# Operator on continuous functions under Alexandroff compactification

Let $X$ be a complete separable non-compact metric space and let us denote $C_b(X)$ to be the space of all continuous bounded real-value functions on $X$. Consider a functional operator $A$ from the space $C_b(X)$ to itself and let $X^* = X\cup\{\infty\}$ be Alexandroff compactification of $X$. Define: $$C^*_b(X^*) = \{f:X^*\to\mathbb R\text{ s.t. }f|_X\in C_b(X)\}$$ to be the space of all extensions of continuous bounded functions from $X$.

Let us extend the operator $A$ as follows: for any $f\in C^*_b(X^*)$ define $$(A^*f)(x) = \begin{cases} \left(Af|_X\right)(x)&\text{ if }x\neq \infty \\ \\ f(\infty)&\text{ if }x = \infty. \end{cases}$$ Clearly, $A^*$ maps $C^*_b(X^*)$ into itself. Does it also map $C_b(X^*)$ into itself?

If it matters, $A$ is a bounded linear operator.

-
If $Af=(1/2)f$, the constant function $1$ is mapped by $A^*$ to the function that is $1/2$ on $X$ and $1$ at $\infty$. Am I missing something? – user31373 May 22 '12 at 19:20
@LeonidKovalev: no, that's true. Even worse, if $X = \mathbb N$ and $Af(x) = \mathrm{sgn}(x)$ then regardless of how the extension $A^*$ is defined on $\infty$ the constant function $0$ to a discontinuous function. Would you put your example as an answer? – Ilya May 22 '12 at 20:17

If $Af=(1/2)f$, the constant function $1$ is mapped by $A^∗$ to the [discontinuous] function that is equal to $1/2$ on $X$ and $1$ at $\infty$.