1
$\begingroup$

Let $\phi:\mathbb R^n\to \mathbb R\in C^2(\mathbb{R^n})$, such that $D^2\phi$ is positive semi-definite and let $c \in\mathbb{R}$ be arbitrary.
Show that $\phi^{-1}((-\infty,c])$ is convex.

Hint: Define $a(t):=v+t(w-v)$ as the line that connects two vectors and show the following:

  • $D^2(\phi\,\circ a)$ is positive semi-definite.

  • Use this to show $a([0,1])\subset\phi^{-1}((-\infty,c])$.

My work so far: I've used the chain rule and got the following: $D(\phi\,\circ a(t))=(D\phi)(a(t)\,\circ(Da)(t))$ and thus $D^2(\phi\,\circ a(t))=D\left((D\phi)(a(t)\,\circ(Da)(t))\right) = ((D^2 \phi)(a(t))\,\circ (Da)(t))\,\circ D^2(a)(t)$ but $D^2a=0$ so the whole expression would equal zero, which isn't a contradiction but still seems strange.

Thank you very much in advance.

$\endgroup$
2
$\begingroup$

Let $f(t) = \phi(a(t))$. Then $f'(t) = D \phi(a(t))D a(t) = D \phi(a(t)) (w-v)$ and $f''(t) = (w-v)^T D^2\phi(a(t)) (w-v)$.

| cite | improve this answer | |
$\endgroup$

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.