# Are Besov spaces $B^{s}_{p,q}$ invariant under Fourier transform?

(This may be very easy question for MO; as I am just trying to understand Besov spaces)

Let $\phi \in C^{\infty}(\mathbb R^{n})$ with $\operatorname{supp} \phi \subset \{\xi \in \mathbb R^{n}: |\xi|\leq 2\} , \phi(\xi)=1$ if $|\xi|\leq 1.$ We put, $$\phi_{j}(\xi)= \phi(2^{-j}\xi)- \phi(2^{-j+1}\xi), (\xi \in \mathbb R^{n}, j \in \mathbb N).$$ Then we have $$\operatorname{supp} \phi_{j} \subset \{\xi\in \mathbb R^{n}: 2^{j-1}\leq |\xi| \leq 2^{j+1} \}, j\in \mathbb N$$ and, with $\phi_{0}=\phi,$ $$\sum_{k=0}^{\infty} \phi_{k}(\xi)=1, \text{if} \ \xi\in \mathbb R^{n}.$$

Perhaps there different ways to introduce Besov spaces; we define in the following way.

Let $0<p\leq \infty, 0 <q \leq \infty$ and $s\in \mathbb R$ then $$B^{s}_{p,q}(\mathbb R^{n})=\{f\in \mathcal{S'}(\mathbb R^{n}):\|f\|_{B^{s}_{p,q}}=\left(\sum_{k=0}^{\infty} 2^{ksq} \|(\phi_{k}\hat{f})^{\vee}\|_{L^{p}}^{q}\right)^{1/q}<\infty \}.$$

Examples. $B^{s}_{2,2}(\mathbb R^{n})= H^{s}(\mathbb R^{n})(=\text{Sobolev spaces}).$

My naive questions are:

(1) Is $B^{0}_{1,1}(\mathbb R^{n})$ can be embedded in $L^{1}(\mathbb R^{n})$ ? (or other $L^{p}$ for $1\leq p \leq \infty$) (2) Is $B^{0}_{1,1}(\mathbb R^{n})$ is invariant under Fourier transform, that is, if $f\in B^{0}_{1,1}(\mathbb R^{n}),$ then $\hat{f} \in B^{0}_{1,1}(\mathbb R^{n})$ ? (3) What about $B^{s}_{p,q}(\mathbb R^{n})$ except for $p=q=2$ ? (4) What does this definition tells us intuitively ? (5) Is there some thing special about dyadic decompositions ?

Thanks,

• I know that the Fourier transform of a function in a modulation space is at least again in a modulation space. Depending a bit on their definition, the usual 3 parameter definition should be invariant?! I think for $p=q$.
– CPJ
Commented Sep 2, 2014 at 12:25
• What do you mean by "(5) Is there some thing special about dyadic decompositions?" Since some Besov spaces can be embedded in modulation spaces you could write their norms with frequency uniform decompositions. Like $B^s_{2,2} = M_{2,2}^s$ coincides with $H^{s}$ so it is characterizable by dyadic or frequency uniform decompositions.
– CPJ
Commented Sep 2, 2014 at 12:37
• @CPJ; Thanks, but $M^{1,1}\subset B^{0}_{1,1}$; so I don't know how does it help here ?; thanks Commented Sep 2, 2014 at 12:53
• As I said I don't know whether or not the Fourier pictures of Besov spaces are Besov spaces; I think in genral it is not the case. (I think the reason for the introduction of modulation spaces was exactly this question to find Fourier-invariant spaces. univie.ac.at/nuhag-php/bibtex/open_files/…)
– CPJ
Commented Sep 2, 2014 at 13:55
• I think Besov spaces do not contain spaces that are invariant under the Fourier transform. That is at least what Feichtinger says in univie.ac.at/nuhag-php/bibtex/open_files/fe83-1_mod-kiev.pdf and I sort of trust him on this. See page 5.
– CPJ
Commented Sep 2, 2014 at 15:05

Let me note $\phi_k(D)$ the Fourier multiplier $\phi_k(\xi)$, i.e. $\text{Fourier}\bigl(\phi_k(D)u\bigr)(\xi)=\phi_k(\xi)\hat u(\xi).$
$\bullet$ The answer to (1) is yes since $$\Vert{u}\Vert_{L^1}=\Vert{\sum_{k\ge 0}\phi_k(D)u}\Vert_{L^1}\le \sum_{k\ge 0}\Vert{\phi_k(D)u}\Vert_{L^1} =\Vert{u}\Vert_{B^0_{1,1}}.$$
$\bullet$ The answer to (2) is no: take $u=\sum_{k\ge 1} a_k\hat \phi(2^{k}x)2^{kn}$ with $a_k\ge 0$ and $\phi$ a smooth function in $\mathbb R^n$ with support the ring $\{\frac12\le \vert \xi\vert\le 2\}$. We have $$((\text{Fourier}(\phi_k(D) u))(\xi)=a_k\phi(\xi 2^{-k})^2,\quad \phi_k(D) u(x)=a_k \widehat{\phi^2}(2^{k}x)2^{kn},\quad \Vert u\Vert_{B^0_{1,1}}=\Vert\widehat{\phi^2}\Vert_{L^1}\sum_{k\ge 1}a_k.$$ On the other hand, we have $\hat u(\xi)=\sum_{k\ge 1}a_k\phi(\xi/2^k )$ and since the dyadic rings $C_k, C_{k+2}$ are disjoint, assuming $a_k=0$ for $k$ odd, we get with $c>0$ $$\Vert \hat u\Vert_{L^1}\ge c\sum_{l\ge 1}a_{2l} 2^{2ln}.$$ Choosing $a_{2l}=2^{-2ln}$ we find that $u\in B^0_{1,1}$ and $\hat u\notin L^1$, which implies from (1) that $\hat u\notin B^0_{1,1}$.
• thanks; would you please tell me, why $\|u\|_{L^{1}}= \|\sum_{k\geq 0}\phi_{k}(D)u\|_{L^{1}}$?(I guess, I need to use $\sum_{k=0}^{\infty}\phi_{k}(\xi)=1$ but I don,t know how); thanks Commented Sep 4, 2014 at 14:48
• The equality $\sum_{k\ge 0}\phi_k(\xi)=1$ is equivalent to $\sum_{k\ge 0}\phi_k(D)=Id.$ Commented Sep 5, 2014 at 7:32