# How are these two Banach spaces related ? (weighted $L2$ type space involving a logarithm and Besov type space)

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

• Is there a reason you take indicator functions instead of a smooth partition of unity in the definition of $B$? – MaoWao Feb 17 '20 at 12:43
• @MaoWao Since we're talking about $L^2$ and not continuous functions, smoothness of the partition of unity is irrelevant. – David C. Ullrich Feb 17 '20 at 13:15
• @MaoWao That's the way it seems to me, yes. – David C. Ullrich Feb 17 '20 at 13:32
• @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. – MaoWao Feb 17 '20 at 13:34
• @MaoWao "Sorry, I deleted my previous comment.": heh, sorry I agreed with it then. Wasn't paying attention to the details, sorry. – 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 $$L^2_{1/2,1/2}$$.
• thanks a lot Davide! (maybe just add a - in $2^2^N$) – Marc_Adrien Feb 18 '20 at 0:08