It's known that the following function (which is a variant of $L_2$-Regression) is homogeneous of degree 1 for every matrix $A \in \mathbb{R}^{n \times m}$ and $x \in \mathbb{R}^m$: $$\left|\left| Ax \right| \right|_2.$$
Let $A \in \mathbb{R}^{n \times m}$ be a matrix, then homogeneous of degree 1 in our context, means that for every $x \in \mathbb{R}^m$ and every $b \in \mathbb{R}_+$ : $$ f(bx) = \left|\left| Abx \right| \right|_2 = b \cdot \left|\left| Ax \right| \right|_2 = b f(x)$$
My interest is to know how does this property changes when translation is added to the mix, in other words, let $g : \mathbb{R}^m \to [0,\infty)$ be the following function:
$$ g(x) = \left|\left| Ax + v \right| \right|_2,$$ where $v \in \mathbb{R}^{n}$.
Is $g$ homogeneous of some degree? or can we establish that for any $b \geq 1$, there exists some constant $L$ such that for every $x \in \mathbb{R}^m$ the following holds: $$ g(bx) \leq b^L \cdot g(x) \ ?$$
Please advise.
If $g:x\mapsto \left|\left| Ax + v \right| \right|_2$ is homogeneous with degree $\alpha>0$, then for $b\geq 0$, $\|bAx+v\|=b^\alpha \|Ax + v\|$. Letting $b\to0$ yields $\|v\|=0$, hence $v=0$, thus $A=0$ or ($A\neq 0$ and $\alpha=1$).
The last claim written by the OP is false in general. It is trivial if $A=0$ or $b=1$. Suppose for the sake of contradiction that it holds when $A\neq 0$ and $b>1$. Since $A\neq 0$, there exists some $x$ such that $Ax+v=0$, hence $g(x)=0$ and $g(bx)=0$, thus $bAx+v=0$. Since $v\neq 0$ (from context), we must have $b=1$, a contradiction.