uniform convergence of strict contractions Define a function $f:X\to Y$ to be a strict contraction, if there exists a $c\in [0,1[$ such that
$$\forall x,y \in X: d_Y\left( f(x),f(y)\right)\le cd_X(x,y)$$
Now, consider a metric space $(X,d)$ and a sequence $(f_n)_n$ of strict contractions $f_n:X \to X$, which is converging to a function $f$. My question is under what conditions will the limit function be a strict contraction again?

I made the following estimate
$$d(f(x),f(y))\le d(f(x),f_n(x))+d(f_n(x),f_n(y))+d(f_n(x),f(y))$$
Now each $f_n$ has its own contraction factor $c_n$ such that this can be rewritten as
$$d(f(x),f(y))\le d(f(x),f_n(x))+c_nd(x,y)+d(f_n(x),f(y))$$
for all $n\in \mathbb{N}$. Now by taking the limit $n\to\infty$ we obtain
$$d(f(x),f(y))\le c d(x,y)$$
where $c = \sup\{c_n|n\in\mathbb{N}\}$. This $c$ is not necessarily strictly smaller than $1$.

So I was looking for additional conditions such that we are guaranteed the limit function to be a strict contraction (i.e. $c&lt;1$). The easiest way, would be to demand that all $f_n$ have the same contraction factor. But that's quite a strong constrain, so my suggestions would be to demand the convergence $f_n\to f$ to be uniformly and $(X,d)$ to be a complete metric space. 

My sketch of the proof:
Each $f_n$ has fixed point $x_n^*$. In addition we have uniform convergence, so there exists some $n_0$ such that $$d(f_n(x_n^*),f_m(x_m^*))=d(x^*_n,x^*_m)&lt; \epsilon$$
for all $n,m\ge n_0$. So $(x^*_n)_n$ is Cauchy sequence in complete X, so it converges to some $x^*$. Using the fact that uniform convergence, we have
$$x^*=\lim_{n\to\infty}x^*_n = \lim_{n\to \infty}f_n(x_n^*) = f(x^*)$$
This shows that $f$ has a unique fixed point in a complete metric space.

Edit: But the existence of a fixed point does not imply that $f$ will be a strict contraction. So how can I make sure $f$ to be a strict contraction?
$f_c(x) = cx$ is a strict contraction whenever $|c|&lt;1$.
Let $c_n = 1-1/n$ so that $c_n\to 1$.

With $f(x) = x$ being the limit of $f_n = f_{c_n}$, we have $|f_n(x) - f(x)| = |x|/n$ so the convergence is uniform on bounded sets but the limit function is not a strict contraction.
It seems to me you don't need the completeness of $X$. As you pointed out, all you need is uniform convergence of the sequence of contractions mappings $(f_{n})$.
Here's my argument, using much of what you've already written
Since each $c_{i} \in [0,1]$, you may write $c = sup \{ c_{i} | i \in \mathbb{N} \}$; however, $c$ is not necessarily the limit of the sequence $(c_{n})$. I think your initial estimate needs a tweak in the third term on the right hand side as follows:
$d(f(x),f(y)) \leq d(f(x),f_{n}(x)) + d(f_{n}(x),f_{n}(y)) + d(f_{n}(y),f(y))$
Your arguments still holds though. $d(f_{n}(x),f_{n}(y)) \leq cd(x,y)$ and there exists $N \in \mathbb{N}$ sufficiently large as to guarantee that $n \geq N$ implies $d(f(z),f_{n}(z)) \leq \frac{\epsilon}{2}$ for all $z \in X$. So for any $\epsilon &gt; 0$ we have,
$d(f(x),f(y)) \leq cd(x,y) + \epsilon$
This means $f$ is a contraction mapping.