The derivative can be seen as a linear operator $\frac{d}{dx} : C^1(\mathbb{R}) \rightarrow C^0(\mathbb{R})$. Does there exist another linear operator $ D : C^1(\mathbb{R}) \rightarrow C^0(\mathbb{R}) $ such that the composition of $D$ with itself gives the derivative operator? (That is, $D \circ D = \frac{d}{dx}$)
My inclination would be that such an operator would not exist, but I haven't been able to prove this.
