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Let $f_1, \dots, f_n$ be positive functions from $\mathbb R^m \rightarrow \mathbb R$.

How do we show that $$\min_x \sum_{i=1}^n f_i(x) = \sum_{i=1}^n \min_x f_i(x)$$

Actually, I am not sure this is true. Maybe adding convexity of the functions helps ?

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It is not true. Take $f_1,f_2\colon\mathbb R\longrightarrow\mathbb R$ defined by $f_1(x)=(x-1)^2$ and $f_2(x)=(x+1)^2$. Then $\min f_1+\min f_2=0$, but $\min(f_1+f_2)=2$.

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  • $\begingroup$ Thank you. Do you know under what hypothesis the statement becomes true ? $\endgroup$
    – W. Volante
    Jun 12, 2019 at 23:20
  • $\begingroup$ It is true if all the functions attain their minimum at the same point. $\endgroup$
    – José Carlos Santos
    Jun 12, 2019 at 23:22
  • $\begingroup$ Could you provide an idea of a proof ? $\endgroup$
    – W. Volante
    Jun 12, 2019 at 23:38
  • $\begingroup$ If each $f_i$ attains its minimum at a point $a$, then, for every $x\in\mathbb R^m$,$$\left(\sum_{i=1}^nf_i\right)(x)=\sum_{i=1}^nf_i(x)\geqslant\sum_{i=1}^nf_i(a)$$and therefore $\sum_{i=1}^nf_i$ attains its minimum at $a$. $\endgroup$
    – José Carlos Santos
    Jun 13, 2019 at 7:43

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