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?
Here is an argument that works for signed measures. Suppose by contradiction that $\mu$ exists and let us analyse $\mu$ on $\mathbb R\setminus \{0\}$.
If $f$ is any function in $C_c^1(\mathbb R\setminus \{0\})$ then we may extend $f$ to the whole real line by setting $f(0)=0$, and it will follow that $f'(0)=0$, so $$\int f d\mu=0.$$ This implies that the restriction of $\mu$ to $\mathbb R\setminus \{0\}$ vanishes so $\mu$ must be a multiple of the Dirac measure at $0$, but this clearly doesn't work either, hence $\mu$ doesn't exist.