Reverse Morrey Embedding In the theory of Sobolev spaces, we have the inclusion $W^{1,p}(\mathbb{R}^n) \subset C^{0,\alpha}(\mathbb{R}^n)$ where $\alpha = 1-\frac{n}{p}$ and $p>n$. 
What can one say about reverse Morrey type inequality?, i.e given a function $f \in C^{0,\alpha}$, is there any criteria to decide if it is actually in $W^{1,q}$?
 A: There is no criterion based only on $\alpha$. In fact there is no embedding $C^{0,\alpha}(\Omega)\to W^{1,q}(\Omega)$ for open $\Omega$, unless $\alpha=1$ and $\Omega$ is bounded, in which case $C^{0,1}$ coincides with the space of Lipschitz functions and by Rademacher's theorem embeds into $W^{1,\infty}$.
You can have a look at this question on Mathoverflow for an example of a function which is Holder continuous for every exponent $\alpha<1$ but not locally in $W^{1,q}$ for any $q$, therefore by itself disproves the imbedding.
A simpler example comes from constructing a $\alpha$-Holder function which is not in a Sobolev space, where $\alpha<1$ is any fixed exponent. The Cantor function is such an example with $\alpha=\frac{\log 2}{\log 3}$. It does not belong to any Sobolev space basically because it has zero derivative a.e. without being constant. Indeed, if it were in some Sobolev space the corresponding Poincaré inequality 
$$\int_\Omega |u-u_\Omega|^q\leq C\int_\Omega|Du|^q$$
would hold. However you can show that the distributional derivative $Du$ is zero in the complement of the Cantor set, which has full measure, therefore $u$ should be constant.
This proof adapts to higher dimension too, considering the Cantor function of one coordinate, and I believe that it can also be adapted to any exponent $\alpha<1$ with a generalized Cantor function, where the "middle interval" at each step of the usual construction of the Cantor function is of relative measure $\delta$ instead of $\frac13$. The only thing to check is the Holder continuity with exponent close to $1$, and the rest of the proof remins the same. I wold guess that it would be $\frac{\log 2}{\log (2/(1-\delta))}$-Holder, but I don't have a proof of it.
