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Let $C_p$ the class of functions $f\in L^p(\mathbb{R}^d)$ such that $$\lim_{|\mathbf{h}|\to 0} \|f(\mathbf{x+h})-f(\mathbf{x})\|_p = 0 .$$

To prove that $C_p = L^p(\mathbb{R}^d)$ we start proving the result for indicators functions of cubes. The book Measure and Integral of Weeden and Zygmund, just says: "Clearly the characteristic function of a cube belongs to $C_p$." Why is that clear?

The problem can be reduced to prove that $m(A\cap (A+\mathbf{h}))\to m(A)$. I have proved that $m(A)=\sup\{m(A\cap (A+\mathbf{h})):\mathbf{h}\in\mathbb{R}^d\}$. But that still isn't enough to conclude the desired result.

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Let $f^h(x):=f(x+h)$. For your first question, you should notice that if $f=\chi_Q$ ($Q$ a cube), then $f^h=\chi_{Q-h}$ and that $\lVert f^h-f\rVert_p=m((Q-h)\Delta Q)=2\ m(Q\setminus (Q-h))$ (here $\Delta$ is the symmetric difference of sets). – Pacciu Dec 15 '11 at 6:20
up vote 3 down vote accepted

Exploring Pacciu's comment.

Let $Q=[0,1]^d$ and let $\chi_Q$ be the characteristic function (or indicator function) of $Q$.
Moreover, for $h=(h_1,h_2,\ldots,h_d)$ we write $$Q_h=Q + h = [h_1,1+h_1]\times [h_2,1+h_2]\times\cdots\times[h_d,1+h_d].$$ Then $$\int_{\mathbb{R}^d} |\chi_Q(x)-\chi_Q(x+h)|^p dx = \int_{\mathbb{R}^d} |\chi_Q(x)-\chi_Q(x+h)| dx =$$ $$\qquad\qquad\qquad\qquad \int_{Q\setminus Q_h} dx + \int_{Q_h\setminus Q} dx $$ Now if $|h|<1$ we have ${Q\setminus Q_h} = [0,h_1)\times[0,h_2)\times\cdots\times[0,h_d)$ so
$$\int_{Q\setminus Q_h} dx = h_1\cdot h_2\cdots h_d$$ and similarly ${Q_h\setminus Q} = (1,1+h_1]\times(1,1+h_2]\times\cdots\times(1,1+h_d]$ $$\int_{Q_h\setminus Q} dx = h_1\cdot h_2\cdots h_d,$$ that is $$\|\chi_Q-\chi_{Q_h}\|_p^p = 2\cdot h_1\cdot h_2\cdots h_d\le 2\cdot |h|^d$$ Hence $\lim_{|h|\to0}$$\|\chi_Q-\chi_{Q_h}\|_p=0$, that is to say $\chi_Q\in C_p$ (the class in the problem).
Next it is clear that translates and dilations of $\chi_Q$ are also in $C_p$, and since $C_p$ is closed under linear combinations it follows that $C_p$.

Now suppose $f_n\in C_p$ converges to $f\in L^p$ (in the $L^p$-norm), then $$\left(\int|f(x)-f(x+h)|^pdx\right)^{1/p}=\qquad\qquad\qquad\qquad\qquad\qquad$$ $$\left(\int|f(x)-f_n(x) + f_n(x)-f_n(x+h)+ f_n(x-h) - f(x+h)|^pdx\right)^{1/p} \le$$ $$\qquad\qquad\qquad 2\|f-f_n\| + \left(\int|f_n(x)-f_n(x+h)|^pdx \right)^{1/p} \to0$$ as $n\to\infty$ (due to translation invariance)- in other words $C_p=L^p$

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Nice answer. Thanks. – leo Dec 15 '11 at 19:06
Thanks and good luck! – AD. Dec 15 '11 at 19:32

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