Let $f:R^n\rightarrow R$ ($n>1$) be a convex function satisfying condition:

$\forall x\in R^n \ \exists t_0 = \arg\min\limits_{t\in R} f(tx).$

Is it true that there exists point $x_0\in R^n$ being global minimum of $f$?

EDIT: $t_0$ does not need to be unique.

  • 1
    $\begingroup$ The claim is not true if $f$ is not convex. To see this, consider $$ f(x,y) = (x-y^2)(x-3y^2). $$ This function has a minimum on each line passing through the origin. However, it is unbounded below: $f(2t^2,t)=-t^4$. $\endgroup$ – daw Sep 13 '18 at 14:49

It was quite tough to come up with this counterexample.

Take $$f(x,y) = \max( y^2 - x, -y ).$$ It is pretty easy to see that this function is convex and the existence of the "directional minimizers" is straightforward to check. However, $$f( s^2 + s, s) = \max( s^2 - s^2 - s, -s) = -s$$ shows that $f$ is not bounded from below. Hence, it does not have a global minimizer.


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