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Banach Spaces and Preserving Finite Dimensional Theorems

09 Sep 2021 - Tags: analysis-qual-prep

Banach Spaces are ubiquitous in analysis, and they let us rein in analytic objects using algebra (in particular, vector spaces) and completeness. Infinite dimensional vector spaces can be pathological, but by restricting attention to continuous operations for our topology, we can recover analogues for a lot of the finite dimensional theory!

Recall a Banach Space is a vector space equipped with a norm so that the resulting metric space is complete. We have already seen some examples of this, but let’s explicitly list some to get a sense of just how common Banach Spaces are!

Of course, we can also build new Banach Spaces from old, and these work in much the same way as in classical (by which I mean finite dimensional) linear algebra.

⚠ Short exact sequences of banach spaces do not split in general! That is, if $A$ is a closed subspace of $X$, it is not the case that $X \cong A \oplus (X / A)$!

See, for instance, this survey by Mohammad Sal Moslehian8.

As a quick exercise, prove

\[\lVert Tx \rVert \leq \lVert T \rVert \lVert x \rVert\]

This is one of the most important inequalities in the subject.

Now that we have a wide array of examples of banach spaces, we should start asking how much of our intuition from finite dimensional linear algebra carries over. After all, a lot of these constructions looked really familiar, but then we got blindsided by the lack of complements.

Thankfully, there are lots of foundational theorems in banach space theory which tell us that certain things work exactly as we’d like!

For instance, in the finite dimensional case, if we’ve defined a functional on some subspace, then we can always extend it to the whole space. But the proof crucially relies on a choice of basis, so in the infinite dimensional case, we need to be a bit careful to guarantee that the extension is still continuous.

Thankfully, everything works out:

The Hahn-Banach Theorem11

If $f$ is a continuous linear functional defined on a subspace of $X$, then $f$ extends to a continuous linear functional $F$ defined on all of $X$.

Moreover, $\lVert F \rVert = \lVert f \rVert$.

Another piece of intuition from the finite dimensional case is that a bijective linear map is automatically an isomorphism (that is, the inverse map is automatically linear). Is it the case that a continuous bijective linear map is automatically an isomorphism? That is, must its inverse also be continuous? Again, the answer is “yes”!

The Open Mapping Theorem

Let $X$ and $Y$ be banach spaces. If $T \in \mathcal{L}(X,Y)$ is surjective, then it is open.

In particular, if $T$ is bijective, then $T^{-1}$ is continuous.

There’s another nice corollary of the open mapping theorem too. Topologically we expect a quotient map to be open, and we know from the homomorphism theorems that we can factor any surjection $T : X \to Y$ as

\[X \to X / \text{Ker}(T) \cong Y\]

the open mapping theorem says that this quotient map is open, as we would expect. In fact, the projection $\pi : X \to X / A$ (for $A$ closed, of course) always has norm $1$.

Continuing with our examples, in the finite dimensional case, we think of subspaces as being “much smaller” than the ambient space. For instance, the $xy$-plane is measure $0$ inside $\mathbb{R}^3$ (with lebesgue measure) because it has no thickness. One can ask if sub-banach spaces (that is, closed subspaces) must be “small” in some sense. Again, the answer is “yes”12, but we have to use a different notion of “small”13:

The (Strong) Open Mapping Theorem

If $X$ and $Y$ are banach spaces14 and $T \in \mathcal{L}(X,Y)$, then if the image $T[X]$ is nonmeagre in $Y$, we automatically have surjectivity and open-ness.

As a simple corollary, every proper closed subspace is meagre.

Of course, we can’t talk about banach spaces without talking about a theorem which honestly feels like magic.

The Uniform Boundedness Principle

Say ${ T_\alpha }$ is a family of continuous linear maps between banach spaces15 $X$ and $Y$. If, this family is bounded pointwise in the sense that every $x$ satisfies

\[\sup \lVert T_\alpha x \rVert \lt \infty\]

(where the precise bound is allowed to depend on $x$)

then we actually get a uniform bound for free!

\(\sup \lVert T_\alpha \rVert \lt \infty\)

The ability to boost pointwise results to uniform results is often important (this is one of many reasons to care about compactness), and there are innumerable appplications of this theorem. Here’s one that seems to show up on a lot of practice quals:

Show that a weakly convergent sequence $(x_n)$ is bounded.

