Let $f_n:I \to \mathbb{R}$ a sequence of functions,that does not get zero at any point.We suppose that $f_n \to f$ uniformly and that $\exists M>0$ such that $|f(x)| \geq M, \forall x \in I$.Then $\frac{1}{f_n} \to \frac{1}{f}$ uniformly. To show that the condition $|f(x)| \geq M, \forall x \in I$ is necessary,how can I find a counterexample,so that I show that the sentence above is not true if the condition does not stand??
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Here's one example: define $f_n:(0,1) \to \mathbb{R}$ by $f_n(x) = \dfrac{n}{n+1}x$.
Cameron Williams
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Ben Grossmann
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How did you find it? – evinda Mar 24 '14 at 16:40
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I edited your post for readability. I hope you don't mind. – Cameron Williams Mar 24 '14 at 16:40
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@CameronWilliams Not at all, thank you. – Ben Grossmann Mar 24 '14 at 16:40
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@evinda I think the logic is to find a sequence for which $1/f$ behaves "badly". This is a very good example of such. – Cameron Williams Mar 24 '14 at 16:41
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@evinda consider what $1/f_n(x)$ looks like near $x = 0$. Functions with vertical asymptotes are one of the "usual suspects" for issues with uniform convergence. – Ben Grossmann Mar 24 '14 at 16:42
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So you mean that $\frac{1}{f_n}$ or $\frac{1}{f}$ should converge to $\infty$ ? – evinda Mar 24 '14 at 16:44
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Yes, but only at the limit point of the domain. The real idea is that if we don't have that lower bound $M$, $\frac{1}{f_n}$ can be unbounded, which (as you can see with the example) is a problem. Notice that $1/f_n$ is defined over the whole domain of $(0,1)$, as is its pointwise limit. – Ben Grossmann Mar 24 '14 at 16:48