About convex combinations of real-stable multivariable complex polynomials

Say $f: \mathbb{C}^{n+1} \rightarrow \mathbb{C}$ is a real stable multivariable polynomial on the variables $(z,w_1,w_2,...,w_n)$. (a "real-stable" polynomial is one which has no zeroes in the open set where all its arguments have a positive imaginary part)

Now let $f(w_i +1)$ denote the function $f$ with its argument $w_i$ shifted by $1$ and all other $w$s and $z$ held fixed. Let $p_i \in [0,1]$ such that $\sum_{i=1}^n p_i =1$.

• I guess it follows that for all $i \in \{1,2,3...,n\}$, $f(w_i +1)$ is also real-stable. Right?

• Can one say if or when would $f' = \sum_{i=1}^n p_i f(w_i +1)$ would also be real-stable?

If one could go from f to f' by acting on f by some real-stability preserving operator (like the ones classified by Borcea and Branden in their 2009 Annals paper) then this would become obvious. But I can't see such an operator in the situation when f is quadratic in the w's as is in the case of my interest. (...but there is such a candidate operator when f is linear in the w's..)

Shifting $(z,w_1, \ldots, w_n)$ by a real vector $v$ shifts the set of zeros by $-v$, and therefore does not change the property of real stability. Taking a convex combination, however, can spoil real stability. For example, consider $f(w_1, w_2) = (w_1 + 2 w_2)^2$ which is real-stable. However, $\dfrac{f(w_1+1,w_2) + f(w_1,w_2+1)}{2}$ is not real-stable, e.g. it is zero at $w_1 = -3/2 + i/6$, $w_2 = i/6$.
• Like if you $f$ were linear in the $w$s then this would always be true... – guest Feb 25 '15 at 23:52