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I know that if function is uniformly convergent ($ |f_n(x)-f(x)|<\epsilon. \forall n > N(\epsilon)$), it is Cauchy convergent ($ |f_n(x)-f_m(x)|<\epsilon. \forall n,m > N(\epsilon)$)

So my question is: if sequence is Cauchy convergent does this imply uniform convergence? I think the answer is no, but can't figure out example.

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    $\begingroup$ Yes because The set of reals is complete. $\endgroup$ Jun 27, 2020 at 14:25

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Yes it does; suppose $\{f_n\}$ is a series of functions on a subset $E$ of $\mathbb{R}$ such that for all $\epsilon > 0$ there exists a $N(\epsilon)$ such that $m, n > N(\epsilon)$ implies that $|f_m(x) - f_n(x)| < \epsilon$ for all $x \in E$.

Since $f_n(x)$ is a Cauchy sequence for each particular $x \in E$, we know that our sequence $f_n(x)$ has a pointwise limit; we call it $f(x)$. We must now show that the convergence to $f(x)$ is uniform.

So let $\epsilon > 0$. By our assumption on $\{ f_n \}$ there exists an $N(\epsilon)$ such that $m, n > N(\epsilon)$ implies that $$|f_m(x) - f_n(x)| < \epsilon$$ for all $x \in E$. Now comes the interesting step: fix $n$, and take the limit $m \to \infty$ in the above expression. The result is that $$|f(x) - f_n(x)| \leq \epsilon$$ for all $x \in E$ and $n > N(\epsilon)$. This proves the uniform convergence.

(This last step has to be justified; it relies on the fact that the function $\phi(x) = |x - c|$ is continuous.)

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  • $\begingroup$ does this last step use the fact that function is uniformly convergent? Because than we take limit we assume that function $f_m(x)$ converges to $f(x)$ $\endgroup$
    – Jack Havis
    Jun 27, 2020 at 14:44
  • $\begingroup$ @JackHavis no, we use the fact that $\text{lim}_{m \to \infty} f_m(x) = f(x)$ for each $x \in E$, that is, pointwise convergence, which I proved above. $\endgroup$
    – Steven
    Jun 27, 2020 at 14:49
  • $\begingroup$ But what if we take a sequence $1/n$ on $X=(0,1)$ which is Cauchy $|1/n-1/m|<2/N<\epsilon, so N=2/\epsilon$,but it is not pointwise convergent, because the limit is 0 and it is not in the domain of analysis, and it is not uniformly convergent $\endgroup$
    – Jack Havis
    Jun 27, 2020 at 15:50
  • $\begingroup$ @JackHavis I think you're confused about the definition of pointwise convergence; I think it is more useful for you to look it up than for me to explain it. $\endgroup$
    – Steven
    Jun 27, 2020 at 15:52
  • $\begingroup$ I actually can't figure out what's wrong. We write $f_k \rightarrow f \in X$ pointwise if $ \forall x \in X, \lim\limits_{k \to \infty} f_k(x) = f(x)$. So in my case $\lim\limits_{n \to \infty} 1/n = 0$ it is not in our domain. So convergence is not pointwise. Correct me, please, if I wrong $\endgroup$
    – Jack Havis
    Jun 27, 2020 at 16:02

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