# Existence of global minimum of $f(x) = \sum_{i=1}^n e^{x_i}$

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.

• 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
– lcv
Mar 25, 2022 at 1:30
• @lcv Edited the question to try and make it more clear. When $n=2$ there are lots of possible 𝐾, but one is $\{(x,y) \in \mathbb R ^2 : x+y =0 \}$ Mar 25, 2022 at 3:34
• @B.Martin When I see that it screams geometric programming. Have you looked at it?
– KBS
Mar 30, 2022 at 14:50
• @KBS I'll check it out. Apr 1, 2022 at 3:08

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 $$a_j<0 for some $$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...

• Thank you. How did you use the separation theorem to show that considering the maximal case is sufficient? Mar 26, 2022 at 4:48
• @B.Martin: Seems I've stated it in rush. (No, it's not sufficient.) Gonna think more... Mar 29, 2022 at 11:42

(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.

• $K$ 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. Mar 25, 2022 at 3:21
• Edited the question to (hopefully) make it more clear. Mar 25, 2022 at 3:27