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Let $u(x,y) \in L^1(\mathbb{R}^2)$, then $u(x,v)-u(x+\xi,v+\eta)$ considered as function of $(x,y)$ is also belongs to $L^{1}(\mathbb{R}^2)$. We can define a function $$ F(\xi,\eta) = \int\limits_{\mathbb{R}^2}|u(x,y)-u(x+\xi,y+\eta)|dxdy $$ Is it true that $F(\xi,\eta) \to 0$ when $(\xi,\eta) \to 0$? In other words, it is possible to pass to the limit under the integral sign?

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Try studying Lemma 1.4 here: – Siminore Oct 28 '12 at 9:49
@Siminore I have some problem: continuous function $g \in C(\mathbb{R}^2)$ might not be uniformly continuous – Nimza Oct 28 '12 at 9:56
up vote 1 down vote accepted

The key is to use $\{u_n\}$, a sequence of continuous functions with compact support, which approximate $u$ in $L^1(\Bbb R^2)$, say $\lVert u-u_n\rVert_{L^1(\Bbb R^2)}\leq n^{—1}$. We have \begin{align}|F(\xi,\eta)|&\leq \int_{\Bbb R^2}|u(x,y)-u_n(x,y)|dxdy+\int_{\Bbb R^2}|u_n(x+\xi,y+\eta)-u_n(x,y)|dxdy\\&+\int_{\Bbb R^2}|u_n(x+\xi,y+\eta)-u(x+\xi,y+\eta)\\ &\leq 2n^{-1}+\int_{[-A_n,A_n]^2}|u_n(x+\xi,y+\eta)-u_n(x,y)|dxdy, \end{align} as $u_n$ has a compact support, and we can assume $|\eta|+|\xi|\leq 1$. So for a fixed $n$, using uniform continuity of $u_n$ on $[-A_n,A_n]^2$, we get $$\limsup_{(\eta,\xi)\to (0,0)}|F(\xi,\eta)|\leq 2n^{-1},$$ which gives the wanted result as $n$ is arbitrary.

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