Dear all,
I got an assignment to prove or to give a well-explained counter example, however I am a bit stuck solving the problem.
The assignment
A function [MATH]f: F \to \mathbb{R} [/MATH] is convex if and only if every local minimum off is a global minimum.
My try:
If we take a convex set [MATH]D[/MATH]. Therefore for any [MATH]y[/MATH], a feasable direction is [MATH]y - x*[/MATH].
Since [MATH]x*[/MATH] is a local minimum for any [MATH]v \in D[/MATH], we can choose a small enough [MATH]\alpha > 0[/MATH], such that
[MATH]f(x*) \le f(x* + \alpha (y-x*)) [/MATH]Since [MATH]f[/MATH] is convex, we have [MATH]f(x* + (y - x*)) = f(\alpha y + (1 - \alpha )x*) \le \alpha f(y) + (1 - \alpha )f(x*)[/MATH]Combining these to equations gives:
[MATH]f(x*) \le \alpha f(y) + (1 - \alpha )f(x*)[/MATH] which implies [MATH]f(x*) \le f(y).[/MATH]
Since y is an arbitrary point in D, this immediately proves that x is a global minimum.
Is this right? Am I missing something?
Thanks in advance for your reply!!
I got an assignment to prove or to give a well-explained counter example, however I am a bit stuck solving the problem.
The assignment
A function [MATH]f: F \to \mathbb{R} [/MATH] is convex if and only if every local minimum off is a global minimum.
My try:
If we take a convex set [MATH]D[/MATH]. Therefore for any [MATH]y[/MATH], a feasable direction is [MATH]y - x*[/MATH].
Since [MATH]x*[/MATH] is a local minimum for any [MATH]v \in D[/MATH], we can choose a small enough [MATH]\alpha > 0[/MATH], such that
[MATH]f(x*) \le f(x* + \alpha (y-x*)) [/MATH]Since [MATH]f[/MATH] is convex, we have [MATH]f(x* + (y - x*)) = f(\alpha y + (1 - \alpha )x*) \le \alpha f(y) + (1 - \alpha )f(x*)[/MATH]Combining these to equations gives:
[MATH]f(x*) \le \alpha f(y) + (1 - \alpha )f(x*)[/MATH] which implies [MATH]f(x*) \le f(y).[/MATH]
Since y is an arbitrary point in D, this immediately proves that x is a global minimum.
Is this right? Am I missing something?
Thanks in advance for your reply!!