Consider the Holder space $C^{0,1-\frac{n}{p}}(\mathbb{R}^{n})$ and the Sobolev Space $W^{1,p}(\mathbb{R}^{n})$. Take $u_{m} \in C_{c}^{\infty}(\mathbb{R}^{n})$ such that Morrey's Inequality we have $||u_{m}||_{C^{0,1-\frac{n}{p}}(\mathbb{R}^{n})} \leq C||u_{m}||_{W^{1,p}(\mathbb{R}^{n})}$. We are also given that $u_{m} \rightarrow \bar{u}$ in $W^{1,p}(\mathbb{R}^{n})$ and $u_{m} \rightarrow u^{*}$ in $C^{0,1-\frac{n}{p}}(\mathbb{R}^{n})$.
How does it follow that $u^{*} = u \text{ a.e}$? What result is used?
Thanks for assistance. Let me know if anything is unclear.
Since $u_m \to u^*$ in $C^{0,1-\frac np}$ you know that $u_m(x) \to u^*(x)$ for all $x$. On the other hand, since $u_m \to u$ in $W^{1,p}$ (and in particular in $L^p$) there is a subsequence with the property that $u_{m_j}(x) \to \bar u(x)$ almost everywhere. Since $u_{m_j}(x) \to u^*(x)$ for all $x$, you get a.e. equality.