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If $a < b$ then $$((f \wedge b) -a)^-=(f-a)^-$$ where $f$ is a real valued function. I can't figure out why this holds. I know that $f^-=-(f\wedge 0)$ so I tried showing this from definitions. I would greatly appreciate any help.

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    What is $\wedge$ here? – 2017-02-01
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    @MJD minimun of the two – 2017-02-01
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    if $a$b=f$. Other cases hold. – 2017-02-01

1 Answers 1

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It's easy checking cases…

For $f \ge b > a$ it holds: $$((f \wedge b) -a)^- = (b - a)^- = 0 = (f - a)^-$$

For $f < b$ it's $$((f \wedge b) -a)^- = (f - a)^-$$

Nothing wild…