In infinite-dimensional space, does linear mapping always have a fixed point? In infinite-dimensional space, does linear mapping that satisfy the strictly nonexpansive always have a fixed point?
I believe the answer is no and we can construct a counterexample to show that such as in $X=c_0$  or $X=l^\infty$ but I am still working on constructing my example.
Any help would be appreciated
 A: Again, there is no linear mapping that is without a fixed point since trivially the zero vector will always satisfy $T(0) = 0$ for any linear mapping $T:V\to V$.  It may be that a linear map has only this one fixed point, e.g. the zero map that sends every $x\in V$ to zero necessarily has but that one fixed point.  Indeed any strictly contractive linear map, $||T(x-y)|| \lt ||x-y||$ when $x\neq y$, has no fixed point other than $T(0)=0$.
However an interesting question is whether there can be a choice of initial point $x_0$ such that the sequence defined by $x_{n+1} = T(x_n)$ has no convergent subsequence under the restriction that $||T(x-y)|| \le ||x-y||$.
A standard example of this is the shift operator on $\ell^\infty$.  That is, define $T(x)$ to be the result of shifting all the entries of $x$ to the right (and introducing a new zero entry to fill the first component).  For any choice of nonzero $x$, the sequence $T^n(x)$ will neither converge nor have any convergent subsequence using the $\infty$-norm. 
The shift operator on $\ell^\infty$ satisfies $||T(x-y)|| = ||x-y||$.
