A Multiplicative Weight Updates Algorithm for Packing and Covering Semi-infinite Linear Programs

Abstract

We consider the following semi-infinite linear programming problems: $\max$ (resp., $\min$) $c^Tx$ s.t. $y^TA_ix+(d^i)^Tx \le b_i$ (resp., $y^TA_ix+(d^i)^Tx \ge b_i)$, for all $y \in {{\mathcal {Y}}}_i$, for $i=1,\ldots ,N$, where ${{\mathcal {Y}}}_i\subseteq {\mathbb {R}}^m_+$ are given compact convex sets and $A_i\in {\mathbb {R}}^{m_i\times n}_+$, $b=(b_1,\ldots b_N)\in {\mathbb {R}}_+^N$, $d^i\in {\mathbb {R}}_+^n$, and $c\in {\mathbb {R}}_+^n$ are given non-negative matrices and vectors. This general framework is useful in modeling many interesting problems. For example, it can be used to represent a sub-class of robust optimization in which the coefficients of the constraints are drawn from convex uncertainty sets ${{\mathcal {Y}}}_i$, and the goal is to optimize the objective function for the worst case choice in each ${{\mathcal {Y}}}_i$. When the uncertainty sets ${{\mathcal {Y}}}_i$ are ellipsoids, we obtain a sub-class of second-order cone programming. We show how to extend the multiplicative weights update method to derive approximation schemes for the above packing and covering problems. When the sets ${{\mathcal {Y}}}_i$ are simple, such as ellipsoids or boxes, this yields substantial improvements in running time over general convex programming solvers. We also consider the mixed packing/covering problem, in which both packing and covering constraints are given, and the objective is to find an approximately feasible solution.

Appendices

Efficient Implementations of the Oracle Integral

In this section we discuss how to efficiently implement the oracle Integral$(p,f,\epsilon ,\sigma ,{{\mathcal {Y}}})$ for a given compact convex set Y, a log-concave function $p:Y\rightarrow {\mathbb {R}}_+$, a function $f:Y\rightarrow {\mathbb {R}}^k$ which is either $f(y):=1$ or $f(y):=y$, and $\epsilon ,\sigma \in [0,1)$.

Write $F(y):=f(y)p(y)$ for $y\in {{\mathcal {Y}}}$. Then $F_i(y)$ is log-concave for all $i\in [k]$. Thus, the question is to how to efficiently well-approximate $\int _{Y}F_i(y)dy$.

1.1 General Convex Sets

Lovász and Vempala show how to implement the integration oracle in ${\tilde{O}}(m^4)$ time. More precisely, they prove the following result.

Theorem 4

[44] Let $F:Y\rightarrow {\mathbb {R}}$ be a log-concave function over a compact convex set $Y\subseteq {\mathbb {R}}^m$. Assume also the availability of a point $\tilde{y}$ that (approximately) maximizes F over Y. Given $\epsilon ,\sigma >0$, a number A can be computed such that with probability $1-\sigma$,

$$\begin{aligned} (1-\epsilon )\int _{Y} F(y)dy \le A \le (1+\epsilon )\int _{Y} F(y)dy \end{aligned}$$

(33)

in $O\left( \displaystyle \frac{m^4}{\epsilon ^2}\log ^7\frac{m}{\epsilon \sigma }\right)$ oracle calls, each involving ether a membership oracle for Y and/or an evaluation of F.

1.2 Reduction to Volume Computation

We consider here the particular case that we actually need in our algorithms, where $F_i(y)$ is either $(1\pm \epsilon )^{a^Ty}$ or $y_i(1\pm \epsilon )^{a^Ty}$, for some $a\in {\mathbb {R}}_+^m$, and where⁹$a^Ty\le T/m$ for all $y\in {{\mathcal {Y}}}\subseteq {\mathbb {R}}_+^m$. Standardly, the idea is slice Y along the direction orthogonal to a and approximate the values of the function $a^ty$ by their values at the boundary of the slices.

