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I'm interested in the subset of functions \begin{equation} A := \{f:[0,1]\rightarrow [0,1] | \text{$f$ is non-decreasing}\} \subset L^1[0,1]. \end{equation} Do we know if $A$ is compact? Since $L^1[0,1]$ is a Banach space, I tried to use the result in one of answers in Examples of compact sets that are infinite dimensional and not bounded, namely: $A$ is compact iff,

(1) $A$ is closed and bounded.

(2) For each $\varepsilon > 0$ there exists a finite dimensional subset $F \subset L^1[0,1]$ such that $d(f, F) < \varepsilon$ for all $f \in A$.

I think (2) should hold since any non-decreasing functions are Riemann integrable and so I can approximate any $f \in A$ by points in a space of rectangular functions (essentially Riemann sum approximation of the Riemann integral) to any $\varepsilon$ accuracy.

For (1) I think it is clear that $A$ is bounded (in a unit ball). But do we know if $A$ is closed? In other words, if we have a sequence $\{f_i\} \in A$ such that $f_i\rightarrow_{L^1} f$, do we know if $f \in A$?

I think I can see this if $L^1$ convergence on $A$ implies point-wise convergence, but I'm also not sure if this is true.

user113988
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  • $L^1$ convergence indeed implies almost everywhere convergence. – LL 3.14 Feb 19 '23 at 17:56
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    That's actually what I'm wondering too. It doesn't seem true in general though, even on compact subset: https://math.stackexchange.com/questions/138043/does-convergence-in-lp-imply-convergence-almost-everywhere. But could you suggest a proof for non-decreasing functions? – user113988 Feb 19 '23 at 17:59
  • Hmm indeed, you are right, this is only up to a subsequence – LL 3.14 Feb 19 '23 at 18:05
  • This might be helpful: https://en.wikipedia.org/wiki/Helly%27s_selection_theorem – PhoemueX Feb 19 '23 at 20:56

1 Answers1

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It is true that $A$ is closed, if you include "almost everywhere" in its definition (otherwise, it doesn't even make sense as a subset of $L^1$). Because in that case if $f$ is in the closure of $A$, then $f$ is a limit of non-decreasing functions. As $L^1$-convergence implies almost everywhere pointwise convergence of a subsequence, $f$ is a pointwise almost everywhere limit of non-decreasing functions, and hence almost everywhere non-decreasing.

Also, as defined $A$ cannot be compact because it is not bounded. But even if you consider a "bounded version" like $$ A=\{f\in L^1[0,1]:\ \|f\|_1\leq1\ \text{ and } f\ \text{ non-decreasin a.e.}\}, $$ it is still not compact.

the set of non-decreasing functions, it cannot be compact.

Let $$ f_n=2^n\,1_{\big[1-2^{-n},1]}. $$ Then, for each $n$, $f_n$ is non-decreasing and $\|f_n\|_1=1$. And if we take $m>n$, then \begin{align} \|f_m-f_n\|_1 &=\int_0^1|2^m\,1_{[1-2^{-m},1]}-2^n\,1_{[1-2^{-n},1]}|\\[0.3cm] &=\int_0^1|2^n\,1_{[1-2^{-n},1-2^{-m}]}+(2^m-2^n)\,1_{[1-2^{-m},1]}|\\[0.3cm] &=2^n(2^{-n}-2^{-m})+(2^m-2^n)\,2^{-m}\\[0.3cm] &=2-2^{n-m+1}\geq1 \end{align} This shows that $\{f_n\}$ does not admit a convergent subsequence, and hence $A$ is not compact.

Martin Argerami
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    Thank you for your answer. In my definition, however, I specified that $f : [0,1]\rightarrow [0,1]$. I think in your example $f_n = 2^n \notin [0,1]$ near $x = 1$. – user113988 Feb 20 '23 at 02:34
  • So I should change definition of my subset to $A = {f:[0,1]\rightarrow [0,1] | \text{$f$ is non-decreasing a.e.}}$? Where $f$ is non-decreasing a.e. means where exists $\tilde{f}$ function which is non-decreasing and $f=\tilde{f}$ a.e.? Is that necessary since everything in $L^1$ are quotiented by a.e. relation anyway? – user113988 Feb 20 '23 at 03:11
  • You are right, I totally misread the definition of your $A$. As for "a.e.", if two functions differ up to a nullset they are undistinguishable in $L^1[0,1]$, so if you give a definition that may apply to one function $f$ but does not apply to $g$ if $g=f$ up to a nullset, they you are not working in $L^1$. – Martin Argerami Feb 20 '23 at 03:51
  • Could you show me where to find the theorem/proof of the fact that $L^1$ convergence implies pointwise convergence a.e. subsequence? Do you have any comments on showing compactness by arguing that there exists finite dim subspace distance at most $\varepsilon$ (for any $\varepsilon > 0$) from $A$, since we can approximate $\int f$ with Riemann sum? – user113988 Feb 20 '23 at 05:00
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    I can check books tomorrow. But if you look for the usual proof that $L^1$ is complete, you'll see that the way the proof works is by starting with a Cauchy sequence ${f_n}$ and then choosing a subsequence ${f_{n_k}}$ such that $\sum_k|f_{n_{k+1}}-f_{n_k}|$ converges pointwise a.e. The limit obtained is the $L^1$-limit of the subsequence, which has to agree with the original limit of the sequence. In the end, the subsequence ${f_{n_k}}$ converges pointwise a.e. to $f$. – Martin Argerami Feb 20 '23 at 05:17
  • I think I have found a reference. Thank you very much again for guidences. – user113988 Feb 20 '23 at 07:24