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Extending from If $f$ is strictly convex in a convex set, show it has no more than 1 minimum and Examples of $f$ strictly convex, either with one minimizer or with no minimizer.

Can we have the same minimisers $(x_1^*,x_2^*,\cdots,x_n^*)$ for $$f\left( \frac{1}{n} \sum_{j=1}^{n} x_j\right)$$ and $$ \frac{1}{n} \sum_{j=1}^{n} f\left( x_j\right)?$$ where $f:\mathbb{R}^+ \to \mathbb{R}^+$ is strictly convex and $n$ is finite natural number.

rookie
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1 Answers1

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If you're asking whether the two quantities $$f\left( \frac{1}{n} \sum_{j=1}^{n} x_j\right)\tag1$$ and $$\frac{1}{n} \sum_{j=1}^{n} f\left( x_j\right)?\tag2$$ can have the same minimizers, the answer is yes: Let $a$ be the minimizer of $f$, and set $x_1=x_2=\cdots=x_n=a$.

If you're asking whether the two quantities (1) and (2) must have the same minimizers, the answer is no. The quantity (1) is minimized by any $x_1,\ldots,x_n$ for which $\sum x_i= na$, whereas $x_1=x_2=\cdots=x_n=a$ is the only minimizer for (2). If $a>0$, then the second solution set is a proper subset of the first.

grand_chat
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  • Agreed. What changes are required to be done if one wants an Yes for the case of must in your answer? – rookie May 09 '17 at 20:33
  • Can you also look at this, https://math.stackexchange.com/questions/2273653/determining-type-of-a-function-to-fulfill-the-requirement, question and try to address the same? Thanks in advance! – rookie May 09 '17 at 20:34
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    @stud_iisc I think a necessary and sufficient condition for must is that the minimizer of $f$ be zero (since by assumption $f$ is defined only for non-negative $x$). – grand_chat May 09 '17 at 20:38
  • @stud_iisc I looked at and wrote an answer to your other question. – grand_chat May 09 '17 at 21:40