At unique critical point, does local max imply global max

calculusmultivariable-calculus

Let $f:(a,b)\to\mathbb{R}$ be a continuous function with continuous derivative. Suppose $f$ has a unique critical point $x_0\in (a,b)$. If $f$ has a local maximum at $x_0$, then $f$ must have a global maximum at $x_0$. Intuitively, if $f$ had a global maximum somewhere else, the graph would have to turn around, leading to a second critical point.

I would like to know if a similar results holds in two dimensions.

Does there exist a connected open subset $U\subset\mathbb{R}^2$ and a function $f:U\to\mathbb{R}$ with continuous first partial derivatives, such that $f$ has a unique critical point $u_0\in U$, which is a local maximum but not a global maximum?

Best Answer

No, it doesn't, see example below. The question arises due to the phenomenon that a real valued differentiable function $f$ (defined on an interval) which has only one critical point in which it possesses a maximum that maximum must be global. (It may have some other critical points as well.)

Take $f(x,y)=-x^3-y^2+xy$, defined on $\mathbb R^2$ without the negative $x$-axis, for example. The domain is simply connected. Now the only maximum $f$ has is in $ (1/6, 1/12)$, obviously not a global one.

The behaviour of $f$ becomes clearer if we write $f$f as $$f(x,y)=x^3+\frac14x^2+(y-x/2)^2.$$ This shows that $f$ is a parabolic transformation surface.

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[Math] Unique critical point does not imply global maximum/global minimum

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

[Math] Only one critical point, local minimum but not global

Using your example, one can show that similar examples exists for all $n\geq 3$: Let $F :\mathbb R^{2+n} \to \mathbb R$, $n\geq 1$ be defined by

$$F(x, y, z_1, \cdots, z_n) = f(x, y) + \epsilon(z_1^2 + \cdots + z_n^2)$$

where $f$ is the first example you gave. Then $dF=0$ only at the point $(0,1,0,\cdots, 0)$ and it is a local minimum by the second derivative test. Also $F(0,-3, 0,\cdots, 0) <F(0,-1, 0\cdots, 0)$ for some small $\epsilon$.

Best Answer

Related Solutions

[Math] Unique critical point does not imply global maximum/global minimum

[Math] Only one critical point, local minimum but not global

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