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There is a definition for difference quotients of Sobolev functions I do not understand. Let $U\subset\Omega\subset\mathbb{R}^n$ be open sets such that the closure of $U$ is compact in $\Omega$ and let $h\in\mathbb{R}$ with $0<|h| < \mbox{dist}(U,\partial\Omega)$. Then one defines a function $\partial^h_i u$ on $U$ by

$\partial_i^h u(x) := \frac{u(x+he_i) - u(x)}{h}$,

(where $e_1,\dots,e_n$ is the standard basis of $\mathbb{R}^n$), see e.g. [1].

But how can one even define this? One must be able to evaluate $u$ at $x$ and $x+he_i$, but since $u$ is only an $L^p$-function, point evaluation is not even well defined. One needs higher regularity, I assume - an unique continuous representative of $u$ which can be evaluated. But no mention of such higher regularity is made in [1] (and the other texts on PDEs I'm reading) where such difference quotients are used. So I guess that no such higher regularity is needed after all?

Am I missing something obvious here? Why can one define $\partial_i^h u$ in such a way? Could you please give me a hint why this definition works or recommend me some literature on this?

Thank you in advance!

Reference:

[1] Lawrence C. Evans, "Partial Differential Equations", 5.8.2. There, Evans defines $\partial_i^hu$ for 'locally summable functions' but uses it for $W^{1,p}$ functions.

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2 Answers 2

up vote 5 down vote accepted

To avoid any discussion of representatives and equivalences, one can introduce translation operators $\tau_h f = f(\cdot +he_i)$. This operator is originally defined on continuous functions, but is immediately seen to be bounded on $W^{1,p}$ (and on pretty much every un-weighted function space), though some care needs to be taken with the domain of the function (it's best to consider functions on $\mathbb R^n$ first). Thus, $\tau_h$ uniquely extends to a bounded operator on $W^{1,p}$.

The operator-based approach also allows one to handle the issue of boundary values (traces), where the equivalence classes do not help much.

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Awesome! Thank you. Very elegant, I think :) –  Sh4pe Sep 27 '12 at 15:31

This actually doesn't define $\partial_i^hu$ as an usual function, but as an equivalence class of functions. It is well-defined, as if $u_1=u_2$ almost everywhere, $u_1(x+he_i)=u_2(x+he_i)$ for almost every $x$ which gives that $\partial_i^hu_1(x)=\partial_i^hu_2(x)$ for almost every $x$.

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