For simplicity let us consider the case when $F_i(y)=F(y)=(1+\epsilon )^{a^Ty}$, and we can compute the the maximum and minimum of a linear function over Y exactly. Let $y_{\min }\in {\text {argmin}}_{y\in {{\mathcal {Y}}}}a^Ty$, $y_{\max }\in {\text {argmax}}_{y\in {{\mathcal {Y}}}}a^Ty$, and define the set of points $y_{\min }=y^0,y^1,\ldots ,y^r=y_{\max }$ in Y, such that $a^Ty^i=a^Ty^{i-1}+\eta$ for $i=1,\ldots ,r$, where $\eta :=\frac{\log (1+\epsilon _1)}{\log (1+\epsilon )}$ and $r:=\lceil \frac{a^Ty_{\max }-a^Ty_{\min }}{\eta }\rceil \le \lceil \frac{T}{m\eta }\rceil$. Note that $(1+\epsilon )^{a^Ty^i}=(1+\epsilon _1)(1+\epsilon )^{a^Ty^{i-1}}$. Note also that, once we have $y_{\min }$ and $y_{\max }$, we can easily compute $y^i=y_{\min }+\lambda _i(y_{\max }-y_{\min })$, where $\lambda _i:=\frac{\eta \cdot i}{a^Ty_{\max }-a^Ty_{\min }}$, for $i=1,\ldots ,r$.

For $i=1,\ldots ,r$, define the slice $Y(i):=\{y\in {{\mathcal {Y}}}:~a^Ty^{i-1}\le a^Ty\le a^Ty^i\}$. Then it is easy to see that

$$\begin{aligned} A:=(1+\epsilon )^{a^Ty_{\min }}\sum _{i=1}^r(1+\epsilon _1)^{i}{\text {vol}}(Y(i)) \end{aligned}$$

(34)

satisfies (33). Indeed,

$$\begin{aligned} \frac{A}{1+\epsilon _1}&=\sum _{i=1}^r(1+\epsilon )^{a^Ty_{\min }+\eta \cdot (i-1)}{\text {vol}}(Y(i))\le \sum _{i=1}^r\int _{Y(i)}(1+\epsilon )^{a^Ty}dy=\int _Y(1+\epsilon )^{a^Ty}dy\\&\le \sum _{i=1}^r(1+\epsilon )^{a^Ty_{\min }+\eta \cdot i}{\text {vol}}(Y(i))=A. \end{aligned}$$

If $F(y)=y$, then ${\text {vol}}(Y(i)$ in (34) should be replaced by $\int _{Y(i)}ydy$.

1.3 Euclidean Balls

We can further show that for the case of Euclidean balls (and similarly for Ellipsoids), one can reduce the integral computation to numerical integration over a single variable.

For $z\in {\mathbb {R}}^m$ and $\rho \in {\mathbb {R}}_+$, let ${\mathbf {B}}_m(y^0,\rho ):=\{y\in {\mathbb {R}}^m:~\Vert y-y^0\Vert \le \rho \}$ be the ball of radius $\rho$ centered at $y^0$. Denote by ${\text {vol}}_m({\mathbf {B}}_m(y^0,\rho ))$ the volume of the ball ${\mathbf {B}}_m(y^0,\rho )$ in ${\mathbb {R}}^m$.

Lemma 24

Let $Y:={\mathbf {B}}_m(y^0,\rho )$, $a\in {\mathbb {R}}^m$ be a given vector, and $F(y):=y(1+\epsilon )^{a^Ty}$ for $y\in {\mathbb {R}}^m$. Let ${\hat{y}}:={\text {argmin}}\{a^Ty:~y\in {{\mathcal {Y}}}\}$. Then

$$\begin{aligned} \int _YF(y)dy= (1+\epsilon )^{a^T{\hat{y}}}\displaystyle \int _{0}^{2\rho } (1+\epsilon )^{\Vert a\Vert h}v_{m-1} (h) ({\hat{y}}+h\frac{a}{\Vert a\Vert })dh, \end{aligned}$$

(35)

where $v_{m-1} (h)$ is the volume of an $(m-1)$-dimensional ball of radius $\sqrt{h(2\rho -h)})$.

Proof

To simplify matters, we use the change of variable $z=Uy$, where $U\in {\mathbb {R}}^{m\times m}$ is a rotation transformation (i.e., $U^TU=I$) such that $Ua = \Vert a\Vert \varvec{1}_m$. Hence, $dy = \prod _i dy_i = \det U \prod _i dz_i = dz$. Note that this transformation maps the ball ${\mathbf {B}}_m(y^0,\rho )$ to to the ball $Z:={\mathbf {B}}_m(Uy^0,\rho )$, that is, $U {{\mathcal {Y}}}=Z$.

