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I am stuck with the following problem. I can find the absolute minimum or maximum of a function with 2 variables or more, but I can't prove that absolute values exist.

Assume we have the following function:

$$f(x,y)=\frac{x^2-y^2}{\pi^{x^2-y^2}}$$

I need to show that the absolute maximum and minimum values exist

My idea: For a function to have absolute min/max value, the domain set should be compact: that is, closed and bounded and the function be continuous. But I am really stuck here. I dont know how to show that the domain set is compact. Could anyone please help me with this matter?

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    $\begingroup$ This is a one variable problem. $\endgroup$ Oct 27 '14 at 0:11
  • $\begingroup$ So how should I approach to the problem then? Could you please elaborate a little? $\endgroup$ Oct 27 '14 at 0:12
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We are maximizing/minimizing $\frac{t}{\pi^t}$. The derivative is $(1-t\ln \pi)/\pi^t$. So the function is increasing for a while then decreasing. It follows that the max exists.

There is no minimum, since when $t$ is large negative, then the top is large negative and the bottom is close to $0$, so our function is very large negative.

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a note: the domain need not be compact for a function to have an absolute minimum or maximum. Take the trivial function $g(x)=0$. This function has an absolute minimum equal to its absolute maximum, which is 0.

As Andre Nicolas mentions, you can make this into a one variable problem. Let $r=x^2+y^2$. Then let $\tilde{f}(r)=\frac{r}{\pi^r}$. Try taking the derivative and see where this leads you.

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  • $\begingroup$ Then how can I know whether the function has absolute min or max? I mean is there a general category then? Because that was the definition I know $\endgroup$ Oct 27 '14 at 0:16
  • $\begingroup$ @primenumber57 I hope this doesn't sound too harsh, but that wasn't a definition. You were misquoting a theorem. The definition of a function that is bounded above is one that is finite for all elements in its domain. The theorem that you were misquoting is: a continuous function attains its maximum and its minimum on a compact domain. There is no "general category". But a good thing to do is plot the function in Wolfram Alpha and see what you get. One thing to note is that, for the way we've defined r, r is always non-negative. $\endgroup$
    – NicNic8
    Oct 27 '14 at 0:18
  • $\begingroup$ I think $r$ can be negative if $x<y$ $\endgroup$ Oct 27 '14 at 0:22
  • $\begingroup$ @primenumber57 I'm sorry, I got confused. You're absolutely right. $\endgroup$
    – NicNic8
    Oct 27 '14 at 0:30

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