Suppouse $A \in \mathbb{R^n}$ is convex. If $f:A\to\mathbb{R^n}$ 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


  • 2
    $\begingroup$ A function is strictly convex if and only if $f((1-t)x_0 + tx_1) \lt (1-t)f(x_0) + tf(x_1)$ for all $t \in (0,1)$. Now take two minimizers... $\endgroup$ – Martin Mar 21 '13 at 17:51

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.


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