For a given $h\in [0,2\rho ]$, let ${\mathbf {B}}(h):={\mathbf {B}}_{m-1}({\hat{z}}+h\varvec{1}_m,\sqrt{h(2\rho -h)})$ be the lower-dimensional ball centered at $U \left( {\hat{y}}+ha/||a||\right) = {\hat{z}}+h\varvec{1}_m$ with radius $\sqrt{\rho ^2-(\rho -h)^2}=\sqrt{h(2\rho -h)}$. For any vector $z \in {\mathbb {R}}^m$, we denote by $z_{{\overline{m}}}$ the vector of the first $m-1$ components of y, with $y_m$ being the mth component.

$$\begin{aligned} \int _YF(y)dy&= \displaystyle U^T\int _Z z(1+\epsilon )^{a^TU^Tz}dz\\&=U^T\displaystyle \int _{{\hat{z}}_m}^{{\hat{z}}_m+2\rho } \int _{B(z_m-{\hat{z}}_m)} z(1+\epsilon )^{||a||\varvec{1}_m^Tz}dz_{\overline{m}}dz_m\\&= U^T\displaystyle \int _{0}^{2\rho } \int _{{\mathbf {B}}(h)} \left[ \begin{array}{c}z_{{\overline{m}}}\\ \hat{z}_m+h\end{array}\right] (1+\epsilon )^{\Vert a\Vert ({\hat{z}}_m+h)}dz_{\overline{m}}dh\\&= (1+\epsilon )^{\Vert a\Vert {\hat{z}}_m}U^T\displaystyle \int _{0}^{2\rho } (1+\epsilon )^{||a||h}\int _{{\mathbf {B}}(h)}\left[ \begin{array}{c}z_{\overline{m}}\\ {\hat{z}}_m+h\end{array}\right] dz_{{\overline{m}}}dh\\&= (1+\epsilon )^{\Vert a\Vert {\hat{z}}_m}U^T\displaystyle \int _{0}^{2\rho } (1+\epsilon )^{||a||h}{\text {vol}}_{m-1} ({\mathbf {B}}(h)) ({\hat{z}}+h\varvec{1}_m )dh\\&=(1+\epsilon )^{a^T{\hat{y}}}\displaystyle \int _{0}^{2\rho } (1+\epsilon )^{\Vert a\Vert h}{\text {vol}}_{m-1} ({\mathbf {B}}(h)) ({\hat{y}}+h\frac{a}{\Vert a\Vert })dh. \end{aligned}$$

The equality before the last is due to the fact that for $i=1,\cdots ,m-1$, $\int _{{\mathbf {B}}(h)} z_idz_{\overline{m}}={\hat{z}}_i{\text {vol}}_{m-1} ({\mathbf {B}}(h))$ due to symmetry, while $\int _{{\mathbf {B}}(h)} ({\hat{z}}_m+h)dz_{{\overline{m}}}=({\hat{z}}_m+h){\text {vol}}_{m-1} ({\mathbf {B}}(h))$. $\square$

Corollary 5

Suppose that there is an algorithm that approximates the integral $\int _{h_1}^{h_2}\tau (h)dh$ to within an additive error $\epsilon$, in time $q(\tau _{max},\frac{1}{\epsilon })$, where $\tau _{\max }:=\max _{h\in [h_1,h_2]}\log \tau (h)$. Then there is an algorithm that approximates the integral in (35) within an additive error of $\epsilon$ in time $q(O(m+\frac{T}{m}+H)\log m,\frac{1}{\epsilon })$, where H is the maximum number of bits needed to represent any of the components of a, $\rho$, and $y^0$.

Proof

The function $\tau (h)$ inside the integral on the R.H.S. of (35) can be written as

$$\begin{aligned} \tau (h):=\frac{\pi ^{\frac{m-1}{2}}}{\varGamma (\frac{m-1}{2}+1)}(1+\epsilon )^{\Vert a\Vert h}\left( h(2\rho -h)\right) ^{\frac{m-1}{2}} ({\hat{y}}+h\frac{a}{\Vert a\Vert }). \end{aligned}$$

where $\varGamma$ is Euler’s gamma function, and $\Vert a\Vert h\le T/m$. It can be easily verified that $\tau _{\max }=O((m+\frac{T}{m}+H)\log m)$. $\square$

Proofs Omitted from Section 5.1

Lemma 25

$\varPhi (t+1) \le \displaystyle \varPhi (t) \exp \Big ( -{\epsilon }\delta (t)\sum _{i\in I(t)} \int _{{{\mathcal {Y}}}_i(t)} \frac{p_i(y,t)}{\varPhi (t)} y^TA_i\mathbf {1}_{j(t)}dy \Big )$.

