Prove that a distribution has its primitive distribution. Given a distribution $F$ in $\mathbb{R}$,prove that there exists a distribution $F_1$ such that $$\frac{d}{dx}F_1=F$$,and it is unique up to an additive constant.
 A: Existence.
Let $T\in\mathcal{D}'(\mathbb{R})$, to get insight let us show the following:

Proposition. If $T$ is compactly supported, then there exists $S\in\mathcal{D}'(\mathbb{R})$ such that $S'=T$.

Proof. Since $T$ is compactly supported, $\{\textrm{supp}(T),\textrm{supp}(H)\}$ is convolutive and $T$ and $H$ are convolable. Let define $S:=T\ast H\in\mathcal{D}'(\mathbb{R})$, one has: $$S'=(T\ast H)'=T\ast H'=T\ast\delta_0=T.$$
Whence the result. $\Box$
Let $\chi\in\mathcal{C}^{\infty}(\mathbb{R})$ such that $0\leqslant\chi\leqslant 1$, $\chi_{\vert(-\infty,0]}=0$ and $\chi_{\vert [1,+\infty)}=1$. One has the:

Proposition. 
  
  
*
  
*$H$ and $\chi$ are convolable. 
  
*$(1-H)$ and $(1-\chi)$ are convolable.
  

Proof. One has $\textrm{supp}(H)=[0,+\infty)$ and $[1,+\infty)\subseteq\textrm{supp}(\chi)\subseteq[0,+\infty)$. Let define the following map: $$\Sigma:\left\{\begin{array}{ccc}\textrm{supp}(H)\times\textrm{supp}(\chi)&\rightarrow&\mathbb{R}\\(x,y)&\mapsto&x+y\end{array}\right..$$
Let $K$ be a compact of $\mathbb{R}$ there exists $A\in[0,+\infty)$ such that $K\subseteq[-A,A]$. Therefore, one has: $$\Sigma^{-1}(K)\subset\Sigma^{-1}([-A,A]).$$ Let $(x,y)\in\Sigma^{-1}(K)$, one has: $$|x+y|\leqslant A.$$ Therefore, since $(x,y)\in [0,+\infty)^2$, one has: $$\left\{\begin{array}{ll}0\leqslant x\leqslant x+y\leqslant A\\0\leqslant y\leqslant x+y\leqslant A\end{array}\right..$$
Hence, one has $(x,y)\in[0,A]^2$ and $\Sigma^{-1}(K)\subseteq[0,A]^2$. Therefore, $\Sigma^{-1}(K)$ is bounded and since $\Sigma$ is continuous and $K$ is closed, $\Sigma^{-1}(K)$ is closed. Finally, since $\mathbb{R}$ is finite-dimensional, $\Sigma^{-1}(K)$ is compact and $\Sigma$ is proper. By definition, $\{\textrm{supp}(H),\textrm{supp}(\chi)\}$ is convolutive and $H$ and $\chi$ are convolable.
Similary, since $\textrm{supp}(1-H)=(-\infty,0]$ and $\textrm{supp}(1-\chi)\subseteq(-\infty,1]$, $H$ and $\chi$ are convolable. $\Box$
Let define $S:=\chi T\ast H-(1-\chi)T\ast(1-H)\in\mathcal{D}'(\mathbb{R})$, one has: $$S'=(\chi T\ast H)'-((1-\chi)T\ast(1-H))'=\chi T\ast\delta_0-(1-\chi T)\ast(-\delta_0)=T.$$
Whence the result.
Uniqueness up to an additive constant. Let $S_1,S_2\in\mathcal{D}'(\mathbb{R})$ such that ${S_1}'=T$ and ${S_2}'=T$, hence : $$(S_1-S_2)'=0.$$
There exists $C\in\mathbb{R}$ such that: $$S_1-S_2=C.$$
Whence the result.
Remark. I assume you have learnt that the solutions of $T'=0$ in $\mathcal{D}'(\mathbb{R})$ are the regular distributions associated with the constant functions. Otherwise, notice that $\varphi\in\mathcal{C}_0^{\infty}(\mathbb{R})$ is the derivative of a function in $\mathcal{C}_0^{\infty}(\mathbb{R})$ if and only if $\displaystyle\int_{\mathbb{R}}f(x)\,\mathrm{d}x=0$.
A: Lemma
Suppose that $\phi\in C^\infty_c(\mathbb{R})$ such that $\langle 1, \phi \rangle := \int_{-\infty}^{\infty} \phi(x) \, dx = 0.$ Then there exists a unique $\psi\in C^\infty_c(\mathbb{R})$ such that $\psi'=\phi.$ Furthermore, $\operatorname{supp}\psi \subseteq [\min\operatorname{supp}\phi, \max\operatorname{supp}\phi].$
Proof
Set $\psi(x) = \int_{-\infty}^{x} \phi(t) \, dt.$ It's left to the reader to show that 1) $\psi'=\phi,$ 2) $\psi \in C^\infty_c(\mathbb{R}),$ 3) there is no other $\xi \in C^\infty_c(\mathbb{R})$ such that $\xi'=\phi$ (uniqueness), and 4) $\operatorname{supp}\psi \subseteq [\min\operatorname{supp}\phi, \max\operatorname{supp}\phi].$

