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Consider the Besov spaces $B_{p,q} ^s$ consisting of those functions $f$ such that

$$f \in W^{n,p} (\mathbb{R}), \quad \int_{0} ^{\infty} \Big | \frac{\omega_p ^2(f^{(n)}, t)}{t^{a}} \Big |^q \frac{dt}{t} < \infty$$

where $n \in \mathbb{N}$ and $a \in \mathbb{R}$ are such that $s = n + a$ with $0< a\leqslant 1$, $W^{n,p} (\mathbb{R})$ is the integer Sobolev space and $\omega_p ^2(f, t)$ is the modulus of continuity defined by

$$\omega_p ^2(f, t) := \sup \limits_{|h| \leqslant t} \| f(x-2h) + f(x) - 2f(x-h) \|_p.$$ (see e.g. https://en.wikipedia.org/wiki/Besov_space).

Let us also consider weighted $L^2$ spaces with respective obvious norms :

$$L^2_{\alpha} := \Big \{ f \in L^2(\mathbb{R}) : \int_{\mathbb{R}} | (1+x^2)^{\alpha/2} f(x) |^2 dx < \infty \Big \},$$

$$L^2_{\alpha, \beta} := \Big \{ f \in L^2(\mathbb{R}) : \int_{\mathbb{R}} | (1+x^2)^{\alpha/2} \log^{\beta}(2 + |x|) f(x) |^2 dx < \infty \Big \}.$$

Does the following inclusion hold ?

$$L^2_{\alpha} \subset L^2_{1/2,\beta} \subset B_{2,1} ^{1/2}, \quad \forall \alpha > 1/2, \beta > 1/2.$$ If so, how to prove the second one using this or other constructions of the Besov spaces ?

Thanks in advance

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1 Answer 1

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No, the last inclusion is false! The first exponent $s=1/2$ in the Besov space corresponds to regularity. The other spaces are just weighted $L^2$ spaces. The weight just measures the behaviour of the function at infinity, and not the local regularity. For example, if $f$ is compactly supported then $f\in L^2_{\alpha,\beta} \iff f\in L^2$. And actually the reverse inclusion holds $$ B^{1/2}_{2,1} \subset L^2 $$ and this inclusion is strict.

If you want a counterexample, take $\varphi$ smooth and compactly supported and $f(x) = \varphi(x)/|x|^{1/2-\varepsilon}$ for $\varepsilon>0$ small enough.

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  • $\begingroup$ thank you! much appreciated for the input $\endgroup$ Aug 10, 2020 at 4:01

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