Why the sequence of functions $f_n(x)=\frac{1}{nx+1}$ is not uniform convergent in (0,1)? I've already prove the pointwise convergence but I can't justify why this sequence is no uniform convergent.
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1$\begingroup$ math.stackexchange.com/questions/561737/… $\endgroup$– user42761Commented Jun 6, 2017 at 18:27
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3$\begingroup$ Possible duplicate of Uniform convergence of a family of functions on $(0,1)$ $\endgroup$– user42761Commented Jun 6, 2017 at 18:27
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1$\begingroup$ Sorry, you are right. $\endgroup$– user42761Commented Jun 6, 2017 at 18:31
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1$\begingroup$ I don't know that I'd characterize the convergence as "punctual" - it seems rather slow, all things considered! (The English term for this is "pointwise".) $\endgroup$– ChrisCommented Jun 6, 2017 at 19:09
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1$\begingroup$ @Chris Well, slow does not imply not punctual. You just need to get there on time. $\endgroup$– zhw.Commented Jun 7, 2017 at 22:39
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2 Answers
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HINT:
What happens when we take $x=1/n\in (0,1)$?
answered Jun 6, 2017 at 18:27
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$\begingroup$ I'm not specially good with analisys but are you suggesting that can I take a subsequence in the domain such that the familiy doesn't cpnverges to zero? $\endgroup$ Commented Jun 6, 2017 at 18:31
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2$\begingroup$ For negating uniform convergence we need to find an $\epsilon>0$ (here we take $\epsilon=\frac12$) such that for all $N$ there exists an $n>N$ (here take any $n>N$) and an $x\in (0,1)$ (take $x=1/n$) such that $|f_n(x)-f(x)|\ge \epsilon$ (here $\frac{1}{1+nx}=\frac12 \ge \epsilon=\frac12$). $\endgroup$ Commented Jun 6, 2017 at 18:42
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These functions are uniformly convergent on $(\epsilon, 1)$ for any $\epsilon > 0$ but not on the whole interval $(0,1)$.
answered Jun 6, 2017 at 18:38