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Let $B^{\alpha}_p:=B^{\alpha}_{p,\infty}$ be the Besov space of regularity $\alpha<0$ and integrability $p\ge1$. Recall that a distribution $f$ from the dual Schwarz space is in $B^{\alpha}_p$ if and only if $$ \sup_{t\in(0,1]} t^{-\alpha/2}\|P_t f\|_{L_p(\mathbb{R})}<\infty, $$ where $P_t$ is the heat kernel. It is well-known that following Besov embedding holds for any $\alpha\in\mathbb{R}$, $p\ge1$: $$ B^{\alpha+1/p}_p\subset B^{\alpha}_\infty $$ and thus $$ \bigcup_{p\ge1} B^{\alpha+1/p}_p\subset B^{\alpha}_\infty. $$

My question is whether this embedding is strict. Namely, is it possible (for some $\alpha\in\mathbb{R}$) to find a function $f$ which belongs to $B^{\alpha}_\infty$ but not to $B^{\alpha+1/p}_p$ for any finite $p$?

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  • $\begingroup$ Yes, for sure, the above index indicates the "regularity", and you cannot improve regularity by improving the integrability index $p$. I don't have a precise example, but perhaps the heaviside function works ? Since it is in $L^\infty$, it is in $B^0_{\infty,\infty}$, but the strong discontinuity should destroy the regularity... $\endgroup$
    – LL 3.14
    Oct 6, 2020 at 21:35
  • $\begingroup$ So, this is not a good example, since $\delta_0 ∈ B^0_{1,\infty}$, so the Heaviside function is in $B^1_{1,\infty}$, and so in all $B^{1/p}_{p,\infty}$ ... but I am still sure that it should be possible to find an example $\endgroup$
    – LL 3.14
    Oct 6, 2020 at 22:27

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