Prove that the function $\mathcal{D}:{C[0,1]}\rightarrow\mathbb{R}$ with given properties is such that $\mathcal{D}\equiv 0$. I found this problem in an old math competition, been battling it for the whole day with no significant progress. The properties are:


*

*$\mathcal{D}(\alpha f + \beta g) = \alpha \mathcal{D}(f) + \beta \mathcal{D}(g)$;

*$\mathcal{D}(fg) = g(1/2)\mathcal{D}(f) + f(1/2)\mathcal{D}(g), \, \forall\alpha,\beta\in\mathbb{R}, \, \forall f,g \in C[0,1].$
So far I have found out that $\mathcal{D}(c) = 0$ for any constant $c$, and if there is $f$ such that $\mathcal{D}(f)\neq0$, then we can build $\tilde{f}$ such that:


*

*$\mathcal{D}(\tilde{f}) = \mathcal{D}(f)$

*$\mathcal{D}(\tilde{f})>0$; 

*$\tilde{f}(x)>0,\, \forall x\in[0,1]$; 

*$\tilde{f}(1/2) = 1;$ 

*$\mathcal{D}(\frac{1}{\tilde{f}}) = -\mathcal{D}(\tilde{f})$,
all that in hope to find a contradiction $\mathcal{D}(f)\neq0 \implies \mathcal{D}(f) = 0$.
 A: Despite this problem appearing on a math competition, this is actually a somewhat well-known theorem (see for instance Do "exotic vectorfields" exist? for a closely related result).  Morally, what's going on here is that the given axioms say that $\mathcal{D}$ should be something like "differentiation at $1/2$", but no nontrivial such thing should exist because not all functions in $C[0,1]$ are differentiable at $1/2$.
Here, then, is the key insight.  Notice that $\mathcal{D}(f^2)=2f(1/2)\mathcal{D}(f)$.  In particular, if $f(1/2)=0$ then $\mathcal{D}(f^2)=0$.  But, every nonnegative function $g$ with $g(1/2)=0$ can be written as $f^2$ for some $f$ (just take $f=\sqrt{g}$--here it is crucial that our domain is $C[0,1]$ and not $C^1[0,1]$ since $\sqrt{g}$ will be continuous but not necessarily differentiable!).  So, $\mathcal{D}(g)=0$ for every such $g$.  Moreover, if $g$ is any function with $g(1/2)=0$, we can write it as a difference of two such nonnegative functions, namely $g=\max(g,0)-\max(-g,0)$.  So $\mathcal{D}(g)=0$ for any $g$ such that $g(1/2)=0$.  Finally, we can always subtract a constant from any function to make the value at $1/2$ equal to $0$, so since $\mathcal{D}$ also vanishes on constants, $\mathcal{D}=0$.
