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For now, say that we have a sequence of real-valued $L^2$ functions $(f_n)$, defined on $\mathbb{R},$ converging to $f\in L^2$ pointwise and the sequence is pointwise bounded by an $L^2$ function. Consider the following limit:

$$\lim_{n\rightarrow\infty}\sum\limits_{m=0}^\infty\left|\left(\int_{K_m} |f_n(x)|^2\, dx\right)^{1/2}\right|,$$ where $\{K_m\}$ is just some collection of compact sets with union $\mathbb{R}$. Under what conditions can we move the limit inside both the sum and the integral? I'm sure it's intimately related to dominated convergence, but I'd appreciate it if someone could write it explicitly for me.

I currently have it set up so that I can swap the limit and the integral, so I just have to justify swapping the limit and the sum. Do I just do something like define a sequence $$g_m(x)=\left(\int_{K_m} |f_n(x)|^2\, dx\right)^{1/2},$$ then require that it converges pointwise and is bounded a.e. by an $\ell^1$ function?

The conditions on $(f_n)$ and $f$ are pretty loose, so we can take them to be fairly nice, if needed (Schwartz, for example).

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  • $\begingroup$ what is the domain of the functions $f_n,f$? What are $K_m$? $\endgroup$ Jul 14, 2019 at 18:23
  • $\begingroup$ Let's just say that $f_n,f:\mathbb{R}\rightarrow\mathbb{R},$ for each $n$, and we can take the $K_m$ to be just some collection of compact sets with union $\mathbb{R}.$ I'll edit that in! $\endgroup$
    – user269711
    Jul 14, 2019 at 18:28
  • $\begingroup$ gah.. you added "with union $\mathbb{R}$" too late. lemme know if my answer is good enough though. $\endgroup$ Jul 14, 2019 at 18:31

1 Answer 1

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The interchange need not be justified. Consider domain $(1,\infty) \subseteq \mathbb{R}$, $K_m = [m,m+1], f_n(x) = \frac{1}{nx}, f = 0$. Then $f_n \in L^2$ for each $n$, $f \in L^2$, $f_n \to f$ pointwise, and the $f_n$'s are pointwise bounded by $\frac{1}{x} \in L^2$. Then $\int_{K_m} |f_n(x)|^2dx \approx \frac{1}{n^2m^2}$, so $\sum_{m=1}^\infty (\int_{K_m} |f_n(x)|^2dx)^{1/2} \approx \sum_{m=1}^\infty \frac{1}{nm} = +\infty$, so $\lim_n \sum_m \cdot = +\infty$, whereas $\sum_m \lim_n \cdot \approx \sum_m \lim_n \frac{1}{nm} = \sum_m 0 = 0$.

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Remark: If we instead had $\lim_n \sum_m \int_{K_m} |f_n(x)|^2dx$, then we could interchange the limit and the sum. Indeed, if each $f_n$ is pointwise dominated by some $g \in L^2$, then for each $n$, $\int_{K_m} |f_n(x)|^2dx \le \int_{K_m} |g(x)|^2 dx$, so, for each $n$, the function $m \mapsto \int_{K_m} |f_n(x)|^2dx$ is bounded by $\int_{K_m} |g(x)|^2dx$, which is an $L^1$ function in the relevant sense: $\sum_m \int_{K_m} |g(x)|^2dx = \int_{\mathbb{R}} |g(x)|^2dx < \infty$, so dominated convergence theorem allows one to interchange $\lim_n$ and $\int_m = \sum_m$.

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  • $\begingroup$ Very nice! Do you think that if we took the domain to be the whole real line and the $K_m$'s to be a nicer decomposition, like a dyadic one, then we could get a positive result? You can assume that functions are Schwartz class, if you like. $\endgroup$
    – user269711
    Jul 14, 2019 at 18:57
  • $\begingroup$ @user269711 I'm stupid. I have no idea why I had the $\log$ there. Now the union is all of $(1,\infty)$. It's then of course easy to make the union all of $\mathbb{R}$ by modifying some things. So now the question you posted is answers. As for your comment, I think $\{[m,m+1] : m \ge 1\}$ is as nice as it gets. $\endgroup$ Jul 14, 2019 at 19:13
  • $\begingroup$ @user269711 also, see the added remark. $\endgroup$ Jul 14, 2019 at 19:17
  • $\begingroup$ Thank you very much! $\endgroup$
    – user269711
    Jul 14, 2019 at 19:27

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