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The following Exercise came from the lecture notes "Modern Real Analysis" by Ziemer, W. (Problem 6.34, pp. 213):

If $f\in L^p(\mathbb{R}^n), 1\leq p < \infty$, then prove $$\lim_{|h|\to 0}\| f(x+h)-f(x)\|_p = 0.$$ Also show that this result fails when $p=\infty$.

I don't know if the measure considered in the Lebesgue measure or this statement is valid in general measures for $\mathbb{R}^n$. I couldn't even start. Any hint will be really appreciated.

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    http://math.stackexchange.com/questions/842937/show-that-lim-r-to-0-t-rf%e2%88%92f-l-p-0?rq=1 – 2017-02-03
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    @mathematician So it works only for the Lebesgue measure? – 2017-02-03
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    Note that smooth compactly supported functions are dense in $L^p$ if $p \in [1, \infty)$ but not for $p=\infty$. – 2017-02-04
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    $L^p$ spaces are usually defined with Lebesgue measure. The theorem certainly works for some other measures. I'm not sure off the top of my head exactly which measures satisfy the theorem. – 2017-02-04

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First suppose that $f$ is a continuous function with compact support. If $h < 1$ then $|f(x) - f(x+h)|^p \le 2^p ||f||_{\infty}^p 1_{B(0,R)}$ for sufficiently large $R$. Since $f$ is continuous and has compact support, $||f||_{\infty} < \infty$. Since $|f(x+h) - f(x)| \to 0$ as $h \to 0$, the result holds by the dominated convergence theorem.

For general $f$, choose a sequence $f_k$ of continuous, compactly supported functions so that $f_k \to f$ in $L^p$.

For convenience, write $\tau_h(g)(x) = g(x +h)$

So, $||\tau_hf - f||_p = ||\tau_hf - \tau_hf_k + \tau_hf_k - f_k + f_k - f||_p \le ||\tau_hf - \tau_h f_k||_p + ||\tau_h f_k - f_k||_p +||f_k - f||_p$

It's easy to check that $||\tau_h f - \tau_h f_k||_p = ||f - f_k||_p$. Hence the first and third term tend to $0$ independent of $h$ as $k \to \infty$. Since each $f_k$ is continuous and compactly supported, the second term tends to $0$ as $h \to 0$.

When $p = \infty$ consider the function $f := 1_{\{0\}}$. Then $f(x+h) - f(x) = 1$ when $h = -x$, $-1$, when $x = 0$, and $0$ otherwise. So $||\tau_h f - f||_{\infty} = 1$ for all $h$.

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    I think your counterexample is incorrect. The function $f$ vanish almost everywhere... Instead, take $f=\chi_{(-1,1)}$ in $\mathbb{R}$. Then, given $0<\delta<1$, take $h=\delta/2$ and conclude that $||\tau_h f - f||_{\infty} = 1$. – 2017-02-15