Confirming Stationary Minima & Convexity of a Function

[ahamdiheme]

If the stationary points of a function are minimum points does that qualify the function to be a convex function?
Also, if the function has only one stationary minimum point does that mean that that point is its global minimum?
Can someone please confirm these for me
Thank you

 

[slider142]

If the stationary points of a function are minimum points does that qualify the function to be a convex function?

Not necessarily. Consider the function f, whose graph is the union over all even integers n of the functions f_n(x) = (x + n)^2 each defined over the interval [n-1, n+1]. All of f's stationary points are minimum points, but f is not convex on [n, n + 2] for even n. Perhaps you mean critical points instead of stationary points?

 

[DyslexicHobo]

Also, if the function has only one stationary minimum point does that mean that that point is its global minimum?
Thank you

If "stationary minimum" is the same as "local minimum", then not necessarily.

Look at the graph of [itex]-x^4 + x^2[/tex]. You can see that it has only one local minimum at x=0, but it's global minimum is at ±∞.[/itex]

 

[ahamdiheme]

Considering the shape of the graph, that makes it convex right?

 

[DyslexicHobo]

I may possibly not be very well versed in the subject, but I'm fairly sure that a graph is either "concave up", "concave down", or neither. I have never used the term convex to describe a 2-d Cartesian graph. The graph I supplied is concave for one interval and concave down at two intervals. You can solve at these points of inflection (where the graph changes from concave up to concave down) by using the second derivative test.

http://en.wikipedia.org/wiki/Second_derivative_test