For simplicity, I will consider $u\in W^{1,p}(\Omega)$, where $1<p<\infty$ and $\Omega\subset\mathbb{R}$ is open and bounded. I am able to show that
\begin{equation}
|u(x)-u(y)| \leq |x-y|^{1-1/p} \lVert u'\rVert_{L^p(\Omega)}
\end{equation}
for all $x,y\in\Omega$. Next, I want to show that the set of Holder continuous functions with Holder exponent $1-1/p$ is continuously embedded in $W^{1,p}$. To do so, I was told to prove that there exists a uniformly continuous function $v$ such that $u=v$ in almost every $x,y\in\Omega$.
I am confused what I really want to do to find $v$. Can I say $v$ is a function such that
\begin{equation}
\sup_{x\neq y} \frac{|v(x)-v(y)|}{|x-y|^{1-1/p}} = \lVert u'\rVert_{L^p(\Omega)}
\end{equation}
for all $x,y\in\Omega$? Then I know Holder continuous functions are uniformly continuous, but is $v=u$ a.e?
Thank you.
Best Answer
I think you have the inclusion backwards: For instance the function $f(t)=|t|^{1/2}$ is Holder continuous in, say, $\Omega=(-1,1)$ but $f'(t)=t^{-1/2}\notin L^2(\Omega)$, therefore the inclusion $C^{0,1/2} \to W^{1,2}$ fails.
To prove that every $u\in W^{1,p}$ has a (locally absolutely) continuous representative see here.