v\rVert.
  • For all vectors v and w, we have \lVert v+w\rVert\leq\lVert v\rVert+\lVert w\rVert.
  • The norm \lVert v\rVert is zero if and only if the vector v is the zero vector.

Notice that we need to be working over a field in which we have a notion of absolute value, so we can measure the size of scalars. We might also want to do away with the last condition and use a “seminorm”. In any event, it’s important to note that though our earlier examples of norms all came from inner products we do not need an inner product to have a norm. In fact, there exist norms that come from no inner product at all.

So if we define a norm we get a “normed vector space”. This is a metric space, with a metric function defined by d(u,v)=\lVert u-v\rVert. This is nice because metric spaces are first-countable, and thus sequential. That is, we can define the topology of a (semi-)normed vector space by defining exactly what it means for a sequence of vectors to converge, and in particular what it means for them to converge to zero.

Finally, if we’ve got a normed vector space, it’s a natural question to ask whether or not this vector space is complete or not. That is, we have all the pieces in place to define Cauchy sequences in our vector space, and we would like for all of these sequences to converge under our uniform structure. If this happens — if we have a complete normed vector space — we call our structure a “Banach space”. Most of the spaces we’re concerned with in functional analysis are Banach spaces.

Again, for finite-dimensional vector spaces (at least over \mathbb{R} or \mathbb{C}) this is all pretty easy; we can always define an inner product, and this gives us a norm. If our underlying topological field is complete, then the vector space will be as well. Even without considering a norm, convergence of sequences is just given component-by-component. But infinite-dimensional vector spaces get hairier. Since our algebraic operations only give us finite sums, we have to take some sorts of limits to even talk about most vectors in the space in the first place, and taking limits of such vectors could just complicate things further. Studying these interesting topologies and seeing how linear algebra — the study of vector spaces and linear transformations — behaves in the infinite-dimensional context is the taproot of functional analysis.

May 12, 2010 - Posted by | Algebra, Analysis, Functional Analysis, Linear Algebra, Measure Theory, Topology

9 Comments »

  1. […] We can also turn around and define what it means for a sequence to be uniformly Cauchy almost everywhere — for any there is some so that for all we have . Unpacking again, there is some measurable set so that for all . It’s straightforward to check that a sequence that converges uniformly a.e. is uniformly Cauchy a.e., and vice versa. That is, the topology defined by the essential supremum norm is complete, and the algebra of essentially bounded measurable functions on a measure space is a Banach space. […]

    Pingback by Convergence Almost Everywhere « The Unapologetic Mathematician | May 14, 2010 | Reply

  2. I’d known for a long time that a proper linear subspace (i.e. a subspace in the usual linear algebra sense) of an infinite dimensional normed space can be fairly exotic (e.g. they can be dense), but I had no idea how exotic linear subspaces could be until a few years ago. For example, in a finite dimensional normed space, each proper linear subspace is closed and nowhere dense. On the other hand, in EACH infinite dimensional separable Banach space there exist proper linear subspaces that are dense and **, where ** can be any of the following:

    F_sigma_delta,
    but not F_sigma or G_delta

    G_delta_sigma,
    but not F_sigma or G_delta

    F_sigma_delta AND G_delta_sigma,
    but not F_sigma or G_delta

    F_sigma_delta_sigma,
    but not F_sigma_delta or G_delta_sigma

    G_delta_sigma_delta,
    but not F_sigma_delta or G_delta_sigma

    F_sigma_delta_sigma AND G_delta_sigma_delta,
    but not F_sigma_delta or G_delta_sigma

    Moreover, the 2nd order examples, represented by the first 3, and the 3rd order examples, represented by the last 3, are just the tip of the iceberg. There are analogous 4th order examples, 5th order examples, …, n’th order examples, … In fact, such examples can be found at each of the w_1 (first uncountable ordinal) Borel levels. Also, there exist proper linear subspaces that are not Borel, and worse.

    I did a little literature research on this topic about 6 years ago and finally got around to writing an essay on it about 4 years ago:

    “Exotic normed linear subspaces” (18 August 2006)
    http://tinyurl.com/2fd7gl6

    Comment by Dave L. Renfro | May 14, 2010 | Reply

  3. […] it turns out that there is no metric which gives this sense of convergence, but we still refer to a sequence as being “Cauchy in […]

    Pingback by Convergence in Measure I « The Unapologetic Mathematician | May 19, 2010 | Reply

  4. […] We can now introduce a norm to our space of integrable simple functions, making it into a normed vector space. We […]

    Pingback by The L¹ Norm « The Unapologetic Mathematician | May 28, 2010 | Reply

  5. […] This shows that any mean Cauchy sequence of integrable functions is also mean convergent to some function, and thus the space of integrable functions equipped with the norm is a Banach space. […]

    Pingback by Mean Convergence is Complete « The Unapologetic Mathematician | June 8, 2010 | Reply

  6. […] it so happens that we can define a norm on this space. Indeed, taking the Jordan decomposition, we must have both and , and thus . We […]

    Pingback by The Banach Space of Totally Finite Signed Measures « The Unapologetic Mathematician | June 28, 2010 | Reply

  7. […] functions on a measure space , which we called or . We’ve seen that this gives us a complete normed vector space — a Banach space. This is what we’d like to […]

    Pingback by Hölder’s Inequality « The Unapologetic Mathematician | August 26, 2010 | Reply

  8. […] of an essentially bounded function. We’ll now write this as , suggesting that it’s a norm. And it’s clear that , and that if and only if almost everywhere. Verifying the triangle […]

    Pingback by The Supremum Metric « The Unapologetic Mathematician | August 30, 2010 | Reply

  9. […] Linear Transformations In the context of normed vector spaces we have a topology on our spaces and so it makes sense to ask that maps between them be […]

    Pingback by Bounded Linear Transformations « The Unapologetic Mathematician | September 2, 2010 | Reply


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