Let $f:[a,b]\rightarrow E$ be absolutely continuous and Pettis integrable, i.e. there exists $I_f\in E$ such that $x^*(I_f)=\int x^*\circ f$ for $x^*\in E^*$. Because $f$ is absolutely continuous, $\|f\|$ is integrable, but can we actually conclude $\|I_f\|\le\int\|f\|$?
2026-03-27 22:03:52.1774649032
Absolute value inequality for Pettis integral
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Yes, choose $x^*\in E^*$ with $\|x^*\|=1$ and $|x^*(I_f)|=\|I_f\|$. Then $$\|I_f\|=|x^*(I_f)|=|\int x^*\circ f|\le\int|x^*\circ f|\le\int\|f\|.$$