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:

*

*$K$ is a linear subspace of $\mathbb R^n$

*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.
 A: 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...
A: (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.
