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This is going to be a long post, so I'm giving a description first:

I recently came across the following exercise: Let $(X,\mathcal{A},\mu)$ be a measure space. If $(f_n)\subset L^p(\mu)$ and $f\in L^p(\mu)$, $p\geq 1$, such that $\|f_n-f\|_p\leq n^{-c}$ with $c>1/p$, prove that $f_n\to f$ a.e. on $X$. The solution is not a big deal, but leads to an interesting question: which rates are good enough for convergence in norm to imply convergence a.e.? Recall that with no further assumptions, convergence in norm only implies the existence of a subsequence that converges a.e.

Anyway, the following definition is necessary: Define $\text{gr}^p(\mu)$ as the set $$gr^p(\mu):=\{(a_n)\in c_0|\text{ for all } (f_n)\subset L^p(\mu): \big{(}\forall n\in\mathbb{N}: \|f_n\|_p\leq |a_n|\big{)}\implies f_n\to0\text{ a.e.}\}$$

What I would like is to describe this set. My progress is the following:

1) For any measure space and any $p\geq 1$ it is $(0)\in\text{gr}^p(\mu)$, therefore this set is never empty.

2) If $(a_n)\in\text{gr}^p(\mu)$ and $\lambda\in\mathbb{C}$ then $\lambda\cdot(a_n)\in\text{gr}^p(\mu)$.

Indeed, if $\lambda=0$ it is obvious; otherwise if $(f_n)\subset L^p(\mu)$ with $\|f_n\|_p\leq|\lambda a_n|$ for all $n$ we have that $\displaystyle{\|\frac{1}{\lambda}f_n\|_p\leq|a_n|}$ for all $n$ therefore $\frac{1}{\lambda}f_n\to 0$ a.e. which is true iff $f_n\to 0$ a.e.

3) For any measure space, $\ell^p\subset\text{gr}^p(\mu)$.

Let $(a_n)\in\ell^p$ and $(f_n)\subset L^p(\mu)$ with $\|f_n\|_p\leq|a_n|$ for all $n$. We have $\displaystyle{\int_X|f_n|^pd\mu\leq|a_n|^p}$ for all $n$ and by summing and using the Monotone convergence theorem we have that $\displaystyle{\int_X\sum_{n}|f_n|^pd\mu\leq\|(a_n)\|_{\ell^p}<\infty}$, therefore the series $\sum_n|f_n|^p$ converges a.e. hence $|f_n|^p\to0$ a.e. which implies $f_n\to 0$ a.e.

4) In any measure space, if $(a_n)\in\text{gr}^p(\mu)$ and $(a_{n_k})\subset(a_n)$, we have $(a_{n_k})\in\text{gr}^p(\mu)$.

Let $(a_{n_k})\subset(a_n)\in\text{gr}^p(\mu)$ and $(f_k)\subset L^p(\mu)$ s.t. for all $k$ it is $\|f_k\|_p\leq |a_{n_k}|$; Define $g_n$ as $0$ if $n\not\in\{n_k: k\in\mathbb{N}\}$ and $g_{n_k}=f_k$ for all $k$. Then $\|g_n\|_p\leq |a_n|$ for all $n$, hence $g_n\to 0$ a.e. which of course implies $f_k\to0$ a.e.

5) $\text{gr}^p(\mu)$ is a linear subspace of $c_0$.

We need only to prove that it is closed under addition. Let $(a_n),(b_n)\in\text{gr}^p(\mu)$ and $(f_n)\subset L^p(\mu)$ with $\|f_n\|_p\leq|a_n+b_n|$ for all $n$. We partition $\mathbb{N}$ in $S=\{n: a_n=0\}$ and its complement $\mathbb{N}-S$.

Case 1: $S$ is an infinite set. By 4), $(b_n)_{n\in S}\in\text{gr}^p(\mu)$ and $\|f_n\|_p\leq|b_n|$ for all $n$ in $S$; therefore the subsequence $(f_n)_{n\in S}$ converges a.e. to $0$. Now for $n\not\in S$ we can find $\lambda_n\in\mathbb{C}$ such that $b_n=\lambda_n\cdot a_n$. We have to deal with two sub-cases:

Sub-case 1: There exists $M>0$ s.t. for all $n\in\mathbb{N}-S$ it is $|\lambda_n|\leq M$.

In this sub-case, for $n\not\in S$ we have $\|f_n\|_p\leq |a_n|+|b_n|\leq (1+M)|b_n|$. But $(b_n)_{n\in\mathbb{N}-S}\in\text{gr}^p(\mu)$ by 4), and by 2) we have $((1+M)b_n)_{n\in\mathbb{N}-S}\in\text{gr}^p(\mu)$. Hence $(f_n)_{n\in\mathbb{N}-S}$ converges to $0$ a.e.

Sub-case 2: $|\lambda_n|\to\infty$ as $n\to\infty$ through $\mathbb{N}-S$ (note that if $\mathbb{N}-S$ is finite we are automatically in sub-case 1).

