Is local minimum/maximum necessarily global when it’s the only stationary point of a continuous & differentiable function

Question

Couldn't find this theorem even though it feels very intuitive to me.

If the $f:R^n \to R$ is continuous, and has only one stationary point – a local minimum/maximuma. Doesn't it necessarily makes it global?

If not – can you please give an example?

If yes – where is it proven?

Answer 1

For $n=1$: you need $f \in C^1$ (the function is not only continuous but continuously differentiable). (a counterxample: $f(x) = e^x - |x + 1|$)

For $n>1$ things are more complicated. See here

A function with a single critical point, which is a local minimum but not a global minimum

For a function of a single variable $f(x)$, if $f$ is continuous on an interval $I$, has only one critical point in $I$, and that critical point is a local minimum, then it is the absolute (or global) minimum. However, for functions of more than one variable things are not so nice.

Let $f(x, y)=e^{3 x}+y^3-3 y e^x$. This is differentiable for all $(x,y)$. In looking for critical points, we solve \begin{aligned} & f_x=3 e^{3 x}-3 y e^x=0 \\ & f_y=3 y^2-3 e^x=0 \end{aligned} The only point solving both of these equations is $(0,1)$, and the second derivative test shows that this point is a local minimum for $f$. However, $f(0,-3)=-17<f(0,1)=-1$, so that this local minimum is not an absolute minimum. Here's a picture:

(The way I constructed this example was by starting with $x^3+y^3-3 x y$, which has a local minimum at $(1,1)$ and a saddle point at $(0,0)$, and then sent the saddle point off to infinity by replacing $x$ by $e^x$.)

Henry Wente (of the University of Toledo) has given me a polynomial with the same property: $f(x, y)=x^2+y^2(1+x)^3$, which he found in the Math Monthly, vol. 100, no. 3, March 1993. The article is ``Counting Critical Points of Real Polynomials in 2 Variables", by Durfee et. al., pp. 255-271.