Q1: How do we show that the function $f:\mathbb{R}^2 \to \mathbb{R}$, $(x,y) \mapsto xy$ is continuous?

Work: The open sets on $\mathbb{R}$ are the open intervals $(a,b)$. We need to show that the preimage of the open interval $(a,b)$ is also open. Would $f^{-1}(a,b)= (a,b) \times (a,b)$ the cartesian product of the interval $(a,b) \subseteq \mathbb{R}$ with itself? Since the cartesian product of two open sets is open, the preimage of $(a,b)$ is open, so $f$ is continuous.

Q2: Is there a systematic method for finding the maximum and minimum values of a function $g$ on some subset $\Omega$ of the domain? I think I remember learning something about it in calculus, but I can't find anything about it in my textbook "Comple Variables" by Flanigan.

Suppose $\overline{D} \subseteq \mathbb{R}^2$ is the closed unit disk centered at the origin $(0,0)$ and that $g$ is function $f$ defined above. How do we find the maximum and minimum values of the function $f(x,y)=xy$ on the closed disk?

Since the unit disk satisfies $x^2+y^2 \leqslant 1$ and the function $f$ gets larger for increasing values of $x$ and $y$, I know that the maximum values are determined by $(1/\sqrt{2},1\sqrt{2})$ and $(-1/\sqrt{2},-1\sqrt{2})$ and the mimimum values are determined by $(-1/\sqrt{2},1\sqrt{2})$ and$(1/\sqrt{2},-1\sqrt{2})$, but I would like to learn a concrete method for finding these values, in case $\Omega$ and $g$ are more complicated.


Sketch of answers:

Q1: you only need to show $p_1(x,y)=x, \, p_2(x,y)=y$ are continuous.

Since $\bar D$ is compact, max and min are indeed attained. Clearly neither is attained in the interior (check why), so there is only the boundary $S^1$ left to examine. Then let $x=\sin \phi,\, y=\cos\phi$.

Q2: Yes. But, in practice, usually we need to presume differentiability of $g$, and $\Omega$ a subset whose boundary is an easily identifiable curve (or a set of such curves). We first work on $\text{int}\Omega$: let $D g=0$ to obtain all the stationary points, then calculate the values assumed by $g$ at these points. Then we move to $\partial \Omega$: if the curve is readily parametrisable, then we are happy because all we need to do is to plug the curve equation into the goal function, then optimise this one-variable function; if the curve appears in an implicit form, say $r(x,y)=0$, use Lagrange multiplier method. Finally, compare all candidate extreme values against each other to find out the max and min.

  • $\begingroup$ Thanks for the help. One more quick question: What do you mean by the equation "$Dg=0$? For the part about the boundary $\partial \Omega$, parametrizing the circle $C_1(0)$ by $x(t)=(\cos t, \sin t)$, would the function be $f(\cos t, \sin t )= \cos t \sin t$? I'll make sure to review the Lagrange multiplier method. $\endgroup$ – user167857 Apr 10 '16 at 6:34
  • $\begingroup$ @user167857 The first one: $D g=0$ is an abbreviation for $(\partial g/\partial x, \partial g/\partial y)=(0,0)$. The second one: yes, you're right. $\endgroup$ – Vim Apr 10 '16 at 6:37

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