The classic Milnor-Thom upper bound on the sum of the Betti numbers of real algebraic sets (for a nice exposition and references, see e.g. N.R. Wallach, On a Theorem of Milnor and Thom, in S. Gindikin (ed.), Topics in Geometry: In Memory of Joseph D'Atri, Progress in Nonlinear Differential Equations and their Applications 20 (Birkhäuser, 1996), pp. 331-348) allows one to give an upper bound on the number of connected components of the complement $$ U_P=\{\xi\in\mathbb{R}^n\ |\ P(\xi)\neq 0\} $$ of the zero set of a real polynomial function $P:\mathbb{R}^n\rightarrow\mathbb{R}$, since $U_P$ is homeomorphic to the real algebraic subset $\tilde{U}_P=\{(\xi,\lambda)\in\mathbb{R}^{n+1}\ |\ \lambda P(\xi)=1\}$ of $\mathbb{R}^{n+1}$ through the map $$ U_P\ni\xi\mapsto\left(\xi,\frac{1}{P(\xi)}\right)\in\tilde{U}_P\ , $$ whose inverse is simply the (restriction of the) projection $$\mathbb{R}^{n+1}\ni(\xi_1,\ldots,\xi_n,\xi_{n+1})\mapsto(\xi_1,\ldots,\xi_n)\in\mathbb{R}^n$$ (to $\tilde{U}_P$).

The aforementioned bound tells us that if the degree of $P$ is $r$, then the sum of the Betti numbers of $U_P$ (and, therefore, its number of connected components) is bounded above by $(r+1)(2r+1)^n$.

Question: is there an upper bound on the number of convex connected components of $U_P$ which is sharper than the Milnor-Thom bound but also only depends on $n$ and $r$?

My intuition is that the Milnor-Thom bound is too crude to this end, even in the exceptional case when all connected components of $U_P$ are convex - take, for instance, $P(\xi)=\prod^n_{j=1}\xi_j$, in which case $r=n$ but the connected components of $U_P$ are precisely the $2^n$ orthants $$ \mathbb{R}^n_I\doteq\{\xi\in\mathbb{R}^n\ |\ \xi_j>0\text{ if }j\in I\ ,\,\xi_j<0\text{ if }j\not\in I\}\ ,\quad I\subset\{1,\ldots,n\} $$ of $\mathbb{R}^n$, which are clearly convex. More generally, if $P(\xi)$ is the elementary symmetric polynomial of order $1\leq r\leq n$ $$P(\xi)=\sum_{1\leq j_1<\cdots<j_r\leq n}\xi_{j_1}\cdots\xi_{j_r}\ ,$$ then $U_P$ has $2^r$ connected components, all of them convex.

My motivation for the above question is related to Lars Gårding's theory of hyperbolic polynomials (see e.g. L. Gårding, An Inequality for Hyperbolic Polynomials, J. Math. Mech. 8 (1959) 957-965). Recall that an homogeneous polynomial $P:\mathbb{R}^n\rightarrow\mathbb{R}$ is said to be hyperbolic with respect to $0\neq\tau\in\mathbb{R}^n$ if $P(\tau)\neq 0$ and the roots of the one-variable polynomial $P_{\xi,\tau}(\lambda)\doteq P(\xi-\lambda\tau)$ are all real. The hyperbolicity cone $C(P,\tau)$ to which $\tau$ belongs is the connected component of $U_P$ to which $\tau$ belongs. Gårding has shown that $$ C(P,\tau)=\{\xi\in\mathbb{R}^n\ |\ \text{all roots of }P_{\xi,\tau}\text{ are positive}\} $$ and, as a consequence, that $C(P,\tau)$ is an open convex cone. Roughly speaking, dual cones of hyperbolicity cones provide upper bounds to the propagation speed of the support of solutions of hyperbolic partial differential equations with principal symbol $i^rP$.

