Is there a nice relationship between $\dot{B}^{-1}_{p,1}$ and $\dot{B}^{0}_{p,1}$?

Question

In particular, I'm hoping there is an embedding/inequality relating elements of the two.

For background, I have a particular function, $f \in W^{\infty,p}(\mathbb{R}^2)$, for all $p \in [1,\infty]$, which is a solution to a particular PDE. I am currently investigating the decay properties fo rthis function, and have reached a point where I have the two norms:

$||f||_{\dot{B}^{-1}_{p,1}}$ and $||\nabla f||_{\dot{B}^{-1}_{p,1}}$

All I wish to do is show that these two norms are finite. As I am very new to Besov spaces, I am not sure how I can do this.

I have been given a hint that the norm should be finite for $p>2$, but I have no idea why this might be the case.

I am aware of the following embedding:

$ \dot{B}^{0}_{p,1} \subseteq L^p \subseteq \dot{B}^{0}_{p,\infty} $

But I am completely unaware of what we can say for Besov spaces of the form $\dot{B}^{s}_{p,\infty}$ where $s < 0$.

The $\dot{B}^{s}_{p,q}$-norm is defined as follows:

$ ||f||_{\dot{B}^{s}_{p,q}} = || \{ 2^{sj}||\phi_j \ast f||_{L^p(\mathbb{R}^2)} \} ||_{\mathcal{l}^q}$, where $\phi_j$ are defined as the inverse FT of terms from the Littlewood-Paley decomposition. That is:

$ \hat{\phi_j}(\xi) := \hat{\phi_0}(2^{-j}\xi)$, $\ \ \text{Supp}(\hat{\phi_j}) \subseteq \{ \xi\in \mathbb{R}^2 \ | \ |2^{j-1}| \leq |\xi| \leq |2^{j+1}| \}$, $\ \ \Sigma_{j \in \mathbb{Z}} \hat{\phi_j}(\xi) = 1$, for all $\xi \neq 0.$

Thank you.

Answer 1

Being in $W^{p,\infty}$, I assume this means having all derivatives in $L^p$, that is, for all $n \ge 0$ we have $$ \left\|\left(\sum_{k=-\infty}^\infty |2^{kn} \Delta_k f|^2\right)^{1/2} \right\|_p < \infty ,$$ where $\hat \Delta_k f(\xi) = \varphi(2^{-k} \xi) \hat f(\xi)$, where $\varphi$ is a smooth function compactly supported in $\{\xi: \frac14 < |\xi| < 4\}$ satisfying $\sum_{k=-\infty}^\infty \varphi(2^k \xi) = 1_{\xi \ne 0}$.

Also $$ {\|f\|}_{\dot{B}^{-1}_{p,1}} := \sum_{k=-\infty}^\infty 2^{-k} {\|\Delta_k f\|}_p .$$

This is how you do these kinds of arguments.

Let's first consider $1 \le p \le 2$.

Split it into two halves, and suppose $\epsilon>0$: \begin{align} \sum_{k=0}^\infty 2^{-k} {\|\Delta_kf\|}_p &\le \left(\sum_{k=0}^\infty 2^{-\epsilon p/(p-1)}\right)^{(p-1)/p} \left(\sum_{k=0}^\infty 2^{-kp(1-\epsilon)} {\|\Delta_kf\|}_p^p \right)^{1/p} \\ &\le C_1 \left\| \left(\sum_{k=0}^\infty 2^{-kp(1-\epsilon)} |\Delta_kf|^p \right)^{1/p} \right\|_p \\ &\le C_1 \left(\sum_{k=0}^\infty 2^{-\epsilon 2p(2-p)}\right)^{(2-p)/(2p)} \left\| \left(\sum_{k=0}^\infty 2^{-2k(1-2\epsilon)} |\Delta_kf|^2 \right)^{1/2} \right\|_p \\ &\le C_2 \left\| \left(\sum_{k=0}^\infty 2^{-2k(1-2\epsilon)} |\Delta_kf|^2 \right)^{1/2} \right\|_p \\ &\le C_2 {\|f\|}_{W^{-1+2\epsilon, p}}, \end{align} Do a similar thing for the other half, which will be bounded by a constant times ${\|f\|}_{W^{-1-2\epsilon, p}}$.

And a similar, but slightly simpler, argument works for $2 \le p < \infty$.

However, for the other half (the low frequencies), you don't get exactly what you need. Because $$ {\|f\|}_{\dot B^{-1}_{p,1}} \ge 2^{-k} {\|\Delta_k f\|}_p ,$$ and for $-k$ large, we simply don't have any kind of estimate coming from the Sobolev norm.

Answer 2

I think one of your questions is whether $W^{\infty,p} \subset \dot B^0_{p,2}$. This would imply: \begin{align} \sum_{k=-\infty}^{-1} {\| \Delta_kf \|}_p &\le C \left\| \left(\sum_{k=-\infty}^{-1} |\Delta_kf|^2 \right)^{1/2}\right\|_p &(*) \end{align} This is definitely not the case if $p = 2$, because this would imply that $\ell^2 \subset \ell^1$. Considering the case $$ f(x) = \sum_{k=-\infty}^{-1} a_k 2^{-dk}\varphi(2^k x) , \quad x \in \mathbb R^d,$$ for suitable $a_k$ should work to show this is not true for any other $1<p<\infty$ either, because the left hand side of (*) will be approximately the $\ell^1$ norm of $(a_k)$, and the right hand side will be approximately the $\ell^p$ norm of $(a_k)$.