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Say we have a sequence of functions $(f_n):(0,1)\rightarrow \mathbb{R}$ which convergence point-wise to a function $f$, and converge uniformly to $f$ on every compact sub-interval of $(0,1)$. Is there anything about the sequence of functions $(f_n)$ that is not preserved in the limit that would be if $(f_n)$ converged uniformly to $f$ on all of $(0,1)$?

Edit: I should clarify, I'm also interested in if the limit can be 'passed' through the integral (assuming it exists) $\int f_n$, incase that wasn't clear.

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Maybe think about the functions $f_n(x) = x^{-1/n}$. The sequence $f_n(x)$ converges uniformly to $1$ on every compact subinterval of $(0,1)$, thus the limit clearly doesn't see the behavior of $f_n$ near $0$. –  treble Mar 6 '12 at 3:54
    
Just to be clear: are you assuming the functions are integrable on $(0,1)$? --- Okay, I see. Perhaps not. –  Patrick Mar 6 '12 at 4:49
    
Well, you can, what I'm interested in is if you do assume they are, does the limit pass through the integral? –  Zaubertrank Mar 6 '12 at 4:51

2 Answers 2

Consider the function $f_n$ whose graph coincides with the $x$-axis on $[0,1-{1/ n }] $. On the interval $[1-1/n,1]$ the graph is comprised of the straight line segments from $[1-{1 / n },0]$ to $[1-{1 /(2 n) },2n]$ and from $[1-{1 /(2 n) },2n]$ to $[1,0]$ (over $[1-{1 / n },1]$, the graph is a "triangular spike" of height $2n$).

Then the sequence $(f_n)$ converges pointwise to $f=0$ on $[0,1]$ and converges uniformly on $[0,1-\epsilon]$ for each $1>\epsilon>0$. But, $\int_0^1 f_n=1$ for each $n$ while $\int_0^1 f=0$.

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Ah yes you are correct, although I'm suspicious that you couldn't well define an antiderivative in this case. That is to say one that would allow you to take n to infinity without the output at a certain point going to infinity as well. –  Zaubertrank Mar 6 '12 at 4:16
    
The integrals certainly exist. $\int_0^1 f_n(x)\,dx$ is simply the area of a triangle of base length $1/n$ and height $2n$ (draw the graph). By the way I changed the example slightly so that it's a bit simpler. –  David Mitra Mar 6 '12 at 4:31
    
Yes the values of the integrals $\int_0^1f_n(x)dx$ form the constant sequence 1,1,1,... and thus converge to 1, but the antiderivatives do not converge to a well defined function. –  Zaubertrank Mar 6 '12 at 4:36
    
@Zaubertrank Maybe I misunderstood your question. In my example, if the $f_n$ converged uniformly on $[0,1]$, then $\int_0^1 f_n$ would would converge to $\int_0^1 f$. So, here, the limit can't be passed through definite integrals. –  David Mitra Mar 6 '12 at 4:47
    
Actually I misspoke, I was concerned with if the limit of your f_n was well defined, not the limit of their antiderivatives, but now I see I was wrong in either case since in your limit the tip of your triangle drops to zero and does not go to infinity, mea culpa. –  Zaubertrank Mar 6 '12 at 5:00

A similar answer, based on what David said: consider the sequence of functions $(f_n):[0,1]\rightarrow\mathbb{R}$ defined as: $$f_n(x)=n^2x(1-x)^n$$ It's easy to check $f_n$ converges pointwise to $f=0$, but: $$1=\lim_{n\to\infty}\int_0^1 f_n(x)\,\mathrm dx\neq\int_0^1 \left(\lim_{n\to\infty}f_n(x)\right)\,\mathrm dx=0$$

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