Showing Uniform Convergence of Cauchy Sequence of Functions

[fishturtle1]

Homework Statement

Let $X \subset \mathbb{C}$, and let $f_n : X \rightarrow \mathbb{C}$ be a sequence of functions. Show if $f_n$ is uniformly Cauchy, then $f_n$ converges uniformly to some $f: X \rightarrow \mathbb{C}$.

Homework Equations

Uniform convergence: for all $\varepsilon > 0$ there exists $N(\varepsilon)$ such that for any $x \epsilon X$ if $n > N(\varepsilon)$ then $\vert f_n(x) - f(x) \vert < \varepsilon$.

Uniformly Cauchy: for all $\varepsilon > 0$ there exists $N(\varepsilon)$ such that for any $x \epsilon X$: if $n, k > N(\varepsilon)$ then $\vert f_n(x) - f_k(x) \vert < \varepsilon$.

The Attempt at a Solution

Proof: Suppose $f_n$ is uniformly Cauchy. First we show $f_n$ converges point wise to some $f : X \rightarrow \mathbb{C}$.

Fix some $x \epsilon X$. Then $f_n(x)$ is a Cauchy sequence of complex values. By completeness, $f_n(x)$ converges to some $f(x): X \rightarrow \mathbb{C}$. Thus, $f_n$ converges point wise to some $f$ on $X$.

Let $\varepsilon/2 > 0$. By definition of uniformly Cauchy, there exists some $N_1 = N_1(\varepsilon/2)$ such that for any $x \epsilon X$: if $l, k > N_1$ then $\vert f_l(x) - f_k(x) \vert < \varepsilon / 2$. By point wise convergence, there exists $N_2 = N_2(\varepsilon, x)$ such that if $n > N_2$ then $\vert f_n(x) - f(x) \vert < \varepsilon / 2$.

..
Not sure, I think I'm on the wrong track, I was going to try to pick max(N_1, N_2) and use triangle inequality but I don't see how to do this/how this would help to show uniform conv.

Any hints?, please

 

[wrobel]

Try to assume the converse: there exist a constant $c>0$ , a sequence $x_n$ and a subsequence $f_{i_n}$ such that
$$|f(x_n)-f_{i_n}(x_n)|\ge c$$
......
Contradiction.

 

[fishturtle1]

Try to assume the converse: there exist a constant $c>0$ , a sequence $x_n$ and a subsequence $f_{i_n}$ such that
$$|f(x_n)-f_{i_n}(x_n)|\ge c$$
......
Contradiction.

Thanks for the reply,

The converse would be if $f_n$ converges uniformly to some $f: X \rightarrow \mathbb{C}$, then $f_n$ is uniformly Cauchy. The text proved this case so I can use it, but I'm not sure how this statement implies, or why we assume $\vert f(x_n) - f_{i_{n}}(x_n) \vert \ge c$ for some constant $c \epsilon \mathbb{C}$.

From Wikipedia(https://proofwiki.org/wiki/Limit_of_Subsequence_equals_Limit_of_Sequence): if $a_n \rightarrow L$ then for any subsequence of $a_n$ we have $a_{n_k} \rightarrow L$.

From there though, we can say $\lim_{n\rightarrow\infty} f_{i_{n}}(x_n) \neq f(x_n)$. By contrapositive of wikipedia statement, we have $\lim_{n\rightarrow \infty} f(x_n) \neq f(x_n)$, a contradiction. We can conclude...

 

[wrobel]

I meant proof by contradiction

 

[fishturtle1]

I meant proof by contradiction

Thanks,

I have: Assume (the converse) if $f_n$ converges uniformly to some $f: X \rightarrow \mathbb{C}$, then $f_n$ is uniformly Cauchy. Furthermore, assume $f_n$ is uniformly Cauchy.

Assume by contradiction, there exists some $c \epsilon \mathbb{C}$ and subsequence $f_{i_n}$ such that for all $N(c)$ if $n > N(c)$ then $\vert f(x_n) - f_{i_n}(x_n) \vert \ge c$. But this implies $\lim_{n\rightarrow\infty} f(x_n)$ does not exist, a contradiction?. We can conclude for all $\varepsilon > 0$ there exists $N(\varepsilon, x_n)$? such that if $n > N(\varepsilon, x_n)$ then $\vert f(x_n) - f_{i_n}(x_n) \vert < \varepsilon$.

But I don't know where to get a contradiction because we only know $f_n$ converges point wise. Since we have a sequence of $x$'s and not a fixed $x$ I'm not sure if we can use point wise convergence as a contradiction?

 

[wrobel]

we can write
$$|f(x_n)-f_{i_n}(x_n)|\le |f(x_n)-f_k(x_n)|+|f_k(x_n)-f_{i_n}(x_n)|$$
Take $\varepsilon=c/2$ and choose $n$ big such that if $j>i_n$ then $|f_j(x)-f_{i_n}(x)|<\varepsilon$ for any $x\in X.$
Choose $k>i_n$ such that for this fixed $n$ (and fixed $x_n$) it follows that $|f(x_n)-f_k(x_n)|<\varepsilon$
we have got $|f(x_n)-f_{i_n}(x_n)|<c$

 

[wrobel]

Actually the proof can be simplified a lot.

You have: for any $\varepsilon>0$ there is a number $N$ such that if $i,j>N$ then
$$|f_i(x)-f_j(x)|<\varepsilon\qquad (*)$$ for each $x\in X$
now fix $i$ and for each $x$ pass to the limit as $j\to \infty$ in formula (*)
you will have $|f_i(x)-f(x)|\le\varepsilon$