Recall $(x_n)$ is weakly convergent if $(T x_n)$ is convergent in $\mathbb{C}$ for every continuous functional $T$.

And here’s one that says a certain pathology you might remember from an undergraduate analysis class doesn’t happen in the banach space setting:

Show that if $X$, $Y$, and $Z$ are banach spaces, then every separately continuous bilinear map $X \times Y \to Z$ is automatically jointly continuous16.

Here $X \times Y$ is meant as merely the product of topological spaces, rather than as the product of banach spaces defined earlier.

As with the open mapping princple, it’s actually enough to know that you’re pointwise bounded on some set that isn’t small:

The (Strong) Uniform Boundedness Principle

If $X$ and $Y$ are normed vector spaces17 and $(T_\alpha)$ is a sequence of continuous linear maps from $X$ to $Y$, then as long as the family is pointwise bounded on a nonmeagre set, it’s actually uniformly bounded!

More formally, if $E \subseteq X$ is nonmeagre, and we have a bound

\[\sup \lVert T_\alpha x \rVert \lt \infty\]

for each $x \in E$

then we actually get a uniform bound for free!

\(\sup \lVert T_\alpha \rVert \lt \infty\)

This is fairly indicative of working with meagre and nonmeagre sets. We frequently get a kind of dichotomy where things are either bad almost everywhere or good almost everywhere (by which I mean on a comeagre set).

So if you can show that something good/bad happens on a set that isn’t small (meagre) then you often get for free that good/bad things happen almost everywhere!

For instance, let’s look at the contrapositive of the above theorem. It says that if $\sup \lVert T_\alpha \rVert = \infty$, then actually for comeagrely many choices of $x$, we must have $\sup \lVert T_\alpha x \rVert = \infty$!

Next time18, we’ll talk about Hilbert Spaces, where we’ll require even more algebraic structure, and in exchange we’ll gain better structure theorems telling us about our spaces. These structure theorems will lead us into the beautiful world of Fourier Analysis, which we’ll discuss afterwards.

In the meantime, you should definitely read Terry Tao’s post about banach spaces here.

See you soon! ^_^

  1. In fact, this accounts for all banach spaces!

    The Banach-Mazur Representation Theorem says that every banach space is isometric to a closed subspace of $C(K)$ for some compact space $K$ (in fact, the unit ball in the dual with the weak-* topology works).

    In case your banach space is separable, we can do better – it is a closed subspace of $C([0,1])$! 

  2. Intuitively this makes sense, but formally it is far from obvious (at least to me!). That said, you can find a smattering of proofs here. I like the proof by Radon-Nikodym, if you want to do it directly.

    The most conceptual way to see this is by citing the Riesz Representation Theorem, which says that this space of measures is actually isometric to $C_0^*$, and thus is banach. Of course, that only works when $X$ is locally compact hausdorff. The theorem as proven in that mse link works more generally. 

  3. The category $\mathsf{Ban}$ of banach spaces with all continuous linear maps turns out to be somewhat badly behaved. But if we restrict to contracting maps, we get a much nicer category, $\mathsf{Ban}_1$.

    Notice we can rescale any linear transformation $T$ by a constant to make it a contraction, so this is not really a limitation. 

  4. In fact, in the finite case we can take $\lVert (x,y) \rVert$ to be any of

    • $\max { \lVert x \rVert_X, \lVert y \rVert_Y }$
    • $\sqrt{ \lVert x \rVert_X^2 + \lVert y \rVert_Y^2 }$
    • $\lVert x \rVert_X + \lVert y \rVert_Y$

    and we’ll get the same banach space up to isomorphism (but NOT isometry!).

    If you’ve not seen this before, you should prove it! It’s a fairly quick exercise to show these are all equivalent norms, and that they are complete. 

  5. Notice the coproduct you might expect

    \[\{ (x_\alpha) \mid x_\alpha \neq 0 \text{ for finitely many $\alpha$ } \}\]

    is not complete, and thus not banach (do you see why?). The actual definition of the coproduct is exactly the completion of this space in the coproduct norm, though. 

  6. Notice we need $A$ to be closed so that it is itself complete, and is thus a sub-banach space. 

  7. Again, we need $A$ to be closed so that we’re working with sub-banach spaces. Notice this is not a serious issue – as a quick exercise, you might show that the kernel of a continuous linear map is always a closed subspace. 

