I really wish I could just say in post titles.
Anyway, I want to investigate the continuous dual of for . That is, we’re excluding the case where either (but not its Hölder conjugate ) is infinite. And I say that when is -finite, the space of bounded linear functionals on is isomorphic to .
First, I’m going to define a linear map . Given a function , let be the linear functional defined for any by
It’s clear from the linearity of multiplication and of the integral itself, that this is a linear functional on . Hölder’s inequality itself shows us that not only does the integral on the right exist, but
That is, is a bounded linear functional, and the operator norm is at most the norm of . The extremal case of Hölder’s inequality shows that there is some for which this is an equality, and thus we conclude that . That is, is an isometry of normed vector spaces. Such a mapping has to be an injection, because if then , which implies that .
Now I say that is also a surjection. That is, any bounded linear functional is of the form for some . Indeed, if then we can just pick as a preimage. Thus we may assume that is a nonzero bounded linear functional on , and . We first deal with the case of a totally finite measure space.
In this case, we define a set function on measurable sets by . It’s straightforward to see that is additive. To prove countable additivity, suppose that is the countable disjoint union of a sequence . If we write for the union of through , we find that
Since is continuous, we conclude that , and thus that is a (signed) measure. It should also be clear that implies , and so . The Radon-Nikodym theorem now tells us that there exists an integrable function so that
Linearity tells us that
for simple functions , and also for every , since each such function is the uniform limit of simple functions. We want to show that .
If , then we must show that is essentially bounded. In this case, we find
for every measurable , from which we conclude that a.e., or else we could find some set on which this inequality was violated. Thus .
For other , we can find a measurable with so that . Setting and defining , we find that on , , and so
We thus find
Applying the monotone convergence theorem as we find that .
Thus in either case we’ve found a so that .
In the -finite case, we can write as the countable disjoint union of sets with . We let be the union of the first of these sets. We note that for every measurable set , so is a linear functional on of norm at most . The finite case above shows us that there are functions on so that
We can define if , and let be the sum of all these . We see that
for every , and since we find that . Then Fatou’s lemma shows us that . Thus the -finite case is true as well.
One case in particular is especially worthy of note: since is Hölder-coonjugate to itself, we find that is isomorphic to its own continuous dual space in the same way that a finite-dimensional inner-product space is isomorphic to its own dual space.
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 continuous. In the finite-dimensional case, all linear functions are continuous, so this hasn’t really come up before in our study of linear algebra. But for functional analysis, it becomes much more important.
Now, really we only need to require continuity at one point — the origin, to be specific — because if it’s continuous there then it’ll be continuous everywhere. Indeed, continuity at means that for any there is a so that implies . In particular, if , then this means implies . Clearly if this holds, then the general version also holds.
But it turns out that there’s another equivalent condition. We say that a linear transformation is “bounded” if there is some such that for all . That is, the factor by which stretches the length of a vector is bounded. By linearity, we only really need to check this on the unit sphere , but it’s often just as easy to test it everywhere.
Anyway, I say that a linear transformation is continuous if and only if it’s bounded. Indeed, if is bounded, then we find
so as we let approach — as approaches — the difference between and approaches zero as well. And so is continuous.
Conversely, if is continuous, then it is bounded. Since it’s continuous, we let and find a so that for all vectors with . Thus for all nonzero we find
Thus we can use and conclude that is bounded.
The least such that works in the condition for to be bounded is called the “operator norm” of , which we write as . It’s straightforward to verify that , and that if and only if is the zero operator. It remains to verify the triangle identity.
Let’s say that we have bounded linear transformations and with operator norms and , respectively. We will show that works as a bound for , and thus conclude that . Indeed, we check that
and our assertion follows. In particular, when our base field is itself a normed linear space (like or itself) we can conclude that the “continuous dual space” consisting of bounded linear functionals is a normed linear space using the operator norm on .
We will soon need to know that Hölder’s inequality is in a sense the best we can do, at least for finite . That is, not only do we know that for any and we have , but for any there is some for which we actually have equality. We will actually prove that
That is, not only is the integral bounded above by — and thus by — but there actually exists some in the unit ball which achieves this maximum.
Hölder’s inequality tells us that
so must be at least as big as every element of the given set. If , then it’s clear that the asserted equality holds, since a.e., and so is the only element of the set on the right. Thus from here we can assume .
We now define a function . At every point where we set as well. At all other we define
In the case where we will verify that . That is, the essential supremum of is . And, indeed, we find that at points where , and at points where .
If , then we check
In either case, it’s easy to see that
To complete what we were saying about the spaces, we need to show that they’re complete. As it turns out, we can adapt the proof that mean convergence is complete, but we will take a somewhat different approach. It suffices to show that for any sequence of functions in so that the series of -norms converges
the series of functions converges to some function .
