Let \begin{equation} \lVert f \rVert_{B_{q,2}^\sigma} = \lVert P_0 f \rVert_p + ( \sum_{j> 0} 2^{2 \sigma j} \lVert P_j f \rVert_p^2 )^{\frac{1}{2}}. \end{equation} For $\sigma \ge0$ and $p \ge 2$, by Littlewood-Paley theorem and Bernstein's inequality, we can easily get that $B_{q,2}^\sigma \subset L^p$ for some $p\ge1$.
But my question is that: as for $\sigma \ge 0$ and $1 \le p \le 2$, can we get the analogous result such like $L^p \subset B_{q,2}^\sigma$?
In fact, I wonder can the following inequality
\begin{equation} \lVert f \rVert_{B_{\frac{6}{5},2}^{\frac{5}{6}}(\mathbb{R}^3)} \lesssim \lVert f\rVert_{L^p(\mathbb{R}^3)} \text{ holds for some p }\ge1? \end{equation} (However, in my opinion it may be wrong, since any $L^p$ space will not be imbedded in any Sobolev space with positive index of regularity.)