Proof

First, we note that

$$\begin{aligned} \varPhi (t+1)&= \sum _{i\in I(t+1)} \int _{{{\mathcal {Y}}}_i(t+1)} p_i(y,t+1)dy \\&= \sum _{i\in I(t+1)} \int _{{{\mathcal {Y}}}_i(t+1)} (1-\epsilon )^{g_i(y)^Tx(t+1)}dy \\&= \sum _{i\in I(t+1)} \int _{{{\mathcal {Y}}}_i(t+1)} (1-\epsilon )^{g_i(y)^T\left[ x(t)+\delta (t)\mathbf {1}_{j(t)}\right] }dy \\&= \sum _{i\in I(t+1)} \int _{{{\mathcal {Y}}}_i(t+1)} p_i(y,t)(1-\epsilon )^{\delta (t)g_i(y)^T\mathbf {1}_{j(t)}}dy \\&\le \sum _{i\in I(t)} \int _{{{\mathcal {Y}}}_i(t)} p_i(y,t)(1-\epsilon )^{\delta (t)g_i(y)^T\mathbf {1}_{j(t)}}dy. \end{aligned}$$

The last inequality is due to the fact that updates of x(t) are non-negative, and hence, $g_i(y)^Tx(t)\le g_i(y)^Tx(t)+\delta (t) g_i(y)^T\varvec{1}_{j(t)}=g_i(y)^Tx(t+1)$, implying that ${{\mathcal {Y}}}_i(t+1) \subseteq {{\mathcal {Y}}}_i(t)$ and $I(t+1)\subseteq I(t)$.

By the definition of the oracle MaxVec, the exponent of $(1-\epsilon )$ satisfies

$$\begin{aligned} \delta (t)g_i(y)^T\mathbf {1}_{j(t)} \le \frac{g_i(y)^T\mathbf {1}_{j(t)}}{\max _i \max _{z \in {{\mathcal {Y}}}_i(t)} g_i(z)^T\mathbf {1}_{j(t)}} \le 1. \end{aligned}$$

(36)

Recalling that for $z \in [0,1]$ and for $a\ge 0$, $(1-a)^z \le 1-az$, we have

$$\begin{aligned}&\sum _{i\in I(t)} \int _{{{\mathcal {Y}}}_i(t)} p_i(y,t)(1-\epsilon )^{\delta (t)g_i(y)^T\mathbf {1}_{j(t)}}dy\nonumber \\&\quad \le \sum _{i\in I(t)} \int _{{{\mathcal {Y}}}_i(t)} p_i(y,t)\Big (1-\epsilon \delta (t)g_i(y)^T\mathbf {1}_{j(t)}\Big )dy\nonumber \\&\quad = \varPhi (t)\left( 1-\epsilon \delta (t)\sum _{i\in I(t)} \int _{{{\mathcal {Y}}}_i(t)} \frac{p_i(y,t)}{\varPhi (t)} g_i(y)^T\mathbf {1}_{j(t)}dy\right) . \end{aligned}$$

(37)

Finally, using $1-z \le e^{-z}$ for all z, we get:

$$\begin{aligned} \varPhi (t+1)&\le \varPhi (t) \exp \left( -{\epsilon }\delta (t)\sum _{i\in I(t)} \int _{{{\mathcal {Y}}}_i(t)} \frac{p_i(y,t)}{\varPhi (t)} g_i(y)^T\mathbf {1}_{j(t)}dy\right) . \end{aligned}$$

(38)

$\square$

Define $1-{\bar{\epsilon }}:=\frac{(1-\epsilon _2)(1-\epsilon _1)}{1+\epsilon _1}$.

Lemma 26

Let $\kappa (t):=\sum _{t'=0}^{t-1} \delta (t')\sum _{i\in I(t)} \int _{{{\mathcal {Y}}}_i(t')} \frac{p_i(y,t')}{\varPhi (t')} g_i(y)^T\mathbf {1}_{j(t')}dy$. Then with probability at least $1-2N\sigma t$, $\kappa (t) \ge \frac{(1-{\bar{\epsilon }})\varvec{1}^Tx(t)}{z^*}$ for all t.

