Is following function $\exp(x) + \exp(y)$ super-coercive?

Question

Consider the function $f(x,y) = \exp(x) + \exp(y)$, $(x, y) \in \mathbb{R}^2$.

I want to know if it is true that,$\exp(x) + \exp(y)$ super-coercive, that is,

$(\exp(x) + \exp(y))/\|(x,y)\|_2 \to \infty$ as $\|(x,y)\|_2 \to \infty $

This seems intuitively true, but I am having some trouble.

Let $x = r\cos(t), y = r\sin(t)$.

Then the left hand side becomes, $$(\exp(x) + \exp(y))/\|(x,y)\|_2 = (\exp(r\cos(t)) + \exp(r\sin(t)))/r$$

Using the approximation, $\exp(x) \geq 1+ x$, we have, $$(\exp(x) + \exp(y))/\|(x,y)\|_2 \geq (1 + r\cos(t) + 1 + r\sin(t))/r$$

Which does not go to $\infty$ as $r \to \infty$.

But the gap between $\exp(x)$ and $1+x$ is huge.

How can I prove my original statement?

Answer 1

As $x \to -\infty$ and $y \to -\infty$ is is clear that the limit is $0$, not $\infty$.

If you are considering only positive values of $x$ and $y$ then the inequalities $e^{x} > \frac {x^{2}} 2$ and $e^{y} > \frac {y^{2}} 2$ make the result quite evident.

Answer 2

For $x$ and $y$ large enough you have that

$e^{|x|}>x^2$ and $e^{|y|}>y^2$ so

$e^{|x|}+e^{|y|}>x^2+y^2=||(x,y)||^2$ that means

$\frac{e^{|x|}+e^{|y|}}{||(x,y)||}>||(x,y)||\to \infty$

In your case you have a problem because for $x\to-\infty$ and $y\to-\infty$ you have that the limit is zero.