0
$\begingroup$

So I am studying convex optimization and I came across this theorem regarding the minimizer of a function in a space $\mathcal{X}$. In particular, the theorem states that if $f:\mathcal{X} \rightarrow \mathbb{R}$ is a convex and differentiable function then $x^*$ minimizes $f$ on $\mathcal{X}$ if and only if:

\begin{equation} \langle\nabla f(x^*),x-x^*\rangle \geq 0,\text{ for all x}\in \mathcal{X}. \end{equation}

However, I remember when studying $\mathbb{R}^n$, $\nabla f(x^*) = 0$ was a sufficient condition for $x^*$ to be a minimizer. So my question is in which cases does $x^*$ being a minimizer imply $\nabla f(x^*) = 0$. Thank you.

$\endgroup$
  • $\begingroup$ $\nabla f(x^*)=0$ applies only if $x^*$ is in the interior of the set. Otherwise you have boundary conditions to deal with. $\endgroup$ – Michael Grant Nov 9 '15 at 16:32
1
$\begingroup$

Assume that $\mathcal{X}$ is a vector space. Your condition easily implies $\nabla f(x^*) = 0$ (in $\mathcal{X}^*$). Indeed, for any $x \in \mathcal{X}$, you have $x + x^* \in \mathcal{X}$. Hence, $$\langle \nabla f(x^*), x\rangle = \langle \nabla f(x^*), (x + x^*) - x^* \rangle \ge 0.$$ Similarly, you can show $$\langle \nabla f(x^*), -x\rangle \ge 0.$$ This gives $\langle \nabla f(x^*), x \rangle = 0$ for all $x \in \mathcal{X}$, thus, $\nabla f(x^*) = 0$ in $\mathcal{X}^*$.

If the set $\mathcal{X}$ is a proper subset, this is no longer the case.

$\endgroup$
  • $\begingroup$ This should hold only for vector spaces? $\endgroup$ – SpawnKilleR Nov 9 '15 at 8:05
  • $\begingroup$ Yes. Because you need that $x \in \mathcal{X}$ implies $x^* \pm x \in \mathcal{X}$. $\endgroup$ – gerw Nov 9 '15 at 9:57

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.