Lipschitz Selectors May Not Yield Competitive Algorithms for Convex Body Chasing

Abstract

The current best algorithms for the convex body chasing (CBC) problem in online algorithms use the notion of the Steiner point of a convex set. In particular, the algorithm that always moves to the Steiner point of the request set is O(d) competitive for nested CBC, and this is optimal among memoryless algorithms [Bubeck et al.: Chasing nested convex bodies nearly optimally. In: 31st Annual ACM–SIAM Symposium on Discrete Algorithms (Salt Lake City 2020), pp. 1496–1508. SIAM, Philadelphia (2020)]. A memoryless algorithm coincides with the notion of a selector in functional analysis. The Steiner point is noted for being Lipschitz with respect to the Hausdorff metric, and for achieving the minimal Lipschitz constant possible. It is natural to ask whether every selector with this Lipschitz property yields a competitive algorithm for nested CBC. We answer this question in the negative by exhibiting a selector that yields a non-competitive algorithm for nested CBC but is Lipschitz with respect to Hausdorff distance. Furthermore, we show that being Lipschitz with respect to an $L_p$-type analog of the Hausdorff distance is sufficient to guarantee competitiveness if and only if $p=1$.

Appendix

1.1 Omitted Proofs

1.2 Steiner Selector is Lipschitz for $D_1$

Fact 4.1

The Steiner selector $\textrm{st}:\mathcal {K}^d \rightarrow \mathbb {R}^d$ is d-Lipschitz with respect to the metric $D_1$.

Proof

For any two sets $A,B \in \mathcal {K}^d$, by the definition of Steiner point we have

$$\begin{aligned} \Vert \text {st}(A)-\text {st}(B)\Vert&=\left\| \, d\int _{y\in S^{d-1}} y\cdot h_{A}(y) \, d\sigma (y)-d\int _{y\in S^{d-1}} y\cdot h_{B}(y) \,d\sigma (y)\right\| \\&\le d\int _{y\in S^{d-1}}\Vert (h_{A}(y)- h_{B}(y))\cdot y\Vert \, d\sigma (y) \qquad \qquad \qquad \qquad (\hbox {Jensen's})\\&= d\int _{y\in S^{d-1}}|h_A(y)- h_B(y)|\ d\sigma (y) \qquad \qquad \qquad \qquad \qquad \quad ({\Vert y\Vert =1})\\&= d\cdot D_1(A,B), \end{aligned}$$

which concludes the proof.$\square$

1.3 Being Lipschitz for $D_1$ Implies Competitiveness

Fact 4.2

If $s:\mathcal {K}^d \rightarrow \mathbb {R}^d$ is a L-Lipschitz selector with respect to $D_1,$ then s is an L-competitive selector.

Proof

Let $B \supseteq K_1\supseteq K_2 \supseteq \ldots$ be any nested sequence. Then since s is L-Lipschitz with respect to $D_1$,

$$\begin{aligned}&\sum _{t=1}^\infty \Vert s(K_t)-s(K_{t+1})\Vert \le \sum _{t=1}^\infty L\int _{y\in S^{d-1}} \left| h_{K_t}(y)- h_{K_{t+1}}(y)\right| \,d\sigma (y) \\&\quad = L\sum _{t=1}^\infty \int _{y\in S^{d-1}} \left( h_{K_t}(y)- h_{K_{t+1}}(y)\right) \,d\sigma (y)\qquad \qquad ({K_t\supseteq K_{t+1}\hbox { so } h_{K_t}\ge h_{K_{t+1}}})\\&\quad = L\int _{y\in S^{d-1}} h_{K_1}(y) \, d\sigma (y)-d\cdot \lim _{t\rightarrow \infty } \int _{y\in S^{d-1}}h_{K_t}(y)\, d\sigma (y)\le L, \end{aligned}$$

where the last inequality follows from $h_{K_1}(y) \le 1$ and $h_{K_t}(y) + h_{K_t}(-y) \ge 0$. Hence the proof.$\square$

1.4 No Lipschitz Selector Extension for $D_p$

Fact 4.3

Fix $p \in [1,\infty )$ and $d > 2p+2$. Let $\hat{s}$ be the partial selector defined only on the unit ball B with $\hat{s}(B) = e_1$ (which is trivially Lipschitz). Then $\hat{s}$ cannot be extended to a selector which is Lipschitz with respect to $D_p$.

Proof

Let s be an arbitrary extension of $\hat{s}$. For $\theta \in [0, {\pi }/{2}]$, let $K_\theta := B - \overline{C}(e_1, \theta )$. We claim that

$$\begin{aligned} \lim _{\theta \rightarrow 0^+}\frac{\Vert s(K_\theta ) - s(B)\Vert }{D_p(K_\theta , B)} = \infty . \end{aligned}$$

(10)

First, we have $\Vert s(K_\theta ) - s(B)\Vert \ge h_{B}(e_1) - h_{K_\theta }(e_1) = 1 - \cos \theta \ge \varOmega (\theta ^2)$ (the asymptotic $\varOmega (\,{\cdot }\,)$ is as $\theta \rightarrow 0^+$). Also, for $y\in S^{d-1} - C(e_1, \theta )$ we have $h_B(y) = h_{K_\theta }(y)$, and for $y\in C(e_1,\theta )$ we have the bounds $0\le h_{K_\theta }(y) \le h_B(y) = 1$. In particular, $|h_{K_\theta }(y)-h_B(y)|^p \le 1$. Therefore,

$$\begin{aligned} D_p(K_{\theta }, B)= & {} \left( \,\int _{y\in S^{d-1}} |h_{K_\theta }(y) - h_B(y)|^p\,d\sigma (y)\right) ^{\!1/p}\\ {}\le & {} [\sigma (C(e_1,\theta ))]^{1/p}\le O(\theta ^{(d-2)/p}), \end{aligned}$$

where the last bound follows from Lemma 2.3 (again the asymptotic $O(\,{\cdot }\,)$ is with $\theta \rightarrow 0^+$). Choosing $d > 2p+2$ proves (10), and hence the fact. $\square$