Let $p>n$ and $\alpha\in(0,1)$.
If $$ \sup_{x,y\in\mathbb{R}^{n}}\frac{|u(x)-u(y)|}{|x-y|^{\alpha}}\leq C\|Du\|_{L^{p}(\mathbb{R}^{n})} $$ for $u\in W^{1,p}(\mathbb{R}^{n})$ with $C$ : independent of $u$,
then want to show that $$ \alpha=1-\frac{n}{p} $$ This statement looks like a inverse of Morrey's inequailty.
Next is what I've done for this question : By Fundamental theorem of Calculus, $$ u(x)-u(y)=\int_{C_{x,y}}(Du) $$ where $C_{x,y}$ is a line from $x$ to $y$.
Then, by Jensen's Inequality, $$ |u(x)-u(y)|^{p}=\left(\int_{C_{x,y}}(Du)\right)^{p}\leq\left(\int_{C_{x,y}}(Du)^{p}\right). $$ And then what can I do next?
You need a dilation argument here: let $u$ be a smooth cutoff, supported in $B(0,1)$, with $u(0)=1$ and $$\int_{B(0,1)}|Du|^p=1.$$ For $\lambda>0$, set $$u_{\lambda}(x)=u(\lambda x).$$ Note then that $$\sup_{x,y\in\mathbb R^n}\frac{|u_{\lambda}(x)-u_{\lambda}(y)|}{|x-y|^{\alpha}}\geq\frac{|u_{\lambda}(e_1/\lambda)-u_{\lambda}(0)|}{|e_1/\lambda|^{\alpha}}=\frac{|u(e_1)-u(0)|}{\lambda^{-\alpha}}=\lambda^{\alpha},$$ and $$\int_{\mathbb R^d}|Du_{\lambda}|^p=\int_{\mathbb R^d}|\lambda Du(\lambda x)|^p\,dx=\lambda^{p-d}\int_{\mathbb R^d}|Du|^p=\lambda^{p-d}.$$ Then, your inequality applied to $u_{\lambda}$ shows that $$\lambda^{\alpha}\leq \sup_{x,y\in\mathbb R^n}\frac{|u_{\lambda}(x)-u_{\lambda}(y)|}{|x-y|^{\alpha}}\leq C\|Du_{\lambda}\|_p=C\lambda^{1-\frac{d}{p}},$$ or, equivalently, $$\lambda^{\alpha-1+\frac{d}{p}}\leq C.$$ Letting $\lambda\to \infty$ and $\lambda\to 0$, the last inequality shows that $\alpha=1-\frac{d}{p}$.