Proof

Let $x^*$ be an optimal solution for (Covering). Then by the feasibility of $x^*$ for (Covering), $g_i(y)^Tx^* \ge 1$ for all $y \in {{\mathcal {Y}}}_i$ and for all $i\in [N]$. Multiplying both sides by $p_i(y,t')$, integrating over the respective active sets, and summing over all $i\in I(t')$, we get

$$\begin{aligned} \sum _{i\in I(t')} \int _{{{\mathcal {Y}}}_i(t')} p_i(y,t') g_i(y)^Tx^*dy \ge \sum _{i\in I(t')}\int _{{{\mathcal {Y}}}_i(t')} p_i(y,t')dy = \varPhi (t'), \qquad \forall t'. \end{aligned}$$

Therefore,

$$\begin{aligned} \sum _{i\in I(t')} \int _{{{\mathcal {Y}}}_i(t')} \frac{p_i(y,t')}{\varPhi (t')} g_i(y)^Tx^*dy&\ge 1, \qquad \forall t'. \end{aligned}$$

(39)

For ${i\in I(t')}$, let ${\bar{y}}^i(t)$ and ${\bar{\phi }}^i(t)$ be the outputs of the oracles ${\textsc {Integral}}(p_i(y,t'),f(y):=y,\epsilon _1,\sigma ,{{\mathcal {Y}}}_i(t'))$ and ${\textsc {Integral}}(p_i(y,t'), f(y):=1,\epsilon _1,\sigma , {{\mathcal {Y}}}_i(t'))$ in steps 15 and 16 of iteration $t'$, respectively. Then we have by the union bound,

$$\begin{aligned}&\Pr \left[ ~\left| \bar{y}^i(t')-\int _{{{\mathcal {Y}}}_i(t')}p_i(y,t')ydy\right| \le \epsilon _1\int _{{{\mathcal {Y}}}_i(t')}p_i(y,t')ydy, \text { for all }i \in I(t')\right] \ge 1-N\sigma ,\\&\Pr \left[ ~\left| \bar{\phi }^i(t')-\int _{{{\mathcal {Y}}}_i(t')}p_i(y,t')dy\right| \le \epsilon _1\int _{{{\mathcal {Y}}}_i(t')}p_i(y,t')dy, \text { for all }i \in I(t')\right] \ge 1-N\sigma , \end{aligned}$$

which imply

$$\begin{aligned} \Pr \Big [\left| \sum _{i\in I(t')} (A_i^T{\bar{y}}^i(t')+\bar{\phi }^i(t')d^i)^T\varvec{1}_j-\sum _{i\in I(t')}\int _{{{\mathcal {Y}}}_i(t')}p_i(y,t')g_i(y)^T\varvec{1}_jdy\right| \\ \le \epsilon _1\sum _{i\in I(t')}\int _{{{\mathcal {Y}}}_i(t')}p_i(y,t')g_i(y)^T\varvec{1}_jdy, \text { for all }j\in [n]\Big ] \ge 1-2N\sigma . \end{aligned}$$

These together with the setting of $j(t')$ in step 18 give that with probability at least $1-2N\sigma$,

$$\begin{aligned}&\sum _{i\in I(t')}\int _{{{\mathcal {Y}}}_i(t')}p_i(y,t')g_i(y)^T\varvec{1}_{j(t')}dy\\&\quad \ge \frac{1}{1+\epsilon _1}\sum _{i\in I(t')}(A_i^T{\bar{y}}^i(t')+\bar{\phi }^i(t')d^i)^T\varvec{1}_{j(t')}\\&\quad \ge \frac{1-\epsilon _2}{1+\epsilon _1}\max _j\sum _{i\in I(t')}(A_i^T{\bar{y}}^i(t')+{\bar{\phi }}^i(t')d^i)^T\varvec{1}_j\\&\quad \ge \frac{(1-\epsilon _2)(1-\epsilon _1)}{1+\epsilon _1}\max _j\sum _{i\in I(t')}\int _{{{\mathcal {Y}}}_i(t')}p_i(y,t')g_i(y)^T\varvec{1}_jdy. \end{aligned}$$

It follows that with probability at least $1-2N\sigma$,

$$\begin{aligned}&\sum _{i\in I(t)}\int _{{{\mathcal {Y}}}_i(t')}p_i(y,t')g_i(y)^T\varvec{1}_{j(t')}dy\nonumber \\&\quad \ge (1-{\bar{\epsilon }}) \sum _{i\in I(t')}\int _{{{\mathcal {Y}}}_i(t')}p_i(y,t')g_i(y)^T\varvec{1}_jdy,\quad \text {for all }j\in [n]. \end{aligned}$$

(40)

