The generalized mean can be given using the following equation:

$ M_p(x_1, \dots, x_n) = (\frac{1}{n}\sum_i x_i^p)^{1/p} $

Is it convex /concave when $p<1$ ?

  • $\begingroup$ We have that $|x|^p$ is concave, rather than convex, so we have the following inequality: $\frac{1}{n}\sum_i x^p_i\le (\frac{1}{n}\sum_i x_i)^p$. Raising to the $\frac{1}{p}$, we have $M_p(x_1,...,x_n)\le \frac{1}{n}\sum_i x_i$. Is this what you are asking? If so I will post it as an answer. $\endgroup$
    – A. S.
    Jun 10, 2016 at 17:48
  • $\begingroup$ The function $M_p : \mathbb{R}^n \to \mathbb{R}$ is not convex for $p < 1$. $\endgroup$
    – gerw
    Jun 10, 2016 at 18:37
  • $\begingroup$ The function as you wrote it can only be defined as $M_p: \mathbb{R}^n_{+} \rightarrow \mathbb{R_+}$. This function is concave for $0 < p \le 1$. On the contrary, $M_p: \mathbb{R}^n \rightarrow \mathbb{R_+}$ defined as $M_p(x_1,\dots,x_n)=(\frac{1}{n}\sum_i |x_i|^p)^{1/p}$ (notice the absolute value) is neither concave nor convex. $\endgroup$
    – Luca Citi
    Jun 12, 2016 at 21:22


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