Two-stage convex relaxation approach to least squares loss constrained low-rank plus sparsity optimization problems

Abstract

This paper is concerned with the least squares loss constrained low-rank plus sparsity optimization problems that seek a low-rank matrix and a sparse matrix by minimizing a positive combination of the rank function and the zero norm over a least squares constraint set describing the observation or prior information on the target matrix pair. For this class of NP-hard optimization problems, we propose a two-stage convex relaxation approach by the majorization for suitable locally Lipschitz continuous surrogates, which have a remarkable advantage in reducing the error yielded by the popular nuclear norm plus $\ell _1$-norm convex relaxation method. Also, under a suitable restricted eigenvalue condition, we establish a Frobenius norm error bound for the optimal solution of each stage and show that the error bound of the first stage convex relaxation (i.e. the nuclear norm plus $\ell _1$-norm convex relaxation), can be reduced much by the second stage convex relaxation, thereby providing the theoretical guarantee for the two-stage convex relaxation approach. We also verify the efficiency of the proposed approach by applying it to some random test problems and some problems with real data arising from specularity removal from face images, and foreground/background separation from surveillance videos.

Appendix

Lemma 6.1

Let ${\mathcal {T}}$ be the subspace given by (3). Then, for any $Z\in {\mathbb {R}}^{n_1\times n_2}$, it holds that

$$\begin{aligned}&{\mathcal {P}}_{{\mathcal {T}}}(Z) = U_1^*(U_1^*)^{{\mathbb {T}}}Z + ZV_1^*(V_1^*)^{{\mathbb {T}}}-U_1^*(U_1^*)^{{\mathbb {T}}} ZV_1^*(V_1^*)^{{\mathbb {T}}} \ \ \mathrm{and} \\&{\mathcal {P}}_{{\mathcal {T}}^{\perp }}(Z) = U_2^*(U_2^*)^{{\mathbb {T}}}ZV_2^*(V_2^*)^{{\mathbb {T}}}. \end{aligned}$$

Proof

Let ${\mathcal {B}}:{\mathbb {R}}^{n_1\times n_2}\rightarrow {\mathbb {R}}^{n_1\times n_2}$ be defined by ${\mathcal {B}}(X)= U_1^*(U_1^*)^{{\mathbb {T}}}XV_2^* (V_2^*)^{{\mathbb {T}}}+XV_1^*(V_1^*)^{{\mathbb {T}}}$ for $X\in {\mathbb {R}}^{n_1\times n_2}$. Then, from the definition of the subspace ${\mathcal {T}}$, it follows that

$$\begin{aligned} {\mathcal {P}}_{{\mathcal {T}}}(Z) = \mathop {\arg \min }_{X\in {\mathbb {R}}^{n_1\times n_2}}\left\{ \frac{1}{2} \big \Vert X-Z\big \Vert _F^2:\ X={\mathcal {B}}(X)\right\} . \end{aligned}$$

(20)

By the necessary optimal conditions of (20), there exists $Y\in {\mathbb {R}}^{n_1\times n_2}$, such that

$$\begin{aligned} {\mathcal {P}}_{{\mathcal {T}}}(Z)-Z-Y+{\mathcal {B}}^*(Y)=0\ \ \mathrm{and}\ \ {\mathcal {P}}_{{\mathcal {T}}}(Z) = {\mathcal {B}} ({\mathcal {P}}_{{\mathcal {T}}}(Z)). \end{aligned}$$

By the expression of ${\mathcal {B}}$, it is easy to check that ${\mathcal {B}}$ is self-adjoint and ${\mathcal {B}}({\mathcal {B}}(X))={\mathcal {B}}(X)$ for any $X\in {\mathbb {R}}^{n_1\times n_2}$. Together with the last equation, we deduce that

$$\begin{aligned}&{\mathcal {B}}({\mathcal {P}}_{{\mathcal {T}}}(Z)) - {\mathcal {B}}(Z)-{\mathcal {B}}(Y)+{\mathcal {B}}({\mathcal {B}}^*(Y))=0 \Longleftrightarrow {\mathcal {P}}_{{\mathcal {T}}}(Z) - {\mathcal {B}}(Z){-}{\mathcal {B}}(Y)+\,{\mathcal {B}}(Y)=0, \end{aligned}$$

which implies the expression of ${\mathcal {P}}_{{\mathcal {T}}}(Z)$. Using the same arguments, one may easily obtain the expression of ${\mathcal {P}}_{{\mathcal {T}}^{\perp }}(Z)$. Thus, we complete the proof. $\square$