  8. In fact, there’s more to say! A (nontrivial) theorem says that if a banach space satisfies “every closed subspace has a complement” then it must actually be a hilbert space!

    So every non-hilbert banach space must contain a noncomplemented closed subspace!

    See here (as well as the linked article), for more. 

  9. Recall a linear function is continuous if and only if it is bounded in the sense that the norm we’re defining is finite. 

  10. In fact, $X^*$ is complete even when $X$ isn’t. One slick corollary of this is the construction of the completion of a normed space.

    Since $X$ (isometrically!) embeds into its double dual $X \hookrightarrow X^{**}$, we can define the completion of $X$ to be the closure of $X$ under this embedding. 

  11. The proof of this fact makes use of the fact that any vector space is the filtered colimit of its finite dimensional subspaces. We show how to extend by one basis element at a time in a way that preserves the norm, then we apply Zorn’s Lemma to the partial order of these extensions.

    It turns out this appeal to Zorn’s Lemma is somewhat unavoidable. There are models of $\mathsf{ZFC}$ where Hahn-Banach fails in full generality, so we need some amount of choice to prove it. However it’s strictly weaker than full AC (see here for more discussion).

    Thankfully, in many concrete situations, we don’t need choice! If $X$ is separable, then we can extend one dimension at a time, making sure we eventually choose each element of our countable dense subset. At the end of this (countable length!) process, we can extend to the whole space by continuity. 

  12. Interestingly, we can ask about more general subspaces (which are necessarily not closed). It turns out the answer here is a firm “no”.

    Every infinite dimensional banach space has a proper nonmeagre subspace (which is necessarily not closed). It turns out such a subspace must be dense and cannot have BP.

    These subspaces arise as kernels of discontinuous functionals, so the next question is “does every discontinuous functional work?”, and the answer here is “it’s subtle”.

    Using Martin’s Axiom one can show that every separable banach space has a discontinuous functional whose kernel is still meagre. See here for more info. 

  13. The post about the baire category theorem is coming up! 

  14. Actually we only need $X$ to be banach – a priori $Y$ can be any normed vector space.

    Interestingly, though. As soon as we know that $T[X] = Y$, we also know that $Y \cong X / \text{Ker}(T)$ is banach. 

  15. Again, we only really need $X$ to be banach. 

  16. It might be helpful to recall that a bilinear operator is jointly continuous if and only if it is “jointly bounded” in the sense that \(\sup \{ \lVert f(x,y) \rVert \ \mid \ \lVert x \rVert = 1, \lVert y \rVert = 1 \} \lt \infty\).

    As a bigger hint, you might try applying the uniform boundedness principle to the family of maps $f(x,-) : Y \to Z$.

    Also, as a fun ~ bonus game ~ for particularly enthusiastic readers, it turns out we don’t need all of $X$, $Y$, and $Z$ to be banach spaces! How weak can you make the assumptions and still prove this theorem? 

  17. In particular neither needs to be complete. 

  18. I didn’t forget about the closed graph theorem, I just couldn’t find a way to make it fit into the narrative of this blog post (comparing finite dimensional and infinite dimensional banach spaces).

    I have some interesting stuff to say, though, since it’s analogous to a theorem in universal algebra. Terry Tao talks some about this in one of his blog posts (and gives a very interesting application in another), but the idea behind the closed graph theorem is true in very high generality:

    Show that $f : A \to B$ is a homomorphism of algebras if and only if its graph is a subalgebra of $A \times B$.

    Moreover, if $Y$ is compact and hausdorff, then $f : X \to Y$ is continuous if and only if its graph is closed in $X \times Y$. This should make some vague sense, since compact hausdorff spaces behave a lot like algebras (in fact, they are a category of algebras for the ultrafilter monad. See here, for instance), so $f$ is a “homomorphism” (is continuous) exactly when its graph is a subalgebra (sub-compact-hausdorff space) of $X \times Y$. Of course, to be sub-compact-hausdorff, it suffices to check closedness.

    Now the closed graph theorem says that this is true of banach spaces as well. $f : X \to Y$ is a continuous linear map if and only if its graph is a sub-banach space of $X \times Y$. Linearity comes from the “space” part of “sub-banach space” and continuity comes from the “banach” part. Of course, as in the compact hausdorff case, it suffices to check closedness.

    Unlike the case of compact hausdorff spaces, though, I don’t know of any categorical justification for this theorem! If anyone happens to have one, I would love to hear about it!