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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 ?

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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.

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  • $\begingroup$ Very interesting and well written! +1 $\endgroup$ – Fimpellizieri May 29 '17 at 19:43
  • $\begingroup$ Very nice and square root trick is very smart +1 $\endgroup$ – JJR May 29 '17 at 21:18
  • $\begingroup$ @Eric Wofsey : why did you write $|q|$ in the one before the last line ? $\endgroup$ – Souvik Dey May 30 '17 at 6:36
  • $\begingroup$ @Eric Wofsey: btw .. how do you get $F(q)=q$ ? Are you assuming $F(1)=1$ in the defn. of ring homomorphism ? Without this definition , can a ring homomorphism still be continuous ? $\endgroup$ – Souvik Dey May 30 '17 at 10:41
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    $\begingroup$ Oops, the $|q|$ was an error (left over from an earlier draft). I am assuming homomorphisms are unital. If you don't assume that though, the result is still true, since $F(1)$ must be an idempotent, which must be the characteristic function of some clopen subset $Z\subset Y$. The entire image of $F$ is then contained in a subring isomorphic to $C_b(Z)$, and so you can consider $F$ as a unital homorphism $C_b(X)\to C_b(Z)$. $\endgroup$ – Eric Wofsey May 30 '17 at 15:21
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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.


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.

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    $\begingroup$ As far as I know the Gelfand-Kolmogorov-Theorem holds if $C(X)$ and $C(Y)$ are isomorphic $\endgroup$ – JJR May 29 '17 at 19:20
  • $\begingroup$ @TomekKania: The point is that you need a ring isomorphism to apply the theorem. At the very least, you need an epimorphism. Otherwise, the preimage of a maximal ideal may not be a proper ideal. As for your second paragraph, that is not needed at all: by the universal property, $C_b(X)=C(\beta X)$ even if $X$ is not completely regular. It's just that in this case, $X\to \beta X$ is not injective, but you don't use that anywhere. $\endgroup$ – tomasz May 29 '17 at 19:48
  • $\begingroup$ That said, the criterion you cited is very interesting, although it is very much an overkill in this case. $\endgroup$ – tomasz May 29 '17 at 19:52
  • $\begingroup$ @JJR, in the non-compact case the Gelfand-Kolmogorov theorem fails. Take $X=\omega_1$ and $Y=\omega_1+1$ with order topologies. $\endgroup$ – Tomek Kania May 29 '17 at 20:13
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    $\begingroup$ If you're just talking about ring-homomorphisms, you can't immediately use the closed graph theorem, since you don't know that $h$ is $\mathbb{R}$-linear. $\endgroup$ – Eric Wofsey May 29 '17 at 20:25

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