Derivative of delta distribution I'm reading  Reed & Simon's book on Functional Analysis. In the chapter of locally convex spaces they say:
"consider the tempered distribution $\delta'(f)=-f'(0)$, which doesn't come from a measure".
Why is that true? I've tried to prove that claim but it's been unsuccessful.
 A: The support of the $\delta'$ is $\{0\}$; for any $f$ that is $0$ on a neighborhood of $\{0\}$, $\delta'(f)=0$.
If $\delta'$ were described by a measure, $\mu$, then that measure must also be supported on $\{0\}$. Then
$$
\mu(f)=\int f\,\mathrm{d}\mu\tag{1}
$$
and $(1)$ is dependent only on the value of $f$ on $\{0\}$.  Both $1+x$ and $1$ are functions which on $\{0\}$ have the same value, but $\delta'(1+x)=1$ and $\delta'(1)=0$.
A: Suppose that $\delta'$ is given by a mesure $\mu$. Then we have for any compact $\mathrm K$ $$ \int_{\mathrm K} d\mu = 0$$
A: According to Riesz's representation theorem, a measure is a continuous linear functional on the Fréchet space of continuous functions with compact support. The functional you have here is not even well-defined on that space. 
A possible objection is that the definition should be considered on the subspace of sufficiently regular functions and then extended by density. But the assignment $$f\in C^1(\mathbb{R})\to f'(0)$$ is not continuous with respect to the topology of $C(\mathbb{R})$. For example, the sequence 
$$f_n(x)=\frac{\sin(nx)}{n}\zeta(x), $$
where $\zeta$ is a smooth cutoff function, is such that $f_n \to 0$ in $C(\mathbb{R})$ (that is, uniformly on compact sets) but $f'_n(0)=1$ for all $n$.
