Let $ f:[0,1]\rightarrow R $ a continuous function. It is true that, if $ f $ is also convex, than $$ f\left(\frac{a +b}{2}\right) \leq \frac{1}{b-a} \int_{a}^{b} f.$$ How to prove this? It holds a reverse of this statement, i.e. f is convex if $$ f\left(\frac{a +b}{2}\right) \leq \frac{1}{b-a} \int_{a}^{b} f ?$$ Thank you for the help.
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What does "a "viceversa" " mean? – kimchi lover Jun 01 '19 at 16:49
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I would say a sort of reverse of the statement, i.e. f is conves if.... – C. Bishop Jun 01 '19 at 16:50
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You need to elaborate what you mean by the reverse. Explicitly, not with ellipses. – copper.hat Jun 01 '19 at 16:53
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What does ... mean?????? – copper.hat Jun 01 '19 at 16:56
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Are you on $[0,1]$ or $[a,b]?$ – zhw. Jun 01 '19 at 17:07
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The function is defined on $ [0,1]$ – C. Bishop Jun 01 '19 at 17:10
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So then you might want to say for all $a,b\in [0,1].$ – zhw. Jun 01 '19 at 17:22
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1It looks unlikely that the converse holds. Take a look here – Lázaro Albuquerque Jun 02 '19 at 18:39
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Since ${1 \over 2}(a+b) = {1 \over b-a} \int_a^b x dx $, Jensen gives $f({1 \over 2}(a+b)) \le {1 \over b-a} \int_a^b f(x) dx $.
If you take $f_n$ to be the function defined by the graph (with $m={1 \over 2} (a+b)$): $(a,1), (m-{1 \over n}, 1), (m,0), (m+{1 \over n},1), (b,1)$ then it should be clear that $f_n$ is not convex and $f_n(m) = 0$ for all $m$ and $\int f_n >0 $.
copper.hat
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2Your function $f_n$ satisfies the inequality for fixed values of $a$ and $b$ while the converse problem assumes it does so for every pair $a<b$. The constant functions, for example, are convex and satisfies the inequality for every pair $a<b$. – Lázaro Albuquerque Jun 01 '19 at 18:08
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@lzralbu: It was not clear at the time of writing (and still needs to be clarified) that that is what the OP meant. – copper.hat Jun 01 '19 at 18:18