TITLE: If $f$ is strictly convex in a convex set, show it has no more than 1 minimum
QUESTION [20 upvotes]: Suppouse $A \in \mathbb R^n$ is convex. If $f:A\to\mathbb R$ is strictly convex, show that the set of minimizers if either a singleton or empty.
Ok, Suppose there exist more than one minimizer, then $f(x_i)\le f(x)\: \forall x\in B_r(x_i),\: r\gt 0$ where $x_i$ is a minimizer. Therefore there exist $x_j\:s.t\:f''(x_j)\lt0$, which is a contradiction. I understand intuitively why this is so but i think my proof is wrong
Any help would be appreciated
Thanks

REPLY [25 votes]: Suppose that f has a local minimum at $x_1$ and also another local minimum at $x_2$ with the condition that $$f(x_1)\le f(x_2) \quad x_1\ne x_2 $$
The definition of strict convexity is
$$f\big(h\ x_1+(1-h)x_2\big)\lt h\ f(x_1)+(1-h)f(x_2)\qquad 0\lt h\lt 1$$
Since h is positive
$$f(x_1)\le f(x_2) \Rightarrow h\ f(x_1)\le h\ f(x_2)$$
which justifies below condition
$$h\ f(x_1)+(1-h)f(x_2)\le h\ f(x_2)+(1-h)f(x_2)$$
$$\Rightarrow h\ f(x_1)+(1-h)f(x_2)\le f(x_2)$$
Replacing this condition to the definition of convexity
$$f\big(h\ x_1+(1-h)x_2\big)\lt f(x_2)$$
If $x_2$ is a local minimum the neighborhood must be defined such as $f(x)\gt f(x_2)$ which is a contradiction with the above condition. To satisfy both conditions it must be that $x_1=x_2$ which shows that f has at most one local minimum.