Here is question I tried to answer for some time - it seems to be straightforward, but I have trouble figuring it out.

Let $\Omega$ be a compact domain in $\mathbb{R}^n$. For any signed Borel measure $\mu$ on $\Omega$ let $\mu_+$ denote its positive part (obtained by Hahn-Jordan decomposition). My question is:

Is taking the positive part a continuous operation, i.e. does $\mu^n\to \mu$ in the space signed Borel measures imply $\mu^n_+\to \mu_+$ in the same space?

Of course, the answer would be positive if $$\|w^n_+-w_+\|_{\mathfrak{M}}\leq \|w_n-w\|_{\mathfrak{M}}$$ where $\|\cdot\|_{\mathfrak{M}}$ denotes the variation norm, but I could not see this.


If $\mu =f\cdot \lambda$ for a positive measure $\lambda$ (i.e., $\mu(A)=\int_A fd\lambda$), isn't then $\mu_+= f_+ \cdot\lambda$ (where $f_+$ is the positive part $\max\{f,0\}$ of $f$) and $\|\mu\|=\int|f|d\lambda$? Then $\|\mu_+-\nu_+\| \le \|\mu-\nu\|$ just follows from Radon-Nikodym (applied to $\lambda=|\mu|+|\nu|$) and $|f_+-g_+|\le |f-g|$.

  • $\begingroup$ Thanks! That was indeed not too hard. Clever use of Radon Nikodym... $\endgroup$ – Dirk Oct 18 '18 at 13:26
  • $\begingroup$ Here you tacitly assume that both $|\mu|$ and $|\nu|$ are absolutely continuous with respect to the same positive measure $\lambda$ so you can invoke the densities $f$ and $g$. $\endgroup$ – Liviu Nicolaescu Oct 18 '18 at 13:27
  • $\begingroup$ I think he just takes $\lambda =|\mu| + |\nu|$. $\endgroup$ – Dirk Oct 18 '18 at 13:40
  • $\begingroup$ OK. I get now. Nice $\endgroup$ – Liviu Nicolaescu Oct 18 '18 at 14:06

The Borel measures on $\Omega$ can be identified with continuous linear functionals $\newcommand{\bR}{\mathbb{R}}$ $\mu:C(\Omega)\to\bR$. Assume that $\Omega$ is compact so $C(\Omega)$ is Banach space. Denote by $\Vert-\Vert$ the sup norm on $C(\Omega)$ and by $\newcommand{\eM}{\mathscr{M}}$ $\eM(\Omega)$ the dual of $C(\Omega)$ equipped with the dual norm $$ \Vert \mu\Vert_*:=\sup_{\Vert f\Vert\leq 1}|\mu(f)|. $$ Set $$ C(\Omega)_+:=\big\{ f\in C(\Omega):\;\; f(x)\geq 0,\;\;\forall x\in\Omega)\big\}. $$Let $\mu\in \eM(\Omega)$ and $f\in C(\Omega)$. Then, for any $f\in C(\Omega)_+$ we have (see Theorem 4.3.2. of R.E. Edwards: Functional Analysis) $$ \mu_+(f)=\sup_{0\leq g\leq f} \mu(g). $$ Let $\mu,\nu\in \eM(\Omega)$, and $f\in C(\Omega)_+$. Then for any $0\leq g\leq f$ we have

$$ \mu(g)-\nu(g)\leq \Vert\mu-\nu\Vert_*\Vert g\Vert \leq \Vert\mu-\nu\Vert_*\Vert f\Vert. $$ Hence, for any $0\leq g\leq f$ $$ \mu(g)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\nu(g), $$ and, symmetrically, $$ \nu(g)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\mu(g). $$ Taking the sup on both sides of the above inequalities we deduce $$ \mu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\nu_+(f)\implies \mu_+(f)- \nu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert, $$ $$ \nu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert+\mu_+(f)\implies \nu_+(f)- \mu_+(f)\leq \Vert\mu-\nu\Vert_*\Vert f\Vert. $$ Hence $$ \big\vert (\mu_+-\nu_+)f\big\vert\leq \Vert\mu-\nu\Vert_*\Vert f\Vert. $$ This implies $$ \Vert \mu_+-\nu_+\Vert_*\leq \Vert\mu-\nu\Vert_*. $$


Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.