Any ring homomorphism from $C_b(X)$ to $C_b(Y)$ is continuous?

Question

For topological space $X$ , let $C_b(X)$ denote the set of all bounded real valued continuous functions on $X$ . It is a ring w.r.t. pointwise addition and multiplication of functions and it is a metric space with respect to the supremum metric (basically a Banach space w.r.t. the sup norm ) . If $X,Y$ are topological spaces then is any ring homomorphism from $C_b(X)$ to $C_b(Y)$ continuous ?

Answer 1

Yes; in fact, any such homomorphism is (nonstrictly) norm-decreasing. Note that for any $r\in\mathbb{R}$ and $f\in C_b(X)$, $f(x)\leq r$ for all $x$ iff for any rational number $q>r$, the function $q-f$ has a square root in $C_b(X)$. Similarly, $f(x)\geq r$ for all $x$ iff for any rational number $q<r$, $f-q$ has a square root in $C_b(X)$. This gives a ring-theoretic definition of the sup norm: $\|f\|$ is the smallest real number $r\geq 0$ such that for any $q\in\mathbb{Q}$ with $q>r$, $q-f$ and $q+f$ have square roots in $C_b(X)$.

Now suppose $F:C_b(X)\to C_b(Y)$ is a homomorphism and $f\in C_b(X)$. Let $r=\|f\|$, so $q-f$ and $q+f$ have square roots in $C_b(X)$ for any $q\in\mathbb{Q}$ with $q>r$. It follows that $F(q-f)=q-F(f)$ and $F(q+f)=q+F(f)$ have square roots in $C_b(Y)$ for any $q\in\mathbb{Q}$ with $q>r$. Since $\|F(f)\|$ is the smallest number with this property, $\|F(f)\|\leq r$. That is, $\|F(f)\|\leq\|f\|$. It follows that $F$ is continuous.

Answer 2

For $x\in X$ the map $\delta_x(f)= f(x)$ is linear, multiplicative and continuous. Moreover, such maps separate points in $C_b(X)$. I claim that for any Banach algebra $A$, every algebra homomorphism $h\colon A\to C_b(X)$ is continuous.

Proof. By the closed graph theorem, we only need to show that $h$ has closed graph. Suppose that $(a_n)$ is a null sequence in $A$ and $h(a_n)\to f$. Then $\delta_x\circ h$ is linear and multiplicative, hence continuous. Thus, $\delta_x(h(a_n))\to 0$ for any $x$. Thus, $h(a_n)\to 0$, and so $h$ is continuous.

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_{Old answer for isomorphisms. In the case where $X$ is completely regular, $C_b(X)$ is isometrically isomorphic to $C(\beta X)$, where $\beta X$ denotes the Stone-Čech compactification of $X$. Thus, if $X$ and $Y$ are completely regular, you may apply the Gelfand-Kolmogorov theorem.}

_{If $X$ and $Y$ are not necessarily completely regular, you may use the Albiac-Kalton criterion to see that $C_b(X)$ and $C_b(Y)$ are isometrically isomorphic as algebras to $C(X^\prime), C(Y^\prime)$, respectively, for some compact Hausdorff spaces $X^\prime, Y^\prime$. Now you are in a position to apply the Gelfand-Kolmogorov theorem.}