# Convergence. Cauchy and uniform

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.

• Yes because The set of reals is complete. Jun 27 '20 at 14:25

## 1 Answer

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.)

• 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 $f(x)$ Jun 27 '20 at 14:44
• @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. Jun 27 '20 at 14:49
• 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 Jun 27 '20 at 15:50
• @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. Jun 27 '20 at 15:52
• 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 Jun 27 '20 at 16:02