Hölder’s inequality is a basic inequality in analysis, used to prove that if the sum of positive numbers equals their product, then the Banach spaces are Banach duals of one another.
Let be a measure space, and for let denote , the Banach space of complex-valued functions on with finite p-norm considered modulo almost everywhere equality. Suppose are positive real numbers such that (that is, ). Then Hölder’s inequality states that for any we have
(in particular, is an function).
The “nPOV” meaning is this: in this situation there is a canonical pairing between and ,
which gives a bounded linear map between Banach spaces. The point of Hölder’s inequality is that this pairing is a short map, i.e., a map of norm bounded above by . In other words, this is morphism in the symmetric monoidal closed category Ban consisting of Banach spaces and short linear maps between them. Accordingly, the map induces (by currying) a map from to the Banach dual of :
(again a short map of course), and reciprocally a map .
It is a short step to prove that in fact the norm of the pairing is exactly , and even better that the maps and are in fact isometric embeddings. With a little more work (with the help of the Radon-Nikodym theorem; see for example here), one sees these maps are surjective and thus isomorphisms in .
Throughout we are working in the range . We also have a Hölder inequality in the extreme case where , , something which is easily seen directly, and it is true also that , but it is not true that is isomorphic to . Or, it is at least not true in ZFC, although it may be true in dream mathematics.
The proof is remarkably simple. First, if and , then we have Young’s inequality, viz. for
with equality precisely when . This is quickly derived from the (strict) convexity of the exponential function, that implies
where equality holds iff . All one has to do is put and arrange so that and .
Then, to prove , we may assume nonzero (so their norms are positive) and normalize them to unit vectors , so that now the object is to prove
But since we are dealing with unit vectors, we have and , and now what we want follows straightaway from Young’s inequality applied to integrands:
and so the proof of Hölder’s inequality is complete.
To prove that the norm of the pairing is exactly (is not less than ), it’s enough to take any of norm , so is a nonnegative function of norm , and then put . We then have (almost) everywhere, where we then have , and now
Actually these calculations do a little better: they show that upon currying, the map
preserves the norm, so that isometrically embeds into .
Recall that Minkowski's inequality is just the triangle inequality in the context of L-p space. There is a well-known trick, covered in just about every functional analysis text, that allows one to deduce Minkowski’s inequality as a corollary of Hölder’s inequality. You can look it up for instance in the English Wikipedia, here.
What seemingly most such presentations lack is motivation for the trick, so let us try to say something about this.
First, Minkowski’s inequality can be restated as asserting the convexity of the unit ball of . If we place ourselves for a moment in the context of real-valued functions, then it suffices to show that is the intersection of a collection of affine half-spaces, say where is a (continuous) linear functional. But with hindsight into the meaning of Hölder’s inequality, seen as paving the way to characterizing linear functionals on as those of the form for some , it’s only natural to see whether we can find a sufficiently large collection of such such that , and in fact the intuition is that the unit ball in ought to work.
Thus the idea is clear, and it’s just a matter of technique from here. We let the relation on set up a Galois connection between subsets of and subsets of . The connection takes the unit ball in to
which is clearly convex, being an intersection of convex sets , one for each . Hölder’s inequality itself just asserts the containment . If we show the other inclusion , then is convex. So we want to show that if whenever , then . But we already did that calculation when we proved is an isometry. Explicitly: take (with where ). Then , so whence . Put ; since , it follows by the hypothesis on that . But this gives
as was to be shown.
The standard derivation of Minkowski’s inequality from Hölder’s inequality is nothing more than a very tidied-up rendering of this argument, but without the additional conceptual explanation given here.
Let be a convex (e.g., affine) space. We say a function is log-convex if is a convex function.
Hölder’s inequality is closely related to the notion of log-convexity. On the one hand, we saw that the inequality follows from the convexity of the exponential function, which is the most basic log-convex function of all. On another hand, we have the following result which uses Hölder’s inequality.
The collection of log-convex functions on a convex domain is closed under pointwise multiplication, pointwise addition, and pointwise max.
The statement for multiplication is clear since and any sum of convex functions is convex.
Similarly, is an isomorphism of partially ordered sets and so . It thus suffices to show that if are convex on , then so is . For and such that , we must show
letting denote the right side, this holds iff and (by definition of ). But
and similarly .
Finally, for the sum , in order to show is convex, it suffices to show that
for such that . But setting
the right side of (1) is . By Hölder’s inequality, this is greater than or equal to
where the last inequality is by log-convexity of and .
Last revised on April 5, 2018 at 14:43:20. See the history of this page for a list of all contributions to it.