TITLE: Any ring homomorphism from $C_b(X)$ to $C_b(Y)$ is continuous?
QUESTION [7 upvotes]: 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 ?

REPLY [7 votes]: 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.