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I was looking at the question whether $$ \int_{\mathbb{R}^N} \left|f(x+h)-f(x)\right|^p dx \leq \left( |h| \cdot \left\Vert f\right\Vert_{W^{1,p}(\Omega)} \right)^p $$ holds for a function $f\in W^{1,p}(\Omega)$ for $1\leq p < \infty$. In Evans' book on PDE, chapter 5 on Sobolev spaces, difference quotients are defined in section 5.8.2. Then follows theorem 3:

$(i)$ Suppose $1\leq p < \infty$ and $u\in W^{1,p}(U)$. Then for each $V\subset \subset U$ $$\left\Vert D^hu\right\Vert_{L^{p}(V)} \leq C \left\Vert Du\right\Vert_{L^{p}(U)} $$ for some constant $C$ and $0<|h| < \frac{1}{2} \text{dist}(V,\partial U)$.

I am pretty sure that there is no neead to assume the bound on $h$ and $V\subset\subset U$.

My question for this post is: Can I use the above theorem to prove my initial statement?

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If $f\in W^{1,p}(\Omega)$, then you cannot write $\int_{\mathbb{R}^N} \left|f(x+h)-f(x)\right|^p dx$ since $f$ is not defined outside $\Omega$. Instead of $\mathbb{R}^N$ you can use the set $$\Omega_h=\{x\in\Omega:\, x+th\in\Omega\text{ for all }t\in [0,1]\}.$$ You need the segment to be contained in $\Omega$ because the proof relies on the fundamental theorem of calculus along that segment. This is why Evans was restricting to the sets $U$ and $V$, to make sure $x+h$ stays inside $\Omega$.

PS: Extending $f$ to be zero outside $\Omega$ will not do because you will lose the property that $f$ is a Sobolev function and won't be able to use the FTC.

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