Uniformly cauchy sequences

Question

A sequence of functions $f_n$ is said to be uniformly cauchy if $$\forall \varepsilon > 0 \ \exists N > 0 :\forall z , \forall r, s > N: |f_r(z) - f_s(z)| < \varepsilon$$

How can I show that if a sequence is uniformly cauchy then $f_n$ converge uniformly to some funciton $f$? We can assume that the metric space is complete.

Thanks!

Answer 1

First, we can show that $f_n$ converges pointwisely to some $f$. Then $\exists N_1$ $|f(z)-f_s(z)|<\epsilon/2 $ $ \forall s>N_1$. (1)

Second, let $\epsilon>0$, $\exists N$ s.t. $\forall z$, we have $|f_s(z) - f_r(z)| < \epsilon/2$. (2)

Now, we want to show that $\forall r>N$ $\forall z$, $|f(z)-f_r(z)|<\epsilon$. Then we are done.

Now choose arbitrary $r>N$, arbitrary $z$.

Using (1), we can find $N_1>N$ s.t. $\forall s>N_1$ and $|f(z)-f_s(z)|<\epsilon/2$. Then $-\epsilon/2+f_s(z) < f(z)<\epsilon/2+f_s(z)$.

Using (2), since $s>N_1>N$ and $r>N$, we can use (2). Then $|f_s(z)-f_r(z)|<\epsilon/2$. Then $-\epsilon/2+f_r(z) < f_s(z) <\epsilon/2+f_r(z)$

Then $|f(z)-f_r(z)|<\epsilon$.

Answer 2

Assume that we are given a sequence of functions $f_n:Z\to Y$, and that the metric space $Y$ is complete. According to assumption, for each fixed $z\in Z$ the sequence $\bigl(f_n(z)\bigr)_{n\geq0}$ is a Cauchy sequence in $Y$. It follows that for each $z\in Y$ the limit $$\lim_{n\to\infty} f_n(z)=: f(z)\in Y$$ exists. I claim that the $f_n$ converge uniformly in $Z$ to the function $f:\>Z\to Y$.

Proof. Let an $\epsilon>0$ be given. Then there is an $N\in{\mathbb N}$ such that for all $z\in Z$ we have $$\bigl|f_{n+m}(z)-f_n(z)\bigr|<\epsilon\tag{1}$$ whenever $n>N$ and $m\geq0$. Letting $m\to\infty$ in $(1)$ we conclude that for all $z\in Z$ we have $$\bigl|f(z)-f_n(z)\bigr|\leq\epsilon\ ,$$ as soon as $n>N$. Since $\epsilon>0$ was arbitray the claim follows.

Here we have made use of the following Lemma: If $a_m\to a\in{\mathbb R}$ $\>(m\to \infty)$ and $a_m\geq 0$ for all $m\geq 1$ then $a\geq 0$.