2
$\begingroup$

Consider a function $f:\Theta \subset \mathbb{R}\rightarrow \mathbb{R}$. Let $\Theta$ be a convex and open set and $f$ a strictly convex function.

If the function has a global minimum, is it unique? I'm confused by the fact that $\Theta$ is not assumed to be compact. Is this a problem for guaranteeing uniqueness?

$\endgroup$
  • 1
    $\begingroup$ Strict convexity should be enough: what if $f$ has a global min attained at two distinct points $x_1$ and $x_2$? how about the value of $f$ on the line segment $[x_1,x_2]$? $\endgroup$ – gniourf_gniourf Feb 15 '16 at 19:25
1
$\begingroup$

Strict convexity is enough: assume that $f$ has a global min $m$ that is attained at two distinct points $x_1$ and $x_2$. Then, since $f$ is convex: $$\forall t\in[0,1],\ f\bigl(tx_1+(1-t)x_2\bigr)\leq tf(x_1)+(1-t)f(x_2)=tm+(1-t)m=m.$$ Since $m$ is a global min of $f$ we must also have $$\forall t\in[0,1],\ m\leq f\bigl(tx_1+(1-t)x_2\bigr),$$ hence $$\forall t\in(0,1),\ f\bigl(tx_1+(1-t)x_2\bigr)=m=tf(x_1)+(1-t)f(x_2)$$ which is impossible since $f$ is strictly convex.

Hence, if $f$ has a global minimum, this global minimum is attained at only one point.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.