Distributional derivative of one-parameter family of distributions

1. Jun 16, 2014

Only a Mirage

Suppose, for a suitable class of real-valued test functions $T(\mathbb{R}^n)$, that $\{G_x\}$ is a one-parameter family of distributions. That is, $\forall x \in \mathbb{R}^n, G_x: T(\mathbb{R}^n) \to \mathbb{R}$.

Now, suppose $L$ is a linear differential operator. That is, $\forall g \in T(\mathbb{R}^n)$ makes sense in terms of the normal definitions of derivates (assuming, of course, that $g$ is sufficiently smooth). $L$ also has meaning when acting on distributions by interpreting all derivatives as distributional derivatives. For example, the derivative of the distribution $\frac{\partial}{\partial x_i} G_{x_0}$ is defined by: $\forall g \in T(\mathbb{R}^n), \frac{\partial}{\partial x_i} G_{x_0}(g) = G_{x_0}(- \frac{\partial}{\partial x_i}g)$.

Note that smooth functions can multiply distributions to form a new distribution in the following way. Suppose $f:\mathbb{R}^n \to \mathbb{R}$ is a smooth function. Then $f G_x$ is defined by: $\forall g \in T(\mathbb{R}^n), (f G_x) (g) = G_x(f g)$

These facts give $L G_x$ meaning.

Also, for fixed $g \in T(\mathbb{R}^n)$, $(x \mapsto G_x (g))$ is a possibly (non-smooth?) function. Define the function $\psi_g : \mathbb{R}^n \to \mathbb{R}$ by $\psi_g (x) = G_x(g)$

Now, here is my question: When is the following equality true?

$$L (\psi_g) (x_0) = (L G_{x_0}) (g), \forall x_0 \in \mathbb{R}^n$$

2. Jul 10, 2014