Multiplying both sides of (40) by $x_j^*/\varPhi (t')$ (a non-negative quantity), summing over all $j\in [n]$, and recalling that $z^* = \varvec{1}^Tx^* = \sum _j x_j^*$, we get that with probability at least $1-2N\sigma$,

$$\begin{aligned}&{z^*\sum _{i\in I(t')} \int _{{{\mathcal {Y}}}_i(t')} \frac{p_i(y,t')}{\varPhi (t')} g_i(y)^T\varvec{1}_{j(t')}}dy\nonumber \\&\quad \ge (1-{\bar{\epsilon }}){\sum _j x_j^*\sum _{i\in I(t')} \int _{{{\mathcal {Y}}}_i(t')} \frac{p_i(y,t')}{\varPhi (t')} g_i(y)^T\varvec{1}_{j}}dy \nonumber \\&\quad = (1-{\bar{\epsilon }}){\sum _{i\in I(t')} \int _{{{\mathcal {Y}}}_i(t')} \frac{p_i(y,t')}{\varPhi (t')} g_i(y)^T {\left( \sum _j x_j^*\varvec{1}_{j}\right) }}dy \nonumber \\&\quad = (1-{\bar{\epsilon }}){\sum _{i\in I(t')} \int _{{{\mathcal {Y}}}_i(t')} \frac{p_i(y,t')}{\varPhi (t')} g_i(y)^Tx^*}dy\nonumber \\&\quad \ge (1-{\bar{\epsilon }}), \end{aligned}$$

(41)

where the last inequality follows from (39). Using this result, we can conclude that with probability at least $1-2N\sigma$,

$$\begin{aligned} z^* \kappa (t)&= z^* \sum _{t'=0}^{t-1} \delta (t'){\sum _{i\in I(t')} \int _{{{\mathcal {Y}}}_i(t')} \frac{p_i(y,t')}{\varPhi (t')} g_i(y)^T\mathbf {1}_{j(t')}} dy \\&= \sum _{t'=0}^{t-1} \delta (t') \left( {z^*\sum _{i\in I(t')} \int _{{{\mathcal {Y}}}_i(t')} \frac{p_i(y,t')}{\varPhi (t')} g_i(y)^T\mathbf {1}_{j(t')}}dy\right) \\&\ge (1-{\bar{\epsilon }})\sum _{t'=0}^{t-1} \delta (t')\\&=(1-{\bar{\epsilon }})\mathbf {1}^Tx(t). \end{aligned}$$

Here, the last equality is a result of the update formula for $x(t+1)$ in step 23 of Algorithm 4. $\square$

Lemma 27

For all t, with probability at least $1-2N\sigma t$, it holds that

$$\begin{aligned} \varPhi (t) \le \varPhi (0) \exp \left( -\displaystyle \frac{\epsilon (1-{\bar{\epsilon }})\mathbf {1}^Tx(t)}{z^*}\right) . \end{aligned}$$

(42)

Proof

By repeated application of (38) and using Lemma 9 we can bound $\varPhi (t)$ as follows:

$\square$

Lemma 28

Suppose $\varPhi (t) \le \gamma$ for some $\gamma >0$ and some iteration t of the algorithm. Assume also that there is $v_i>0$ such that ${\text {vol}}({{\mathcal {Y}}}_i(t))\ge v_i$ for all $i\in I(t)$. In addition, suppose that for given $T>0$ and $\epsilon _3 >0$ there exists an $\alpha$ such that

$$\begin{aligned}&0<\alpha <\alpha _0 \end{aligned}$$

(43)

$$\begin{aligned}&(1-\epsilon )^{-\alpha /2}\min _{i\in I(t)}\left\{ \left( \frac{\alpha }{\alpha _0}\right) ^{m_i}v_i\right\} >1, \end{aligned}$$

(44)

where $\alpha _0:=2T$. Then $(1-\epsilon )^{g_i(y)^Tx(t)} \le \gamma (1-\epsilon )^{-\alpha }$ for all $y \in {{\mathcal {Y}}}_i(t)$, for all $i \in I(t)$.