Theorem
Suppose that $v \in \mathcal{D}'(\mathbb{R}).$ Then there exists $u \in \mathcal{D}'(\mathbb{R})$ such that $u'=v.$
Proof
Construction of $u$:
Fix $\rho\in C^\infty_c(\mathbb{R})$ such that $\langle 1, \rho \rangle = 1.$ Given $\phi \in C^\infty_c(\mathbb{R})$ set $\tilde\phi = \phi - \langle 1, \phi \rangle \rho.$ Then $\langle 1, \tilde\phi \rangle = 0$ so there exists $\psi \in C^\infty_c(\mathbb{R})$ such that $\psi' = \tilde\phi.$ Now set
$\langle u, \phi \rangle := -\langle v, \psi \rangle.$
We need to show that $u \in \mathcal{D}'(\mathbb{R})$ and that $u'=v.$
Proof of $u \in \mathcal{D}'(\mathbb{R})$:
Let $\phi_j \to 0$ in $C^\infty_c(\mathbb{R}).$ We need to show that $\langle u, \phi_j \rangle \to 0.$ But this will follow from $v$ being a distribution if we can only show that $\psi_j \to 0$ in $C^\infty_c(\mathbb{R}),$ where $\psi_j(x)$ is constructed from $\phi_j$ as above.
Since $\phi_j \to 0$ in $C^\infty_c(\mathbb{R})$ there exists a compact set $K$ such that $\operatorname{supp}\phi_j \subseteq K$ for all $j.$ Then $\operatorname{supp}\tilde\phi_j \subseteq K \cup \operatorname{supp}\rho$ and $\operatorname{supp}\psi_j \subseteq [\min(K \cup \operatorname{supp}\rho), \max(K \cup \operatorname{supp}\rho)] =: K'.$ Thus there exists a compact set $K'$ such that $\operatorname{supp}\psi_j \subseteq K'$ for all $j$.
Now note that
$$
\sup|\tilde\phi_j|
= \sup|\phi_j - \langle 1, \phi_j\rangle \rho|
\leq \sup|\phi_j| + |\langle 1, \phi_j\rangle| \sup|\rho|
\leq \sup|\phi_j| + \mu(\operatorname{supp}\phi_j) \sup|\phi_j| \sup|\rho|
\leq (1 + \mu(K)\sup|\rho|) \sup|\phi_j|
$$
so
$$
\sup|\psi_j^{(0)}|
= \sup_{x\in\mathbb{R}} \big| \int_{-\infty}^{x} \tilde\phi_j(t) \, dt \big|
\leq \sup_{x\in\mathbb{R}} \int_{-\infty}^{x} \big| \tilde\phi_j(t) \big| \, dt
= \int_{-\infty}^{\infty} |\tilde\phi_j(t)| \, dt
\leq \mu(\operatorname{supp}\tilde\phi_j) \sup|\tilde\phi_j|
\leq \mu(K+\operatorname{supp}\rho) \sup|\tilde\phi_j|
\to 0
$$
and for $k=1,2,3,\ldots$
$$
\sup|\psi_j^{(k)}|
= \sup|\tilde\phi_j^{(k-1)}|
\leq (1+\mu(K)) \sup|\rho|) \sup|\phi_j^{(k-1)}|
\to 0.
$$
Thus, $\psi_j \to 0$ in $C^\infty_c(\mathbb{R})$ and
$$
\langle u, \phi \rangle
= -\langle v, \psi \rangle
\to 0
$$
so $u$ is a distribution.
Proof of $u'=v$: First note that $\langle 1, \phi' \rangle = 0$ so $\tilde{\phi'} = \phi'$ and the "$\psi$" for $\phi'$ is just $\phi$. Therefore,
$$
\langle u', \phi \rangle
= -\langle u, \phi' \rangle
= \langle v, \phi \rangle,
$$
i.e. $u'=v.$
