Finding maxima and minima of two dimensional function

Question

The procedure for doing this usually goes as follows:

1. Find the critical points of the function,
2. Check which ones are in the region,
3. Find critical points of the function on thr boundary,
4. Compare the value of the function at each one of these points and choose the largest to be the maximum and smallest to be the minimum.

My question is: why do we claim this method works? It could very well be that the function restricted to the boundary has a minimum, but this point is neither a local minimum nor maximum on the region. Take for example $f(x,y)=y^2+1−\frac{1}{1+x}$on the right half plane. It is obvious that this function will have a minimum on the y-axis, but this point will not be a minimum on the region. 

My question arose from reviewing some multivariable calculus. The problem in question was to find the maxima and minima of $f(x,y)=x^2+y^2+2x+4y-1$ on the closed region above the diagonal $y=-x$. It is easy to find that $\nabla f =0$ at $(-1,-2)$. Since this point is not within the region, we restrict $f$ to the diagonal and find that its minimum over this boundary occurs at $(1/2,-1/2)$. In the answer sheets, the instructor claims this is a minimum for the function over the whole region, however I find this claim rigorously unjustified. How do we know whether or not there is some other point on which $f$ is smaller?

I tried to prove this formally by assuming that such a point, say $\vec x$ exists, then defining a "gradient descent" sequence $S=\{\vec x_n\}$ as follows:

$$ \vec x_0=\vec x$$ $$ \vec x_{n+1}= \vec x_n-\nabla f(x_n)$$ Since $f$ grows to infinity in all directions, it seems intuitive that $S$ should be bounded, at which point the result is proven.

It is also very possible I am missing something obvious. Any help is appreciated.

Answer 1

If a function $f : D \to \Bbb R$ from a closed and bounded domain $D$ and is everywhere differentiable, then a maxima or minima must occur at a point where $\nabla f = 0$, for if not then for $x$ with $\nabla f(x) \ne 0$ we would have $f(x + \epsilon \nabla f) = f(x) + \epsilon \nabla f \cdot \nabla f + O(\epsilon^2)$, which is $\ge f(x)$ for sufficiently small $\epsilon$. Thus any minimizer $x$ of $f$ must either satisfy $\nabla f(x) = 0$ or $x$ must be on the boundary. We may thus enumerate the critical points in the region and on the boundary, and since the absolute maxima and minima must be a local maxima or minima, among them is our absolute maximum and minimum.

The problem with rigor comes when the function is defined on a region which is unbounded. In this case, the function is not guaranteed to have an absolute minimum. If it does have an absolute minimum, then the minimizer $x$ must satisfy $\nabla f(x) = 0$ or be on the boundary, as is argued above. If there is no absolute minimum, then this method tells you nothing.