Appendix
In this section, we prove Proposition 3.2, Theorem 3.1, and Lemma 3.1. Firstly, we prove Proposition 3.2.
Proof
(Proof of proposition 3.2) From (6), we know that
$$\begin{aligned} \nabla f(x^{k+1})= & {} \mathcal{A}^*y^k - \sigma \mathcal{A}^*(\mathcal{A}(x^{k+1})-b)) + \sigma (x^k-x^{k+1})\nonumber \\&\quad + \Pi _{K^*}(z^k-\sigma x^k) + d^k. \end{aligned}$$
(32)
And from the MNAL method, we have
$$\begin{aligned} y^{k+1}= & {} y^k-\tau \sigma (\mathcal{A}(x^{k+1})-b),\nonumber \\ z^{k+1}= & {} z^k-\tau \sigma (x^{k+1}-s^{k+1})=z^k-\tau (\sigma x^{k+1}-\Pi _{K}(\sigma x^{k+1}-z^k))\\= & {} z^k+\tau (\Pi _{K^*}(z^k-\sigma x^{k+1})-z^k).\nonumber \end{aligned}$$
(33)
Then (32) and (33) imply that
$$\begin{aligned} \nabla f(x^{k+1})= & {} \mathcal{A}^*(y^{k+1}-(1-\tau )\sigma (\mathcal{A}(x^{k+1})-b))+ \sigma (x^k-x^{k+1})\nonumber \\&\quad + \Pi _{K^*}(z^k-\sigma x^k) + d^k. \end{aligned}$$
(34)
Meanwhile, from the KKT condition (1), we have
$$\begin{aligned} \nabla f(\overline{x}) - \mathcal{A}^*\overline{y} - \overline{z}=0. \end{aligned}$$
(35)
By (34), (35) and (36),
we obtain that
$$\begin{aligned} \langle \mathcal{A}^*(y^{k+1}-(1-\tau )\sigma (\mathcal{A}(x^{k+1})-b)+ \sigma (x^k-x^{k+1})\\ + \Pi _{K^*}(z^k-\sigma x^k) + d^k - \mathcal{A}^*\overline{y} - \overline{z}, \, x^{k+1}-\overline{x}\rangle \\ \ge \Vert x^{k+1}_{e}\Vert ^2_{\Sigma _{f}}. \end{aligned}$$
Furthermore, we obtain that
$$\begin{aligned}&\langle y^{k+1}_{e}, \, \mathcal{A}(x^{k+1}-\overline{x})\rangle -(1-\tau )\sigma \langle \mathcal{A}(x^{k+1})-b, \, \mathcal{A}(x^{k+1})-\mathcal{A}(\overline{x})\rangle \\&\quad +\sigma \langle x^{k}_{e}{-}x^{k+1}_{e}, \, x^{k+1}_{e}\rangle {+} \langle \Pi _{K^*}(z^k{-}\sigma x^k)-\overline{z}, \, x^{k+1}_{e}\rangle + \langle d^k, \, x^{k+1}_{e}\rangle \ge \Vert x^{k+1}_{e}\Vert ^2_{\Sigma _{f}}, \end{aligned}$$
i.e.,
$$\begin{aligned}&\frac{1}{\tau \sigma }\langle y^{k+1}_{e}, \, y^{k}_{e}-y^{k+1}_{e}\rangle -(1-\tau )\sigma \Vert \mathcal{A}(x^{k+1})-b\Vert ^2+\sigma \langle x^{k}_{e}-x^{k+1}_{e}, \, x^{k+1}_{e}\rangle \nonumber \\&\qquad +\langle \Pi _{K^*}(z^k-\sigma x^k)-\overline{z}, \, x^{k+1}_{e}\rangle + \langle d^k, \, x^{k+1}_{e}\rangle \ge \Vert x^{k+1}_{e}\Vert ^2_{\Sigma _{f}}. \end{aligned}$$
(36)
Here we note that the term
$$\begin{aligned} \langle \Pi _{K^*}(z^k-\sigma x^k)-\overline{z}, \, x^{k+1}_{e}\rangle= & {} \langle \Pi _{K^*}(z^k-\sigma x^k)-\Pi _{K^*}(z^k-\sigma x^{k+1}), \, x^{k+1}_{e}\rangle \nonumber \\&+ \langle \Pi _{K^*}(z^k-\sigma x^{k+1})-\overline{z}, \, x^{k+1}_{e}\rangle \nonumber \\= & {} \langle \Pi _{K^*}(z^k-\sigma x^k)-\Pi _{K^*}(z^k-\sigma x^{k+1}), \, x^{k+1}_{e}\rangle \nonumber \\&+ \langle z^{k+1}+(1-\tau )(\Pi _{K^*}(z^k-\sigma x^{k+1})-z^k)\nonumber \\&-\overline{z}, \, x^{k+1}_{e}\rangle . \end{aligned}$$
