Please mark this as duplicate if this has been asked before.
I started a deep dive in distribution theory on $\mathbb{R}^n$ I want to understand the singular support better. I know that there are regular distributions $T_u$ for $u \in L^1_{loc}(\mathbb{R}^n)$ and Delta distributions and its derivatives $\delta^{(\alpha)}_x$ with $\delta^{(\alpha)}_x (\varphi) = (\partial^\alpha \varphi) (x)$
My question is if every Distribution can be split up like that. So is the following true? $$ \text{Let}\ T \in \mathcal{D}'(\mathbb{R}^n)\text{.} \\ \text{Then there is}\ u \in L^1_{loc}(\mathbb{R}^n)\ \text{and for all}\ i\in \mathbb{N} \ \text{there is} \ b_i \in \mathbb{R} , x_i \in \mathbb{R}^n \ \text{and}\ {_i\alpha} \ \text{Multiindices such that} \\ T = T_u + \sum_{i=0}^\infty b_i \ \delta^{(_i\alpha)}_{x_i} $$ I have read of the Schwartz-Kernel-Theorem and know the notation of singular support but I could not derive this statement from anything I know nor find a counterexample. If you have any suggestions, i am grateful, as I am completely out of ideas.
This is a worthwhile issue to consider, after all! In addition to @md2perpe's example of principal value of $1/x$:
It may be illuminating to look at the analogous question on the circle, where we can show that every distribution has a Fourier expansion with coefficients of polynomial growth, and, conversely, every Fourier expansion with polynomial-growth coefficients gives a distributions. (These expansions do converge in suitable Sobolev spaces, though naturally not pointwise...)
The Fourier expansion of (periodic) Dirac $\delta$ (also known as "Dirac comb"...) has coefficients that are all $1$'s (up to some normalizing constant(s)). $k$th derivatives of $\delta$ have coefficients that are (uniform constant multiples of) $n^k$. Translation twists these by exponentials of absolute value $1$.
So the $n$th Fourier coefficients of (finite) linear combinations of derivatives of Dirac $\delta$s are finite linear combinations of terms of the form $e^{itn}n^k$ for real $t$ and non-negative integer $k$.
$L^1$ functions have Fourier coefficients that go to $0$ at infinity (though not every such sequence gives an $L^1$ function).
So, for example, $\sum_n \sqrt{|n|} e^{inx}$ cannot be expressed as a sum of an $L^1$ function and derivatives of $\delta$. :)
But, of course, the reason we have trouble thinking of distributions other than relatively elementary ones is that our descriptive apparatus and our experience has certain limitations. :) And there's the eternal hidden question about "describe in terms of what?" In terms of simpler things? Oop, but most things admit no simpler description than... themselves. :)