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 $g_j(x) \rightarrow h(x), j\rightarrow\infty$ pointwisely,and if $ f_{ij}(x)\rightarrow g_j(x), i\rightarrow \infty, \forall j$ pointwisely, can we find a subsequence $f_{ij_{(i)}}$ from $f_{ij}$ such that $f_{ij_{(i)}}\rightarrow h, i\rightarrow\infty$ pointwisely?

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

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.

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 $\{f_{ij_{(i)}}\}$ from $\{f_{ij}\}$ 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 took into account.

Now my question is: what if the convergence is pointwise? Can we still do that, namely, if $g_j(x) \rightarrow h(x), j\rightarrow\infty$ pointwisely,and if $ f_{ij}(x)\rightarrow g_j(x), i\rightarrow \infty, \forall j$ pointwisely, can we find a subsequence from sequence $f_{ij_{(i)}}$ from $f_{ij}$ such that $f_{ij_{(i)}}\rightarrow h, i\rightarrow\infty$ pointwisely.

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

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 $g_j(x) \rightarrow h(x), j\rightarrow\infty$ pointwisely,and if $ f_{ij}(x)\rightarrow g_j(x), i\rightarrow \infty, \forall j$ pointwisely, can we find a subsequence $f_{ij_{(i)}}$ from $f_{ij}$ such that $f_{ij_{(i)}}\rightarrow h, i\rightarrow\infty$ pointwisely?

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

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 $\{f_{ij_{(i)}}\}$ from $\{f_{ij}\}$ 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(lazy actually), so that more examples or counterexamples can be took into account.

Now my question is: what if the convergence is pointwise? Can we still do that, namely, find a subsequence from sequence $f_{ij_{(i)}}$ from $f_{ij}$ such that $f_{ij_{(i)}}\rightarrow h, i\rightarrow\infty$.

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

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 $\{f_{ij_{(i)}}\}$ from $\{f_{ij}\}$ 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 took into account.

Now my question is: what if the convergence is pointwise? Can we still do that, namely, if $g_j(x) \rightarrow h(x), j\rightarrow\infty$ pointwisely,and if $ f_{ij}(x)\rightarrow g_j(x), i\rightarrow \infty, \forall j$ pointwisely, can we find a subsequence from sequence $f_{ij_{(i)}}$ from $f_{ij}$ such that $f_{ij_{(i)}}\rightarrow h, i\rightarrow\infty$ pointwisely.

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