We can find $n_0\in\mathbb{N}$ such that for all $n\geq n_0$ and $n\not\in S$ it is $|\lambda_n|>1$. For those $n$ it is $a_n=\frac{1}{\lambda_n}b_n$ therefore $\|f_n\|_p\leq|1+1/\lambda_n|\cdot|b_n|\leq2|b_n|$; now since $(b_n)_{n\geq n_0, n\in\mathbb{N}-S}\in\text{gr}^p(\mu)$ it is $(f_n)_{n\geq n_0, n\in\mathbb{N}-S}\to0$ a.e. and we are done.

Case 2: S is finite; we can do exactly what we did in the two sub-cases above for $\mathbb{N}-S$ and we are done.

Anyway, my questions to the community are these:

  1. Are these spaces any interesting in your opinion?

  2. What would be a good norm for these spaces? I can't think of anything that is of interest.

In this post I prove that for a series of Dirac point-mass measures the space $\text{gr}^p(\mu)$ is the entire $c_0$ for all $p$ and that for the measure space $(\mathbb{R}^d, \mathcal{L}^d, \lambda_d)$ the space $\text{gr}^p(\mu)$ is only $\ell^p$.

EDIT: An easy argument that shows that $\text{gr}^p(\mu)$ is not closed under the supremum norm, unless $\text{gr}^p(\mu)=c_0$: obviously $c_{00}\subset\text{gr}^p(\mu)$, where $c_{00}$ denotes the subspace of sequences that are eventually $0$. Thus $\text{gr}^p(\mu)$ is dense in $c_0$ with the supremum norm.

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  • $\begingroup$ PS: $gr^p$ is short for "p-good rates", i didnt know what to call this $\endgroup$ Nov 20, 2018 at 21:49
  • $\begingroup$ You have a good norm on $c_0$, so your subspace inherits this norm. Have you already established if your subspace is closed or not? $\endgroup$
    – J. De Ro
    Jun 20, 2020 at 23:01
  • $\begingroup$ @ε-δ Check out my edit on why the space is not closed! $\endgroup$ Jul 6, 2020 at 8:53

1 Answer 1

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Per your profile, this is so you can sleep at night. Your space is exactly $\ell^p$. Here is a proof:

Suppose that $\|a\|_p<\infty$ and $\|f_n\|_p\le |a_n|$ for each $n$. Then $\|f_n^p\|_1\le |a_n|^p$ for each $n$, so that $\|\sum |f_n|^p\|_1<\infty$. In particular, $\sum_n |f_n|^p$ is finite almost everywhere, so that $|f_n|^p$ converges to 0 almost everywhere and hence $f_n$ converges to 0 almost everywhere.

For the converse, suppose that $\|a\|_p=\infty$. Then we construct a sequence of functions $f_n$ on $[0,1]$ with $\|f_n\|_p\le a_n$ so that $f_n(x)$ does not converge for every $x\in [0,1]$. Let $b_n=\min(|a_n|^p,\frac 12)$ and notice that if $f_n$ is the indictator function of an interval of length $b_n$, then $\|f_n\|_p=b_n^{1/p}\le |a_n|$. By assumption, $\sum b_n=\infty$.

Now we just use the standard construction of a sequence of functions that converges to zero in $L^1$, but not pointwise. Let $t_0=0$ and $t_{n}=(t_{n-1}+b_n)\bmod 1$ for each $n$. Then let $f_n=\mathbf 1_{[t_{n-1},t_n]}$ (where if $t_n<t_{n-1}$, this means $\mathbf 1_{[t_{n-1},1]} + \mathbf 1_{[0,t_n]}$. Now it is known that $\limsup f_n(x)=1$ for all $x$ and $\liminf f_n(x)=0$ for all $x$. (This has been described to me as the "Goodyear blimp": the support of the function continually moves to the right, each one starting where the previous one finished. The blimp passes overhead infinitely often because the $b_n$ are not summable).

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  • $\begingroup$ You are only proving that the space is equal to $\ell^p$ for the space of the real line equipped with the lebesgue measure. This is true, I have proved this for higher dimensions too as you can see in my examples. However the construction is for any measure space. I have also proved as you can see in my examples that if the measure is a point mass then the space is $c_0$. $\endgroup$ Jan 23, 2021 at 8:25
  • $\begingroup$ OK - right so you need to distinguish between atomic spaces; and spaces that have sets of arbitrarily small measure $\endgroup$ Jan 23, 2021 at 8:26
  • $\begingroup$ so you claim that the space is c_0 for atomic spaces and ell^p for non-atomic spaces? $\endgroup$ Jan 23, 2021 at 8:29
  • $\begingroup$ Right. I think this argument shows that the space is precisely $c_0$ if the space is a union of atoms; and $\ell^p$ if the space has anything else. (I may be thinking of slightly nice measure spaces here like Polish spaces) $\endgroup$ Jan 23, 2021 at 8:29
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    $\begingroup$ So if there are no atoms and the space has some subsets of finite measure, let's say the measure is at least 1. You can (I think) pick out two disjoint subsets of measure 1/2; then you can split each of those into disjoint subsets of measure 1/4 etc. Now for any dyadic interval in [0,1], it corresponds to a union of your subsets and it has the same measure as the Lebesgue measure of the interval. Now you can extend this to all intervals by taking limits.(or just work with dyadic intervals, which are rich enough to run the above argument). $\endgroup$ Jan 23, 2021 at 8:36

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