Let $f$ be a generalized function in $\mathcal{D}'(\mathbb{R})$ such that its derivative $\mathrm{D} f = \mu \in \mathcal{M}(\mathbb{R})$, the space of signed Radon measure with finite total variation, endowed with the total variation norm
$ \| \mu \|_{\mathcal{M}} = |\mu|(\mathbb{R}) < \infty.$
Is there any simple argument to say that $f \in L_\infty(\mathbb{R})$ is (a.e.) bounded.
Best Answer
Suppose $f\in L^{loc}_1(\mathbb{R})$ and $\nu_f:=Df\in\mathcal{M}(\mathbb{R})$. Let $F(x)=\nu_f(-\infty,x]$. It follows from Lebesgue integration by parts that for any $\phi\in\mathcal{D}(\mathbb{R})$, $$\int_{\mathbb{R}}\phi(s)\nu_f(ds)=-\int_{\mathbb{R}}F(s-)\phi'(s)\,ds=-\int_{\mathbb{R}} F(s)\,\phi'(s)\,ds$$ On the other hand, $$\int_\mathbb{R} \phi(s)\nu_f(ds)=-\int_\mathbb{R} f(s)\phi'(s)\,ds$$
Hence $$\int_{\mathbb{R}}F(s)\,\phi'(s)\,ds=\int_\mathbb{R} f(s)\,\phi'(s)\,ds$$ for all $\phi\in\mathcal{D}(\mathbb{R})$. The problem reduces to
This would imply that $f(t)=\nu_f(-\infty,t]+c$ (Lebesgue almost surely) for some $c$ and hence, $f\in L_\infty$.
Problem 0 has been discussed in MSE. See here for example.
Edit: This is to address the more general setting of the OP.
Suppose $u\in \mathcal{D}'(\mathbb{R})$ and $\mu:=Du\in\mathcal{M}(\mathbb{R})$. Define $f(x)=\mu(-\infty,x]$. $f$ is a bounded function of finite bounded variation. By Lebesgue's integration by parts, for any $\phi\in\mathcal{D}(\mathbb{R})$ $$Df(\phi):=-\int f(x)\phi'(x)\,dx=\int\phi(x)\,\mu(dx)=Du(\phi)$$ Hence $f$, as an element in $\mathcal{D}'(\mathbb{R})$, and $u$ have the same derivative. Hence, they differ by a constant $c$. This means that $u$ admits a representation in $L^{loc}_1(\mathbb{R})$, namely $$u(\phi)=\int (f(x)+c)\phi(x)\,dx,\qquad \phi\in\mathcal{D}(\mathbb{R})$$ As a consequence, $u\in L_\infty$.
Thanks to @MaoWao for his observation.