Show that $f \in L_\infty(\mathbb{R})$ as soon as $f'$ is a Radon measure of finite total variation.

Question

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.

You are right, I work with the convention that a measure in $\mathcal{M}(\mathbb{R})$ has a finite total variation but I should specify it explicitely. — Goulifet, Feb 21 '22 at 18:16
When you differentiate, I guess you should take into account the jumps of the indicator functions. If I am correct, this gives $f'(x) = \sum_{n \neq 0} n 1_{(n,n+n^{-3})}(x) + (n+n^{-3}) \delta(x-n-n^{-3}) - n\delta(x-n)$. And the Dirac impulses imply that $| f' |_{TV} = \infty$. Do you agree? — Goulifet, Feb 21 '22 at 21:37

Mittens · Accepted Answer · 2022-02-23T18:03:50.190

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

Problem 0: Prove that if $u\in\mathcal{D}'(\mathbb{R})$ is such that $Du(\phi)=-u(\phi')=0$ for all $\phi\in\mathcal{D}(\mathbb{R})$, then $u\equiv constant$, that is, for some $c\in\mathbb{R}$, $u(\phi)=c\int_\mathbb{R}\phi(s)\,ds$ for all $\phi\in\mathcal{D}(\mathbb{R})$.

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.

Thanks. I guess the proof also works without the assumption that $f$ is locally integrable, simply observing that $\int_{\mathbb{R}} \varphi(s) \nu_f(ds) = \langle \nu_f , \varphi \rangle = \langle f', \varphi \rangle = - \langle f, \varphi' \rangle$. Why did you need this assumption? — Goulifet, Feb 23 '22 at 14:50
And you indeed need it in the different steps, including the integration by part. So I guess to be complete I should prove first that $f \in L_1^{loc}$. — Goulifet, Feb 23 '22 at 16:32
I don't assume initially that $f$ is a function, only that it's a generalized function. Hence the weak derivative $f'$ is well-defined in $\mathcal{D}'(\mathbb{R})$, and assumed to be a finite Radon measure. Hence, I cannot write a priori that $f(\phi) = \int \phi f d \lambda$. — Goulifet, Feb 23 '22 at 16:49
@Goulifet: It seems that the problem is not complicated. MaoWao commented that $f(x):=Du(-\infty,x]$ and $u$ have the same derivative (in $\mathcal{D}'(\mathbb{R})$) hence, they differ by a constant. — Mittens, Feb 23 '22 at 17:44

Show that $f \in L_\infty(\mathbb{R})$ as soon as $f'$ is a Radon measure of finite total variation.

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