The proof of Lemma 2.2

It suffices to consider that the right-hand side is positive (otherwise the inequality is trivial), which implies that $\langle H,{\mathcal {A}}^*{\mathcal {A}}(G)\rangle >0$, and then $\langle H,{\mathcal {A}}^*{\mathcal {A}}(H)\rangle >0$. Without loss of generality, we assume that $n_1\le n_2$, $\Omega$ takes the form

$$\begin{aligned} \Omega =\left\{ (i,j)\ |\ i+(j-1)n_1\le s^*\ \ \mathrm{with}\ i\in \{1,\ldots ,n_1\}, j\in \{1,\ldots ,\lfloor \frac{s^*}{n_1}\rfloor +1\}\right\} ,\nonumber \\ \end{aligned}$$

(21)

where $\lfloor \frac{s^*}{n_1}\rfloor$ means the integer not more than $\frac{s^*}{n_1}$, and all components $G_{ij}$ for $i+(j-1)n_1>s^*$ are arranged in a descending order by the column index, i.e.,

$$\begin{aligned} |G_{i_0+1,j_0}|\ge & {} \cdots \ge |G_{n_1,j_0}|\ge |G_{1,j_0+1}|\ge \cdots \ge |G_{n_1,j_0+1}| \\\ge & {} \cdots \ge |G_{1,n_2}|\ge \cdots \ge |G_{n_1,n_2}|, \end{aligned}$$

where $i_0$ and $j_0$ are positive integers such that $i_0+n_1(j_0-1)=s^*$. For $k=1,2,\ldots$, let

$$\begin{aligned}&\Omega _k:=\Big \{(i,j)\ |\ s^*+(k-1)t<i+(j-1)n_1\le s^*+kt\ \ \mathrm{for}\ i\in \{1,\ldots ,n_1\}, \\&\qquad \quad j\in \big \{\big \lfloor \frac{s^*}{n_1}\big \rfloor + 1,\ldots ,\big \lfloor \frac{s^*+kt}{n_1}\big \rfloor +1\big \}\Big \}, \end{aligned}$$

except that the largest column index in the last block stops at $n_2$. By comparing with (21), it is immediate to see that $\Omega =\Omega _0$ and $\Lambda =\Omega _1$, and consequently $\Gamma =\Omega _0\cup \Omega _1$. In addition, from the order of $|G_{ij}|$ for $i+(j-1)n_1>s^*$, we have $\Vert {\mathcal {P}}_{\Omega _k}(G)\Vert _{\infty }\le \frac{\Vert {\mathcal {P}}_{\Omega _{k-1}}(G)\Vert _{1}}{t}$ when $k>1$, which implies that $\sum _{k>1}\Vert {\mathcal {P}}_{\Omega _k}(G)\Vert _{\infty }\le \frac{\Vert {\mathcal {P}}_{\Omega ^{c}}(G)\Vert _{1}}{t}$. Then, we have that

$$\begin{aligned}&\big \langle H,{\mathcal {A}}^*{\mathcal {A}}\big \rangle -\big \langle H,{\mathcal {A}}^*{\mathcal {A}} ({\mathcal {P}}_{\Gamma } (G-H))\big \rangle \\&\quad =\big \langle H,{\mathcal {A}}^*{\mathcal {A}}(H)\big \rangle + \big \langle H, {\mathcal {A}}^*{\mathcal {A}}({\mathcal {P}}_{\Gamma ^c}(G))\big \rangle \\&\quad =\big \langle H,{\mathcal {A}}^*{\mathcal {A}}(H)\big \rangle + \sum _{k>1}\big \langle H, {\mathcal {A}}^*{\mathcal {A}}({\mathcal {P}}_{\Omega _k}(G))\big \rangle \\&\quad =\big \langle H,{\mathcal {A}}^*{\mathcal {A}}(H)\big \rangle \left[ 1+\frac{\sum _{k>1}\big \langle H, {\mathcal {A}}^*{\mathcal {A}}{\mathcal {P}}_{\Omega _k}(G)\big \rangle }{\big \langle H,{\mathcal {A}}^*{\mathcal {A}}(H)\big \rangle }\right] \\&\quad \ge \big \langle H,{\mathcal {A}}^*{\mathcal {A}}(H) \big \rangle \left[ 1-\varpi (s^*+t,t) \sum _{k>1}\frac{\Vert {\mathcal {P}}_{\Omega _k}(G)\Vert _{\infty }}{\Vert H\Vert _F}\right] \\&\quad \ge \langle H,{\mathcal {A}}^*{\mathcal {A}} (H)\rangle \left[ 1-\frac{\varpi (s^* + t,t)}{t}\frac{\Vert {\mathcal {P}}_{\Omega ^c}(G)\Vert _1}{\Vert H\Vert _F}\right] \\&\quad \ge \chi _{-}(s^*+t)\left[ \Vert H\Vert _F - \frac{\varpi (s^*+t,t)}{t} \Vert {\mathcal {P}}_{\Omega ^c}(G)\Vert _1\right] \Vert H\Vert _F. \end{aligned}$$

