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Let $I \subset \mathbb{R}$ be an interval, $f: I \rightarrow \mathbb{R}$ be a function and let $n \geq 2, n \in \mathbb{N}$ be fixed number.

Let's consider the following conditions:

  1. $\displaystyle f\left(\frac{x_1+x_2}{2}\right) \leq \frac{f(x_1)+f(x_2)}{2} \textrm{ for } x_1, x_2 \in I$;

  2. $\displaystyle f\left(\frac{x_1+...+x_n}{n}\right) \leq \frac{f(x_1)+...+f(x_n)}{n} \textrm{ for } x_1,...,x_n \in I$.

By https://math.stackexchange.com/a/83398/22907 1. implies 2.

Does 2. imply 1. ?

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Alex
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  • isn't (2) a special case of (1), when you set n to 2? – NoChance Jan 21 '12 at 14:10
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    If $n$ is even, $n=2p$ just take $x_1=\ldots=x_p=a$ and $x_{p+1}=\ldots,=x_{2p}=b$. If $n=2p+1$, put $x_1=\ldots=x_p=a$, $x_{p+1}=\ldots=x_{2p}=b$ and $x_{2p+1}=\frac{a+b}2$. – Davide Giraudo Jan 21 '12 at 14:11
  • @Emmad: That's what I thought when I looked at the question at the first sight. But I looked at it again, I found that here $n$ is fixed. – Paul Jan 21 '12 at 14:16
  • Offtop: Is the implication (2 to 1) also true if we demand $x_i \lt x_{i+1}$ and do not assume continousity / measurability of $f$? – savick01 Jan 21 '12 at 14:22
  • @Davide: You might copy this as an answer. – Did Jan 21 '12 at 14:34
  • @DidierPiau Done, I took me time, sine I tough was I told for the odd case was wrong, but now I am convinced it's not the case. – Davide Giraudo Jan 21 '12 at 14:40
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    @Davide: I liked the short version better... Ma non importa. – Did Jan 21 '12 at 14:43

1 Answers1

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Let $a,b\in I$. If $n$ is even, $n=2p$ just take $x_1=\ldots=x_p=a$ and $x_{p+1}=\ldots,=x_{2p}=b$. We get $$f\left(\frac 1{2p}\left(\sum_{j=1}^pa+\sum_{j=1}^pb\right)\right)\leq \frac{\sum_{j=1}^pf(a)+\sum_{j=1}^pf(b)}{2p},$$ so $$f\left(\frac{pa+pb}{2p}\right)\leq \frac{pf(a)+pf(b)}{2p}\Rightarrow f\left(\frac{a+b}2\right)\leq \frac{f(a)+f(b)}2.$$

If $n=2p+1$, put $x_1=\ldots=x_p=a$, $x_{p+1}=\ldots=x_{2p}=b$ and $x_{2p+1}=\frac{a+b}2\in I$. We get, using 2. that $$f\left(\frac{a+b}2\right)=f\left(\frac{pa+\frac a2+pb+\frac b2}{2p+1}\right)\leq \frac{pf(a)+pf(b)+f\left(\frac{a+b}2\right)}{2p+1}$$ so $$f\left(\frac{a+b}2\right)\left(1-\frac 1{2p+1}\right)\leq p\frac{f(a)+f(b)}{2p+1}$$ hence $2pf\left(\frac{a+b}2\right)\leq p(f(a)+f(b))$, QED.

Davide Giraudo
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