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:


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?

  • 2
    $\begingroup$ 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

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.


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.

  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.