# 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 $\overline{c}$ we will denote constant function at $c$, i.e. $\overline{c}:X\to \mathbb{R}$ is defined by $\overline{c}(x)=c$.

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

Proof. Step one. We will prove that $D(\overline{c})=\overline{0}$ for any $c\in \mathbb{R}$. 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 $f:X\to \mathbb{R}$ is continuous and $c\in \mathbb{R}$, 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 $f:X\to \mathbb{R}$ 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 $D(f)=\overline{0}$ 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

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

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^{} ${C}^{\mathrm{\infty}}(M,\mathbb{R})$ on a smooth manifold^{} $M$, because ${f}^{+}$, ${f}^{-}$ and $\sqrt{f}$ need not be smooth.

Title | derivations on a ring of continuous functions |
---|---|

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 |