Timeline for Is the total variation of a vector measure $\mu$ a (classical) measure, even when $\mu$ is not of bounded variation?
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11 events
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| Mar 31 at 20:57 | answer | added | Salix Liu | timeline score: 0 | |
| Jan 20, 2018 at 18:08 | answer | added | Pietro Majer | timeline score: 2 | |
| Jan 20, 2018 at 17:48 | comment | added | Pietro Majer | @NateEldredge Also, in any Hilbert space with an orthogonal basis $\{e_\lambda\}_{\lambda\in\Lambda}$, an element $u\in H$ is the sum of the summable family $\{(u,e_\lambda)e_\lambda\}_{\lambda\in\Lambda}$, (which of course in general is not absolutely summable). | |
| Jan 20, 2018 at 17:34 | comment | added | Nate Eldredge | Ah, I see. Thank you. Now I can think about the question. | |
| Jan 20, 2018 at 17:31 | comment | added | 0xbadf00d | @NateEldredge I'm sorry, you're right. Here is a working counterexample: Let $(\Omega,\mathcal A)=(\mathbb N,2^{\mathbb N})$, $(\alpha_n)_{n\in\mathbb N}\in c_0$ and $$\mu(I):=\sum_{i\in I}\alpha_ie^i\;\;\;\text{for }I\subseteq\mathbb N\;,$$ where $e^i\in C_0$ is such that $e^i_j=\delta_{ij}$. Then, $$|\mu|(I)=\sum_{i\in I}|\alpha_i|\;\;\;\text{for all }I\subseteq\mathbb N$$ and hence $\mu$ has bounded variation iff $\alpha\in\ell_1$. | |
| Jan 20, 2018 at 17:22 | comment | added | Nate Eldredge | @0xbadf00d: I don't know, is it? If not, what's a counterexample? Also, what's an example of a sequence in a Banach space that is summable but not absolutely summable? | |
| Jan 20, 2018 at 17:00 | comment | added | 0xbadf00d | @JeanDuchon $(1)$ is summability in $E$: $(x_i)_{i\in I}\subseteq E$ is called summable iff there is a $x\in E$ with $$\forall\varepsilon>0:\exists J\subseteq I\text{ with }|J|\in\mathbb N:\forall K\subseteq I\text{ with }|K|\in\mathbb N\text{ and }J\subseteq K:\left\|x-\sum_{k\in K}x_k\right\|_E<\varepsilon\;.$$ If $E$ is complete (which is the case in the question) $\sum_{i\in I}\left\|\mu(A_i)\right\|_E<\infty$ is a sufficient, but not necessary, condition. | |
| Jan 20, 2018 at 15:18 | comment | added | Nate Eldredge | @JeanDuchon: Yes, that's what was worrying me. | |
| Jan 20, 2018 at 15:10 | comment | added | Jean Duchon | What does (1) mean exactly? A priori the sum is defined only if $\sum ||\mu(A_i)||<\infty$, right? So the $\sigma$-additivity of $\mu$ is not general. What then can follow for $|\mu|$ ? | |
| Jan 20, 2018 at 15:03 | comment | added | Nate Eldredge | Maybe this is silly, but I'm having trouble seeing how a vector measure of unbounded variation can exist at all. Is there a standard example? | |
| Jan 19, 2018 at 22:55 | history | asked | 0xbadf00d | CC BY-SA 3.0 |