Why is any distribution is the derivative of another distribution

distribution-theoryfunctional-analysisreal-analysistopological-vector-spaces

Let $\mathcal{D}(\mathbb{R})$ be the space of compactly-supported smooth complex-valued functions on $\mathbb{R}$ (also called test functions), and let $\mathcal{D}'(\mathbb{R})$ be its dual, so that elements of $\mathcal{D}'(\mathbb{R})$ are distributions. For an arbitrary distribution $f$, how can we prove that there exists another distribution $u$ with $u'=f$?

My approach is as follows. The derivative map $d/dx: \mathcal{D}\to\mathcal{D}$ is continuous (#), and its range has codimension one, since it's given precisely by the functions whose integral over $\mathbb{R}$ is 0. Define $u$ on $Ran(d/dx)$ by $u(\phi') = -f(\phi)$ for $\phi\in\mathcal{D}$. Then $u: Ran(d/dx)\to\mathbb{C}$ is continuous (#) and it has a (non-unique) extension to $u:\mathcal{D}\to\mathbb{C}$ (#), as desired. My main issue is with the three points highlighted with (#) above.

  1. Why is $\frac{d}{dx}:\mathcal{D}\to\mathcal{D}$ continuous? I'm familiar with Banach spaces and Fréchet spaces, but $\mathcal{D}$ is something even more general, so I'm unsure if the usual semi-norm argument is enough to prove this.
  2. Why is $u: Ran(d/dx)\to\mathbb{C}$ continuous?
  3. Why does the extension exist? If we were dealing with Fréchet spaces, then we could use Hahn-Banach. But in this full generality setting, I'm unsure how to proceed.

Thank you for your help!

Best Answer

For an arbitrary distribution $f$, how can we prove that there exists another distribution $u$ with $u'=f$?

We can construct a primitive distribution.

First fix $\rho \in C_c^\infty(\mathbb R)$ such that $\langle 1, \rho \rangle = \int_{-\infty}^{\infty} \rho(t)\,dt = 1.$

Given an arbitrary $\varphi \in C_c^\infty(\mathbb{R})$ set $\tilde{\varphi} = \varphi - \langle 1, \varphi \rangle \rho$. Then the primitive function $\tilde\Phi(x) :=\int_{-\infty}^{x} \tilde\varphi(t) \, dt$ is in $C_c^\infty(\mathbb R)$ so $\langle f, \tilde\Phi \rangle$ is defined. Let $F$ be the distribution given by $\langle F, \varphi \rangle = - \langle f, \tilde\Phi \rangle$. Then $F' = f$:

$$ \langle F', \varphi \rangle = - \langle F, \varphi' \rangle = \langle f, \int_{-\infty}^{x} \left(\varphi'(t) - \langle 1, \varphi' \rangle \rho(t) \right) \, dt \rangle \\ = \langle f(x), \int_{-\infty}^{x} \varphi'(t) \, dt \rangle - \langle 1, \varphi' \rangle \langle f(x), \int_{-\infty}^{x} \rho(t) \, dt \rangle = \langle f, \varphi \rangle $$ since $\langle 1, \varphi' \rangle = 0$.

Related Question