## Archive for August 2010

### Functions with exactly one stationary point

August 9, 2010

Let $f:\mathbb{R}^n\rightarrow \mathbb{R}$ be a continuous differentiable function, with exactly one stationary point $x$, and suppose that this point is a local minimum. If $n=1$ it is easy to see that $x$ must be a global minimum, but what if $n\geq 2$?

The answer is written in white. Highlight it to read it.

No, x doesn’t have to be a global minimum. Consider the function f(x,y)=(e^y+e^{-y^2})(-2x^3+3x^2)-e^{-y^2} (I don’t know how to make latex white, so I wrote it in plain text instead, sorry). We see that (e^y+e^{-y^2}) is positive so df/dx=0 only if x=0 or x=1. For x=1 the function is e^y and doesn’t have any stationary points. For x=0 the function is -e^{-y^2}, and (0,0) is a stationary point. The point is a local minimum, since for x < 1 we have f(x,y)>=f(0,y)>=f(0,0)=-1, but it is not a global minimum since f(2,0)=-9. Now the next question is: What if f is a polynomium? I don’t know the answer.