# Are Besov embeddings strict?

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 $$p$$?

• 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... Oct 6, 2020 at 21:35
• 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 Oct 6, 2020 at 22:27