From: ullrich@math.okstate.edu (David C. Ullrich) Subject: Re: Questionable inequality for Holder seminorms Date: Wed, 12 Apr 2000 15:50:14 GMT Newsgroups: sci.math Summary: Besov normed spaces Sorry if this appears twice - mysterious server errors... On Tue, 11 Apr 2000 13:09:11 -0700, "lester l. helms" wrote: >Your last sentence describes the Holder norms I am writing about, as >excerpted below. Well fine then. That woulda been my first guess except you called it a "seminorm"; seems like an actual norm. Anyway, the result seems to be true, and the proof is straightforward if you look at it right. The vast majority of these Holder spaces are also Besov spaces, and the corresponding inequality for Besov spaces is trivial. Turns out you can also get the inequality for the Holder spaces that are not Besov spaces by the same sort of argument, I think: We're going to have more than one _family_ of norms. This is bad enough when it's actually typeset - here let's abandon the traditional "norm" notation and write them as functions instead. So let's agree that U(f) (for "uniform") is the sup of the absolute value of f, while H(a,f) is the norm of f in the Holder space H_a: If a = n + alpha with n a non-negative integer and 0 < alpha <= 1 then f is in H_a if f has n continuous derivatives and the n-th derivative satisfies a Holder condition of order alpha; H(a,f) is the corresponding norm H(a,f) = U(f) + ... + U(D^n f) + sup(|f(x)-f(y)|/|x-y|^alpha) . Now we need a little setup to give the right defintiion of the Besov space B_a: If f is a bounded function on the line we're going to write f = c sum(f_n) (where n ranges over all positive and negative integers), where the support of the Fourier transform of f_n is contained in the set where 2^(n-1) <= |x| <= 2^(n+1) . You do this by using a smooth partition of unity - you should also arrange things so that the various multipliers in this partition of unity are dilates of each other. (See [reference] for details on how this can be done. c is the "constant term", coming from the part of the Fourier transform of f supported on {0}. We will pretend c = 0 below.) We define the Besov norm B(a, f) by B(a, f) = sup(2^(na) U(f_n)) (n ranges over the integers), and then B_a is the class of all f with B(a, f) finite. It's extremely clear that if c = ta + (1-t)b is a convex combination of a and b then (*) B(c, f) <= B(a, f)^t B(b, f)^(1-t) . And this gives what you want, almost: It's "well-known" that if a is not an integer then B_a = H_a (with equivalent norms). Not quite true when a is an integer: If a = n + 1 then f is in H_a if D^n f lies in Lip_1, while f is in B_a if and only if D^n f lies in the "Zygmund class" Lambda: (phi is in Lambda if phi is continuous and (phi(x - h) - 2 phi(x) + phi(x + h)) / h is bounded.) In any case H_a is a subset of B_a, with the corresponding inequality of norms. So if c is not an integer you have H(c, f) <= C B(c, f) <= C B(a, f)^t B(b, f)^(1-t) <= C H(a, f)^t H(b, f)^(1-t) and you're done. If c is not an integer this doesn't quite work. Some would say you shouldn't want to get the inequality for H_a; the Besov spaces work better so you should use them instead. (Krantz has a long paper devoted to showing that Lip_1 is bad.) But if you want the inequality for H_a I believe it follows by the same approach, and it's just as straighforward: In the proof of (*) you throw a lot away, using the fact that the sup of a product is less than the product of the sups. If you use the same estimates but don't throw anything away I'm pretty sure you get the stronger inequality (**) B(c, 1, infinity; f) <= C B(a, f)^t B(b, f)^(1-t) . I checked this carefully last night in the case a = 1/2, b = 3/2, t = 1/2, which is exactly the first case where the answer to your original question wasn't obvious to me - seems clear it's true in general, haven't written it down. If you know (**) the result follows: It's easy to see that (at least when c is an integer) H(c, f) <= C B(c, infinity, 1; f), so when c is an integer you have H(c, f) <= C B(c, infinity, 1; f) <= C B(a, f)^t B(b,f)^(1-t) <= C H(a, f)^t H(b, f)^(1-t). I suppose I should at least explain the notation. The Besov spaces are actually indexed by three parameters; what I call B_a above is really B_{a, infinity, infinity} (if you look closely you can see a "p = infinity" and also a "q = infinity" in the definition of B(a, f) above.) The norm B(a, infinity, 1; f) is defined by B(a, infinity, 1; f) = sum(2^(na) U(f_n)). The same as previously except with a sum instead of a sup. If you don't find a proof you like better I can give details or references for all the well-known facts above - some of these things are easier to prove than to find in the literature, but I believe most of what we need is in the paper of Krantz alluded to above; there's another paper by Joel Shapiro et al, and lots of references in [reference]. We can give a proof that H(c, f) <= C B(c, infinity, 1; f) right here - this is the one I'm not certain I know a reference for, and the main lemma in the proof is also the main ingredient in the proof of the other well-known facts: Lemma. If N is a translation-invariant norm then N(D(f_n)) is equivalent to 2^n N(f). (Ie the ratio of the two is bounded above and below; here D is the derivative.) Proof: This is just like the proof of Bernstein's inequality: First note that we need only show that N(D(f_0)) is equivalent to N(f_0); the general case follows by applying this case to the function f(2^n x). We can find a measure mu such that the Fourier transform satisfies mu^(x) = ix (1/2 < |x| < 2). (Indeed we can find a Schwarz function that does this.) Since the Fourier transform of f_0 is supported in this (pair of) interval(s) it follows that D f_0 = mu * f_0, and hence N(D f_0) <= V(mu) N(f_0) (where V(mu) is the norm of the measure mu). The other inequality follows similarly, with a different choice of mu. QED. The lemma makes it clear that H(a, f) <= C B(a, infinity, 1; f) if a is an integer: Say a = N + 1. It's easy to see that the "other" terms in H(a, f) are bounded by C B(a, infinity, 1; f) (if not it should be easy to see given the following argument for the last term). The last term is just U(D^(N+1) f), or rather it's equivalent to that (where the final derivative is a weak one: a function is in Lip_1 if and only if it has a bounded measurable function for a derivative); applying the lemma N+1 times you see that U(D^(N+1) f) <= sum(U(D^(N+1) f_n) <= C sum(2^((N+1)n) U(f_n)) = B(N+1, infinity, 1; f).