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We know that for a concave function $f\left(\frac{x_1+x_2}{2}\right) \geq \frac{f(x_1)+f(x_2)}2$.

Is it true that $ f\left(\frac{\sum x_i}n\right) \geq \frac{\sum f(x_i)}n$ for finite $x_1, ... x_n$? I tried to prove it by myself using mathematical inductions, I have not sucessfully proved it yet. If it is possible, I would like prove only using $f\left(\frac{x_1+x_2}{2}\right) \geq \frac{f(x_1)+f(x_2)}2$ without other concave function properties.

I do not know if I miss something important. I thought it'd be a lot easier... Thanks in advance.

Sayan Dutta
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Lab
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  • This video might help you – DatBoi Jul 27 '21 at 03:34
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    Yes, what you write is true for concave functions, not convex functions. Be aware that the inequality $f(\frac{x_1 + x_2}{2}) \ge \frac{f(x_1) + f(x_2)}{2}$ is the definition for midpoint-concavity, and is weaker than concavity in general (though not if the function is continuous). – Theo Bendit Jul 27 '21 at 03:43
  • Fixed, thank you :) It "Is weaker than concavity", but concavity implies midpoint-concavity(correct me if wrong). In that case, does midpoint-concavity implies the inequality in the post? – Lab Jul 27 '21 at 03:46
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    @Lab No it doesn't. It will if $n$ is a power of $2$, but not otherwise. – Theo Bendit Jul 27 '21 at 03:47
  • Can you give me a counterexample? – Lab Jul 27 '21 at 03:49

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Suppose $f : D \to \Bbb{R}$, where $D$ is a convex subset of some vector space (could just be $\Bbb{R}$ if you prefer). We say $f$ is concave if the inequality $$f(\lambda x + (1 - \lambda)y) \ge \lambda f(x) + (1 - \lambda)f(y) \tag{$\star$}$$ holds whenever $x, y \in D$, and $\lambda \in [0, 1]$. If $(\star)$ holds just for $\lambda = 1/2$, we say $f$ is midpoint-concave. Note that trivially, concavity implies midpoint-concavity.

Concavity implies the inequality you wrote. We can prove it for all $n$ by induction. For $n = 1$, the inequality becomes an equality, and is trivially true. If it holds for some $n \ge 1$, then \begin{align*} f\left(\frac{x_1 + \ldots + x_n + x_{n+1}}{n + 1}\right) &= f\left(\frac{x_1 + \ldots + x_n}{n + 1} + \frac{x_{n+1}}{n + 1}\right) \\ &= f\left(\frac{n}{n+1}\frac{x_1 + \ldots + x_n}{n} + \frac{1}{n + 1}x_{n+1}\right) \\ &\ge \frac{n}{n+1}f\left(\frac{x_1 + \ldots + x_n}{n}\right) + \frac{1}{n + 1}f(x_{n+1}) \\ &\ge \frac{n}{n+1} \cdot \frac{f(x_1) + \ldots + f(x_n)}{n} + \frac{1}{n + 1}f(x_{n+1}) \\ &= \frac{f(x_1) + \ldots + f(x_n)}{n+1} + \frac{1}{n + 1}f(x_{n+1}) \\ &= \frac{f(x_1) + \ldots + f(x_n) + f(x_{n+1})}{n+1}. \end{align*}

We now look at midpoint-concavity. Let $$I = \{x \in \Bbb{Q} \cap [0, 1] : \exists p, q \in \Bbb{Z}, \text{ such that }x = p/2^q\}.$$ Note that, if $x, y \in I$, then $\frac{x + y}{2} \in I$ as well.

Midpoint-concavity implies that $(\star)$ holds for $\lambda \in I$. This set is dense in $[0, 1]$, so if $f$ is continuous, then midpoint-concavity implies concavity, and hence the average calculation above holds true. If $f$ is discontinuous, then we can find counterexamples. Check out here for an example (though note that this is for convexity, not concavity; just take the negative).

Theo Bendit
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