Revision as of 00:30, 20 April 2025 by Bot (Created page with "<div class="d-none"><math> \newcommand{\mathds}{\mathbb}</math></div>Given a convex function <math>f:\mathbb R\to\mathbb R</math>, prove that we have the following Jensen inequality, for any <math>x_1,\ldots,x_N\in\mathbb R</math>, and any <math>\lambda_1,\ldots,\lambda_N > 0</math> summing up to <math>1</math>, <math display="block"> f(\lambda_1x_1+\ldots+\lambda_Nx_N)\leq\lambda_1f(x_1)+\ldots+\lambda_Nx_N </math> with equality when <math>x_1=\ldots=x_N</math>. In pa...")
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Apr 20'25

Exercise

[math] \newcommand{\mathds}{\mathbb}[/math]

Given a convex function [math]f:\mathbb R\to\mathbb R[/math], prove that we have the following Jensen inequality, for any [math]x_1,\ldots,x_N\in\mathbb R[/math], and any [math]\lambda_1,\ldots,\lambda_N \gt 0[/math] summing up to [math]1[/math],

[[math]] f(\lambda_1x_1+\ldots+\lambda_Nx_N)\leq\lambda_1f(x_1)+\ldots+\lambda_Nx_N [[/math]]

with equality when [math]x_1=\ldots=x_N[/math]. In particular, by taking the weights [math]\lambda_i[/math] to be all equal, we obtain the following Jensen inequality, valid for any [math]x_1,\ldots,x_N\in\mathbb R[/math],

[[math]] f\left(\frac{x_1+\ldots+x_N}{N}\right)\leq\frac{f(x_1)+\ldots+f(x_N)}{N} [[/math]]

and once again with equality when [math]x_1=\ldots=x_N[/math]. Prove also that a similar statement holds for the concave functions, with all the inequalities being reversed.