Uniform continuity of a function and cauchy sequences 
So I'm pretty sure this is almost immediate from the definitions, please tell me if I am incorrect..
Consider two cauchy sequences in D, $\{x_n\}$ and $\{y_m\}$. Since $f$ is uniform continuous we know that there exists $\delta > 0$ for all $     \epsilon > 0$ such that for $x_n , y_m$ ($\forall n,m)$
$|x_n - y_m| < \delta  \implies |f(x_n) - f(y_m)| < \epsilon$
$f(x_n), f(y_n)$ are clearly sequences, and since we have this inequality for all $n,m$ we conclude that $f(x_n) , f(y_n)$ are cauchy sequences.
 A: No, it's not. 
You need to start with one Cauchy sequence, say $(x_n)$. Then you need to show that $\bigl(f(x_n)\bigr)$ is a Cauchy sequence. 
To do this, first fix an $\epsilon>0$. You now need to show that there is an $N$ so that $\bigl|f(x_n)-f(x_m)\bigr|<\epsilon$ whenever $n,m\ge N$. 
To do this, first use the uniform continuity of $f$ to  find a $\delta>0$ so that $|x-y|<\delta$ implies $\bigl|f(x)-f(y)\bigr|<\epsilon$. Now use the Cauchyness of $(x_n)$ to find your desired $N$ (you can make $x_n$ arbitrarily close to $x_m$ for $n,m$ sufficiently large; make them closer than $\delta$).
A: Since $f$ is uniformly continuous, by definition, given $\epsilon > 0$, there exists some $\delta > 0$ (only depending on $\epsilon$) such that 
   $$ \vert f(x) - f(y) \vert < \epsilon $$
whenever $x, y\in D$ satisfies $\vert x - y \vert < \delta$.
Let $(x_n)$ be a Cauchy sequence in $D$, then for $\delta > 0$ (as above), there exists some $N\in\mathbb{N}^\ast$ such that $\vert x_n - x_m \vert < \delta$ as $n,m \geq N$. Hence 
as $n,m \geq N$, we have $$ \vert f(x_n) - f(x_m) \vert < \epsilon .$$
We can conclude that $f(x_n)$ is Cauchy.
For the converse statement, we can consider the example below:
$f(x)= x^2$ 
$x\in\mathbb{R} \longmapsto x^2 \in \mathbb{R}$
this function sends Cauchy sequences to Cauchy sequences but it is not uniformly continuous.
Let $(x_n)$ be a Cauchy sequence in $\mathbb{R}$, clearly it is bounded. In fact, given $\epsilon_0 = 1$, there exists $N_0\in\mathbb{N}$ such that 
  $$\big\vert x_n - x_m \big\vert < 1\quad\text{as $n,m\geq N_0$} $$
this implies that for any $m\geq N_0$, $\vert x_m \vert \leq 1 + \vert x_{N_0} \vert$. Then the whole sequence is bounded by 
 $$\max\big\{  x_1,x_2, \ldots, x_{N_0}, 1 + \vert x_{N_0} \vert \big\} = : M \, . $$
Now we want to show that $\big( x_n^2 \big)$ is also Cauchy, see below:
$$\big\vert x_n^2 - x_m^2 \big\vert = \vert x_n - x_m \vert \cdot \vert x_n + x_m \vert \leq 2 M \vert x_n - x_m \vert \xrightarrow{n,m\to+\infty} 0 \,\, .  $$
Hence $\big\vert x_n^2 - x_m^2 \big\vert$ goes to zero as $n,m$ tend to $+\infty$. 
Q.E.D.
A: $f(x)=x^2$ is not uniformly continuos over $\mathbb C$ and maps Cauchy sequences into Cauchy sequences.
