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$$\lim_{n\to\infty} \lim_{N\to\infty} a(N,n)$$

the reason I want to know is

I have a sequence of functions $\{f_n\}$ which converges uniformly to a function $f$

and I want to say that

$$\lim_{n\to\infty} \lim_{N\to\infty} \sum_{k=1}^N \frac{f_k(x_n)}N = \lim_{N\to\infty} \lim_{n\to\infty} \sum_{k=1}^N \frac{f_k(x_n)}N $$

here my $a(N,n)$ is that sum ($\sum_{k=1}^N f_k(x_n)/N$)

and somehow I want to use that $f_n$ converges uniformly to do this


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put on hold as unclear what you're asking by Jonas Meyer, avid19, Claude Leibovici, Yes, Tom-Tom 2 days ago

Please clarify your specific problem or add additional details to highlight exactly what you need. As it's currently written, it’s hard to tell exactly what you're asking. See the How to Ask page for help clarifying this question. If this question can be reworded to fit the rules in the help center, please edit the question.

(1)As written, your terms don't depend on $N$ at all, and both limits are f(x). Do you mean something else? If so, please edit to clarify. (2) If you are only considering limits at a particular value of $x$, then it is irrelevant whether the convergence is uniform. –  Jonas Meyer Oct 23 '10 at 6:16
After your edit: you do not have the same sequence on each side of the equation. And is $(x_n)$ a sequence converging to $x$, or what? –  Jonas Meyer Oct 23 '10 at 6:28
I still cannot fathom your question, alas. –  Robin Chapman Oct 23 '10 at 10:45
In the 2nd equation on the RHS you have originally written $f_n(x)$. I think you mean $f_k(x_n)$ — is it so? –  kennytm Oct 23 '10 at 13:58
This general theorem may be of use. –  joriki Oct 4 '11 at 16:47

3 Answers 3

If the total sum converges absolutely, you can reorder the terms however you want. Can you prove that?

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I'm not seeing how this is relevant to the question. Could you please be more precise? –  Jonas Meyer Aug 11 '14 at 16:00

Is the limit of the $f_j$'s continuous? If so I think your statement is true. Let $g_N(x) = 1/N \ \sum f_j(x)$. Now if $f_j \to f$, then $g_N \to f$ as well (if each $f_j$ is within an epsilon of $f$, then $g_N$ is within an epsilon of $f$ as well; go far enough out in the sequence to make this happen).

Now say $x_n \to x$. Then the RHS is $$ \lim_{N \to \infty} g_N (x) = f(x)$$, and the LHS is $$ \lim_{n \to \infty} f(x_n) = f(x)$$ if $f$ is continuous.

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If they commute with each other we can interchange the limits........

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