Can we infer from the support function of a convex domain $Q$ when the boundary $\partial Q$ admits nonvanishing curvature? Let $Q$ be a compact, convex domain in the plane, with smooth boundary $\partial Q$. We further assume that the origin is contained in $Q$. For a concrete example, let's take an ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}\le1$.
The support function of $Q$ is given by $\displaystyle h(\theta)=\sup_{\mathbf{q}\in Q}\;\mathbf{q}\cdot \langle\cos\theta,\sin\theta\rangle$.
We can recover a convex domain from its support function by tracking the caustics of the supporting lines $L_\theta=\{h(\theta)\langle\cos\theta,\sin\theta\rangle+t\langle-\sin\theta,\cos\theta\rangle:t\in\mathbb{R}\}$ (Edit: More precisely, this gives the boundary of $Q$, not $Q$ itself.). So $h$ should be able to characterize $Q$. My questions are:


*

*is there some kind of characterization of $h$ such that the domain $Q$ is convex (strictly convex, respectively)?

*Assume $Q$ is already strictly convex. Is there any characterization of $h$ such that the boundary $\partial Q$ admits nonvanishing curvature?
Thank you! :)

Edit: as Harald's comment below, the resulting domain $Q=Q(h)$ is always convex, since it is the intersection of a family of convex domains--half planes.  Thank you Harald!
Let's start with arbitrary $2\pi$-periodic, positive function $h$, and obtain a convex domain $Q=Q(h)$. Then we can reconsider the support function $g=g(Q)=g(Q(h))$ of $Q$, which might be different from $h$. This happens if $Q$ does not touch the boundary of some half plane. 
So we may restrict our attention to the class of faithful support functions, those satisfy the condition $g(Q(h))=h$.
 A: 
is there some kind of characterization of $h$ such that the domain $Q$ is convex

No, because the support function of $Q$ is equal to the support function of the convex hull of $Q$.  

is there some kind of characterization of $h$ such that the domain $Q$ is strictly convex

Yes: if and only if $h$ is differentiable. In other words, if and only if the polar set of $Q$ has smooth boundary. It's worth noticing that the polar of polar set is the original set. 
A corner in a convex set creates a line segment in the boundary of its polar, because the value of support function comes from the same point (the corner) for some interval of   $\theta$. Conversely, suppose the boundary of a convex set contains a vertical segment to the right of origin. Then $h$ is not differentiable at $\theta=0$, because a small change of $\theta$ in either direction increases $h$ at a linear rate. (So, the graph of $h$ has a nonsmooth minimum like $|\theta|$.)   
The general form of the above is known as duality of smoothness and rotundity, and it holds in all finite dimensional spaces. 

Is there any characterization of $h$ such that the boundary $\partial  Q$ admits nonvanishing curvature?

Yes: if and only if $h\in C^2$.  I recommend Convex Bodies: The Brunn-Minkowski Theory by Rolf Schneider, specifically section 2.5 Higher regularity and curvature.  For the two-dimensional case, see also page 2 of Lectures on Mean Curvature Flows by Xi-Ping Zhu. 
Sketch. Introduce the Gauss map $\nu:\partial Q\to S^1$ which gives exterior unit normal vector at every boundary point. This map is defined when $Q$ has $C^1$ boundary. If  $\partial Q$ is also strictly convex, then $\nu$ is injective. And if $\partial Q$ has nonvanishing curvature, then $\nu$ is a diffeomorphism -- indeed, this is if and only if, because the derivative of $\nu$ is the curvature of $\partial Q$.
It is convenient to define $h_Q$ as a function on $\mathbb R^2$, via $h_Q(x)=\sup_{q\in Q} x\cdot q$. Then one can check  that $ h_Q(x)= x\cdot \nu^{-1}(x/|x|) $ and 
$ \nabla h_Q(x)=  \nu^{-1}(x/|x|)$. 
Thus, the equivalence is: nonvanishing curvature $\iff$ $\nu $ is a diffeomorphism $\iff$ $h_Q\in   C^2(\mathbb R^2\setminus \{0\})$. 