Proof

Towards a contradiction, suppose that there exist $i\in I(t)$ and $y^* \in {{\mathcal {Y}}}_i(t)$ such that $(1-\epsilon )^{g_i({y^*})^Tx(t)} > \gamma (1-\epsilon )^{-\alpha }$ in iteration t. Then define $\mu ^* = \frac{\alpha }{\alpha _0}\in (0,1)$. Also, define the following sets:

$$\begin{aligned}&{{\mathcal {Y}}}_i^+(t) := \left\{ y \in {{\mathcal {Y}}}_i(t): (y-y^*)^TA_ix(t) \le \frac{\alpha }{2}\right\} ,\nonumber \\&{{\mathcal {Y}}}_i^{++}(t) :=\left( 1-\frac{1}{\mu ^*}\right) y^*+\frac{1}{\mu ^*}{{\mathcal {Y}}}_i^+(t) =\left\{ y^*+\frac{1}{\mu ^*}(y-y^*) : y \in {{\mathcal {Y}}}_i^+(t) \right\} . \end{aligned}$$

(45)

$\square$

Claim 7

$y^*+\mu '(y-y^*) \in {{\mathcal {Y}}}_i^{++}(t)$ for all $\mu ' \in [0,\frac{1}{\mu ^*}]$ and $y \in {{\mathcal {Y}}}_i^+(t)$. In particular, ${{\mathcal {Y}}}_i^+(t) \subseteq {{\mathcal {Y}}}_i^{++}(t)$.

Proof

$y^*$ is in ${{\mathcal {Y}}}_i^+(t)$ since $(y^*-y^*)^TA_ix(t)=0\le \alpha /2$. ${{\mathcal {Y}}}_i^+(t)$ is the intersection of a half-space and ${{\mathcal {Y}}}_i(t)$, which are both convex. Thus, ${{\mathcal {Y}}}_i^+(t)$ is convex. Therefore, for all $\mu ' \in [0,\frac{1}{\mu ^*}]$ and for any $y \in {{\mathcal {Y}}}_i^+(t)$, we have

$$\begin{aligned}&\mu '\mu ^*y+(1-\mu '\mu ^*)y^* \in {{\mathcal {Y}}}_i^+(t), \\&\therefore y^*+\mu '\mu ^*(y-y^*) \in {{\mathcal {Y}}}_i^+(t). \end{aligned}$$

Consider the point $y^*+\mu '\mu ^*(y-y^*)$. Since this point is in ${{\mathcal {Y}}}_i^+(t)$, we can substitute it into the definition of ${{\mathcal {Y}}}_i^{++}(t)$ to get that

$$\begin{aligned}&y^*+\frac{1}{\mu ^*}\big (y^*+\mu '\mu ^*(y-y^*)-y^*\big ) \in {{\mathcal {Y}}}_i^{++}(t), \\&\therefore y^*+\mu '(y-y^*) \in {{\mathcal {Y}}}_i^{++}(t). \end{aligned}$$

In particular for $\mu '=1$, we have $y\in {{\mathcal {Y}}}_i^{++}(t)$, implying the claim. $\square$

Claim 8

${\text {vol}}({{\mathcal {Y}}}_i^{++}(t)) = \displaystyle \left( \frac{1}{\mu ^*}\right) ^{m_i} {\text {vol}}({{\mathcal {Y}}}_i^+(t))$

Proof

Immediate from the definition in (45). $\square$

Claim 9

${{\mathcal {Y}}}_i(t) \subseteq {{\mathcal {Y}}}_i^{++}(t)$.

Proof

Suppose for a contradiction that there exists a point $y \in {{\mathcal {Y}}}_i(t)\backslash {{\mathcal {Y}}}_i^{++}(t)$. Then by Claim 7, y is also outside ${{\mathcal {Y}}}_i^+(t)$. Now define:

$$\begin{aligned} \mu ^+ = \max \left\{ \mu :y^*+\mu (y-y^*) \in {{\mathcal {Y}}}_i^+(t)\right\} . \end{aligned}$$

(Note that the maximum exists since ${{\mathcal {Y}}}_i^+(t)$ is closed and bounded.) Define also $y^+ = y^* + \mu ^+(y-y^*)$. By the definition of $\mu ^+$, $y^+$ is on the boundary of ${{\mathcal {Y}}}_i^+(t)$, and the segment joining $y\not \in {{\mathcal {Y}}}_i^+(t)$ and $y^*\in {{\mathcal {Y}}}_i^+(t)$ crosses the hyperplane $\{z\in {\mathbb {R}}^{m_i}: (z-y^*)^TA_ix(t)=\frac{\alpha }{2}\}$ at $z=y^+$. This implies that $(y^+-y^*)^TA_ix(t) = \frac{\alpha }{2}$. Therefore,

$$\begin{aligned} {y^*}^TA_ix(t) + \alpha /2&= {y^+}^TA_ix(t)\\&=(\mu ^+y+(1-\mu ^+)y^*)^TA_ix(t).\\ \therefore \mu ^+{y^*}^TA_ix(t) +\alpha /2&= \mu ^+y^TA_ix(t). \end{aligned}$$

