After defining convolutions of two tempered distributions $T,S \in \mathcal{S}'$; and convolutions of functions $f \in \mathcal{S}$ with distributions $T \in \mathcal{S}'$, my textbook goes on to discuss their application to solving linear PDEs, in particular it claims that for the PDE:
$$ D u = f $$
The solution is, in the distribution sense, $u = f * g$ with $g : \ Dg = \delta$. (here the notation is "abused")
My question: How does one convert a "normal" PDE to a "distribution" PDE ? What did my textbook mean to say ? Is $Du =f$ :
$$ \langle D T_u, \phi \rangle = \langle T_f , \phi \rangle \ ? $$
or is it:
$$ \langle D T_u, \phi \rangle = f \ ? $$
Your textbook means that $u$ is a distribution such that $\langle Du, \phi \rangle = \langle T_f, \phi \rangle$ for every test function $\phi$.
What would the formula $\langle DT_u, \phi \rangle = f$ mean? The left hand side is a number, while the right hand side is a function.
An example
In 3 dimensions, one has $$ \Delta G(x) = \delta $$ where $\Delta$ is the Laplace operator, $G(x)=\frac{1}{4\pi\|x\|},$ and $x\in\mathbb{R}^3$ with $\|x\| = \sqrt{x_1^2+x_2^2+x_3^2}.$
A solution to $\Delta u = \rho,$ where $u$ and $\rho$ are distributions, is then given by $$ u = G*\rho . $$ Often both $u$ and $\rho$ are induced by ordinary functions and the convolution is then defined in both a classical sense (i.e. using an integral) and in the distributional sense. To be more precise, if $f$ and $g$ are functions so that $f*g$ is defined using the integral definition, then $T_f*T_g = T_{f*g}$ in the distributional sense.