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Prove $f:\mathbb{R} \to \mathbb{R}$ is convex if and only if $f\left(\frac{a+b}{2}\right)\leq \frac{1}{b-a}\int_a^b f(x) dx$ for any $a,b \in \mathbb{R}$.

The context of this result is proving that in $\mathbb{R}$, a function is $C^0$-subharmonic if and only if it is convex. We have a few equivalent definitions of $C^0$-subharmonic but I thought the above would be the most logical to try to prove. But I'm not sure how to prove either direction. One can interpret the RHS as the average value of $f$ on $[a,b]$ and then the forward direction is at least visually clear, but I don't know how to express this mathematically. Help/hints would be appreciated.

AlephNull
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    For the other direction, see https://math.stackexchange.com/q/1827772/42969. – Martin R Mar 03 '19 at 14:20
  • Thanks, I didn't see these in the 'similar questions' or the search. – AlephNull Mar 03 '19 at 14:21
  • If you modify a constant function in on a measure 0 set in such a way that it becomes lower semicontinuous, then it will obey your inequalities but not be convex. – kimchi lover Mar 03 '19 at 14:24
  • FYI: there’s more intuitive proof that doesn’t have to yse proof by contradiction. Since it is convex, we can find a line that passes through $((a+b)/2,f((a+b)/2))$ and support $f$ (i.e. below $f$), and you get the inequality when you integrate both $f(x)$ and the line on $[a,b]$. – Seewoo Lee Mar 03 '19 at 14:31
  • @MartinR The linked question is has the assumption that $f$ is continuous; this question does not. – kimchi lover Mar 03 '19 at 15:04
  • This is false as stated: If $f(0)=-1$ and $f(t)=0$ for $t\ne0$ then $f$ satisfies the inequality but is not convex. Are you also given that $f$ is continuous? – David C. Ullrich Mar 03 '19 at 15:30
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    Oh yes, $f$ is continuous, forgot to mention that sorry (and that is why this is equivalent to the $C^0$-subharmonic property). – AlephNull Mar 03 '19 at 15:52

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