Showing a $\mathbb{R}^2 \rightarrow \mathbb{R}$ function attains a global maximum

Question

Given $f:\mathbb{R}^2 \rightarrow \mathbb{R}$ defined by $$f(x,y) = (ax^2+by^2)\exp(-x^2-y^2)$$

where $a > b > 0$, how can I show $f$ attains a global maximum? It is easy to show that it attains a local maximum via the Hessian matrix and I suspect its local maxima coincide with its global maxima. This function is reminiscent of the $\mathbb{R} \rightarrow \mathbb{R}$ function $x\exp(-x)$ and I am wondering if there is a simple change of variables to simplify the problem to showing that this $\mathbb{R} \rightarrow \mathbb{R}$ function attains its global maximum. I am looking for some help in this regard.

The function is always greater than or equal to zero, f(0,0)=0, and f approaches zero as r approaches infinity. That should be enough to deduce the existence of a global max — Alex Pavellas, Aug 11 '15 at 00:04

score 2 · Accepted Answer · answered Aug 11 '15 at 00:00

2

Make the polar transformation $x = r\cos \theta, y = r\sin\theta$, $r > 0, \theta \in [0, 2\pi).$ Then \begin{align*} f(x, y) = g(r, \theta) = r^2(a\cos^2\theta + b\sin^2 \theta) e^{-r^2} \end{align*} I think you may take it from here.

answered Aug 11 '15 at 00:00

Zhanxiong

14,040

Jack's wasted life · Answer 2 · 2015-08-11T00:05:46.927

0

Hint: $$ \because \lim_{n\to\infty}n^pe^{-n^2}=0\;\forall\;p>0\\ |f(x,y)|\to 0 \text{ as } ||(x,y)||\to\infty $$

edited Aug 11 '15 at 00:05

answered Aug 11 '15 at 00:01

Jack's wasted life

8,849

Showing a $\mathbb{R}^2 \rightarrow \mathbb{R}$ function attains a global maximum

2 Answers2