Joint convexity of the following function Let $x,x_0 \in \mathbb{R}^n$ and $Y \in \mathbb{R}^{n \times n}$ Is the following function convex as a function of $(x,Y)$?

$f(x,Y) = ||(x-x_0)-Y(x-x_0) ||^2_2$

I know that this function is convex w.r.t. each argument alone.
First of all, if $E$ is a (Banach) space and $g\,\colon\,E\to\mathbb{R}$ is a linear function, then $g^2$ is convex since

$$g^2\left(tx + (1-t)y\right) = g^2(tx) + 2g(tx)g\left((1-t)y\right) + g^2\left((1-t)y\right) = t^2g^2(x) + (1-t)^2g^2(y) + 2t(1-t)g(x)g(y) = tg^2(x) + (1-t)g^2(y) + t(1-t)\left(g(x) + g(y)\right)^2.$$

Then, $\|Yx\|_2^2$ is a sum of squares of linear functions hence a sum of convex functions hence convex. Finally,

$$f(x, Y) = \left\|(Y-I)(x-x_0)\right\|_2^2$$

is the same function shifted by $(x_0, I)$ hence also convex (here $I$ stands for the identity matrix).
First of all, if $E$ is a (Banach) space and $g\,\colon\,E\to\mathbb{R}$ is a linear function, then $g^2$ is convex since

$$g^2\left(tx + (1-t)y\right) = g^2(tx) + 2g(tx)g\left((1-t)y\right) + g^2\left((1-t)y\right) = t^2g^2(x) + (1-t)^2g^2(y) + 2t(1-t)g(x)g(y) = tg^2(x) + (1-t)g^2(y) + t(1-t)\left(g(x) + g(y)\right)^2.$$

Then, $\|Yx\|_2^2$ is a sum of squares of linear functions hence a sum of convex functions hence convex. Finally,

$$f(x, Y) = \left\|(Y-I)(x-x_0)\right\|_2^2$$

is the same function shifted by $(x_0, I)$ hence also convex (here $I$ stands for the identity matrix).