Why is any distribution is the derivative of another distribution? 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.


*

*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.

*Why is $u: Ran(d/dx)\to\mathbb{C}$ continuous? 

*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!
 A: I can answer 1:
We have to understand convergence in $\mathscr{D}$ first in order to see why $\frac{d}{dx}$ is continuous. 
Given a net $\{\phi_\beta\}\subset \mathscr{D}$ we say $\phi_\beta \rightarrow \phi$ iff for some $K\subset\mathbf{R}^n $ compact $$\|\partial^{\alpha}\phi_\beta-\partial^{\alpha}\phi\|_u\rightarrow0 $$ $\forall \alpha\in \mathbf{N}^n$ on $K$  (in our case n=1). 
So now with this definition we can consider a net $\{\phi_\beta\}\subset \mathscr{D}$ with  $\phi_\beta \rightarrow \phi$. 
We can apply the operator and see what happens. 
$$\|\partial^{\alpha}\frac{d}{dx}\phi_\beta-\partial^{\alpha}\frac{d}{dx}\phi\|_u$$
If we use the same compact set $K$ that we have from convergence of the sequence, the derivative operator is going to preserve the limit since it's only going to shift the index up by one. So we will have $$\frac{d}{dx}\phi_\beta \rightarrow \frac{d}{dx}\phi$$ and hence $\frac{d}{dx}$ is  continuous linear operator on the space of test functions. 
A: 
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 $\langle 1, \varphi' \rangle = 0$.
