Proof of Besov space embeddings: $B_{p, \infty}^{s+\varepsilon}$ is continuously embedded in $B_{p, 1}^s$

150 Views Asked by At

Let's consider the Besov spaces $B_{p,q}^s$, $p,q \in [1,\infty]$, $s\in \mathbb{R}$ and $\varepsilon >0$. I keep coming across that $B_{p, \infty}^{s+\varepsilon}$ is continuously embedded in $B_{p, 1}^s$, i.e. there exists an constant $C$ with $\|\cdot\|_{B_{p,1}^s} \leq C\|\cdot\|_{B_{p, \infty}^{s+\varepsilon}}$, but unfortunately I have no idea how to prove it. Could anybody give me a hint (or proof)?

1

There are 1 best solutions below

2
On BEST ANSWER

After some thought, I came to the following proof. I hope firstly that it is correct and secondly that it may be able to help others. $$ \begin{align*} \|f\|_{B_{p,1}^s} &= \sum_{j \geq -1} 2^{js} \|\Delta_j f\|_{L^p} \\ &= \sum_{j \geq -1} 2^{-j(s+\varepsilon- s)} 2^{j(s+\varepsilon)} \|\Delta_j f\|_{L^p} \\ &= \sum_{j\geq -1} 2^{-j\varepsilon} 2^{j(s+\varepsilon)} \|\Delta_j f\|_{L^p} \\ &\leq \sum_{j\geq -1} 2^{-j\varepsilon} \sup_j 2^{j(s+\varepsilon)} \|\Delta_j f\|_{L^p} \\ &= \sum_{j\geq -1} 2^{-j\varepsilon} \|f\|_{B_{p, \infty}^{s+\varepsilon}} \\ &= C_{\varepsilon} \|f\|_{B_{p, \infty}^{s+\varepsilon}} \end{align*}.$$