Quasiconvex fucntions: local minimum is equal to global minimum A function is called strictly quasiconvex if, for all $x,y \in M$ with $f(x) \neq f(y)$ and all $\lambda \in (0,1)$ we can say that
$f(\lambda x + (1-\lambda)y) < max( f(x),f(y) )$.
How can i prove that if a function $f$ is strictly quasiconvex any local minimum of $f$ on $M$ is also a global minimum of $f$ on $M$?
 A: Suppose $x$ is a local minimum: that is, for some $\epsilon>0$, we have
$$f(x) \leq f(z) \quad \forall z\in M, ~\|x-z\|< \epsilon.$$
Now suppose that there is at least one value of $y$ satisfying $f(y)<f(x)$. (Note that $y$ need not be a global minimum itself.) Obviously, this means means that $\|x-y\|\geq \epsilon$.
Consider the segment between $x$ and $y$:
$$z_\lambda = \lambda x + (1-\lambda) y \quad \lambda\in(0,1)$$
At least some portion of this line segment satisfies $\|x - z_\lambda\|<\epsilon$, since
$$\|x-z_\lambda\| = \|x-\lambda x -(1-\lambda)y\| = (1-\lambda)\|x-y\|$$
For such values of $z_{\lambda}$, our local minimum assertion guarantees that $f(z_\lambda)\geq f(x)$. But strict quasiconvexity gives us
$$f(z_\lambda) = f(\lambda x + (1-\lambda)y) < \max(f(x),f(y)) = f(x) \quad\Longrightarrow\quad f(z_\lambda) < f(x).$$
This is a contradiction. Therefore, there cannot be another point $y$ satisfying $f(y)<f(x)$; $x$ is a global minimum.
EDIT: I should add that you can prove a stronger point, which is that not only is $x$ a global minimum, but it must be unique. After all, suppose there is a second point $y\neq x$ satisfying $f(x)=f(y)$. Then strict quasiconvexity says
$$f(\lambda x + (1-\lambda)y) < \max\{f(x),f(y)\} = f(x) \quad \forall \lambda\in(0,1);$$
that is, the entire line segment between the two points obtains a smaller function value. So any minimum must be unique. This is, of course, not necessarily the case for non-strict quasiconvexity.
