Let $1<p<\infty$, $\mathrm{L}^p_s(\mathbb{R}^n)=J_s(\mathrm{L}^p(\mathbb{R}^n))$, where $J_s=(I-\Delta)^{-\frac{s}{2}}$, or $\mathscr{F}(J_sf)(\xi)=(1+|\xi|^2)^{-\frac{s}{2}}\hat{f}(\xi)$. And we use the norm $\Vert f\Vert_{\mathrm{L}^p_s(\mathbb{R}^n)}=\Vert J_{-s}f\Vert_{\mathrm{L}^p(\mathbb{R}^n)}$. On can prove the Trudinger inequality thanks of Young's inequality ; there exists two constants $a,C>0$ such that if $s=\frac{n}{p}$, for all $u\in\mathrm{L}^p_s(\mathbb{R}^n)$, $$ \int_{\mathbb{R}^n}\left(\exp\left({a\frac{|u(x)|^p}{\Vert u\Vert_{\mathrm{L}^p_s}^p}}\right)-1\right)dx\leq C. $$ And for $s=\frac{n}{p}$, we have the Sobolev embedding $\mathrm{L}^p_s(\mathbb{R}^n)\hookrightarrow \mathrm{BMO}(\mathbb{R}^n)$. So I would like to know if from the Trudinger inequality, one could recover this embedding (as a function in $\mathrm{BMO}$ is locally exponentially integrable, I thought that an exponentially integrable function could be in $\mathrm{BMO}$, and that we could recover a norm estimate from the inequality).

## 1 Answer

This is not the case. Functions can have a high level of integrability and still not have bounded mean oscillation. For example, let $u : B_1 \to \mathbb{R}$ be defined for $x \in B_1 \setminus \{0\} \subset \mathbb{R}^n$ by $$ u (x) = \sin \Bigl(\frac{1}{\vert x \vert} \Bigr) \Bigl(\log \frac{1}{\vert x \vert}\Bigr)^\alpha. $$ If $\alpha$ is small enough, then the function $u$ is exponentially integrable. However, it can be checked (by taking small balls centered around the points $x$ at which $\sin 1/\vert x \vert = 0$) that $u$ does not have bounded mean oscillation.