12
$\begingroup$

I have actually two questions here, but both are very much related so I decided to put them both in this question.

From Wikipedia I found the following example of a function that has a single critical point which is a local minimum, yet is not a global minimum:

$f(x,y) = x^2 + y^2(1-x)^3$

Indeed the only critical point of $f$ is $(0,0)$ which is a local minimum by the second derivative test, but $(0,0)$ is not a global minimum because $f(4,1) = -11$. Why does this happen? I am having some trouble understanding intuitively why this is true. I know that $f$ has a minimum when restricted to a compact set. Then $(0,0)$ is a global minimum in any closed ball around $(0,0)$, and thus $(0,0)$ is a global maximum?? This is not true because $f$ is a counterexample, but I can't quite understand why.

In some cases we do have a global minimum at the unique critical point. For example, let

$f(x,y) = (-x+y)^2 + (x-1)^2 + (x+y-1)^2$

Then $f$ has a unique critical point at $(\frac{2}{3}, \frac{1}{6})$. The critical point turns out to be a global minimum (according to the graph and wolframalpha). How would I prove that the point is a global minimum? And how does one in general decide that a function has a global minimum/maximum on a critical point when the function is defined on all of $\mathbb{R}^2$, particularly when there is only one critical point?

$\endgroup$
8
$\begingroup$

Your intuitive picture is probably as follows: Imaging filling water into the hollow at the origin. As the water rises, since there are points lower than the origin, you expect the water to start spilling over into the lower lying areas at some point. And that point would need to be a critical point! The way to resolve the paradox is to note that before any spilling over occurs, the lake will in fact extend off to infinity. Look at a cross section of your function graph for fixed $y$, as $y$ starts at $0$ and then grows. In other words, study the function $x\mapsto x^2+y^2(1-x)^3$ as $y^2$ grows. You will see a local minimum, starting at the origin and approaching $x=1$ from below, and a local maximum, starting at infinity and approaching $x=1$ from above. As $y^2$ grows, the value at the local minimum grows, while the value at the local maximum decreases. So the aforementioned lake will extend out to $(1,\pm\infty)$ and then start to spill over the edge (the local maximum) at infinity before it spills over anywhere else.

A picture is tremendously helpful. If you have access to maple, try this:

plots[animate](plot,
  [x^2+y^2*(1-x)^3, x = 0 .. 2, view = [0 .. 2, -1 .. 2]], y = 0 .. 10);

Or you can take a look at this static plot of $f(x,y)$. It is easy to see how $(0,0)$ is a local, but not a global, maximum:

x^2+y^2(1-x)^3

$\endgroup$
3
  • $\begingroup$ @Adrian: Thanks for adding the picture. I must admit, though, that I find it a bit hard to read, with the whole region around the origin being so very flat, which makes it difficult to see the important features. After a bit of experimenting with Maple, however, I find that plotting the arctangent of a multiple of the given function is quite effective. $\endgroup$ – Harald Hanche-Olsen Mar 18 '12 at 0:18
  • 1
    $\begingroup$ Plot the function tanh(x^2+y^2*(1-x)^3) on the square [-3,3]X[-3,3]. That will give you a much clearer picture of the shape of the graph. $\endgroup$ – wdacda Sep 6 '18 at 8:25
  • $\begingroup$ @HaraldHanche-Olsen If you don't mind, could you please take a look at my question here? Thank you so much. math.stackexchange.com/questions/3987955/… $\endgroup$ – user56202 Jan 16 at 20:35
5
$\begingroup$

I know that f has a minimum when restricted to a compact set. Then (0,0) is a global minimum in any closed ball around (0,0)

This is where your reasoning fails. A global minimum must be attained on a critical point... or at the boundary the set! A much simpler and clearer example could be the function $f(x) = x^3$. It has only one critical point, 0, but when you take a closed ball containing 0, the minimum and maximum that $f$ attains is not at a critical point but at a boundary of the ball.

$\endgroup$
1
  • $\begingroup$ Ah, of course. I forgot about the boundary completely.. Thanks. $\endgroup$ – spin Mar 17 '12 at 20:30
3
$\begingroup$

There is a nice result showing that it is the case for polinomials of low degree.

If f(x,y) is a polynomial of degree $\leq 4$ having a local extrema as the unique critical point, then f(x,y) has a global extrema at that point.

This was proven in Calvert and Vamanamury "Local and global extrema for functions of several variables" J. Austral. Math. Soc. 29, 362-368 (1980)

$\endgroup$
1
$\begingroup$

Consider the following one dimensional example $$ f: \Bbb{R} \to \Bbb{R},\ f(x)=\begin{cases} x^2 & x\geq 0 \\ -x^2 & x <0 \end{cases}$$

This function is differentiable, and it is easy to see that the unique critical point is $x_0=0$. But $x_0=0$ is not a global (not even local) minimum/maximum.

In general, finding critical points is not enough to find the local extrema. You need to use the second derivative test to see whether the critical points are local extrema points.

$\endgroup$
2
  • $\begingroup$ This might also help: tutorial.math.lamar.edu/Classes/CalcI/ShapeofGraphPtII.aspx $\endgroup$ – Beni Bogosel Mar 17 '12 at 19:06
  • $\begingroup$ Thanks, but I already know that a function can have critical points which are not local/global maximum/minimum (saddle points). In the case of $\mathbb{R}$ I think it is actually true by that if a differentiable function has an unique critical point that is a local maximum/minimum, then the critical point is a global maximum/minimum. My question was about determining global max/min for functions $\mathbb{R}^2 \rightarrow \mathbb{R}$ that have a unique critical point which is a local maximum/minimum. $\endgroup$ – spin Mar 17 '12 at 20:29
-1
$\begingroup$

A critical point can be either a minimum, a maximum, a saddle point or none of these three, i.e. undecidable. For a two-variable function f(x,y), if the determinant of its Hessian is positive at the critical point (x*,y*), and fxx(x*,y*) is positive, then the critical point(x*, y*) is a local minimum. Refer to the list of tests in: https://en.wikipedia.org/wiki/Second_partial_derivative_test

$\endgroup$
5
  • $\begingroup$ Your answer of this old and solved post does not seem to add any information to the ones already in place. $\endgroup$ – Bcpicao Nov 4 '20 at 23:22
  • $\begingroup$ The new information is in the test to be applied for deciding whether the critical point is or is not a minimum. $\endgroup$ – Hass Saidane Nov 4 '20 at 23:29
  • $\begingroup$ The same link you add is already on Beni's answer. $\endgroup$ – Bcpicao Nov 5 '20 at 10:53
  • $\begingroup$ The link is the same, but the information I provided is different and answers the outstanding question "about determining global max/min for functions R2→R that have a unique critical point which is a local maximum/minimum" . The test I cited was not reported and answers the question and it is helpful to report. Here is a different link which reports the test in terms of the determinant of the Hessian matrix associated with optimality of multivariate functions: en.wikipedia.org/wiki/Hessian_matrix $\endgroup$ – Hass Saidane Nov 7 '20 at 16:09
  • $\begingroup$ Transcribing the first sentences of a Wikipedia article whose link has already been presented in an earlier answer is not "new and helpful" information. Even more so when the post in question has been marked by the OP as answered for 8 (eight!) years. The purpose of my comment to remind you that this answer is not useful according to this platform's guidelines and explain why that is so. Let us not waste time arguing about minutiae. Have a good day. $\endgroup$ – Bcpicao Nov 7 '20 at 20:09

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.