(37)
Since
$$\begin{aligned} x^{k+1}_{e}= & {} \frac{1}{\sigma }(\sigma x^{k+1}-z^k-\sigma \overline{x}+z^k)\\= & {} \frac{1}{\sigma }(\Pi _{K}(\sigma x^{k+1}-z^k)-\Pi _{K^*}(z^k-\sigma x^{k+1})-\sigma \overline{x}+z^k)\\= & {} \frac{1}{\sigma }(\Pi _{K}(\sigma x^{k+1}-z^k)-\sigma \overline{x}) + \frac{1}{\tau \sigma }(z^k-z^{k+1}), \end{aligned}$$
where the third equality is according to (33), thus
$$\begin{aligned}&\langle \Pi _{K^*}(z^k-\sigma x^{k+1})-\overline{z}, \, x^{k+1}_{e}\rangle \nonumber \\&= \frac{1}{\tau \sigma }\langle z^{k+1}_{e}, \, z^{k}_{e}-z^{k+1}_{e}\rangle \nonumber \\&\quad -\frac{1-\tau }{\tau \sigma }\langle \Pi _{K^*}(z^k-\sigma x^{k+1})-z^k, \, z^{k+1}-z^{k}\rangle \nonumber \\&\quad + \frac{1}{\sigma }\langle \Pi _{K}(\sigma x^{k+1}-z^k)-\sigma \overline{x}, \, \Pi _{K^*}(z^k-\sigma x^{k+1})-\overline{z}\rangle \nonumber \\&=\frac{1}{\tau \sigma }\langle z^{k+1}_{e}, \, z^{k}_{e}-z^{k+1}_{e}\rangle - \frac{1-\tau }{\sigma }\Vert \Pi _{K^*}(z^k-\sigma x^{k+1})-z^k\Vert ^2\nonumber \\&\quad -\frac{1}{\sigma }\langle \Pi _{K}(\sigma x^{k+1}-z^k), \, \overline{z}\rangle -\langle \Pi _{K^*}(z^k-\sigma x^{k+1}), \, \overline{x}\rangle . \end{aligned}$$
(38)
By substituting (38) into (37), and then substituting (37) into (36), we obtain that
$$\begin{aligned}&\frac{1}{\tau \sigma }\langle y^{k+1}_{e}, \, y^{k}_{e}-y^{k+1}_{e}\rangle -(1-\tau )\sigma \Vert \mathcal{A}(x^{k+1})-b\Vert ^2\\&\qquad +\sigma \langle x^{k}_{e}-x^{k+1}_{e}, \, x^{k+1}_{e}\rangle +\frac{1}{\tau \sigma }\langle z^{k+1}_{e}, \, z^{k}_{e}-z^{k+1}_{e}\rangle \\&\qquad -\frac{1-\tau }{\sigma }\Vert \Pi _{K^*}(z^k-\sigma x^{k+1})-z^k\Vert ^2+\langle \Pi _{K^*}(z^k-\sigma x^k)\\&\qquad -\Pi _{K^*}(z^k-\sigma x^{k+1}), \, x^{k+1}_{e}\rangle + \langle d^k, \, x^{k+1}_{e}\rangle \ge \Vert x^{k+1}_{e}\Vert ^2_{\Sigma _{f}}\\&\qquad +\langle \Pi _{K^*}(z^k-\sigma x^{k+1}), \, \overline{x}\rangle +\frac{1}{\sigma }\langle \Pi _{K}(\sigma x^{k+1}-z^k), \, \overline{z}\rangle . \end{aligned}$$
Using the elementary relations \(\langle u, \, v\rangle =\frac{1}{2}(\Vert u\Vert ^2+\Vert v\Vert ^2-\Vert u-v\Vert ^2)=\frac{1}{2}(\Vert u+v\Vert ^2-\Vert u\Vert ^2-\Vert v\Vert ^2)\), we further obtain the following inequality
