Consider the delta dirac distribution $\delta (\varphi) = \varphi (0), \varphi \in \mathcal{S}(\mathbb{R}^n)$ (the Schwartz space). I know that $\delta ^{'} (\varphi) = – {\varphi }^{'} (0)$. How can I prove $\delta^{'}$ is not given by a measure, that is , doesn't exists a measure $\mu$ such that
$$\delta^{'} (\varphi) = \displaystyle\int_{R^n} \varphi (x)\,d \mu (x) $$
I have no idea how to proceed. Someone can give me a hint ?
Thanks in advance
Best Answer
Choose a test function $\varphi$ with $\frac{\partial \varphi}{\partial x_i}(0) = 1$. For $n \in \mathbb{Z}^+$, let
$$f_n(x) = \frac1n\varphi\left(nx\right).$$
For every measure $\mu$, we have
$$\lim_{n\to\infty} \int f_n(x)\,d\mu = 0$$
by the dominated convergence theorem. But
$$\frac{\partial\delta}{\partial x_i}[f_n] = - \frac{\partial f_n}{\partial x_i}(0) = -1$$
for all $n$.