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