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 $\bar{c}(x)=c$.

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

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

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:

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

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

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

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

and thus

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.