Fix $a \in \mathbb R^n$, $r \ge 0$, and $\epsilon \ge 0$, and consider the functions given by $f(x) := \|x-a\|_2$ and $g(x) = f(x) + r\|x\|_2$, for $x \in \mathbb R^n$. Let $S(f) := \{x \in \mathbb R^n \mid f(x) \le \epsilon\}$.
Question. As a function of $r$, $\epsilon$, and $a$, what are some generally good lower-bounds for $\inf_{x \in S(f)} g(x)$ and $\inf_{x \in S(g)} f(x)$ ?
An obvious lower-bound
- It is clear that $$ \begin{split} \inf_{x \in S(f)} g(x) &\ge \epsilon + \min_{\|x-a\|_2 \le \epsilon} r\|x\|_2 = \epsilon + r\max(\|a\|_2-\epsilon,0)\\ &= \begin{cases}\epsilon,&\mbox{ if }\|a\|_2\le \epsilon,\\(1-r)\epsilon + r\|a\|_2,&\mbox{ else.} \end{cases} \end{split} $$
Here is a partial answer, computing $\inf_{x \in S(f)} g(x)$ exactly.
Let $C = \operatorname{conv}\{0, a\} \subseteq L$ and $p_C(x)$ be the projections of $x$ onto $C$. Since projection onto a convex set is non-expansive, we have \begin{align*} g(p_C(x)) &= \|p_C(x) - a\| + r\|p_C(x)\| \\ &= \|p_C(x) - p_C(a)\| + r\|p_C(x) - p_C(0)\| \\ &\le \|x - a\| + r\|x - 0\| \\ &= g(x). \end{align*} Now, using this same non-expansiveness, we can also conclude that if $\|x - a\| \le \varepsilon$, then $$\|p_C(x) - a\| = \|p_C(x) - p_C(a)\| \le \|x - a\| \le \varepsilon.$$ So, if $x \in S(f)$, then $p_C(x) \in S(f)$. Thus, assuming for the moment that $a \neq 0$, \begin{align*} \inf_{\|x - a\| \le \varepsilon} g(x) &= \inf_{\substack{\|x - a\| \le \varepsilon \\ x \in C}} g(x) = \inf_{\substack{\|x - a\| \le \varepsilon \\ x = \lambda a \\ 0 \le \lambda \le 1}} g(x) \\ &= \inf_{\substack{\max\{0, 1 - \varepsilon/\|a\|\} \le \lambda \le 1}} g(\lambda a) \\ &= \inf_{\substack{\max\{0, 1 - \varepsilon/\|a\|\} \le \lambda \le 1}} (1 - \lambda + r\lambda)\|a\| \\ &= \|a\| + \|a\|\inf_{\substack{\max\{0, 1 - \varepsilon/\|a\|\} \le \lambda \le 1}} \lambda(r - 1) \\ &= \|a\| + \|a\|\begin{cases} (r - 1)\max\left\{0, 1 - \frac{\varepsilon}{\|a\|}\right\} & \text{if } r > 1 \\ r - 1 & \text{if }r \le 1\end{cases} \\ &= \begin{cases} \|a\| + (r - 1)\max\{0, \|a\| - \varepsilon\} & \text{if } r > 1 \\ r\|a\| & \text{if }r \le 1\end{cases}. \end{align*} If we relax the $a \neq 0$ assumption, then the above conclusion clearly still holds, by considering the $a = 0$ case separately.