Suppose $X\subseteq\mathbb{R}^{l\times m}$ and $Y\subseteq\mathbb{R}^{m\times n}$ are sets of matrices with real entries. Moreover, $X$ is the set of all $l\times m$ stochastic matrices (each column sums up to 1, and each entry is non-negative), so $X$ is convex. Moreover, $Y$ is a (possibly strict) subset of the set of all $m\times n$ stochastic matrices, and assume $Y$ is also convex.
Is it true that $Z:=\{xy\mid x\in X\text{ and }y\in Y\}$ is also a convex set?
Intuitively I think this may not be true in general, but I had no luck to produce a counter example. Any help would be appreciated.
The proposition is likely not true in general. Here is a way to produce a numerical counterexample:
Let $X\subseteq\mathbb{R}^{2\times2}$ and $Y\subseteq\mathbb{R}^{2\times5}$. Moreover, let $Y=\{\lambda Y_u+(1-\lambda)Y_{v}\,|\,0\leq\lambda\leq1, Y_u\in\mathbb{R}^{2\times5} \text{ and }Y_v\in\mathbb{R}^{2\times5} \text{ are some column-stochastic matrices}\}$.
The question then becomes, for any $X_1,X_2\in X$, $Y_1,Y_2\in Y$ and $0\leq\mu_0\leq1$, can we find an $X_3\in X$ and $0\leq\mu_1\leq1$ such that \begin{equation}\tag{1} \mu_0X_1Y_1+(1-\mu_0)X_2Y_2 = X_3(\mu_1Y_u+(1-\mu_1)Y_v). \end{equation}
Since we just need a counterexample, we can fix $Y_1=Y_u$, $Y_2=Y_v$, $\mu_0=0.2$ (for example) and pick $X_1\in X$ and $X_2\in X$ arbitrarily. In this case, (1) becomes \begin{equation} 0.2X_1Y_u+0.8X_2Y_v = X_3(\mu_1Y_u+(1-\mu_1)Y_v). \end{equation} In the above equation, $X_1$, $X_2$, $Y_u$ and $Y_v$ are all known (fixed or picked). There are $10$ equations in total (because $X\subseteq\mathbb{R}^{2\times2}$ and $Y\subseteq\mathbb{R}^{2\times5}$), but there are only $5$ variables ($X_3$ and $\mu_1$), this give a hint that the solution might not exist.