I am having trouble coming up with such an example.
How do I go about starting this?
EDIT:
I have asked the wrong question. Please refer to this question instead:
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asked Feb 26 '17 at 22:00
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$\begingroup$ Hint: try to approximate the function $x \mapsto 1/x$. $\endgroup$ – Nigel Overmars Feb 26 '17 at 22:03
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$\begingroup$ @NigelOvermars I've thought about that one, but the thing is that $f(x)=1/x$ is not bounded. Is there a way around this? $\endgroup$ – user46372819 Feb 26 '17 at 22:11
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$\begingroup$ What about $f_n \equiv (-1)^n $? $\endgroup$ – PhoemueX Feb 26 '17 at 22:20
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$\begingroup$ @PhoemueX What would be the function $f$? $\endgroup$ – user46372819 Feb 27 '17 at 0:05
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Let $f(x) = 0$ on $(0,1)$ and $f_n(x) = (1- 1/n)^n 1_{[1- 1/n, 1)}.$
We have $f_n(x) \to 0 = f(x)$ pointwise but not uniformly since $(1 - 1/n)^n \to e^{-1}.$
With fixed $x \in (0,1)$ there exists $N \in \mathbb{N}$ such that $x < 1 - 1/N$. Thus for any $\epsilon > 0$ we have $|f_n(x ) - f(x)| = |f_n(x)| < \epsilon$ for all $n > N$ trivially since $f_n(x) = 0$ for such $n$.
However, $\lim_{n \to \infty}\sup_{x \in (0,1)}|f_n(x)-f(x)| = \lim_{n \to \infty}(1-1/n)^n = e^{-1} \neq 0.$
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$\begingroup$ How did you go about thinking of such a function? What was the thought process? $\endgroup$ – user46372819 Feb 27 '17 at 0:04
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1$\begingroup$ @Al Jebr: An easier example is $f_n(x) = 1_{[0,1/n]}$. I first noticed you have an open interval. I just needed a sequence that ultimately converges to $0$ at any fixed point but where the maximum of $f_n$ occurs at a point $x_n$ that converges to an endpoint and $f_n (x_n)$ stays bigger than some $\epsilon_0$. This pushes the discontinuity to the endpoint -- so the limit function is continuous on the open interval -- and for any $n$ no matter how big we always have $f_n(x_n) \geqslant \epsilon_0$. $\endgroup$ – RRL Feb 27 '17 at 0:29
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$\begingroup$ Step functions are not continuous, but the standard example of a sequence that fails to converge uniformly on $(0,1)$ is $f_n(x) = x^n.$ On $[0,1]$ it converges to $\begin{cases} 1, \,\,\, x = 1 \\ 0, \,\,\, x \neq 1 \end{cases}$. This is discontinuous so convergence cannot be uniform. However the discontinuity is only at the endpoint. I just modified this idea. $\endgroup$ – RRL Feb 27 '17 at 0:34
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$\begingroup$ Finally, Dini's theorem states that a monotone sequence of continuous functions converging to a continuous function on a compact set is uniformly convergent. To find non-uniformly convergent sequences you have to remove at least one condition: monotone, continuous or compact. $\endgroup$ – RRL Feb 27 '17 at 0:43
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$\begingroup$ I have actually asked the wrong question. Please help me here: math.stackexchange.com/questions/2163242/… $\endgroup$ – user46372819 Feb 27 '17 at 3:37
