Algebra of bounded functions on a completely regular space

Question

Let $T$ be a completely regular topological space, i.e. a topological space satisfying axiom $T_1$ and such that for any closed set $F\subset T$ and any $t_0\in T\setminus F$ there is a continuous real function $f:T\to\mathbb{R}$ such that $f(t_0)=0$, $\forall t\in F \quad f(t)=0$ and $\forall t\in T \quad f(t)\in[0,1]$.

Let $B_T$ the algebra of all continuous complex bounded functions defined on $T$ equipped with the norm $\|x\|=\sup_{t\in T}|x(t)|$.

I want to understand why:

• $B_T$ is a Banach algebra. I find it easy to prove that it is a normed algebra, so proving that it is complete is all that remains to do;

• the points of $T$ are homeomorphically included in the space $\mathscr{M}$ of the maximal ideals of the algebra $B_T$ and that, in this inclusion, the image of $T$ in $\mathscr{M}$ is an everywhere dense subset (quoting the book). Here I know that $\mathscr{M}$, which I think to be intended with the weak$^\ast$ topology, is the same as the set of all the complex non-null continuous multiplicative linear functionals $B_T\to\mathbb{C}$, but I cannot go further...

• any complex bounded function on this image of $T$ admits a unique continuous prolongation to $\mathscr{M}$.

My book does not give many details about Banach algebras, but I would be very interested in learning more... Can anybody help me with a proof or a link to one?

I heartily thank you!

Edit: added continuous thanks to the comments by Daniel.

Answer 1

I find it easy to prove that it is a normed algebra, so proving that it is complete is all that remains to do;

You need to see that

• if $(f_n)$ is a Cauchy sequence in $B_T$, then for every $t\in T$, the sequence $\bigl(f_n(t)\bigr)$ is a Cauchy sequence in $\mathbb{C}$, which follows directly from $\lvert f_n(t) - f_m(t)\rvert \leqslant \lVert f_n - f_m\rVert$.

So the pointwise limit $f(t) = \lim_{n\to\infty} f_n(t)$ exists. Then you need to know that a (locally) uniform limit of continuous functions is continuous, and finally, you need to show that $\lVert f_n - f\rVert \to 0$ (but that is very easy).

the points of $T$ are homeomorphically included in the space $\mathscr{M}$ of the maximal ideals of the algebra $B_T$ and that, in this inclusion, the image of $T$ in $\mathscr{M}$ is an everywhere dense subset (quoting the book). Here I know that $\mathscr{M}$, which I think to be intended with the weak$^\ast$ topology, is the same as the set of all the complex non-null continuous multiplicative linear functionals $B_T\to\mathbb{C}$, but I cannot go further...

Right, it is intended to carry the (subspace topology of) the weak$^\ast$ topology, and $\mathscr{M}$ is identified with the space of unital complex algebra homomorphisms on $B_T$. (The correspondence between a maximal ideal $\mathfrak{M}$ and a unital algebra homomorphism $\varphi$ is $\varphi \mapsto \ker \varphi$ in the one direction, in the other, it is given by $\mathfrak{M} \mapsto (x \mapsto \operatorname{can} (x + \mathfrak{M}))$, where $\operatorname{can}$ denotes the canonical isomorphism between a Banach algebra in which every nonzero element is invertible and $\mathbb{C}$ as implied by the Gelfand-Mazur theorem).

The inclusion map $\varepsilon\colon t \mapsto \varepsilon_t$, where $\varepsilon_t(f) = f(t)$ is clear, and one needs to see that it is an embedding for completely regular $T$.

Fix an arbitrary $t_0 \in T$. First, we show that $\varepsilon$ is continuous at $t_0$. A neighbourhood basis of $\lambda \in B_T^\ast$ in the weak$^\ast$ topology is given by sets of the form

$$V(\lambda;\delta; f_1,\dotsc, f_n) = \left\{\mu\in B_T^\ast : \bigl(\forall 1 \leqslant i \leqslant n\bigr)\bigl(\lvert \mu(f_i) - \lambda(f_i)\rvert < \delta\bigr) \right\}$$

for $\delta > 0$ and finitely many $f_i$. Now

$$\varepsilon^{-1}(V(\varepsilon_{t_0}; \delta; f_1,\dotsc, f_n)) = \left\{ t\in T : \bigl(\forall 1 \leqslant i \leqslant n\bigr)\bigl( \lvert f_i(t) - f_i(t_0)\rvert < \delta\right\} = \bigcap_{i=1}^n f_i^{-1}\left(D_\delta(f_i(t_0))\right)$$

is a finite intersection of (open) neighbourhoods of $t_0$, hence a(n open) neighbourhood of $t_0$. Thus $\varepsilon^{-1}(U)$ is a neighbourhood of $t_0$ for every neighbourhood $U$ of $\varepsilon_{t_0}$, i.e. $\varepsilon$ is continuous at $t_0$.

Next we see that $\varepsilon$ is open at $t_0$, that means that the image of every neighbourhood of $t_0$ is a neighbourhood of $\varepsilon_{t_0}$ [in $\varepsilon(T) \subset \mathscr{M}$]. So, given a neighbourhood $U$ of $t_0$, we need to find a $\delta > 0$ and finitely many $f_i \in B_T$ such that

$$\varepsilon(T) \cap V(\varepsilon_{t_0};\delta; f_1,\dotsc,f_n) \subset \varepsilon(U),$$

which would follow from

$$\varepsilon^{-1}\bigl(V(\varepsilon_{t_0};\delta;f_1,\dotsc,f_n)\bigr) \subset U.$$

Now, by the complete regularity, there is an $f \in B_T$ with $f(t_0) = 1$ and $f(T\setminus U) \subset \{0\}$. Then $V\left(\varepsilon_{t_0};\frac{1}{2};f\right)$ does the job.

