A distribution that is not a Radon measure

Question

I need help with this question:

Let $N=2$, $\Omega=\mathbb{R}^2$ and $T:\cal{D}(\Omega)\to\mathbb{C}$, defined as:

$<T,\phi>=\frac{\partial^2\phi}{\partial x\partial y}(0)=D^{(1,1)}\phi(0)$

I have to show that it is a distribution, but not a Radon measure.

Any ideas?

Thanks a lot.

Answer 1

Hint: If $\mu$ is a radon measure such that $$ \langle T, \phi \rangle = \int \phi \,d\mu$$ Then for any $\phi$ such that $0 \not\in \operatorname{supp}(\phi)$, it must be that $\int \phi \,d\mu =0$. So, it must be that $\mu$ is a point mass (dirac-type measure) at $0$ (why!? [think inner regularity]). So $\mu = \lambda \delta_0$, meaning that $\int \phi \,d\mu = \lambda \phi(0)$. Now, why is this impossible?