where the second equality is using $\Gamma =\Omega _0\cup \Omega _1$, the third one is due to $\langle H,{\mathcal {A}}^*{\mathcal {A}}(H)\rangle >0$, and the first inequality is by the definition of $\varpi (\cdot ,\cdot )$, Combining the last inequality with

$$\begin{aligned} \langle H,{\mathcal {A}}^*{\mathcal {A}}({\mathcal {P}}_{\Gamma }(G-H))\rangle \ge -\chi _{+}(s^*+t,t)\Vert H\Vert _F\Vert {\mathcal {P}}_{\Gamma }(G-H)\Vert _F, \end{aligned}$$

we immediately obtain the desired result. Thus, we complete the proof. $\square$

The proof of Theorem 4.1

Let the subspaces ${\mathcal {H}}$ and ${\mathcal {K}}$ and the index sets $\Gamma$ and $\Lambda$ be defined as in Lemma 4.2. Using Lemma 2.4 with ${\mathcal {J}}_1={\mathcal {H}}, {\mathcal {I}}={\mathcal {K}},G=\Delta L^{k}$ and $H={\mathcal {P}}_{{\mathcal {K}}}(\Delta L^k)$ yields that

$$\begin{aligned}&\max \left( 0,\frac{\langle {\mathcal {P}}_{{\mathcal {K}}} (\Delta L^k), {\mathcal {Q}}(\Delta L^k)\rangle }{\Vert {\mathcal {P}}_{{\mathcal {K}}}(\Delta L^k)\Vert _F}\right) \\&\quad \ge \vartheta _{-}(2r^*+l)\Big (\Vert {\mathcal {P}}_{{\mathcal {K}}} (\Delta L^k)\Vert _F -\frac{\pi (2r^*+l,l)}{l}\Vert {\mathcal {P}}_{{\mathcal {T}}^\bot } (\Delta L^k)\Vert _*\Big ). \end{aligned}$$

Using Lemma 2.2 with $H={\mathcal {P}}_{\Gamma }(\Delta S^k)$ and $G=\Delta S^k$ then yields that

$$\begin{aligned}&\max \left( 0,\frac{\big \langle {\mathcal {P}}_{\Gamma }(\Delta S^k),{\mathcal {Q}}(\Delta S^k)\big \rangle }{\Vert {\mathcal {P}}_{\Gamma }(\Delta S^k)\Vert _F}\right) \\&\quad \ge \chi _-(s^*+t)\Big (\Vert {\mathcal {P}}_{\Gamma }(\Delta S^k)\Vert _F - \frac{\varpi (s^*+t,t)}{t}\Vert {\mathcal {P}}_{\Omega ^c}(\Delta S^k)\Vert _1\Big ). \end{aligned}$$

In addition, from the definitions of $\vartheta _{+}(\cdot )$ and $\chi _{+}(\cdot )$, it follows that

$$\begin{aligned} \big \langle {\mathcal {P}}_{{\mathcal {K}}}(\Delta L^k), {\mathcal {Q}}(\Delta L^k )\big \rangle&\le \Vert {\mathcal {A}}{\mathcal {P}}_{{\mathcal {K}}}(\Delta L^k)\Vert _F\Vert {\mathcal {A}}(\Delta L^k )\Vert _F \\&\le \sqrt{\vartheta _+(2r^*+l)}\Vert {\mathcal {P}}_{{\mathcal {K}}}(\Delta L^k)\Vert _F\Vert {\mathcal {A}}(\Delta L^k )\Vert _F,\\ \big \langle {\mathcal {P}}_{\Gamma }(\Delta S^k), {\mathcal {Q}}(\Delta S^k )\big \rangle&\le \Vert {\mathcal {A}}{\mathcal {P}}_{\Gamma }(\Delta S^k)\Vert _F\Vert {\mathcal {A}}(\Delta S^k )\Vert _F \\&\le \sqrt{\chi _+(s^*+t)}\Vert {\mathcal {P}}_{\Gamma }(\Delta S^k)\Vert _F\Vert {\mathcal {A}}(\Delta S^k )\Vert _F. \end{aligned}$$

