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Some very pedestrian questions about Besov spaces. Just to fix notation:

1.Let $f \in \mathcal{S}'$, the space of tempered distributions.

2.$\Psi, \{ \Phi_n \}_{n \geq 0} \subset \mathcal{S}$ such that their Fourier transforms $\hat{\Psi}, \{ \hat{\Phi}_n \}$ form a partition of unity subordinate to the cover $A_0 = (-1, 1)$, $ A_n = \{2^{n-1} < |\xi| < 2^{n+1} \}$.

3.So $f = \Psi * f + \sum_{n \geq 0} \Phi_n * f$ in $ \mathcal{S}'$.

4.Say (my impression) $f$ lies in the inhomogeneous Besov space $B^{\alpha}_{p,q}$ if

$$ \| \Psi * f\|_p + (\sum_{n \geq 0} (2^{n \alpha} \|\Phi_n * f\|_p)^q)^{\frac{1}{q}} < \infty. $$

Questions

What exactly does the indices signify in terms of smoothness properties of the function? How does the frequency content of $f$ in the dyadic frequency bands, as summarized by the Besov norm, reflect its regularity properties?

For example, if I want to find $f \in L^2(\mathbb{R})$ that's $\beta$-times differentiable in the Sobolev sense, I can just look for the condition

$$ \| \hat{f}(\xi) \xi^{\beta} \|_2 < \infty $$

What would be a corresponding statement for $B^{\alpha}_{p,q}$ for, say, a piecewise constant function? What if different pieces have different degrees of smoothness?

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  • $\begingroup$ I don't understand why you are thinking about piecewise constant functions. Roughly speaking the $\alpha$ index indicates that a $B^\alpha_{p,q}$ function is $\alpha$ times differentiable (in a appropriate sense). The difference between Besov and Sobolev spaces is basically how features at different scales are compared. See also mathoverflow.net/a/17906/3948 $\endgroup$ – Willie Wong Sep 3 '14 at 13:34
  • $\begingroup$ That's part of what I am asking---what would the Besov parameters be, explicitly, for a trivial/degenerate case? If that's not a good question, why not? $\endgroup$ – Michael Sep 3 '14 at 14:35
  • $\begingroup$ I don't know what you mean by "what would the Besov parameters be, explicitly, for a trivial/degenerate case?" But regarding the question in your post: Besov spaces are not microlocal spaces. Physical space localisation ("different pieces have different degrees of smoothness") cannot be seen in its definition. This is the same as for Lebesgue class: if a function is not $L^p_{loc}$ in a neighbourhood of $x_0$, it cannot be $L^p(\mathbb{R}^n)$. So I find the questions you asked in the last sentence entirely not the right one for understanding Besov spaces. $\endgroup$ – Willie Wong Sep 3 '14 at 14:45
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    $\begingroup$ Perhaps the following answers a little bit of your question: first observe the trivial inclusion $B^s_{p,q_1} \subset B^s_{p,q_2}$ if $q_1 \leq q_2$. We further have that $B^s_{p,1} \subset W^{s,p} \subset B^s_{p,\infty}$. (This can be sharpened to say that $W^{s,p}$ always sits between $B^s_{p,2}$ and $B^s_{p,p}$.) This is indicative that the regularity indicated by the indices gives you not much more information than Sobolev regularity. But in any case, none of this has anything to say about "spatial inhomogeneity" as you indicated in your title. $\endgroup$ – Willie Wong Sep 3 '14 at 14:55
  • $\begingroup$ Ok. Thanks for replying. By "trivial/degenerate cases", I mean locally constant functions with compact support, say. Surely they are Besov, no? I see what you're saying regarding the definition. The reason for my question is that, in the applied literature, Besov spaces are usually said to have "descriptive power". It's very common to read that they contain "spatially inhomogeneous", in the sense that having different smoothness properties at different locations, functions. I am trying to see what they mean. (In this context, Sobolev functions are "spatially homogeneous".) $\endgroup$ – Michael Sep 3 '14 at 14:56
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This answer has a couple of parts.

Part I: A Computation

Consider the simplest case in $\mathbb{R}$. Let $\chi$ be the indicator function of $[-1,1]$. You have that $\|\Phi_n * \chi\|_{L^2} \approx 2^{- \frac{n}2}$. So in which Besov spaces does $\chi$ belong? First clearly we need $\alpha \leq \frac12$. When $\alpha = \frac12$ you have that $q = \infty$ necessarily. In general you want $$ \sum_{n = 0}^{\infty} 2^{n (\alpha - \frac12) q } < \infty$$ But as long as $\alpha < \frac12$, this is a geometric series and converges. And so for $\alpha < \frac12$ we have that $\chi \in B^{\alpha}_{2,q}$ for any $q \in [1,\infty]$.

Note also that $B^{\alpha}_{2,2} = H^{\alpha}$, so for your simple example the Besov space description does not tell you any more than the Sobolev regularity, except for the end-point $q = \infty$ which gives you infinitesimally more.

In fact, from the formula for the Besov norm, if the norm $\|\Phi_n*f\|_p$ has a polynomial decay rate $2^{-\beta n}$ then we see that the Sobolev description gives more or less the same regularity behaviour as the Besov one. Where Besov space shines is when there's a further logarithmic correction term. Imagine the norm now decays like $2^{-\beta n} f(n)$. In this situation you can win a little bit: the Sobolev estimate will not work with $\beta$ derivatives, since you lose convergence of the geometric series, but it will work with $\beta-\epsilon$ derivatives for any $\epsilon > 0$. In the Besov case you have hope to recover the regularity up to $\beta$ derivatives provided $f(n)$ is summable in some $\ell^q$. But generally speaking: that's it. In most practical situations the additional descriptive power of Besov over Sobolev is precisely that: one logarithmic divergence.

Part II: Spatially inhomogeneous smoothness

After doing some light reading on this subject [1,2, 3, 4], what I came to realize is that

  1. The studies are mostly within the field of estimation theory; and
  2. The stark contrast between spatially inhomogeneous smoothness versus spatially homogeneous smoothness is not so much between Sobolev and Besov classes, but between Besov classes and Holder classes.
  3. Furthermore, the contrast does not concern so much the membership of the relevant classes, but rather in the whether one can find estimators reconstructing a function based on (noisy) observations.

To explain: suppose we have a function $f$ on $\mathbb{R}$. We know that if $f$ has Holder regularity $\alpha$ for $\alpha \in (0,1)$ this means that locally we have the bound $$ |f(y) - f(x)| \leq C |y - x|^\alpha $$ when $|y-x| < 1$. Note that the regularity $\alpha$ gives a uniform upper-bound to the local scaling behaviour of the function near a point.

Now suppose $f$ satisfies the following conditions:

  • $f(0) = 0 = f(1)$
  • When $x < \frac14$ we have $f(x) = |x|^{\alpha}$
  • When $|x - 1| < \frac14$ we have $f(x) = |x-1|^{\beta}$.
  • $\alpha < \beta$.

The problem is that now because of the presence of the singularity at $x = 0$, the function cannot belong to any space better than $C^\alpha$. But measured relative to the $C^\alpha$ semi-norm, the singularity at $x = 1$ is invisible since $$ \limsup_{r \to 0} \sup_{y\in (1-r,1+r) \setminus \{1\}} \frac{|f(y) - f(x)|}{|x-y|^\alpha} = 0 $$ For this we see that a $C^\alpha$ function can easily have spatially inhomogeneous smoothness. (Same for Sobolev and Besov classes: most functions in both Sobolev and Besov spaces have really bad spatial inhomogeneity when it comes to pointwise smoothness, see [5].)

(An aside: Please note that the Besov spaces $B^s_{\infty,\infty}$ coincides with the Holder space $C^s$ for $s$ not an integer. (When $s$ is an integer $B^s_{\infty,\infty}$ is the slightly better Zygmund space.) So in a very real sense Besov spaces include among them both the $L^2$-Sobolev and the Holder spaces. So that the constraint that a function lies in a Besov space should be regarded as a relaxation of the constraint that a function lies in a Holder space.)

In any case, when in the literature people write that Besov spaces are adapted to considering spatially inhomogeneous smoothness, what they mean is that assuming that the signal lies in some Besov space (so relaxing the constraint that it lies in a Holder or Sobolev space), one can get very good estimators. And one of the justifications for this is that there exists very powerful wavelet based techniques [6, 7], and Besov (for that matter, also Triebel-Lizorkin) spaces are naturally adapted to wavelet characterisations.

That wavelet techniques are useful for spatially inhomogeneous regularity should not be surprising, as a basic idea behind wavelets is that of microlocalisation: that one decomposes a function simultaneously in physical and Fourier space (of course, the decomposition cannot be simultaneously sharp due to the uncertainty principle).

To finish, let me just say that even with Besov and Sobolev scales, the scales tell you something about "averaged" regularity. By this I mean that it is not possible to pin-point where various degrees of smoothness occur or even their spatial distribution. (If you have the wavelet decomposition sitting in front of you, on the other hand, you can get some additional information.) As indicated in [5], what you typically get by knowing that a function is in Besov class is some bounds on how "big" are the sets of the singularities of various degree.

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  • $\begingroup$ There seems to be something unfinished with your $C^{\alpha}$ $f$...so what's the contrast with Besov? $\endgroup$ – Michael Sep 4 '14 at 13:19
  • $\begingroup$ "...when in the literature people write that Besov spaces are adapted to considering spatially inhomogeneous smoothness, what they mean is that assuming that the signal lies in some Besov space"---that's what I meant when I said spatial inhomogeneity is almost synonymous with Besov. In view of your answer, however, spatial inhomogeneity is present already in the subfamilies of Sobolev and Holder spaces. $\endgroup$ – Michael Sep 4 '14 at 13:24
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    $\begingroup$ The contrast, as far as I can tell from my brief reading, seems to be only that "it is more profitable to work by assuming your signal lies in a Besov space than assuming that it lies in a Holder space". In other words, from the few papers I have looked at the association of Besov spaces with spatial inhomogeneity is that of a heuristic. On the other hand, there is a good reason why Holder spaces are bad: they are defined by spatially uniform pointwise norms. So it can only really see the regularity at the worst point. Using "averaged" norms like Sobolev or Besov intuitively allows you to $\endgroup$ – Willie Wong Sep 4 '14 at 15:05
  • $\begingroup$ probe a bit further. Using an anisotropic example, let $f(x,y) = \sqrt{|x|}$ and $g(x,y) = (x^2 + y^2)^\frac14$. They have the same Holder exponent ($1/2$). But they belong in different (local) Sobolev spaces. $f$ does not belong to $H^1_{loc}$ as $\frac{1}{\sqrt{|x|}}$ is not $L^2$ near the origin. But $g$ does, as $|\nabla g|^2$ is proportional to $(x^2 + y^2)^{-\frac12}$ and is integrable. So you see that while the two functions cannot be distinguished in the Holder sense, the fact that one has a larger "singular set" can be seen from the Sobolev sense of regularity. Besov allows you to $\endgroup$ – Willie Wong Sep 4 '14 at 15:15
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    $\begingroup$ @Shalop: yeah, poor choice of word on my part. I think I may have been influenced by the other things I am writing for this answer and meant "better" in the sense of "admitting more functions".... $\endgroup$ – Willie Wong May 16 '19 at 1:33

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