Let $A: \mathbb{R}^n \to \mathbb{R}^d$ be a matrix.
Let
$$\max_{\|y\|_2 \leq 1 } \min_{\|x\|_2 \leq R}\|Ax-y\|_2 = \epsilon\;.$$ (where $R$ is the radius of the norm ball that $x$ is constrained to).
Can we say the following $$\max_{\|y\|_2 \leq t } \min_{\|x\|_2 \leq R}\|Ax-y\|_2 = O(t\epsilon) \;?$$ (where $O()$ hides some additive and mulitplicative constants).
EDIT: As mentioned in the comments the problem can be written as $$\max_{y \in \mathcal{B}(1)}\|ty - \mathrm{Proj}_{Ax:x\in \mathcal{B}(R)}(ty)\|$$ Note the projection operation is on a closed convex set $\{Ax:x\in \mathcal{B}(R)\}$, which we define for brevity by $C$. Thus the problem reduces to $$\max_{y \in \mathcal{B}(1)}\|ty - \mathrm{Proj}_C(ty)\|\; .$$ Thus we have $\max_{y \in \mathcal{B}(1)}\|y - \mathrm{Proj}_C(y)\| = \epsilon$ which by usual definition of the distance function $d(y, C) \leq \epsilon$ for all $y \in \mathcal{B}(1)$. So if the triangle inequality holds true, then we have $d(ty, C) \leq t\epsilon$ for all $y$ and the it is proved.
Using this link General Triangle Inequality, distance from a point to a set
we have $$d(ty, C) \leq d(ty, (t-1)y) + d((t-1)y, (t-2)y)+ \ldots+ d(y, C)$$ $$\leq O(t\epsilon)$$
EDIT is this correct? Adding an edit to increase visibility