Are bounded linear functionals on $L^{\infty}$ of “bounded variation?”

Question

Let $(X,\mathscr M,\mu)$ be a measure space and let $L^{\infty}$ be the set of (equivalence classes of) essentially bounded measurable functions on it. Suppose that $\phi\in (L^{\infty})^*$; that is, $\phi:L^{\infty}\to\mathbb C$ is a bounded linear functional on $L^{\infty}$.

Conjecture: There exists some $C>0$ such that $$\sup\left\{\sum_{j=1}^n|\phi(\chi_{E_j})|\,\Bigg|\,n\in\mathbb N,\,E_1,\ldots,E_n\text{ are disjoint and measurable}\right\}\leq C,$$ where $\chi_{E}$ is the characteristic function of $E\in\mathscr M$. In a sense, this conjecture amounts to saying that the (finitely additive) complex measure induced by $\phi$, $v(E)\equiv\phi(\chi_E)$, on $\mathscr M$ is of bounded variation.

Do you think if the conjecture is true, at least if $\mu$ is finite?

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More generally, is the following conjecture true? “If

• $\nu:\mathscr M\to \mathbb C$,
• $\nu(\varnothing)=0$,
• $n\in\mathbb N,\,E_1,\ldots,E_n\text{ are disjoint and measurable}$ $\Longrightarrow$ $\nu(\cup_{j=1}^n E_j)=\sum_{j=1}^n\nu(E_j)$,
• $\sup_{E\in\mathscr M}|\nu(E)|<\infty$,

then $$\sup\left\{\sum_{j=1}^n|\nu({E_j})|\,\Bigg|\,n\in\mathbb N,\,E_1,\ldots,E_n\text{ are disjoint and measurable}\right\}<\infty.\text{”}$$ I.e., are all finitely additive, absolutely bounded complex measures of bounded variation?

Answer 1

Good question.

Actually, the supremum is equal to $\lVert \phi \rVert$.

Two hints:

1. finite linear combinations of characteristic functions are dense in $L^\infty$.

2. Fix pairwise disjoint $E_1,\dots,E_n$ and consider linear combinations $f = \sum a_k \chi_{E_k}$ with $|a_k| \leq 1$. How should you pick the $a_k$ in order to ensure that $\phi(f)$ is real and as large as possible?

Answer 2

It turns out the second part is even easier (and it implies the first). Suppose that $n\in\mathbb N$, $E_1,\ldots, E_n$ are measurable and disjoint and that $C>0$ is such that $|\nu(E)|\leq C$ for all $E\in\mathscr M$. Let \begin{align*} N_R^+\equiv&\,\{j\in\{1,\ldots,n\}\,|\,\Re\nu(E_j)\geq0\},\\ N_R^-\equiv&\,\{j\in\{1,\ldots,n\}\,|\,\Re\nu(E_j)<0\},\\ N_I^+\equiv&\,\{j\in\{1,\ldots,n\}\,|\,\Im\nu(E_j)\geq0\},\\ N_I^-\equiv&\,\{j\in\{1,\ldots,n\}\,|\,\Im\nu(E_j)<0\}.\\ \end{align*}

It follows that \begin{align*} \sum_{j=1}^n|\nu(E_j)|\leq&\,\sum_{j=1}^n|\Re\nu(E_j)|+\sum_{j=1}^n|\Im\nu(E_j)|\\ =&\,\sum_{j\in N_R^+}\Re\nu(E_j)-\sum_{j\in N_R^-}\Re\nu(E_j)+\sum_{j\in N_I^+}\Im\nu(E_j)-\sum_{j\in N_I^-}\Im\nu(E_j)\\ =&\,\Re\nu\bigg(\bigcup_{j\in N_R^+} E_j\bigg)-\Re\nu\bigg(\bigcup_{j\in N_R^-} E_j\bigg)+\Im\nu\bigg(\bigcup_{j\in N_I^+} E_j\bigg)-\Im\nu\bigg(\bigcup_{j\in N_I^-} E_j\bigg)\\ =&\,\bigg|\Re\nu\bigg(\bigcup_{j\in N_R^+} E_j\bigg)\bigg|+\bigg|\Re\nu\bigg(\bigcup_{j\in N_R^-} E_j\bigg)\bigg|+\bigg|\Im\nu\bigg(\bigcup_{j\in N_I^+} E_j\bigg)\bigg|+\bigg|\Im\nu\bigg(\bigcup_{j\in N_I^-} E_j\bigg)\bigg|\\ \leq&\,\bigg|\nu\bigg(\bigcup_{j\in N_R^+} E_j\bigg)\bigg|+\bigg|\nu\bigg(\bigcup_{j\in N_R^-} E_j\bigg)\bigg|+\bigg|\nu\bigg(\bigcup_{j\in N_I^+} E_j\bigg)\bigg|+\bigg|\nu\bigg(\bigcup_{j\in N_I^-} E_j\bigg)\bigg|\leq4 C. \end{align*}

Reference: Lemma III.1.5 in Dunford–Schwartz (1958, p. 97).