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Let $K$ be a subset of $\mathbb R^n$ satistifying the following properties:

  1. $K$ is a linear subspace of $\mathbb R^n$
  2. If $x \in K$ either $x= 0$ or there exists an $i \in \{ 1, \ldots n\}$ such that $x_i >0$.

Does $f(x) = \sum_{i=1}^n e^{x_i}$ have a global minimum over $K$ for all such $K$?

Intituitively to me it seems like it should exist as $x \to \infty$ along any ray contained in $K$ then $f(x) \to \infty$. It I could show that this is uniform then the proof would be easy from there.

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  • $\begingroup$ If I understood correctly for $n=2$, $K$ would be the union of first, second and fourth quadrant of the Cartesian plane (counted anticlockwise). But then the infimum is 1 (obtained when $x\to -\infty$ and $y\to 0^+$ or viceversa) but is not attained. But I'm not sure I understood the question correctly $\endgroup$
    – lcv
    Mar 25, 2022 at 1:30
  • $\begingroup$ @lcv Edited the question to try and make it more clear. When $n=2$ there are lots of possible 𝐾, but one is $ \{(x,y) \in \mathbb R ^2 : x+y =0 \}$ $\endgroup$
    – Ben Martin
    Mar 25, 2022 at 3:34
  • $\begingroup$ @B.Martin When I see that it screams geometric programming. Have you looked at it? $\endgroup$
    – KBS
    Mar 30, 2022 at 14:50
  • $\begingroup$ @KBS I'll check it out. $\endgroup$
    – Ben Martin
    Apr 1, 2022 at 3:08

2 Answers 2

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Сonsider the "maximal" case $\dim K=n-1$. That is, $$K=\{a\}^\perp:=\{x\in\mathbb{R}^n:(a,x)=0\}\quad\text{for some}\quad a\in\mathbb{R}^n,\ a\neq 0.$$

If $a_j=0$ for some $j$, then the choice $x_j=-1$ and $x_i=0$ for $i\neq j$ violates the conditions on $K$. Similarly, if $a_j<0<a_k$ for some $j\neq k$, choose $x_j=-a_k$, $x_k=a_j$, and $x_i=0$ for $i\notin\{j,k\}$. Thus, replacing $a$ by $-a$ if necessary, we may assume that $a_i>0$ for each $i$.

For $\lambda>0$, the function $t\mapsto e^t-\lambda t$ (here $t\in\mathbb{R}$) has a global minimum at $t=\log\lambda$. Hence the function $f(x)-\lambda(a,x)$ has a global minimum (in $\mathbb{R}^n$) at $x=x^\star$, where $x_i^\star=\log(\lambda a_i)$.

Now choose $\lambda$ so that $x^\star\in K$. That is, take $$\lambda=\exp\left(-\frac{\sum_{i=1}^n a_i\log a_i}{\sum_{i=1}^n a_i}\right).$$ Then for $x\in K$ $$f(x)=f(x)-\lambda(a,x)\geqslant f(x^\star)-\lambda(a,x^\star)=f(x^\star).$$

To be continued, to consider the general case...

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    $\begingroup$ Thank you. How did you use the separation theorem to show that considering the maximal case is sufficient? $\endgroup$
    – Ben Martin
    Mar 26, 2022 at 4:48
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    $\begingroup$ @B.Martin: Seems I've stated it in rush. (No, it's not sufficient.) Gonna think more... $\endgroup$
    – metamorphy
    Mar 29, 2022 at 11:42
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(Not a full answer, but merely pointing out a big issue with your question.)

If I understood you correctly, you have that

$$K_n=\{x=(x_1,\dots,x_n)\subseteq\mathbb{R}^n: x=0\text{ or } x_i>0\text{ for some } i\}.$$

You claim that $K_n$ is a subspace of the real vector space $\mathbb{R}^n$, however it is not. Consider the vector

$$x=(1,0,\dots,0)\in K_n.$$

If $K_n$ was a vector space, then it would be closed under scalar multiplication, but notice that

$$(-1)\cdot x=(-1,0,\dots,0)\notin K_n,$$

and so $K_n$ is not a vector space.

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  • $\begingroup$ $K$ is a subset of the set you are talking about. An example in dimension 2 would be $\{(x,y) \in \mathbb R^2: x+ y = 0 \}$, which is a subspace. $\endgroup$
    – Ben Martin
    Mar 25, 2022 at 3:21
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    $\begingroup$ Edited the question to (hopefully) make it more clear. $\endgroup$
    – Ben Martin
    Mar 25, 2022 at 3:27

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