Do open sets have maximums?

Question

I've learned the Extreme Value Theorem, where if a continuous function is contained in a compact set, then it has a global min/max in the set.

And from here, I can see that open sets do not have minimums. But do they have maximums?

Edit: For example, we're currently doing Lagrange optimization in class, and it would be nice to be able to prove the set is open and thus has no max/min to avoid doing problems that have no solution.

Edit 2:

Follow-up question then: why does showing that U is open help then in showing that f doesn't attain a max/min there? (that's the end of the solution my prof provided. I found it weird that he stopped there)

Answer 1

It seems to me you are a bit confused: you start by minimizing a function, and then you switch to minima of a subset of $\mathbb{R}$. Although these things are obviously linked, you should be careful. Indeed, a function can attain a maximum on an open set, think of $f(x)=1-x^2$ in $(-1,1)$.

So no, you cannot stop looking for maxima of a function as soon as you realize that its domain is an open set.

As regards your edit, please notice that your professor writes: first show that $U$ is open. In other words, you should now complete the reasoning. As a further hint, you are trying to maximize a linear function, and it is rather clear (why?) that a linear function cannot generically attain a maximum inside an open set.

Answer 2

You are told to determine the extreme values of a smooth function $f:\>{\mathbb R}^3\to{\mathbb R}$ on a compact body $B\subset{\mathbb R}^3$. By the "Extremal Value Theorem" (EVT) the function $f$ takes a global minimum and a global maximum on $B$.

The body $B$ is stratified: It consists of an obviously open interior $U$ and a boundary $\partial B$, which is an ellipsoid, hence a smooth surface. In this example there are no lower-dimensional strata (edges or vertices).

An extremal point $\xi$ of $f$ can lie in the interior $U$ or on the boundary $\partial B$. If $\xi\in U$ then $\xi$ necessarily is a (full) critical point of $f$, i.e., $\nabla f(\xi)=0$. As $\nabla f(x,y,z)\equiv(2,1,-1)$ the function $f$ has no critical points, and there is not even a local extremum of $f$ in $U$. If $\xi\in\partial B$ it will be brought to the fore by Lagrange's method applied with the given constraint, which describes $\partial B$. In the course of the computation you will obtain two conditionally critical points $\xi_1$, $\xi_2\in\partial B$. Since we are guaranteed a min and a max by the EVT one of these points will give the global minimum, the other the global maximum of $f$ on $B$.

In more complicated cases you will end up with a larger candidate list $\{\xi_1,\xi_2,\ldots,\xi_m\}\subset B$, coming from the various present strata of $B$. A comparison of the function values in these points then decides which of the $\xi_i$ gives the min, resp., the max.