The Weierstrass theorem suggest that in order for a function $f$ to have a min and max value, the following must hold:
- $f$ must be continous over the domain
- Domain should be closed and bounded (i.e. compact)
Example:
$ \min (x^2 + y^2)$
$ {\rm s.t.}~ x + y = 3 $
In this example it is clear that the set is not bounded since the values for x and y can be basically anthing.
However, for more complex examples it is hard to determine if a set is bounded and also closed. I am working on the following:
$ \max (x + y^2)$
$ {\rm s.t.}~ 3x^2 + 5y \geq -1 , x + y^2 \leq -1 $
$ \max (2x + y)$
$ {\rm s.t.}~ x^2 + y^2 \leq 9 , x + y = 1 $
$ \max (x+ \ln y)$
$ {\rm s.t.}~ x-5y \geq -1 , x + y^2 \leq 1 $
Can someone help to determine whether Weierstrass theorem holds for the above optimization problems and most importantly how one can approach these?
Thanks in advance.
The trick is to play with the inequalities so that you always have a different variable on a different side.
E.g. with the first one we have $3x^2 + 5y \geq -1$ and $x+y^2\leq -1$. We can re-arrange so that $3x^2 \geq -1-5y$ and $x\leq -1-y^2$.
Now, the first inequality means that $3x^2$ is large (depending on $y$), and the second says that $x$ is small (depending on $y$).
Since these two conditions can always be true, this means that the set is not bounded. To prove that formally, just take $y=0$ (or anything really), and now try to build a sequence of $x_n$'s satisfying the equation with $|x_n|\rightarrow \infty$. One way to do so is that $x_n = -n$, then $x_n\leq -1$, but $3x^2 = 3n^2 \geq -1$. Then $(-n,0)$ is a diverging sequence so the set is not compact.
Sometimes however you will have constraints. For instance in the second equation $x^2+y^2 \leq 9$, this is already a bound on $x$ and $y$ (by definitino $\|(x,y)\|\leq 3$ for all $(x,y)$ in the set).
Informally, the bounded situation can "only" happen in this trivial way, so if you see only inequalities and either $x$ and $y$ are not squared, I would try to disprove the claim.