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?



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