# Injection of Besov spaces in $L^p$

I believe that for $$p\ge 2$$, we have the continuous injection (for $$p=2$$, it is an equality), $$B^0_{p,2}(\mathbb R^n)\subset L^p(\mathbb R^n),$$ where $$B^0_{p,2}(\mathbb R^n)$$ is the Besov space. Is there a straightforward proof?

• Here is a straightforward proof." Write the dyadic Littlewood-Paley characterizations of $B^{0}_{p,2}$ and $F^{0}_{p,2} = L^p$ and use Minkowkski's inequality to take the $L^p$ norm inside the $\ell^2$ norm. In any case, this is not really a research-level question. – sharpend Mar 24 '19 at 22:20

Theorem 2.3.2 d) in Triebel says that for $$- \infty < s < \infty$$ we have (continuous inclusions)
$$B^s_{p,2}(\mathbb{R}^n) \subset F^s_{p,2}(\mathbb{R}^n) = H^s_p(\mathbb{R}^n) \subset B^s_{p,p}(\mathbb{R}^n) \quad \text{for}~2 \leq p < \infty, \\ B^s_{p,p}(\mathbb{R}^n) \subset F^s_{p,2}(\mathbb{R}^n) = H^s_p(\mathbb{R}^n) \subset B^s_{p,2}(\mathbb{R}^n) \quad \text{for}~1 < p \leq 2.$$
The proof seems like a direct estimate for the norms in the respective spaces, which are the ones by dyadic decomposition. Maybe that qualifies as straightforward, at least where $$s=0$$?