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

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

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?

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Bazin
Bazin
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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), $$ B^0_{p,2}(\mathbb R^n)\subset L^p(\mathbb R^n). $$ Is there a straightforward proof?