# Class $C^{- \infty }$ functions?

If my understanding is correct, a class $C^{-1}$ function (in terms of smoothness, of course) can be thought of as a function which integrates to a class $C^{0}$ function. And when we differentiate (in the appropriate sense, of course) it, we can construct a class $C^{-2}$ function.

An example would be these three functions, ordered from highest to lowest class: $$f(x) = |x|$$ $$f(x) = \theta (x)$$ $$f(x) = \delta (x)$$

(those are Heaviside step and Dirac delta function).

A natural question that comes to my mind is, is there such a thing as a $C^{- \infty } - smooth$ function? What about discontinuous-everywhere fuctions?

I have some ideas, but I'm not sure what to think of them, so I'd appreciate some constructive answers if they really do exist.

• The dirac is not a function. In fact, for any locally integrable function $f$, its primitive is a continuous functions, so any such function is $C^{-1}$ in your sense (and any $C^{-1}$ function should be locally integrable for the definition to make sense).
• The primitive of a smooth ($C^\infty$) function being still a smooth function, any $C^\infty$ function is a $C^{-\infty}$ function. In fact, it's obvious from your definition that if $f$ is $C^k$ and $p \leq k$, then $f$ is $C^p$.
• The right notions of regularity and derivation for generalized functions (such as the dirac) is the distribution settings. I invite you to read wikipedia's article on distributions. In short, you replace functions by linear forms on a space of test functions. Any locally integrable function gives rise to such a linear form, but some forms such as $\phi \mapsto \phi(0)$ (the dirac) are not obtained from real functions and are thought of as generalized functions. You can always differentiate as much as you want any distribution. There is a notion of order (a function is a distribution of order 0) and the derivative of a distribution of order $n$ is a distribution of order $n+1$.