Is there some sense in which one could write any distribution as a sum of this sort?


$$A(x,y)=\sum\_{n=0}^{\infty}a\_n(x)i^n\frac{\partial^n}{\partial x^n}\delta (x-y)$$


Provided that the rhs acting on a test function is convergent for all $x$.



Derivatives of delta function as a basis for distributions
Seconding @Victor Ivrii's good answer, with a few more points:


First, as Victor noted, a (properly) infinite sum of the sort written has convergence problems. This is already essentially visible if we just ignore the $y$-variable. Then we're asking whether an infinite sum of derivatives of $\delta$ (all just at $0$) can be a distribution. It is not at all obvious that this is not possible, I think. But there are at least two (covertly equivalent) proofs. First, we can recall E. Borel's theorem that an arbitrary sequence of complex numbers can be the Taylor-Maclaurin coefficients at $0$ of a smooth function. Thus, in a purported infinite linear combination, no matter how rapidly we make the coefficients decay, there is a smooth function whose Taylor-Maclaurin coefficients grow faster, and the sum (implied by evaluating the infinite sum on such a function) does not converge. Another way is to use Taylor-Maclaurin expansions with error term to classify distributions supported at $0$: *finite* linear combinations of derivatives of $\delta$.


On another hand, a similar argument *does* show that all distributions on $\mathbb R^{m+n}$ supported on $\mathbb R^m\times \{0\}$ *are* (finite!) sums of transverse derivatives, followed by evaluation on the smaller $\mathbb R^m$. So, for example, the finite version of your expression *does* give *all* distributions on $\mathbb R^2$ supported on the diagonal.


As in my comment, apart from inescapable issues of convergence (that is, it's not about failure to converge "pointwise", but failure to converge even in the weak dual topology, etc.), differentiation never increases support, so a-priori your expressions only give distributions supported on the diagonal.


Using the extended sense of Fourier transform is fine, of course, if one is careful, but/and will *not* truly produce something with larger support, etc.


