Let $\Omega\subset\mathbb{R}^n$ be a open set and write $\Omega=\cup_{i=1}^\infty Q_i$, where $Q_i$ is a compact set. Define in $L_{loc}^p(\Omega)$ the metric $d(f,g)=\sum_{k=1}^\infty\min\Big(\frac{1}{2^k},\|f-g\|_{L^p(Q_k)}\Big)$.

Let $p\in [1,\infty)$, $F\subset L^p_{loc}(\Omega)$ and define $F_k=\{f\chi_{Q_k}:\ f\in F\}$, where $\chi_{A}$ is the characteristic function of $A$. Suppose that for each $k$, the set $F_k$ is totally bounded in $L^p(\Omega)$. How can one show that $F$ is totally bounded?

My try: Fix $\epsilon>0$ and choose $n$ such that $\sum_{k=n+1}^\infty \frac{1}{2^k}<\frac{\epsilon}{2}$. For each $k=1,...,n$ consider the $\frac{\epsilon}{k+1}$-net $\phi_1^k,...,\phi_{n(k)}^k$ of $F_k$, where $n(k)$ is a natural number depending on $k$.

Consider the set $K$ of all "possible" sequences of the form: $(A_1,A_2,...,A_n)$, where each $A_k$ is a element chosen from the $\frac{\epsilon}{k+1}$-net $\phi_1^k,...,\phi_{n(k)}^k$ (for example $(\phi_2^1,\phi_1^2,...,\phi^n_4)$). Note that $K$ is finite. Enumerate $K=\{U_1,...,U_r\}$ and let's say that $f\in L_{loc}^p(\Omega)$ is in $U_i=(A_1,...,A_n)$, if $\|f\chi_{Q_1}-A_1\|_{Q_1}<\frac{\epsilon}{2}$,...,$\|f\chi_{Q_{n}}-A_n\|_{Q_n}<\frac{\epsilon}{n+1}$ $\Big($ for instance, if $U_i=(\phi_2^1,\phi_1^2,...,\phi^n_4)$, then I am asking that $\|f\chi_{Q_1}-\phi_2^1\|_{Q_1}<\frac{\epsilon}{2}$,...,$\|f\chi_{Q_{n}}-\phi^n_4\|_{Q_n}<\frac{\epsilon}{n+1}\Big)$.

For each $i=1,...,r$ choose a $f_i\in F$, such that $f_i\in U_i$.

We claim that the set $f_i$ is an $\epsilon$-net for $F$. Indeed, if $f\in F$, then $f\in U_i$ for some $i$, hence \begin{eqnarray} d(f,f_i) &=& \sum_{k=1}^\infty\min\Big(\frac{1}{2^k},\|f-f_i\|_{L^p(Q_k)}\Big) \nonumber \\ &\leq& \sum_{i=1}^n\|f-f_i\|_{Q_k}+\frac{\epsilon}{2} \nonumber \\ &<& \epsilon \end{eqnarray}

If you have patience to read my proof, please verify if it is correct.

Thank you for your help.

  • $\begingroup$ Are there any restrictions on $p$? $\endgroup$ – copper.hat Mar 18 '13 at 17:14
  • $\begingroup$ Yes, let me add it, I forgot. $\endgroup$ – Tomás Mar 18 '13 at 17:14
  • $\begingroup$ Maybe this is obvious, but I don't see it: If $F_1,F_2$ are totally bounded, is it true that $F_1 \cup F_2$ is totally bounded? $\endgroup$ – copper.hat Mar 19 '13 at 19:05
  • $\begingroup$ I think yes: Given $\epsilon>0$, Let $F_1\subset\cup U_i$ and $F_2\subset \cup V_i$ where $U_i$ and $V_i$ are sets of diameter less then $\epsilon$. Then, $F_1\cup F_2\subset (\cup U_i)\cup (\cup V_i)$. Is this right @copper.hat ? $\endgroup$ – Tomás Mar 19 '13 at 19:34
  • $\begingroup$ You are correct. However, I didn't express myself correctly. I need to think a bit more about what I meant to ask! $\endgroup$ – copper.hat Mar 19 '13 at 19:41

This is not a complete answer because of an additional assumption I made.

I will assume that the $Q_i$ are increasing. The main import of this assumption is that we have $\|x\|_{L^p(Q_{k})} \le \|x\|_{L^p(Q_{k+1})}$, hence an $\epsilon$-net for $F_k$ is (appropriately adjusted) also an $\epsilon$-net for $F_1,...,F_{k-1}$.

Choose $\epsilon>0$ and choose $n$ such that $\sum_{k>n} \frac{1}{2^k} < \frac{\epsilon}{2}$. Since $F_n$ is totally bounded, there is a finite $\frac{\epsilon}{2n}$-net $\phi_1,...,\phi_p$. Each $\phi_i$ has the form $\phi_i = f_i 1_{Q_n}$ for some $f_i \in F$. Then I claim that $f_1,...,f_p$ is an $\epsilon$-net for $F$ (appropriately adjusted).

Let $f \in F$. Choose $\phi_j$ such that $\|f-\phi_j\|_{L^p(Q_n)} < \frac{\epsilon}{2n}$. Let $f_j$ be the corresponding element of $F$. Then \begin{eqnarray} d(f,f_j) &=& \sum_{k=1}^\infty\min ( \frac{1}{2^k},\|f-f_j\|_{L^p(Q_k)} ) \\ &\le& \sum_{k=1}^n \|f-f_j\|_{L^p(Q_k)} + \sum_{k>n} \frac{1}{2^k} \\ &<& \sum_{k=1}^n \|f-\phi_j\|_{L^p(Q_n)} + \frac{\epsilon}{2} \\ &<& \epsilon \end{eqnarray} Hence $f_1,...,f_p$ is an $\epsilon$-net for $F$.

  • $\begingroup$ I have changed my proof based on yours, maybe now it is readable. If you have some time, please try to read it and verify if it is correct. Thank you @cooper. $\endgroup$ – Tomás Mar 20 '13 at 12:32
  • $\begingroup$ Hi @Tomás; I think there is still a gap in the proof. The issue is how to piece together the elements of $F_k$ into a single function $f_i$. That is why I needed nesting above. $\endgroup$ – copper.hat Mar 20 '13 at 14:50
  • $\begingroup$ Yes that's the problem. This issue is giving me on the head. $\endgroup$ – Tomás Mar 20 '13 at 14:54
  • $\begingroup$ Where did the problem arise from? $\endgroup$ – copper.hat Mar 20 '13 at 14:57
  • $\begingroup$ Take a look in the proof of corollary 8. math.ntnu.no/conservation/2009/037.pdf $\endgroup$ – Tomás Mar 20 '13 at 15:05

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