From the above four inequalities, it is immediate to obtain that

$$\begin{aligned}&\big \Vert {\mathcal {P}}_{{\mathcal {K}}}(\Delta L^k)\big \Vert _F +\big \Vert {\mathcal {P}}_{\Gamma }(\Delta S^k)\big \Vert _F -\frac{\pi (2r^*+l,l)}{l}\big \Vert {\mathcal {P}}_{{\mathcal {T}}^\bot }(\Delta L^k)\big \Vert _* \nonumber \\&\qquad -\frac{\varpi (s^*+t,t)}{t} \big \Vert {\mathcal {P}}_{\Omega ^c}(\Delta S^k)\big \Vert _1 \nonumber \\&\quad \le \frac{\sqrt{\vartheta _+(2r^*+ l)}}{\vartheta _{-}(2r^*+l)}\big \Vert {\mathcal {A}}(\Delta L^k)\big \Vert _F + \frac{\sqrt{\chi _+(s^*+t)}}{\chi _-(s^*+t) }\big \Vert {\mathcal {A}}(\Delta S^k)\big \Vert _F. \end{aligned}$$

(22)

Let $\beta ^k\equiv \max \Big (\frac{\pi (2r^*+l,l)}{l(1-\Vert {\mathcal {P}}_{{\mathcal {T}}^\bot }(W^{k-1})\Vert )}, \frac{\varpi (s^*+t,t)}{t\nu _{k-1}T_\mathrm{min}^{k-1}}\Big )$. From Lemma 4.1 and the definition of $\widehat{\gamma }$,

$$\begin{aligned}&\frac{\pi (2r^*+l,l)}{l}\big \Vert \mathcal{P}_{\mathcal{T}^\bot }(\Delta L^k)\big \Vert _* +\frac{\varpi (s^*+t,t)}{t} \big \Vert \mathcal{P}_{\Omega ^c}(\Delta S^k)\big \Vert _1 \\&\quad \le \beta ^k\left( (1-\Vert {\mathcal {P}}_{{\mathcal {T}}^\bot } (W^{k-1})\Vert )\Vert \mathcal{P}_{\mathcal{T}^\bot }(\Delta L^k)\Vert _* + \nu _{k-1}T_\mathrm{min}^{k-1}\Vert \mathcal{P}_{\Omega ^c} (\Delta S^k)\Vert _1\right) \\&\quad \le \beta ^k\left( \big \Vert W^{k-1} - U_1^*(V_1^*)^{\mathbb {T}}\big \Vert \big \Vert {\mathcal {P}}_{{\mathcal {T}}}(\Delta L^k)\big \Vert _* + \nu _{k-1}T_\mathrm{max}^{k-1}\Vert {\mathcal {P}}_\Omega (\Delta S^k) \Vert _1\right) \\&\quad \le \beta ^k\left( \sqrt{2r^*}\big \Vert W^{k-1} - U_1^*(V_1^*)^{\mathbb {T}}\big \Vert \big \Vert {\mathcal {P}}_{{\mathcal {T}}}(\Delta L^k)\big \Vert _F + \nu _{k-1}\sqrt{s^*} T_\mathrm{max}^{k-1} \Vert {\mathcal {P}}_\Omega (\Delta S^k)\Vert _F\right) , \\&\quad \le \widehat{\gamma }\left( \sqrt{2r^*} \xi _{k-1}\big \Vert {\mathcal {P}}_{{\mathcal {K}}}(\Delta L^k)\big \Vert _F + \sqrt{s^*}\eta _{k-1}\Vert {\mathcal {P}}_\Gamma (\Delta S^k)\Vert _F\right) , \end{aligned}$$

