6
$\begingroup$

Given a $\sigma$-additive measure space $(S,\Sigma,\mu)$ and a linear operator $$ U: L^\infty \to L^\infty $$ where $L^\infty$ is the space of essentially bounded measurable functions.

Assume it is know that the operator fulfills the following three properties

  1. $ f\geq 0 \Rightarrow Uf \geq 0$,
  2. $UI = I$ where $I(x)=1$ for all $x \in S$,
  3. $ ||U||\leq 1$,

where $|| \cdot ||$ is the operator norm induced by $L^\infty$.

I would like to know if this implies the following property:

Given a family of disjoint sets $(A_i)_{i \in \mathbb{N}} \in \Sigma$ and define $B = \bigcup_{i \in \mathbb{N}} A_i$ and $B_n = \bigcup_{i =1}^n A_i$, does then the sequence converge $$ U 1_{B_n} \to U 1_B$$ pointwise $\mu$-almost surely?

I know that the sequence $( U 1_{B_n})_{n\in \mathbb{N}}$ converges $\mu$-almost surely, because from condition $1.$ it is clear that the sequence is monotonically increasing, and it is clear that the sequence is bounded $\mu$-almost surely, however I am not sure if it converges towards $1_B$. I wonder if the linearity and condition $2.$ and $3.$ imply this.

Does anyone know?

$\endgroup$
  • $\begingroup$ Are you assuming that $\mu(S)=1$? It isn't stated in the problem, but the terminology/notation (almost surely, $\mathsf 1_{B}$ instead of $\chi_B$, etc. suggests that you're working in a probability space. $\endgroup$ – Math1000 Oct 3 '15 at 12:18
  • $\begingroup$ @Math1000 No its not a probability space, but it is $\sigma$-additive, I forgot to mention that. Thanks for your note. $\endgroup$ – Adam Oct 3 '15 at 12:21
  • $\begingroup$ Also I'm assuming that the norm $\|\cdot\|$ is the $L^\infty$ norm? $\endgroup$ – Math1000 Oct 3 '15 at 12:26
  • $\begingroup$ @Math1000 it is the operator norm induced by $L^\infty$, I will also add this to the question. Thank you. $\endgroup$ – Adam Oct 3 '15 at 12:29
4
$\begingroup$

It does not follow that $U 1_{B_n} \to U 1_B$ pointwise $\mu$-almost everywhere. Consider $S = \mathbb{N},\Sigma = \mathcal{P}(\mathbb{N})$ and $\mu$ the counting measure (or any measure given by positive masses on every point, so $\mu$ could even be a probability measure).

Let $\mathscr{U}$ be a free ultrafilter on $\mathbb{N}$ and define $\lambda_{\mathscr{U}} \colon L^\infty(S,\Sigma,\mu) \to \mathbb{C}$ by

$$\lambda_{\mathscr{U}}(f) = \lim_{\mathscr{U}} f(n)$$

and $U(f) := \lambda_{\mathscr{U}}(f)\cdot I$. Then $\lambda_{\mathscr{U}}$ is a positive linear functional with norm $1$ and $\lambda_{\mathscr{U}}(I) = 1$, so $U$ satisfies the conditions.

But with $A_i = \{i\}$ we have

$$U(1_{B_n}) = 0$$

for all $n\in \mathbb{N}$. Similar constructions can be made on many measure spaces.

| cite | improve this answer | |
$\endgroup$

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.