prove the following linear functional is not induced by measure Let the linear functional $I$ below defined on $C^1_c(\mathbb{R})$. as follows:
$$I(f) = \frac{df}{dx}(0)$$
Prove it's not induced by a measure,or even signed measure.that is no such measure makes $I(f) = \int fd\mu = \frac{df}{dx}(0)$.
I try to use sequence of function $f_n$ such that makes $\frac{df_n}{dx}(0) \to \infty$,but the total mass of $\int f_nd\mu \to 0$ ?If exist such measure it can not be the Lebesgue measure ,since we can construct such function,can it holds for some other measures?

Take $f \in C^1_c(\mathbb{R})$ with $f \geq 0$ but $f'(0)<0.$ Then $I(f)<0$ but $\int f\,d\mu \geq 0.$ Contradiction!
Such an $f$ can be constructed by smoothing
$$
f_0(x) = \begin{cases}
2-x & \text{when $-1<x<1$} \\
0 & \text{otherwise}
\end{cases}
$$
for example by convolution with a narrow enough nascent delta function.