where the last inequality is due to $\beta ^k\le \frac{\widehat{\gamma }}{\min [(1-\Vert {\mathcal {P}}_{{\mathcal {T}}^{\perp }}(W^{k-1})\Vert ),\,\nu _{k-1}T_\mathrm{min}^{k-1}]}$ and $\Gamma =\Omega \cup \Lambda$. Together with equation (22), we obtain that

$$\begin{aligned}&(1-\widehat{\gamma }\sqrt{2r^*}\xi _{k-1})\Vert {\mathcal {P}}_{{\mathcal {K}}}(\Delta L^k)\Vert _F + (1-\widehat{\gamma }\sqrt{s^*}\eta _{k-1})\Vert {\mathcal {P}}_\Gamma (\Delta S^k)\Vert _F \\&\quad \le \frac{\sqrt{\vartheta _+(2r^*+l)}}{\vartheta _{-}(2r^*+l)} \Vert {\mathcal {A}}(\Delta L^k)\Vert _F +\frac{\sqrt{\chi _+(s^*+t)}}{\chi _-(s^*+t) }\Vert {\mathcal {A}}(\Delta S^k)\Vert _F \\&\quad \le \widetilde{\gamma }\big (\Vert {\mathcal {A}}(\Delta L^k)\Vert _F+\Vert {\mathcal {A}}(\Delta S^k)\Vert _F\big ). \end{aligned}$$

Notice that Assumption 4.1 holds and $0\le \xi _{k-1}<\frac{1}{c_1}$ and $0\le \eta _{k-1}<\frac{1}{c_2}$ imply that $1-\widehat{\gamma }\sqrt{2r^*}\xi _{k-1}>0$ and $1-\widehat{\gamma }\sqrt{s^*}\eta _{k-1}>0$. By the definitions of $a_{k-1}$ and $b_{k-1}$, we have

$$\begin{aligned} \Vert {\mathcal {P}}_{{\mathcal {K}}}(\Delta L^k)\Vert _F\le a_{k-1}\big (\Vert {\mathcal {A}}(\Delta L^k)\Vert _F + \Vert {\mathcal {A}}(\Delta S^k)\Vert _F\big )\nonumber \\ \Vert {\mathcal {P}}_\Gamma (\Delta S^k)\Vert _F\le b_{k-1}\big (\Vert {\mathcal {A}}(\Delta L^k)\Vert _F + \Vert {\mathcal {A}}(\Delta S^k)\Vert _F\big ). \end{aligned}$$

(23)

Next we focus on characterizing the bounds of $\Vert {\mathcal {A}}(\Delta L^k)\Vert _F$ and $\Vert {\mathcal {A}}(\Delta S^k)\Vert _F$. Since

$$\begin{aligned} \Vert {\mathcal {A}}(\Delta L^k)\Vert _F^2 + \Vert {\mathcal {A}}(\Delta S^k)\Vert _F^2&=\Vert {\mathcal {A}}(\Delta L^k+\Delta S^k)\Vert _F^2-2\langle {\mathcal {A}}^*{\mathcal {A}}(\Delta L^k),\Delta S^k\rangle , \\&\le 4\delta ^2 -2\langle {\mathcal {A}}^*{\mathcal {A}}(\Delta L^k),\Delta S^k\rangle , \end{aligned}$$

we only need to bound $|\langle \mathcal{Q}(\Delta L^k),\Delta S^k\rangle |$. Notice that $\Vert \Delta L^k\Vert _\infty \le 2\tau$. We have that

$$\begin{aligned}&|\langle \mathcal{Q}(\Delta L^k),\Delta S^k\rangle | \le \Vert \mathcal{Q}(\Delta L^k)\Vert _\infty \Vert \Delta S^k\Vert _1 \le \Vert {\mathcal {Q}}\Vert _\infty \Vert \Delta L^k\Vert _\infty \Vert \Delta S^k\Vert _1\\&\quad \le 2\tau \Vert {\mathcal {Q}}\Vert _\infty \Vert \Delta S^k\Vert _1. \end{aligned}$$

In addition, from Lemma 4.1 and the definitions of $\zeta _{k-1}$ and $\mu _{k-1}$, it follows that

$$\begin{aligned} \Vert \mathcal{P}_{\Omega ^c}(\Delta S^k)\Vert _1 \le \zeta _{k-1}\big \Vert {\mathcal {P}}_{{\mathcal {T}}}(\Delta L^k)\big \Vert _* +\mu _{k-1}\Vert {\mathcal {P}}_\Omega (\Delta S^k)\Vert _1. \end{aligned}$$

