Convexity of product of elements from two convex set

Question

Given two convex set $X\subseteq \mathbb{R}^N$,$Y\subseteq \mathbb{R}^{N\times N}$

Given a $x\in X$, is the set $\{z|z=yx,\forall y \in Y\}$ convex?

If no, by adding what can force it to be convex?

Answer 1

The current and original edition of your question seems to asking for completely different thing. Do you want to ask whether the set $YX = \{ yx : y \in Y, x \in X \}$ is convex or not?

In any event, for fixed $x$, the set $Yx = \{ yx : y \in Y\}$ is convex.

For any $z_1, z_2 \in Yx, \lambda \in [0,1]$, pick $y_1, y_2 \in Y$ s.t. $z_1 = y_1 x, z_2 = y_2 x$. $$\begin{align}Y \text{ convex } &\implies y = \lambda y_1 + (1-\lambda) y_2 \in Y\\ &\implies \lambda z_1 + (1-\lambda) z_2 = (\lambda y_1 + (1-\lambda) y_2) x = y x \in Yx\end{align}$$

In contrast, the set $YX$ need not be convex. For a counter example, conside the case

$$Y = \left\{ \begin{pmatrix}1&p\\0&1\end{pmatrix} : p \in \mathbb{R} \right\} \quad\text{ and }\quad X = \left\{ \begin{pmatrix}1\\q\end{pmatrix} : q \in \mathbb{R} \right\} $$ It is then clear $$YX = \left\{ \begin{pmatrix}1 + pq\\q\end{pmatrix} : p, q \in \mathbb{R} \right \} = \mathbb{R}^{2} \setminus \left\{ \begin{pmatrix}r\\0\end{pmatrix} : r \in \mathbb{R}, r \ne 1 \right \} $$ is not convex.