How can we prove that the below function does not have fixed point?
Define $S:=\{(x_m)\in l^1\mid \sum^\infty |x_i|\leq 1)\}$, and consider the self map $\Phi$ on $S$ defined by $\Phi((x_m)):=(1-\sum^\infty |x_i|,x_1,x_2,,,,)$. Show that $S$ is a nonempty, closed, bounded , and convex subset of $l^1$ and $\Phi$ is a Lipschitz continuous map without a fixed point.
My solution is, as a way of contradiction, assume that it does have a fixed point. Then by definition of the mapping, $x_i=\bar{x} \ \ \forall i$. Moreover, the first coordinate of the mapping implies $\sum |\bar{x}|=1$, but this is clearly a contradiction, because the infinity sum of a constant number can not converge.
You don't get $\sum |x_i|=1$. What you get is $x_1=1-\sum |x_i|, x_1=x_2,x_2=x_3,...$. It follows that $x_n$ is independent of $n$. If $x_n=x_1 \neq 0$ then $\sum |x_i|=\infty$ contradicting the fact that $x_1=1-\sum |x_i|$. But $x_n=0$ for all $n$ also contradicts $x_1=1-\sum |x_i|$ so there can be no fixed point.