First the standard $L2$ space : $$L^2(\mathbb{R}) = \Big \{ f : \| f \|_2 = \left( \int_{\mathbb{R}} | f(x) |^2 dx \right)^{1/2} < \infty \Big \}.$$

Let $s \geqslant 1/2$. Define a weighted $L2$ space as follows :

$$L^2_{s} := \{ f \in L^2(\mathbb{R}) : \| (2+|x|)^s f(x) \|_2 < \infty \}.$$

There is also another Banach space $B$ (in the literature that I have seen it either doesn't have a special name, or is just called a "Besov" space) :

$$B := \{ f \in L^2(\mathbb{R}) : \sum_{n=0} ^{\infty} \sqrt{2^n} \| \mathbb{1}_{\Omega_n} (x) f(x) \|_2 <\infty \}.$$

Here $\mathbb{1}_{\Omega_n}(x)$ is the indicator function onto the sets $\Omega_n := \{ x \in \mathbb{R} : 2^{n-1} \leqslant |x| < 2^n \}$, $n \geqslant 1$, and $\Omega_0 := \{x \in \mathbb{R} : |x| < 1 \}$. Note $\{ \Omega_n \}$ is a partition of $\mathbb{R}$.

Then one can show that the following inclusions hold (I can include a proof if requested) :

$$L^2_s \subsetneq B \subsetneq L^2_{1/2}, \quad \forall s >1/2.$$

Now let us define another type of weighted $L2$ space involving logarithms. For $s,p \geqslant 1/2$,

$$L^2_{s,p} := \{ f \in L^2(\mathbb{R}) : \| (2+|x|)^s \left(\log(2+|x|)\right)^p f(x) \|_2 < \infty \}.$$

Then one can also show that the following inclusions hold :

$$L^2_s \subsetneq L^2 _{1/2,p} \subsetneq B \subsetneq L^2_{1/2}, \quad \forall s,p >1/2.$$

We also have trivially

$$L^2_s \subsetneq L^2 _{1/2,p} \subsetneq L^2_{1/2,1/2} \subsetneq L^2_{1/2}, \quad \forall s,p >1/2.$$

My question is : what is the relationship between $L^2_{1/2,1/2}$ and $B$ ? Is one included in the other ? Thanks for any tips or references

  • $\begingroup$ Is there a reason you take indicator functions instead of a smooth partition of unity in the definition of $B$? $\endgroup$ – MaoWao Feb 17 '20 at 12:43
  • $\begingroup$ @MaoWao Since we're talking about $L^2$ and not continuous functions, smoothness of the partition of unity is irrelevant. $\endgroup$ – David C. Ullrich Feb 17 '20 at 13:15
  • $\begingroup$ @MaoWao That's the way it seems to me, yes. $\endgroup$ – David C. Ullrich Feb 17 '20 at 13:32
  • $\begingroup$ @DavidC.Ullrich Sorry, I deleted my previous comment. The space $B$ is of course not $F^{-1}(B^s_{2,2})$, but $F^{-1}(B^s_{2,1})$, so the claim $B\subsetneq L^2_{1/2}$ makes sense. $\endgroup$ – MaoWao Feb 17 '20 at 13:34
  • $\begingroup$ @MaoWao "Sorry, I deleted my previous comment.": heh, sorry I agreed with it then. Wasn't paying attention to the details, sorry. $\endgroup$ – David C. Ullrich Feb 17 '20 at 13:37

Let $a_n:= \sqrt{\int_{\Omega_n}f^2}$. A function $f$ belongs to $B$ if and only if $\sum_{n\geqslant 0}2^{n/2}a_n$ is finite, and to $L^2_{1/2,1/2}$ if and only if $\sum_{n\geqslant 0} 2^nna_n^2$ is finite.

  • Let $a_n=2^{-n/2}n^{-1}(\log(n+2))^{-3/4}$: then $f$ belongs to $L^2_{1/2,1/2}$ but not to $B$.
  • If $a_{2^N}=N^{-4}2^{-2^N/2}$ and $a_n=0$ if $N$ is not of the form $2^N$ for some $N$, then $f$ belongs to $B$ but not to $L^2_{1/2,1/2}$.
  • 1
    $\begingroup$ thanks a lot Davide! (maybe just add a - in $2^2^N$) $\endgroup$ – Marc_Adrien Feb 18 '20 at 0:08
  • $\begingroup$ I corrected, thanks! $\endgroup$ – Davide Giraudo Feb 18 '20 at 9:14

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