# A distribution that is not a Radon measure

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.

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?
• I can't see clearly this reasoning, Tom... However, I found this question solved for $N=1$ ($\phi'(0)$), using some functions $\varphi_m(x)=sin(mx)\varphi(x)$, where $\varphi\in\cal{D}(\mathbb{R})$,$supp(\varphi)\subseteq[-1,1]$, $0\leq\varphi\leq 1$ and $\varphi(x)=1$ if $|x|\leq\frac{1}{2}$. As $\varphi_m'(0)=m\leq|\mu|([-1,1])<\infty$ $\forall m\in\mathbb{N}$, we get the contradiction. How could be this argument adapted for N=2? Commented Nov 9, 2013 at 14:18
• In the real case, it seems you could use this same sort of argument with $\varphi_m(x) = \cos(mx)\varphi(x)$ so that $$m^2 = |\varphi''_m(0)| = \left| \int \varphi_m \,d\mu \right| \leq \int_{[-1,1]}|\varphi_m| \,d|\mu| \leq |\mu|([-1,1])$$.
• But shouldn't those $\varphi_m$ be functions of two variables? Commented Nov 9, 2013 at 15:00
• Looks like it would work for a judicious choice of $\varphi(x,y)$!
• Considering $\varphi_m(x,y)=cos(m(x+y))\varphi(x,y)$ would be enough? (and preserving $\varphi\in\cal{D}(\mathbb{R}^2)$, $supp(\varphi)\subseteq[-1,1]$x$[-1,1]$, $0\leq\varphi\leq 1$,$\varphi(x)=1$ if $|x|\leq\frac{1}{2}$) Commented Nov 9, 2013 at 15:17