Due to the homogeneity of $P$, hyperbolicity cones always come in opposite pairs - more precisely, if $P$ is hyperbolic with respect to $\tau$, then it also is with respect to $-\tau$, and $C(P,-\tau)=-C(P,\tau)$. Since $P$ in the above examples is hyperbolic with respect to any $\tau$ in $U_P$, we conclude that the number of hyperbolicity cones of an hyperbolic homogeneous polynomial $P:\mathbb{R}^n\rightarrow\mathbb{R}$ is an even number bounded above by the number of convex connected components of $U_P$, and such an upper bound is sharp. Therefore, a positive answer to the above question would provide an upper bound to the number of hyperbolicity cones of $P$ which is sharper than the Milnor-Thom bound.

Update: This is not exactly an answer to my question (hence it does not appear as a separate answer) but rather to the original question that motivated it - to wit, a sharp upper bound on the number of hyperbolicity cones of an (homogeneous) hyperbolic polynomial. Since the result is connected to Aaron Meyerowitz's bounty-awarded answer, I feel it is worthwhile to bring it up here.

Thorsten Jörgens and Thorsten Theobald have just posted (March 15th, 2017) the arXiv preprint Hyperbolicity Cones and Imaginary Projections, arXiv:1703.04988 [math.AG].

The main result of the paper (Theorem 1.1) states (I have retained only the part of the statement that interests us here and added a short explanation on the notation):

Let $f\in\mathbb{R}[\mathbf{z}]$ be an homogeneous $n$-variable polynomial of degree $r$ with real coefficients. Then the number of hyperbolicity cones of $f$ is at most $$\mathcal{H}(n,r)=\begin{cases} 2^r & \text{for }r\leq n\ ,\\ 2\sum^{n-1}_{k=0}\binom{r-1}{k} & \text{for }r>n\ .\end{cases}$$ The maximum is attained if and only if $f$ is a product of linear polynomials.

Notice that the saturation of the bound occurs precisely as conjectured for the number of connected components of the complement of an arrangement of $r$ hyperplanes of $\mathbb{R}^n$ in general position (see Boris Bukh's comment below). A connection between both numbers is reached through the reasoning discussed in Aaron's comment to his answer: let $$f(n,r)=\begin{cases}2^r &\text{for }r\leq n\ ,\\ \sum^{n}_{k=0}\binom{r}{k} &\text{for }r>n\ .\end{cases}$$ be the latter number, so that we clearly have $$\mathcal{H}(n,r)=\begin{cases} f(n,r) & \text{for }r\leq n\ ,\\ 2f(n-1,r-1) &\text{for }r>n\ .\end{cases}$$ Let us consider from now on the case $r>n$ - adding an $(r+1)$-th hyperplane $E$ in general position to the arrangement causes the intersection of the arrangement with $E$ to be an arrangement of $r$ hyperplanes in $\mathbb{R}^{n-1}$ in general position, whose complement has $f(n-1,r)$ connected components, each cutting a connected component of the previous arrangement in half and thus adding $$f(n-1,r)=f(n,r+1)-f(n,r)$$ connected components. This formula entails $$\begin{split} f(n,r)-f(n-1,r-1) &=f(n,r)-f(n-1,r)+f(n-1,r)-f(n-1,r-1) \\ &=f(n-1,r-1)+\binom{r}{n}-\binom{r-1}{n-1}\end{split}$$ and hence $$\begin{split} f(n,r) &=\mathcal{H}(n,r)+\binom{r}{n}-\binom{r-1}{n-1} \\ &=\mathcal{H}(n,r)+\binom{r-1}{n-1}\frac{r-n}{n}\ .\end{split}$$ Particularly, this entails that not all connected components of the complement of the zero level set of a product of $r$ linear polynomials in $\mathbb{R}^n$ are hyperbolicity cones if $r>n$, contrary to what happens for $r\leq n$. This was already to be expected since, as I discussed above, hyperbolicity cones always come in pairs and $f(r,n)$ can clearly fail to be even for $r>n$. This also entails that connected components of the complement of the zero set of an (homogeneous) hyperbolic polynomial may be convex cones without being hyperbolicity cones.

Of course, the bound $\mathcal{H}(n,r)$ is far tighter than the Milnor-Thom bound, but too tight for being the desired answer.