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If we have in certain norm

1). $g_j(x) \rightarrow h(x), j\rightarrow\infty$ and

2). $ f_{ij}(x)\rightarrow g_j(x), i\rightarrow \infty, \forall j$ ,

then we can choose a subsequence $\{f_{ij_{(i)}}, i=1,2,...\}$ from $\{f_{ij}, i=1,2,...;j=1,2,...\}$ such that $f_{ij_{(i)}}\rightarrow h, i\rightarrow\infty$, in that norm sense. And it's just a trivial matter of triangle inequality. (I leave some freedom for function definition, so that more examples or counterexamples can be taken into account.)

Now my question is:

What if the convergence is in pointwise sense? Can we still do that? namely, if $g_j(x) \rightarrow h(x), j\rightarrow\infty$ pointwise,and if $ f_{ij}(x)\rightarrow g_j(x), i\rightarrow \infty, \forall j$ pointwise, can we find a subsequence $f_{ij_{(i)}}$ from $f_{ij}$ such that $f_{ij_{(i)}}\rightarrow h, i\rightarrow\infty$ pointwise?

I constructed a counterexample which seems to be complicated. I wonder if there is any easy answer. Thanks.

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closed as off-topic by Bill Johnson, Andrey Rekalo, Willie Wong, Todd Trimble, Ramiro de la Vega Jul 17 '13 at 11:35

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I love the word pointwisely :-) –  Mariano Suárez-Alvarez Jul 17 '13 at 2:36
2  
It's better than pointfoolishly –  Robert Israel Jul 17 '13 at 2:37
    
I cut your love. ;-) –  pde_bk Jul 17 '13 at 3:07

1 Answer 1

up vote 3 down vote accepted

Let $h$ be the indicator function of the rationals. This is of Baire class 2, i.e. it is not the pointwise limit of a sequence of continuous functions, but it is the pointwise limit of a sequence of pointwise limits of continuous functions. For example, let $r_1, r_2, \ldots$ be an enumeration of the rationals, $g_n(x)$ the indicator function of $S_n = \{r_1, r_2, \ldots, r_n\}$, and $f_{ij}(x) = \exp(-i \;\text{dist}(x, S_j))$.

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It's a wonderful example, thanks a lot. :-) –  pde_bk Jul 17 '13 at 3:38

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