For finite , Minkowski’s inequality allows us to conclude that
The monotone convergence theorem now tells us that the limiting function
is defined a.e., and that . The dominated convergence theorem can now verify that the partial sums of the series are -convergent to :
In the case , we can write . Then except on some set of measure zero. The union of all the must also be negligible, and so we can throw it all out and just have . Now the series of the converges by assumption, and thus the series of the must converge to some function bounded by the sum of the (except on the union of the ).
We can actually extend what we’ve been doing with Hölder’s inequality and Minkowski’s inequality a little further. Given a metric space , we’ve already discussed the idea of an “essentially bounded” function — one for which there is some real constant so that for almost all . We will write for the collection of essentially bounded functions on the measure space. It should be clear that these form a vector space.
We also discussed the “essential supremum” 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 identity is exactly Minkowski’s inequality.
And, indeed, we know that and a.e., so a.e., so whatever the least such essential upper bound is smaller still. That is, .
Now for Hölder’s inequality. For this purpose we consider , and thus , which means that and are Hölder-conjugates. Thus our assertion is that if is integrable and is essentially bounded, then is integrable and . Indeed, we know that , and so — both inequalities holding almost everywhere. From this, we conclude that
as we asserted. From now on, we’ll allow (and ) whenever we’re talking about a Hölder-conjugate pair or -space.
We continue our project to show that the spaces are actually Banach spaces with Minkowski’s inequality. This will allow us to conclude that is a normed vector space. It states that if and are both in , then their sum is in , and we have the inequality
We start by considering Hölder’s inequality in a toy space I’ll whip up right now. Take two isolated points, and let each one have measure ; the whole space of both points has measure . A function is just an assignment of a pair of real values , and integration just means adding them together. Hölder’s inequality for this space tells us that
where and are Hölder-conjugate to each other. We can set , , and and use this inequality to find
Dividing out and raising both sides to the th power, we conclude that . Thus if both and are integrable, then so is . Thus must be in .
Now we calculate
Dividing out by we find that
This lets us conclude that is a vector space. But we can also verify the triangle identity now. Indeed, if , , and are all in , then Minkowski’s inequality shows us that
which is exactly the triangle inequality we want. Thus is a norm, and is a normed vector space.
We’ve seen the space of integrable 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 generalize.
Given a real number , we define the space or to be the collection of all measurable functions for which is integrable. As in the case of , we identify two functions if they’re equal -almost everywhere.
It will turn out that these are Banach spaces. We define the norm
and we write to define a metric. This is clearly non-negative, and we see that if and only if -a.e., just as before. It’s also clear that . What we need to work to check is the triangle inequality. It’s also not quite so apparent a problem, but we actually don’t know yet that this is a vector space at all! That is, how do we know that is integrable if and are?
As a first step in this direction, we prove Hölder’s inequality: if and are real numbers greater than such that , and if and , then the product and . To see this, we will use the function defined for all positive real numbers by
Differentiating, we see that , so the only (positive) critical point of is . Since the limit as approaches and are both positive infinite, must be a local minimum. That is
For any two real numbers and , we can consider the value
and it follows that
which is clearly also true even if we allow or to be zero. This is known as “Young’s inequality”.
Okay, so now we can turn to the theorem itself. If either or , the inequality clearly holds. Otherwise, we define
we can plug these into the above inequality to find
Since the measurability of and implies that of , and the right hand side of this inequality is integrable, we conclude that is integrable. If we integrate, we find
and Hölder’s inequality follows.
The condition relating and is very common in this discussion, so we will say that such a pair of real numbers are “Hölder conjugates” of each other. Given , the Hölder conjugate is uniquely defined by , which is a strictly decreasing function sending to itself (with order reversed, of course). The fact that this function has a (unique) fixed point at will be important. In particular, we will see that this norm is associated with an inner product on , and that Hölder’s inequality actually implies the Cauchy-Schwarz inequality!
Let be the unit interval , let be the class of Borel sets on , and let be Lebesgue measure. If is a sequence of partitions of the maximal element of the measure algebra into intervals, and if the limit of the sequence of norms is zero, then is dense.
If is a positive number, then we can find some so that . If is a subinterval of , then let be the unique interval containing the left endpoint of . If also contains the right endpoint, then we can stop. Otherwise, let be the next interval of to the right of , and keep going (at most a finite number of steps) until we get to an interval containing the right endpoint of . The union of all the can overshoot by at most on the left and the same amount on the right, and so . Thus any interval can be approximated arbitrarily closely by some partition in the sequence. The general result follows, because we can always find a finite collection of intervals whose union is arbitrarily close to any given Borel set.
Now, say that every separable, non-atomic, normalized measure algebra is isomorphic to the measure ring .
