# derivations on a ring of continuous functions

Let $X$ be a topological space and denote by $\mathbb{R}$ the set of reals. Of course the set of all continuous functions $C(X,\mathbb{R})$ is a $\mathbb{R}$-algebra. Let $c\in\mathbb{R}$. By the symbol $\bar{c}$ we will denote constant function at $c$, i.e. $\bar{c}:X\to\mathbb{R}$ is defined by $\bar{c}(x)=c$.

If $D:C(X,\mathbb{R})\to C(X,\mathbb{R})$ is a $\mathbb{R}$-derivation, then $D(x)=\bar{0}$ for any $x\in C(X,\mathbb{R})$.

Proof. Step one. We will prove that $D(\bar{c})=\bar{0}$ for any $c\in\mathbb{R}$. Indeed

 $D(\bar{1})=D(\bar{1}\cdot\bar{1})=\bar{1}\cdot D(\bar{1})+\bar{1}\cdot D(\bar{% 1})=D(\bar{1})+D(\bar{1})=2\cdot D(\bar{1})$

and thus $D(\bar{1})=\bar{0}$. Now from linearity of $D$ we obtain that

 $D(\bar{c})=D(c\cdot\bar{1})=c\cdot D(\bar{1})=c\cdot\bar{0}=\bar{0}.$

Step two. If $f:X\to\mathbb{R}$ is continuous and $c\in\mathbb{R}$, then $f+\bar{c}$ is continuous and obviously $D(f)=D(f+\bar{c}).$ Moreover, if $x\in X$ then $D(f-\overline{f(x)})=D(f)$, but $(f-\overline{f(x)})(x)=0$. Thus we may assume that $f(x)=0$ for fixed $x\in X$.

Let $x_{0}\in X$. Now we will restrict only to such maps $f:X\to\mathbb{R}$ that $f(x_{0})=0$.

Step three. We now decompose $f$ into sum of two nonnegative functions. Indeed, if $f:X\to\mathbb{R}$ is continuous, then define $f^{+},f^{-}:X\to\mathbb{R}$ by the formula:

 $f^{+}(x)=\mathrm{max}(f(x),0);\ \ \ f^{-}(x)=\mathrm{max}(-f(x),0).$

Of course both $f^{-}$ and $f^{+}$ are continous, nonnegative and $f=f^{+}-f^{-}$. Thus

 $D(f)=D(f^{+})-D(f^{-}),$

so it is enough to show that $D(f)=\bar{0}$ only for nonnegative and continuous functions.

Step four. Assume that $f:X\to\mathbb{R}$ is nonnegative, continuous and $f(x_{0})=0$. Then there exists $g:X\to\mathbb{R}$ continuous such that $g^{2}=f$ (indeed $g=\sqrt{f}$ and it is well defined, continuous map, because $f$ was nonnegative). Then we have

 $D(f)=D(g^{2})=g\cdot D(g)+g\cdot D(g)=2\cdot g\cdot D(g).$

Now we have $g(x_{0})=\sqrt{f(x_{0})}=0$ and thus

 $D(f)(x_{0})=2\cdot g(x_{0})\cdot D(g)(x_{0})=0.$

Now we can take any $x\in X$ and repeat steps two, three and four to get that for any $x\in X$ we have

 $D(f)(x)=0$

and thus

 $D(f)=\bar{0},$

which completes the proof. $\square$

Remark. Note that this proof cannot be repeated if we (for example) consider the set of all smooth functions $C^{\infty}(M,\mathbb{R})$ on a smooth manifold $M$, because $f^{+}$, $f^{-}$ and $\sqrt{f}$ need not be smooth.

Title derivations on a ring of continuous functions DerivationsOnARingOfContinuousFunctions 2013-03-22 18:37:25 2013-03-22 18:37:25 joking (16130) joking (16130) 14 joking (16130) Example msc 17A36 msc 16W25 msc 13N15