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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

sorry.edited

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(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
1  
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 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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