Convergence Analysis under Consistent Error Bounds

Abstract

We introduce the notion of consistent error bound functions which provides a unifying framework for error bounds for multiple convex sets. This framework goes beyond the classical Lipschitzian and Hölderian error bounds and includes logarithmic and entropic error bounds found in the exponential cone. It also includes the error bounds obtainable under the theory of amenable cones. Our main result is that the convergence rate of several projection algorithms for feasibility problems can be expressed explicitly in terms of the underlying consistent error bound function. Another feature is the usage of Karamata theory and functions of regular variations which allows us to reason about convergence rates while bypassing certain complicated expressions. Finally, applications to conic feasibility problems are given and we show that a number of algorithms have convergence rates depending explicitly on the singularity degree of the problem.

Appendix A

Proof

The fact that $f^{-}(0) = 0$ follows from $f(0) = 0$ and definition (4.2). We also note that in (4.2), if we increase s, the set after the “$\inf$” potentially shrinks, so $f^{-}$ is monotone nondecreasing. Next, we prove each item.

• (i) Fix any $s\in (0,\, \sup f)$. Suppose that $f^{-}(s) = 0$. By definition (4.2), given any $\epsilon _k > 0$, there exists $t_k\in [0,\, \epsilon _k]$ such that $f(t_k)\ge s$. Consequently, there exists a sequence $t_k\rightarrow 0_+$ with $f(t_k)\ge s > 0$. This together with $f(0) = 0$ contradicts the (right)-continuity of f at 0, and thus proves (i).

• (ii) Let $s\ge 0, t \ge 0$ be such that $s\le f(t)$. Since f is monotone increasing, $\sup f$ is never attained, which implies $0\le s\le f(t) < \sup f$. Furthermore, by definition (4.2), we have $f^{-}(s)\le t$.

• (iii) Let $s\ge 0, t \ge 0$ be such that $s < \sup f$ and $f(t) < s$. By definition, $f^{-}(f(t)):=\inf \left\{ u\ge 0: f(u)\ge f(t)\right\}$, therefore $f^{-}(f(t))\le t$. On the other hand, the strict monotonicity of f implies that there is no $u < t$ with $f(u)\ge f(t)$. This implies $f^{-}(f(t))\ge t$ and thus $f^{-}(f(t)) = t$. Together with the monotonicity of $f^{-}$, we obtain $t = f^{-}(f(t))\le f^{-}(s)$.

• (iv) Suppose that there exists some $\bar{s}\in (0,\, \sup f)$ such that $f^{-}$ is not continuous at $\bar{s}$. Since $f^{-}$ is monotone, both the left-sided limit $f^{-}(\bar{s}-)$ and the right-sided limit $f^{-}(\bar{s}+)$ exist and $f^{-}(\bar{s}-) < f^{-}(\bar{s}+)$. Fix any $t\in (f^{-}(\bar{s}-),\, f^{-}(\bar{s}+))$. From the monotonicity of $f^{-}$, there exist $\epsilon > 0$ such that whenever $s_1,s_2$ satisfy $0< s_1< \bar{s}< s_2 < \sup f$ we have

  $$\begin{aligned} f^{-}(s_1)< t - \epsilon< t + \epsilon < f^{-}(s_2). \end{aligned}$$

  We now show that $f(t) = \bar{s}$. Suppose that $f(t)\ne \bar{s}$. Then either $f(t) < \bar{s}$ or $f(t) > \bar{s}$. If $f(t) < \bar{s}$, let $s_1 = (f(t) + \bar{s})/2 \in (f(t),\, \bar{s})$. Thus, we know from item (iii) that $f^{-}(s_1)\ge t$, which contradicts $f^{-}(s_1) < t - \epsilon$. If $f(t) > \bar{s}$, let $s_2 = (f(t) + \bar{s})/2 \in (\bar{s},\, f(t))$. Then, from item (ii), we have $f^{-}(s_2) \le t$, which contradicts $t + \epsilon < f^{-}(s_2)$. This proves $f(t) = \bar{s}$. The arbitrariness of $t\in (f^{-}(\bar{s}-),\, f^{-}(\bar{s}+))$ contradicts the strict monotonicity of f. Consequently, $f^{-}$ is continuous on $(0,\, \sup f)$.