Let me start with a couple of notational reminders. For $\xi\in \mathbb R^n$, $$ 1=\varphi_{0}(\xi)+\sum_{\nu \ge 1}\varphi_{\nu}(\xi),\quad \varphi_{0}\in C^\infty_c(\mathbb R^{n}),\quad \varphi_{\nu}=\varphi(\frac{\xi}{2^{\nu}}), $$ where $\varphi\in C^\infty_c(\mathbb R^{n}), \text{supp}\varphi=\{1/4\le \vert \xi\vert \le 4\}\supset\varphi^{-1}(\{1\})= \{1/2\le \vert \xi\vert \le 2\}$. We have \begin{align} \Vert{u}\Vert_{B^{s}_{p,q}}=\bigl\Vert{(2^{\nu s}\Vert{\varphi_{\nu}(D) u}\Vert_{L^{p}})_{\nu\ge 0}}\bigr\Vert_{\ell^{q}(\mathbb N)} \quad\text{(Besov spaces)},\\ \Vert{u}\Vert_{F^{s}_{p,q}}=\bigl\Vert{ \Vert(2^{\nu s}\vert{\varphi_{\nu}(D) u}\vert )_{\nu\ge 0}\Vert_{ \ell^{q}(\mathbb N) }}\bigr\Vert_{L^{p}(\mathbb R^n)} \quad\text{(Triebel-Lizorkin spaces).} \end{align} It is easy to see that for $1<q\le p<+\infty$, we have $ B^s_{p,q}\subset F^s_{p,q}\subset B^s_{p,p} $ and thus $F^s_{p,p}=B^s_{p,p}$.

My question: is it true that $$ B^s_{p,p}=W^{s,p}=\{u\in \mathscr S'(\mathbb R^n), (1+\vert D\vert^2)^{s/2} u\in L^p\} ? $$


Your space $W^{s,p}$ is the same as $F^s_{p2}$, not $F^s_{pp}$.

See these lecture notes. The definition of $H^{s,p}$ (which you call $W^{s,p}$) is given on page 34 and the result is on page 52.

| cite | improve this answer | |

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.