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Nov 19, 2017 at 11:57 vote accept Learner
Nov 19, 2017 at 5:18 answer added R W timeline score: 1
Nov 18, 2017 at 22:46 comment added Nate Eldredge I think at a minimum you need $\phi$ to be concave. For instance consider $\phi(t) = t^2$. If we take $X =\mathbb{R}$ with Lebesgue measure, we would have to have $\|\chi_{[0,1)} + \chi_{[1,2]}\| = \|\chi_{[0,2]}\| = 4 > 2 = \|\chi_{[0,1)}\| + \|\chi_{[1,2]}\|$ contradicting the triangle inequality for the norm.
Nov 18, 2017 at 20:02 review First posts
Nov 18, 2017 at 20:05
Nov 18, 2017 at 19:59 history asked Learner CC BY-SA 3.0