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Suppose $1\leq p < \infty$, and that $\mathbb{R}^d$ is equipped with lebesgue measurea. Show that if $f\in L^p (\mathbb{R}^d)$, then

$||f(x+h)-f(x)||_{L^p}\to 0$ as $|h|\to 0$.

I want to use dominated convergence theorem. First note that the question is equivalent to $||f(x+1/n)-f(x)||_{L^p}\to 0$ as $n\to \infty$ and $n\to -\infty$.

But we see that $||f(x+1/n)-f(x)||_{L^p}\leq (\int |f(x+1/n)|^p+|f(x)|^p)^{1/p}=(\int (2|f(x)|^p)^{1/p}<\infty$. So we can use dominated convergence theorem to conclude. But the hint provided in this problem suggests that this argument is naive and wrong. What did I do wrong?

My guess is that since $f$ is not continuous we cannot conclude $f(x+h)-f(x)\to 0$. Is this the only part that I missed?

jk001
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    You can do a little search in MSE and find that there are many (good) postings dealing with this problem. – Mittens Aug 06 '21 at 22:42
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    First, that the family ${f(\cdot+h)-f(\cdot)}_{|h|<\delta}$ is bounded in $L^p$ does not necessarily imply that the DCT is applicable. Indeed, there are examples in which the DCT cannot be applied. For the proof, there are many good postings available on this MSE community, so you would definitely want to give them a look. But to illustrate a idea, first consider the case where $f$ is nice enough (such as continuous and compactly supported) and then use the density argument. – Sangchul Lee Aug 06 '21 at 22:44
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    Here is another one. In any case the idea is to approximate your functions by continuous functions of compact support and use the uniform continuity of such functions. – Mittens Aug 06 '21 at 22:44
  • @Sangchul Lee It's surprising that DCT does not work and I'm trying to find the reason, although I know that there are other solutions.. The only thing I can think is that $f$ is not continuous. Could you please help me with why I cannot use DCT by just simply changing $\delta$ to $1/n$? The reason I asked this question was to make sure where I did do wrong, not necessarily to find solutions. But many thanks – jk001 Aug 07 '21 at 00:07
  • For example, consider the integrable function $f$ defined by $$f(x)=\frac{1}{\sqrt{x}}\mathbf{1}{(0,1)}(x).$$ Then any function $\phi$ that dominates $|f(x+\frac{1}{n})-f(x)|$ for all $n$ must satisfy $$ \frac{1}{\sqrt{x+\frac{1}{n}}} \leq \phi(x) \quad \text{for} \quad -\frac{1}{n} < x < -\frac{1}{n+1}, $$ and so, $$\int{-1}^{0} \phi(x) , \mathrm{d}x \geq \sum_{n=1}^{\infty} \int_{-\frac{1}{n}}^{-\frac{1}{n+1}} \frac{1}{\sqrt{x+\frac{1}{n}}} , \mathrm{d}x = \sum_{n=1}^{\infty} \frac{2}{\sqrt{n^2+n}} = \infty. $$ – Sangchul Lee Aug 07 '21 at 05:01
  • @Sangchul Lee I'm looking for the example where $|f(x+1/n)-f(x)|^p$ is bounded by some integrable function but $f(x+1/n)-f(x)\not\to 0$ in $L^p$ sense as $n\to\infty$, but I can't see how your example counts. Could you explain a bit? – jk001 Aug 07 '21 at 09:14
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    @jk001, Well, that is not what you originally asked to me. My interpretation is that you asked me why DCT may not work, and I provided an example for that. More precisely, my example shows that DCT can fail to apply because you can't find an integrable function that dominates ${|f(x+\frac{1}{n})-f(x)|^p}_{n=1}^{\infty}$. Also, note that if $f$ is in $L^p$ then you always have $f(x+\frac{1}{n})-f(x)\to0$ in $L^p$, which means that you can't find such an example for your last comment. – Sangchul Lee Aug 07 '21 at 09:41
  • Sorry for bad the wording. $||f(x+h)-f(x)||{L^p}\to 0$ as $|h|\to 0$ isn't really equivalent to $||f(x+1/n)-f(x)||{L^p}\to 0$ as $n\to\infty$ and $n\to-\infty$, is it? Or do I miss something?? That's quite surprising! – jk001 Aug 07 '21 at 18:11
  • Also, even if $f\in L^p$ and even if we can use DCT, we cannot conclude $f(x+\frac{1}{n})-f(x)$ in $L^p$ because $f(x+\frac{1}{n})-f(x)$ may not converge to $0$ point-wise since $f$ is not guaranteed to be continuous – jk001 Aug 07 '21 at 18:15

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