Talk:Triangle Inequality for Integrals

A short comment that $\int \text{Im} (\alpha f) = \text{Im} \left(\int \alpha f\right)$ vanishes by definition of $\alpha$ would be great. Correct nontheless.


 * Not quite sure what you mean -- can you elaborate? --prime mover (talk) 13:29, 26 April 2015 (UTC)

The proof is a bit weird, while not wrong. The integral is real so we're only talking about $\alpha = \pm 1$. (and $\map \Re {\alpha f} = \alpha f$) I have a feeling like the proof is for a more general context, maybe $f : X \to \C \cup \set \infty$ or something, but we don't have the setup for complex Lebesgue integrals yet. Proof 2 is a much more natural approach anyway. Caliburn (talk) 17:52, 27 September 2021 (UTC)