Thus,

$$\begin{aligned} \alpha /2&= \mu ^+(y-y^*)^TA_ix(t)\nonumber \\&\le \mu ^+\left( {y}^TA_ix(t)+(d^i)^Tx(t)\right) \nonumber \\&\le \mu ^*\left( {y}^TA_ix(t)+(d^i)^Tx(t)\right) \nonumber \\&\le \mu ^*T, \end{aligned}$$

(46)

where the first inequality is due to the non-negativity of $y^{*T}A_ix(t)$ (by (A1)) and $(d^i)^Tx(t)$, the second is because $\mu ^+ < \mu ^*$ (by Claim 7, as $y^*+\frac{1}{\mu ^+}(y^+-y^*)\not \in {{\mathcal {Y}}}_i^{++}(t)$), and the third is because $y \in {{\mathcal {Y}}}_i(t)$ and thus $g_i(y)^Tx(t) \le T$ by the definition of ${{\mathcal {Y}}}_i(t)$. But then plugging $\mu ^*=\frac{\alpha }{\alpha _0}$ in (46) gives $\alpha _0<2T$, contradicting the definition of $\alpha _0$. Thus, no points $y \in {{\mathcal {Y}}}_i(t) \backslash {{\mathcal {Y}}}_i^{++}(t)$ exist, and ${{\mathcal {Y}}}_i(t) \subseteq {{\mathcal {Y}}}_i^{++}(t)$. $\square$

Recalling that $y^TA_ix(t) \le {y^*}^TA_ix(t)+\alpha /2$, and hence, $g_i(y)^Tx(t)\le g_i(y^*)^Tx(t)+\alpha /2$, for all y in ${{\mathcal {Y}}}_i^+(t)$, we have

$$\begin{aligned} \varPhi _i(t)&= \int _{{{\mathcal {Y}}}_i(t)} (1-\epsilon )^{g_i(y)^Tx(t)}dy \ge \int _{{{\mathcal {Y}}}_i^+(t)} (1-\epsilon )^{g_i(y)^Tx(t)}dy\\&\ge \int _{{{\mathcal {Y}}}_i^+(t)} (1-\epsilon )^{g_i({y^*})^Tx(t)+\alpha /2}dy = (1-\epsilon )^{g_i({y^*})^Tx(t)+\alpha /2}{\text {vol}}({{\mathcal {Y}}}_i^+(t)). \end{aligned}$$

From the claims above, the volumes of the sets ${{\mathcal {Y}}}_i(t)$, ${{\mathcal {Y}}}_i^+(t)$ and ${{\mathcal {Y}}}_i^{++}(t)$ are related by ${\text {vol}}({{\mathcal {Y}}}_i^+(t)) = \left( \mu ^*\right) ^{m_i}{\text {vol}}({{\mathcal {Y}}}_i^{++}(t)) \ge \left( {\mu ^*}\right) ^{m_i}{\text {vol}}({{\mathcal {Y}}}_i(t))\ge \left( {\mu ^*}\right) ^{m_i} v_i$. Therefore,

$$\begin{aligned} \varPhi _i(t)&\ge (1-\epsilon )^{g_i({y^*})^Tx(t)+\alpha /2}\left( {\mu ^*}\right) ^{m_i}v_i. \end{aligned}$$

(47)

The point $y^*$ is one such that $(1-\epsilon )^{g_i({y^*})^Tx(t)} > \gamma (1-\epsilon )^{-\alpha }$. Also, $\alpha$ was chosen such that $(1-\epsilon )^{-\alpha /2}\left( \frac{\alpha }{\alpha _0}\right) ^{m_i}v_i>1$. Using these two facts in (47), we obtain

$$\begin{aligned} \varPhi (t) \ge \varPhi _i(t)&> \gamma (1-\epsilon )^{-\alpha +\alpha /2}(\mu ^*)^{m_i}{\text {vol}}({{\mathcal {Y}}}_i(t))\\&\ge \gamma (1-\epsilon )^{-\alpha /2}\left( \frac{\alpha }{\alpha _0}\right) ^{m_i}v_i > \gamma . \end{aligned}$$

This contradicts the hypothesis of the lemma.