derivations on a ring of continuous functions

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

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

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

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

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

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

Step two. If $f:X\to ℝ$ is continuous and $c\in ℝ$, then $f+\overline{c}$ is continuous and obviously $D(f)=D(f+\overline{c}).$ 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 $f:X\to ℝ$ that $f(x_0)=0$.

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

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

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 $D(f)=\overline{0}$ only for nonnegative and continuous functions.

Step four. Assume that $f:X\to ℝ$ is nonnegative, continuous and $f(x_0)=0$. Then there exists $g:X\to ℝ$ 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

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

which completes the proof. $\mathrm{\square }$

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

Title	derivations on a ring of continuous functions
Canonical name	DerivationsOnARingOfContinuousFunctions
Date of creation	2013-03-22 18:37:25
Last modified on	2013-03-22 18:37:25
Owner	joking (16130)
Last modified by	joking (16130)
Numerical id	14
Author	joking (16130)
Entry type	Example
Classification	msc 17A36
Classification	msc 16W25
Classification	msc 13N15