At unique critical point, does local max imply global max?

Question

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?

Michael Hoppe · Accepted Answer · 2019-11-14T11:47:28.107

2

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.

edited Nov 14 '19 at 11:47

answered Nov 04 '19 at 17:41

Michael Hoppe

18,764

You should mention what $U$ is: a connected open set that contains $(1/6, 1/12)$ and, say, $(-1,-1/2)$, but not $(0,0)$... – Robert Israel Nov 04 '19 at 17:55
@RobertIsrael Thanks, corrected. – Michael Hoppe Nov 04 '19 at 18:42

At unique critical point, does local max imply global max?

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