Sums of contractive maps is still a contractive map Suppose $f_1$, $f_2$ are contractive maps such that each is a mapping from $X \to X$ where $X$ is a metric space. Then does it necessary follow that their average sum $(f_1 + f_2)/2$ is also a contractive map? What if $(f_1 + f_2 + \dots +f_n)/n$?
My idea is yes, but I am rusty on the proof. Since $f_1$ and $f_2$ are contractive maps, we can do 
$$d((f_1+f_2)/2, (f_1+f_2)/2) \leq kd( (x + x)/2, (y + y)/2) = kd(x,y)$$
Does that work?I am not sure about substituting the $x$ and $y$ inside the metric.
 A: This only makes sense for vector spaces over the reals, in which case i am taking the distance as a norm with triangle inequality as usual. S, for two points $x,y \in X,$ let us define
$$  x_j = f_j(x), \; \; y_j = f_j(y).   $$
Each is a contraction, so
$$ \parallel x_j - y_j \parallel <   \parallel x - y \parallel   $$ as long as $x \neq y.$
Furthermore 
$$  \left\| \sum_{j=1}^n x_j - y_j \right\|  \leq  \sum_{j=1}^n  \parallel x_j - y_j \parallel $$
Then
$$ \frac{1}{n} \; \sum_{j=1}^n  \parallel x_j - y_j \parallel <  \frac{1}{n} \; \sum_{j=1}^n  \parallel x - y \parallel  \; = \;   \frac{1}{n} \cdot n \cdot \parallel x - y \parallel \; = \; \parallel x - y \parallel   $$
I see, you are using the stronger contraction with multiplier, $0 < k < 1$ and
$$ \parallel x_j - y_j \parallel \leq \; k   \parallel x - y \parallel   $$
where we no longer care whether $x,y$ are distinct. Then
$$ \frac{1}{n} \; \sum_{j=1}^n  \parallel x_j - y_j \parallel \leq   \frac{1}{n} \; \sum_{j=1}^n k  \parallel x - y \parallel  \; = \;   \frac{k}{n} \cdot n \cdot \parallel x - y \parallel \; = \; k \parallel x - y \parallel   $$ 
Let's see. we are never guaranteed a fixed point with the first, weaker type of contraction, because a vector space is not compact. In finite dimension, the stronger type with multiplier $k$ does give a fixed point as the space is metrically complete. In infinite dimension, I think not.