From the last three inequalities, it follows that

$$\begin{aligned}&\Vert {\mathcal {A}}(\Delta L^k)\Vert _F^2 + \Vert {\mathcal {A}}(\Delta S^k)\Vert _F^2 \le 4\delta ^2 + 4\tau \Vert \mathcal{Q}\Vert _\infty \Vert \Delta S^k\Vert _1\nonumber \\&\quad \le 4\delta ^2 + 4\tau \Vert \mathcal{Q}\Vert _\infty \Vert \mathcal{P}_{\Omega ^c}(\Delta S^k)\Vert _1 + 4\tau \Vert \mathcal{Q}\Vert _\infty \Vert \mathcal{P}_\Omega (\Delta S^k)\Vert _1\nonumber \\&\quad \le 4\delta ^2 + 4\tau \Vert \mathcal{Q}\Vert _\infty \zeta _{k-1}\big \Vert {\mathcal {P}}_{{\mathcal {T}}}(\Delta L^k)\big \Vert _* + 4\tau \Vert \mathcal{Q}\Vert _\infty (1+\mu _{k-1})\Vert {\mathcal {P}}_\Omega (\Delta S^k)\Vert _1\nonumber \\&\quad \le 4\delta ^2 + 4\tau \Vert \mathcal{Q}\Vert _\infty \big (\zeta _{k-1}\sqrt{2r^*}\Vert \Delta L^k\Vert _F + \sqrt{s^*}(1+\mu _{k-1})\Vert \Delta S^k\Vert _F\big ). \end{aligned}$$

(24)

Combining inequality (23) with inequality (24), we obtain that

$$\begin{aligned}&\Vert {\mathcal {P}}_{{\mathcal {K}}}(\Delta L^k)\Vert _F \\&\quad \le a_{k-1}\sqrt{4\delta ^2 + 4\tau \Vert \mathcal{Q}\Vert _\infty \big (\zeta _{k-1}\sqrt{2r^*}\Vert \Delta L^k\Vert _F + \sqrt{s^*}(1+\mu _{k-1})\Vert \Delta S^k\Vert _F\big )}, \\&\Vert {\mathcal {P}}_{\Gamma }(\Delta S^k)\Vert _F \\&\quad \le b_{k-1}\sqrt{4\delta ^2 + 4\tau \Vert \mathcal{Q}\Vert _\infty \big (\zeta _{k-1}\sqrt{2r^*}\Vert \Delta L^k\Vert _F + \sqrt{s^*}(1+\mu _{k-1})\Vert \Delta S^k\Vert _F\big )}. \end{aligned}$$

Now substituting the bounds of $\Vert {\mathcal {P}}_{{\mathcal {K}}}(\Delta L^k)\Vert _F$ and $\Vert {\mathcal {P}}_{\Gamma }(\Delta S^k)\Vert _F$ into (12) gives that

$$\begin{aligned}&\Vert \Delta L^k\Vert _F^2 + \Vert \Delta S^k\Vert _F^2 \\&\quad \le \theta _{k-1}\big [4\delta ^2 + 4\tau \Vert \mathcal{Q}\Vert _\infty \big (\zeta _{k-1}\sqrt{2r^*}\Vert \Delta L^k\Vert _F + \sqrt{s^*}(1+\mu _{k-1})\Vert \Delta S^k\Vert _F\big )\big ]. \end{aligned}$$

Notice that $x^2+y^2\le ax+by +c$ for $a,b,c\in {\mathbb {R}}_{+}$ implies $(x+y)\le \frac{a+b}{2}+\sqrt{2}\sqrt{c+\frac{a^2+b^2}{4}}$. Therefore, the last equation implies that $\Vert \Delta L^k\Vert _F+\Vert \Delta S^k\Vert _F\le \frac{a+b}{2}+\sqrt{2}\sqrt{c+\frac{a^2+b^2}{4}}$ with

$$\begin{aligned} a=4\tau \Vert \mathcal{Q}\Vert _\infty \sqrt{2r^*}\theta _{k-1}\zeta _{k-1},\ b=4\tau \Vert \mathcal{Q}\Vert _\infty \sqrt{s^*}\theta _{k-1}(1+\mu _{k-1}),\ c=4\delta ^2\theta _{k-1}. \end{aligned}$$

A suitable rearrangement yields the desired result. The proof is then completed. $\square$