Let $X$ be a topological space and denote by $\mathbb{R}$ the set of reals. Of course the set of all continous 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 $\bar{c}(x)=c$ .

Proposition. 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 continous and $c\in\mathbb{R}$ , then $f+\bar{c}$ is continous 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 continous, 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 continous functions.

Step four. Assume that $f:X\to\mathbb{R}$ is nonnegative, continous and $f(x_0)=0$ . Then there exists $g:X\to\mathbb{R}$ continous such that $g^2=f$ (indeed $g=\sqrt{f}$ and it is well defined, continous 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.