I am following a course of real analysis and my teacher, while proving the continuity of translation of functions in $L^p$, used the dominated convergence theorem (DCT) in a strange way. I write the first half of proof, for being clear.

Given $f \in L^p(E)$, with $E$ measurable and $1\le p< \infty$, then for every $\varepsilon >0$ exists $\delta=\delta(\varepsilon)$ such that $$ \Vert T_hf-f\Vert_p<\varepsilon \ \ \text{ if}\ \ \vert h \vert < \delta $$ Where the operator $T_hf$ is defined by $$ T_hf(x)= \begin{cases} f(x+h) & \text{if}\ x+h \in E, \\ 0 & \text{if}\ x+h \in \mathbb{R}^N-E \end{cases} $$

Proof.First we may suppose $f$ continuous and compactly supported (and at the end of the proof uses the density of such functions in $L^p$). For all $x \in E$ we have that $$ \vert T_hf(x)- f(x)\vert^p=\vert f(x+h)-f(x) \vert^p \to 0 \ \ \text{if}\ \ \vert h \vert \to 0$$ Now $f$ is continuous and compactly supported, so $\Vert f \Vert_\infty < \infty$, than we can write $$ \vert T_hf(x)- f(x)\vert^p \le \left(\vert f(x+h)\vert+\vert f(x) \vert\right)^p \le 2^p \Vert f \Vert_\infty$$ So, letting $K$ be the compact support of $f$ and $\vert h\vert < \delta$, we have that the function $\vert T_hf-f \vert$ vanishes outside the set $K_\delta=K+B(0,\delta)$ (the ball centered in $0$ with radius $\delta$), that is measurable. So we can dominate $$ \vert T_hf-f \vert \le 2^p \Vert f \Vert_\infty \chi_{K_\delta}$$ Now using the DCT we can say that $\Vert T_hf-f \Vert_p <\varepsilon$ for $|h|<\delta$. For $f \in L^p(E)$ generic we use the density of continuous and compactly supported functions.

The DCT is true for numerable families of functions $\{f_n \}$, but here is used for a non-countable family $\{f_t\}_{t \in I \subseteq \mathbb{R}}$, such $T_hf$. I managed as follow for giving a sense to the proof.

The DCT states that if we have a sequences of measurable function $f_n$ such that $f_n \to f$ pointwise, and all functions $f_n$ are "dominated" by a summable function $g$, i.e $|f_n(x)| \le |g(x)|, \ \forall n, \ \forall x$, so $$\lim_{n \to \infty} \int_{E} f_{n} d \mu= \int_{E} f d \mu$$ Now we have a subset $I$ of real numbers (we suppose for semplicity $I$ as interval), and a family of function $\{f_t\}_{t \in I}$ (dominated by a summable function $g$). Consider the function $$ \begin{aligned} \phi \colon I &\longrightarrow \mathbb{R} \\ t &\longmapsto \int_{E}f_t d \mu \end{aligned} $$ Now suppose we have the property that, fixed $\bar t \in I$, for every sequences $\{t_n\} \subseteq I$ such that $t_n \to \bar t$ as $n$ increase, result that $f_{t_n} \to f_{t}$ poinwise. By DCT we have that $$\lim_{n\to \infty}\phi(t_n) = \phi(\bar t)$$ for every sequence $t_n$ convergent to $\bar t$. So for a well known theorem of topology, we have $$ \lim_{t \to \bar t} \phi(t)=\phi(\bar t) $$

Choosing $\phi(h)=T_hf$ and $\bar t=0$ we have concluded the problem.

Because I don't have practice with $L^p$ space my question is if the above reasoning is right or not. My teacher use that fact easy, without long arguments, so there may be a faster reasoning way?

Thank you a lot.

  • $\begingroup$ Is it not easier to use an $\frac{\epsilon}{3}$ argument? $\endgroup$ – user38404 Dec 19 '12 at 1:12
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    $\begingroup$ @ Chris: judging from the first paragraph, I'd guess $T_h(x) = f(x-h)$. $\endgroup$ – user31714 Dec 19 '12 at 4:29
  • $\begingroup$ @Christipher you are right, indeed I added the definition of $T_hf$ operator. $\endgroup$ – Lorban Dec 19 '12 at 11:12
  • $\begingroup$ @Vivek, Yes the proof ends with a $\varepsilon/3$ argument. $\endgroup$ – Lorban Dec 19 '12 at 11:23
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    $\begingroup$ @Lorban I'm not sure what TCD is. and where exactly is your professor claiming to use DCT? $\endgroup$ – user38404 Dec 19 '12 at 18:49

I think your reasoning was more or less what your teacher had in mind. Actually, was we have to show is that for each sequence $\{h_n\}$ converging to $0$, we have $\lVert \tau_{h_n}f-f\rVert_{L^p}\to 0$. It's enough to reach the conclusion, as if it was not true, there would be an $\varepsilon$ such that if $\delta>0$, we can find $|h|<\delta$ such that $\lVert \tau_{h}f-f\rVert_{L^p}\geqslant \varepsilon$. In particular, taking $\delta=\frac 1n$ for $n$ integer, we find a sequence $\{h_n\}$ converging to $0$ such that $\lVert \tau_{h_n}f-f\rVert_{L^p}\geqslant \varepsilon$, a contradiction.

  • $\begingroup$ Sure, but we need that $\{\tau_{h_n}f \}$ has to converge to $f$ for every sequence $\{h_n\}$ converging to $0$, that is obviously true in this case. In general with a family $\{ f_t\}_{t \in I}$ with $I=[0,1]$ (for example) is used that $f_t$ converges to $f_0$ with continuity respect to $t$, and this is true if for every sequence $t_n$ converging to $0$, we have that $f_{t_n}$ converge to $f_0$, for the same theorem I mentioned above. I hope you understand what I mean. $\endgroup$ – Lorban Apr 28 '13 at 15:26

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