Find supremum of a function Let $x,y\in \mathbb{R}^n$. Find $$\sup_x (<x,y> - e^{<x,1>})$$ or we can rewrite this as $$\sup_x (\sum_i^nx_iy_i - e^{\sum_i^nx_i})$$ We can find the derivative, it gives $y = e^{<x,1>}$, but it looks like it can not possible to get an explicit formula if we put this into the initial one. Moreover, it looks like this function does not have the supremum at all (i.e it's $\infty$). Is this true? How to properly show it then?
P.s if we consider $<x,1> = 0$ then the sup is just $0$, but what if $<x,1> \not =$0 ?
 A: This depends on $y$.

*

*If there is $y_i$<0, then take $x_n=(0,\cdots,0,x_{i,n},0,\cdots,0)$, where $x_{i,n}=-n$. Then
$$\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}n|y_i|-e^{-n}=\infty$$
So $\operatorname{sup}_xf(x)=\infty$.

*Assume $y_i, \forall i$ positive, but there exist $y_j\neq y_k$. Assume w.l.o.g. $y_j>y_k$, take $x_n=(0,\cdots,0,x_{j,n},0,\cdots,0,x_{k,n},0,\cdots,0)$ where $x_{j,n}=n$ and $x_{k,n}=-n$. Then
$$\lim_{n\to\infty}f(x_n)=\lim_{n\to\infty}n(y_j-y_k)-1=\infty$$
So $\operatorname{sup}_xf(x)=\infty$.

*At last assume all $y_i$s are the same and positive, denote $y_i=a>0$. Then the function takes the form
$$f(x)=a<x,1>-e^{<x,1>}$$
Lets look at the function $g(z)=e^z-az$, $z\in\mathbb{R}$.
$$\lim_{z\to-\infty}g(z)=\infty$$
Obviously. Also
$$\lim_{z\to\infty}\frac{az}{e^z}=0,\ \lim_{z\to\infty}\frac{z}{e^z-az}=0$$
So for all $z$ big enough we have
$$z<e^z-az$$
And this means that
$$\lim_{z\to\infty}g(z)=\infty$$
So the function $g(z)$ is coercive, and this implies it has global minimum. Taking the derivative and equating to $0$ we find $e^z-a=0$, so $z^*=\ln a$. This means that
$$a(1-\ln a)\leq e^z-az,\ \forall z\in\mathbb{R}$$
or
$$a(\ln a-1)\geq az-e^z,\ \forall z\in\mathbb{R}$$
This still holds for $<x,1>,\ x\in\mathbb{R^n}$
$$a(\ln a-1)\geq a<x,1>-e^{<x,1>}$$
So we can take $x_1=z^*,\ x_i=0,\ i>1$ and state that $\operatorname{sup}_xf(x)=a(\ln a-1)$.

