# A question about the Stone-Čech compactification and ultrafilter

Let $$X$$ be a Tychonoff space and let $$\beta X$$ is the Stone-Čech compactification of $$X$$. Assume $$f:X\longrightarrow \mathbb{R}$$ is a bounded function. Then there exists a function $$f^{\beta }:\beta X\longrightarrow \mathbb{R}$$ (the map $$f\longrightarrow f^{\beta }$$ is an isomorphism of $$C^{\ast }\left( X\right)$$ onto $$C\left( \beta X\right)$$).

Moreover, every point $$p$$ of $$\beta X$$ is the limit of a unique $$z$$-ultrafilter $$\mathcal{A}^{p}$$ on $$X$$. If $$p\in X$$, then $$\mathcal{A}^{p}=\left\{ Z\in Z\left( X\right) :p\in Z\right\}$$.

My question is that for $$p\in \beta X$$, how to define $$f^{\beta }\left( p\right)$$ in terms of $$z$$-ultrafilter $$\mathcal{A}^{p}$$ on $$X$$?

Actually, my purpose is to check whether $$f^{\beta }$$ also provides a property that $$f$$ provides (for example, $$f$$ is constant on some subsets) using ultrafilters.

Of course $$f^\beta(p)$$ is the limit of $$f(x)$$ along the z-ultrafilter $$\mathcal A^p$$. For $$t \in \mathbb R$$, we write $$\lim_{\mathcal A^p} f(x) = t$$ iff for every neighborhood $$U$$ of $$t$$ there is $$V \in \mathcal A^p$$ so that $$\forall x \in V, f(x) \in U$$.
In your setting, there exists a unique $$t$$ with that property.
The map $$f^\beta$$ is the unique continuous extension of $$f$$ from the dense subset $$X$$ to all of $$\beta X$$.