So, $\varepsilon$ is continuous at every point, and open at every point, hence it is continuous and open, i.e. an embedding.

To see that $\varepsilon(T)$ is dense in $\mathscr{M}$, we suppose there were a $\varphi \in \mathscr{M}\setminus \overline{\varepsilon(T)}$ and derive a contradiction. By definition, $\varphi \in \mathscr{M}\setminus \overline{\varepsilon(T)}$ means there is a (weak$^\ast$) neighbourhood of $\varphi$ not intersecting $\varepsilon(T)$, i.e. there is a $\delta > 0$ and finitely many $f_1,\dotsc,f_n\in B_T$ with $V(\varphi;\delta;f_1,\dotsc,f_n)\cap \varepsilon(T) = \varnothing$, or, in other words,

$$\bigl(\forall t\in T\bigr)\bigl(\exists \nu \in \{1,\dotsc,n\}\bigr)\bigl(\lvert \varphi(f_\nu) - f_\nu(t)\rvert \geqslant \delta\bigr).\tag{$\ast$}$$

Then consider

$$F\colon t \mapsto \sum_{\nu=1}^n \lvert f_\nu(t) - \varphi(f_\nu)\rvert^2 = \sum_{\nu = 1}^n (f_\nu(t)-\varphi(f_\nu))(\overline{f_\nu(t)} - \overline{\varphi(f_\nu)}).$$

Since $B_T$ is an algebra containing the constant functions and $B_T$ is conjugation-invariant [$f\in B_T \iff \overline{f}\in B_T$], we have $F\in B_T$, and by $(\ast)$, we have $F(t) \geqslant \delta^2$ for all $t\in T$, hence $F$ is a unit in $B_T$, whence $\varphi(F) \neq 0$. On the other hand,

$$\begin{align} \varphi(F) &= \varphi\left(\sum_{\nu = 1}^n (f_\nu-\varphi(f_\nu)\cdot \mathbb{1})(\overline{f_\nu} - \overline{\varphi(f_\nu)}\cdot \mathbb{1})\right)\\ &= \sum_{\nu = 1}^n \varphi\left(f_\nu - \varphi(f_\nu)\cdot \mathbb{1}\right)\cdot\varphi\left(\overline{f_\nu} - \overline{\varphi(f_\nu)}\cdot \mathbb{1}\right)\\ &= \sum_{\nu=1}^n \left(\varphi(f_\nu) - \varphi(\varphi(f_\nu)\cdot \mathbb{1})\right)\cdot\varphi\left(\overline{f_\nu} - \overline{\varphi(f_\nu)}\cdot \mathbb{1}\right)\\ &= 0, \end{align}$$

since $\varphi(\mathbb{1}) = 1$, so $\varphi(\varphi(f_\nu)\cdot \mathbb{1}) = \varphi(f_\nu)$.

Finally we treat the last item,

any complex bounded [and continuous] function on this image of $T$ admits a unique continuous extension to $\mathscr{M}$.

Well, $\mathscr{M}$ is a (topological, not linear) subspace of $B_T^\ast$ endowed with the weak$^\ast$ topology. Since $\varepsilon \colon T \to \varepsilon(T)$ is a homeomorphism, every continuous and bounded function $f$ on $\varepsilon(T)$ can be identified with an element $g = f\circ \varepsilon \in B_T$, and with the canonical embedding $J$ of $B_T$ into its bidual, we obtain a continuous map $J(g) = J(f\circ\varepsilon) \colon B_T^\ast \to \mathbb{C}$ (where $B_T^\ast$ is endowed with the weak$^\ast$ topology, since that's the one we're interested in). Then

$$\tilde{f} = J(f\circ\varepsilon)\bigl\lvert_{\mathscr{M}}$$

is a continuous extension of $f$ to $\mathscr{M}$. The continuity is clear, since it's the restriction of a continuous function, and

$$\tilde{f}(\varepsilon_t) = J(f\circ\varepsilon)(\varepsilon_t) = \varepsilon_t(f\circ\varepsilon) =(f\circ\varepsilon)(t) = f(\varepsilon_t)$$

shows it is indeed an extension of $f$.

Now, $\mathscr{M}$ is a compact (quasicompact and Hausdorff) space, since it is a weak$^\ast$-closed subset of the norm-closed unit ball of $B_T^\ast$, hence normal, and in particular completely regular. Thus the uniqueness of the extension is equivalent to the denseness of $\varepsilon(T)$ in $\mathscr{M}$.

Note: This is a functional-analytic construction of the Stone-Čech compactification of a completely regular space. Since compact spaces are completely regular, and subspaces of completely regular spaces are completely regular, only completely regular spaces can have compactifications (a compactification of a topological space $X$ is a compact space $K$ together with an embedding of $X$ into $K$ as a dense subspace). This construction shows that every completely regular space has a compactification, namely the Stone-Čech compactification - many completely regular spaces also have other compactifications, for example locally compact (but not compact) [Hausdorff] spaces have the Alexandrov compactification aka one-point compactification, which is the smallest compactification a non-compact space can have; the two very rarely coincide. The Stone-Čech compactification is in a sense the largest compactification that a space can have, it is characterised by each of the following criteria:

• every continuous map $f \colon X \to K$ where $K$ is compact has a (unique) continuous extension to the compactification.
• every bounded continuous real (or complex) valued function has a (unique) continuous extension to the compactification.

The latter criterion directly translates to an isometric isomorphism between the algebra $C_b(X)$ of bounded continuous functions on $X$ (which is called $B_XT$ here) and the algebra $C(\beta X)$ of continuous functions on the Stone-Čech compactification $\beta X$ of $X$.