Since the metric space has a diameter of — no two sets can differ by more than , and — we can find a dense sequence in the space. For each we can consider sets of the form
where either or . The collection of all such sets for a given is a partition . It should be clear that the sequence is decreasing, and the fact that is dense implies that the sequence of partitions is also dense. We thus conclude that .
The first partition has two elements and . We can define and , so that and . This gives us a partition of that reflects the structure of . Similarly, we can carve up each of these intervals so that the resulting partition reflects the structure of . And we can continue, each time subdividing into smaller intervals so that the next partition reflects the structure of . Since this correspondence preserves measures, it follows that , and the above result then shows that the sequence is dense.
Now, we can extend from partition elements occurring in to finite unions of such elements by sending such a finite union to the finite union of corresponding elements of . This gives an isometry from a dense subset of to a dense subset of . Thus we can uniquely extend it to an isometry . Since preserves unions and differences, and since these operations are uniformly continuous, it follows that gives us an isomorphism of measure algebras.
Let be a totally finite measure algebra, and write for the maximal element. Without loss of generality, we can assume that is normalized so that .
We define a “partition” of an element to be a finite set of “disjoint” elements of whose “union” is . Remember, of course, that the elements of are not (necessarily) sets, so the set language is suggestive, but not necessarily literal. That is, if then and
The “norm” of a partition is the maximum of the numbers . If is a partition of and if is any element of below , then is a partition of .
If and are partitions, then we write if each element in is contained in an element of . We say that a sequence of partitions is “decreasing” if for each . A sequence of partitions is “dense” if for every and every positive number there is some and an element so that , and is exactly the union of some elements in . That is, we can use the elements in a fine enough partition in the sequence to approximate any element of as closely as we want.
Now, if is a totally finite, non-atomic measure algebra, and if is a dense, decreasing sequence of partitions of , then . Indeed, the sequence of norms is monotonic and bounded in the interval , and so it must have a limit. We will assume that this limit is some positive number , and find a contradiction.
So if then at least one of the must be big enough that for all . Otherwise the sequence of norms would descend below and that couldn’t be the limit. Let be just such an element, and consider the sequence of partitions of . The same argument is just as true, and we find another element from the partition , and so on.
Now, let be the intersection of the sequence . By assumption, each of the has , and so as well. Since is non-atomic, can’t be an atom, and so there must be an with . This element must be either contained in or disjoint from each element of each partition .
We can take smaller than either or . Now no set made up of the union of any elements of any partition can have a distance less than from . This shows that the sequence of partitions cannot be dense, which is the contradiction we were looking for. Thus the limit of the sequence of norms is zero.
The Stone space functor we’ve been working with sends Boolean algebras to topological spaces. Specifically, it sends them to compact Hausdorff spaces. There’s another functor floating around, of course, though it might not be the one you expect.
The clue is in our extended result. Given a topological space we define to be the Boolean algebra of all clopen subsets. This functor is contravariant — given a continuous map , we get a homomorphism of Boolean algebras sending the clopen set to its preimage . It’s straightforward to see that this preimage is clopen. Another surprise is that this is known as the “Stone functor”, not to be confused with the Stone space functor .
So what happens when we put these two functors together? If we start with a Boolean algebra and build its Stone space , then the Stone functor applied to this space gives us a Boolean algebra . This is, by construction, isomorphic to itself. Thus the category is contravariantly equivalent to some subcategory of . But which compact Hausdorff spaces arise as the Stone spaces of Boolean algebras?
Look at the other composite; starting with a topological space , we find the Boolean algebra of its clopen subsets, and then the Stone space of this Boolean algebra. We also get a function . For each point we define the Boolean algebra homomorphism that sends a clopen set to if and only if . We can see that this is a continuous map by checking that the preimage of any basic set is open. Indeed, a basic set of is for some clopen set . That is, . Which functions of the form are in ? Exactly those for which . Since is clopen, this preimage is open.
Two points and are sent to the same function if and only if every clopen set containing also contains , and vice versa. That is, and must be in the same connected component. Indeed, if they were in different connected components, then there would be some clopen containing one but not the other. Conversely, if there is a clopen that contains one but not the other they can’t be in the same connected component. Thus this map collapses all the connected components of into points of .
If this map is a homeomorphism, then no two points of are in the same connected component. Thus each singleton is a connected component, and we call the space “totally disconnected”. Clearly, such a space is in the image of the Stone space functor. On the other hand, if , then , and so this is both a necessary and a sufficient condition. Thus the “Stone spaces” form the full subcategory of , consisting of the totally disconnected compact Hausdorff spaces. Stone’s representation theorem shows us that this category is equivalent to the dual of the category of Boolean algebras.
As a side note: I’d intended to cover the Stone-Čech compactification, but none of the references I have at hand actually cover the details. There’s a certain level below which everyone seems to simply assert certain facts and take them as given, and I can’t seem to reconstruct them myself.