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,
 A: 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 
$\hat u\notin B^0_{1,1}$.
