# Does the converse of uniform continuity -> Preservance of Cauchy sequences hold?

We know that if a function $f$ is uniformly continuous on an interval $I$ and $(x_n)$ is a Cauchy sequence in $I$, then $f(x_n)$ is a Cauchy sequence as well.

Now, I would like to ask the following question:

The function $g:(0,1) \rightarrow \mathbb{R}$ has the following property: for every Cauchy sequence $(x_n)$ in $(0,1)$, $(g(x_n))$ is also a Cauchy sequence. Prove that g is uniformly continuous on $(0,1)$.

How do we go about doing it?

-

$g$ is clearly continuous, you really only need to worry about what would happen at the endpoints. some sequence $x_n\to 1$ has $g(x_n)\to g(1)$, similarly for zero. so you can define $g$ on $[0,1]$ where it will be uniformly continuous (basically $g$ doesnt zoom off to infinity at the endpoints, else $g(x_n)$ wouldnt be cauchy)

-

Just a remark: a key point is that the open interval $(0,1)$ is not complete as a metric space. Suppose you have a function $f: X \rightarrow \mathbb{R}$ where $X$ is complete. Then $f$ is Cauchy-continuous -- i.e., if $\{x_n\}$ is a Cauchy sequence in $X$, then $\{f(x_n)\}$ is Cauchy in $\mathbb{R}$ -- iff it is continuous. To see that this is in general much weaker than being uniformly continuous, consider the example $f: \mathbb{R} \rightarrow \mathbb{R}$, $x \mapsto x^2$.

In the other direction, if $X$ is incomplete, but with compact completion, then every Cauchy continuous function $f: X \rightarrow \mathbb{R}$ extends to a continuous function $\overline{f}: \overline{X} \rightarrow \mathbb{R}$ on the completion $\overline{X}$. Since $\overline{X}$ is compact, $\overline{f}$ is uniformly continuous, hence so is its restriction $f$. This latter situation obtains in the question asked by the OP.

-

You have to prove is that if $(x_n)_{n \ge 0} \in (0,1)^{\mathbb{N}}$ and $(y_n)_{n \ge 0} \in (0,1)^{\mathbb{N}}$ converge to the same limit $x \in [0,1]$, then $(g(x_n))_{n \ge 0}$ and $(g(y_n))_{n \ge 0}$ have the same limit (you may define $z_{2n} = x_n$, $z_{2n+1} = y_n$, and use uniqueness of the limit). Then you can the deduce that $g$ is continuous on the compact $[0,1]$ (or can be extended to such a function), hence uniformly continuous.

-

You can also prove it by contradiction.

Suppose that $f$ is not uniformly continuous. Then there exists an $\epsilon >0$ so that for each $\delta>0$ there exists $x,y \in (0,1)$ with $|x-y| < \delta$ and $|f(x)-f(y)| \geq \epsilon$.

For each $n$ pick $x_n, y_n$ so that $|x_n-y_n| < \frac{1}{n}$ and $|f(x_n)-f(y_n)| \geq \epsilon$.

Pick $x_{k_n}$ a Cauchy subsequence of $x_n$ and $y_{l_n}$ a Cauchy subsequence of $y_{k_n}$.

Then the alternaticng sequence $x_{l_1}, y_{l_1}, x_{l_2}, y_{l_2},..., x_{l_n}, y_{ln}, ...$ is Cauchy but

$$\left| f(x_{l_n}) - f(y_{l_n}) \right| \geq \epsilon \,.$$

-