I am trying to solve the following exercise:
Let $U$ be an open set in $\mathbb{R}^n$ and let $V$ be a compact subset with Lipschitz boundary. Assume that $f$ is in $L^p(U)$ with $1<p<\infty$ and there are constant $C_1$ and $h_2$ such that $\|\tau_{h,i}f\|_{L^p(V)}\le C_1$ for all $h<h_2$ and each $i$. Then show
$f$ is in $W^{1,p}(V),\|Df\|_{L^p(V)}\le C_1$ and
$\tau_{h,i}f$ converges to $\frac{\partial f}{\partial x_i}$ in $L^p(V)$ as $h\to 0$.
I have shown $(1)$ by extracting a weak convergence subsequence of $\tau_{h,i}f$. But I don't know how to show $(2)$.
Given $\epsilon>0$ ,since $C^\infty(V)\cap W^{1,p}(V)$ is dense in $W^{1,p}(V)$, we can find a smooth $\phi$ such that $\|\phi-f\|_{W^{1,p}(V)}<\epsilon$, hence $\|\frac{\partial \phi}{\partial x_i}-\frac{\partial f}{\partial x_i}\|_{L^p(V)}<\epsilon$. Moreover, since $\phi$ is smooth, and by compactness of $V$, for small enough $h$, $\|\tau_{h,i}\phi-\frac{\partial \phi}{\partial x_i}\|_{L^p(V)}\le \epsilon$
Hence $$\begin{align} \|\tau_{h,i}f-\frac{\partial f}{\partial x_i}\|_{L^p(V)}&\le \|\tau_{h,i}f-\tau_{h,i}\phi\|_{L^p(V)}+\|\tau_{h,i}\phi-\frac{\partial \phi}{\partial x_i}\|_{L^p(V)}+\|\frac{\partial \phi}{\partial x_i}-\frac{\partial f}{\partial x_i}\|_{L^p(V)}\\ &\le \|\tau_{h,i}f-\tau_{h,i}\phi\|_{L^p(V)}+2\epsilon \end{align}$$
But the problem is I don't know how to argue $\|\tau_{h,i}f-\tau_{h,i}\phi\|_{L^p(V)}<\epsilon$ for small enough $h$ since $f(x+he_i)$ can still goes outside $V$ and there is $h$ in denominator.
Since $V$ has Lipschitz boundary, I know there exists $\bar{f}\in W^{1,p}(\mathbb{R}^n)$ such that $\bar{f}=f$ on $V$. But I still don't know how to handle it since $\bar{f}(x+he_i)$ may not always equal to $f(x+he_i)$ for all $x\in V$.
As Tomas mentioned in the comment, the result may not hold for $L^p(V)$. Hence I currently am trying to prove the $L^p_{loc}(V)$, that is for open $W\subset\subset \textrm{int}(V)$.