Functions with exactly one stationary point

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.

Explore posts in the same categories: Uncategorized

One Comment on “Functions with exactly one stationary point”

  1. gowers Says:

    I was first told this problem about twenty years ago and it is still one of my favourites. I prefer to think about it visually. First of all, it is not hard to see that if you allow yourself just one other stationary point then it is easy to find such a function from \mathbb{R}^2 to \mathbb{R}. You just start with a function like f(x,y)=x, which gives you a sloping plane, and then you stick a finger into that plane and push down, creating a local minimum, but also a saddle point nearby. Then all you need to observe is that you can compose your function by another function that takes the saddle point to infinity. With those thoughts you can then create any number of examples, though they won’t be polynomials since those won’t take the saddle point to infinity. But a rational function shouldn’t be too hard.

Leave a Reply

Fill in your details below or click an icon to log in: Logo

You are commenting using your account. Log Out / Change )

Twitter picture

You are commenting using your Twitter account. Log Out / Change )

Facebook photo

You are commenting using your Facebook account. Log Out / Change )

Google+ photo

You are commenting using your Google+ account. Log Out / Change )

Connecting to %s

%d bloggers like this: