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?
 A: Assume that there is some (positive) measure $\mu$ that induces this functional. Clearly $\mu\ne0$.
Take $f_n(x)$ a function such that

*

*$f_n(x)=1$ if $x\in[-n,n]$.

*$f_n(x)=0$ if $x\notin [-n-1,n+1]$.

*$f_n\in{\cal C}^1(\Bbb R)$.

*$0\le f_n(x)\le 1$ for all $x\in\Bbb R$.

Then $I(f_n)=f_n'(0)=0$, obviously.
On the other hand,
$I(f_n)=\int_{\Bbb R} f_nd\mu \ge \int_{(-n,n)} f_n d\mu=\int_{(-n,n)}d\mu=\mu(-n,n)
$.
So $\mu(-n,n)=0$ for all $n\in\Bbb N$; and since $\Bbb R=\bigcup_n (-n,n)$, by subaditivity, $\mu(\Bbb R)\le \sum_n \mu(-n,n)=0$, so $\mu=0$, wich is a contradiction.
A: 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.
A: 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.
