Banach fixed point theorem for a function $f_k(x) = k(x+1/x)$

Question

Suppose $X = [1,\infty)$. The function $f_k(x) = k(x+\frac{1}{x})$ where $k\in(0,1)$ is a contraction on $X$, furthermore, $X$ is complete and $f:X\rightarrow X$. So all the requirements for the Banach fixed point theorem are satisfied. However, for say $k = 1/3$ the function has no fixed point in $X$, yet we are guarenteed from the theorem that there should be a unique fixed point in $X$, what went wrong here?

Answer 1

If $k = 1/3$, then $f(x) = k(x+\tfrac{1}{x})$ is not a function from $X = [1,\infty)$ into $X$.

For instance, $f(1) = \dfrac{1}{3}(1+\tfrac{1}{1}) = \dfrac{2}{3} \not\in [1,\infty)$.