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

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?

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.
• You should mention what $U$ is: a connected open set that contains $(1/6, 1/12)$ and, say, $(-1,-1/2)$, but not $(0,0)$... Commented Nov 4, 2019 at 17:55