$$\begin{aligned}&\frac{1}{2\tau \sigma }\Vert y^{k}_{e}\Vert ^2-\frac{1}{2\tau \sigma }\Vert y^{k+1}_{e}\Vert ^2\\&\qquad -\frac{1}{2\tau \sigma }\Vert y^{k}_{e}-y^{k+1}_{e}\Vert ^2-(1-\tau )\sigma \Vert \mathcal{A}(x^{k+1})-b\Vert ^2+\frac{\sigma }{2}\Vert x^{k}_{e}\Vert ^2-\frac{\sigma }{2}\Vert x^{k+1}_{e}\Vert ^2\\&\qquad -\frac{\sigma }{2}\Vert x^{k}_{e}-x^{k+1}_{e}\Vert ^2+\frac{1}{2\tau \sigma }\Vert z^{k}_{e}\Vert ^2\\&\qquad -\frac{1}{2\tau \sigma }\Vert z^{k+1}_{e}\Vert ^2-\frac{1}{2\tau \sigma }\Vert z^{k}_{e}-z^{k+1}_{e}\Vert ^2-\frac{1-\tau }{\sigma }\Vert \Pi _{K^*}(z^k-\sigma x^{k+1})-z^k\Vert ^2\\&\qquad +\frac{\sigma }{2}\Vert x^{k}_{e}-x^{k+1}_{e}\Vert ^2+\frac{\sigma }{2}\Vert x^{k+1}_{e}\Vert ^2+\langle d^k, \, x^{k+1}_{e}\rangle \\&\ge \Vert x^{k+1}_{e}\Vert ^2_{\Sigma _{f}}+\langle \Pi _{K^*}(z^k-\sigma x^{k+1}), \, \overline{x}\rangle +\frac{1}{\sigma }\langle \Pi _{K}(\sigma x^{k+1}-z^k), \, \overline{z}\rangle , \end{aligned}$$
then we have
$$\begin{aligned}&\frac{1}{\tau \sigma }\Vert y^{k}_{e}\Vert ^2-\frac{1}{\tau \sigma }\Vert y^{k+1}_{e}\Vert ^2\nonumber \\&\qquad -(2-\tau )\sigma \Vert \mathcal{A}(x^{k+1})-b\Vert ^2+\sigma \Vert x^{k}_{e}\Vert ^2-\sigma \Vert x^{k+1}_{e}\Vert ^2+\frac{1}{\tau \sigma }\Vert z^{k}_{e}\Vert ^2-\frac{1}{\tau \sigma }\Vert z^{k+1}_{e}\Vert ^2\nonumber \\&\qquad -\frac{2-\tau }{\sigma }\Vert \Pi _{K^*}(z^k-\sigma x^{k+1})-z^k\Vert ^2+\sigma \Vert x^{k+1}_{e}\Vert ^2+2\langle d^k, \, x^{k+1}_{e}\rangle \nonumber \\&\ge 2\Vert x^{k+1}_{e}\Vert ^2_{\Sigma _{f}}+2\langle \Pi _{K^*}(z^k-\sigma x^{k+1}), \, \overline{x}\rangle +\frac{2}{\sigma }\langle \Pi _{K}(\sigma x^{k+1}-z^k), \, \overline{z}\rangle . \end{aligned}$$
(39)
Finally, we reformulate (39) to get the required inequality (20). \(\square \)
Secondly, we prove Theorem 3.1.
Proof (Proof of Theorem 3.1)
By Proposition 3.3, we obtain the following relation
$$\begin{aligned} \frac{1}{\tau \sigma }\Vert \overline{y}^{k+1}_{e}\Vert ^2+\frac{1}{\tau \sigma }\Vert \overline{z}^{k+1}_{e}\Vert ^2+\sigma \Vert \overline{x}^{k+1}_{e}\Vert ^2\le & {} \frac{1}{\tau \sigma }\Vert y^{k}_{e}\Vert ^2+\frac{1}{\tau \sigma }\Vert z^{k}_{e}\Vert ^2+\sigma \Vert x^{k}_{e}\Vert ^2\nonumber \\ \end{aligned}$$
(40)
Now we define the sequences \(\{\xi ^k\}\) and \(\{\overline{\xi }^k\}\) with
$$\begin{aligned} \xi ^k:= & {} \left( \frac{1}{\sqrt{\tau \sigma }}y^k_e,\frac{1}{\sqrt{\tau \sigma }}z^k_e,\sqrt{\sigma } x^k_e\right) ,\\ \overline{\xi }^k:= & {} \left( \frac{1}{\sqrt{\tau \sigma }}\overline{y}^k_e,\frac{1}{\sqrt{\tau \sigma }}\overline{z}^k_e,\sqrt{\sigma } \overline{x}^k_e\right) . \end{aligned}$$
Thus by (40), we get
$$\begin{aligned} \Vert \overline{\xi }^{k+1}\Vert\le & {} \Vert \xi ^{k}\Vert . \end{aligned}$$
Furthermore, we have
$$\begin{aligned} \Vert \xi ^{k+1}\Vert\le & {} \Vert \xi ^{k}\Vert +\Vert \overline{\xi }^{k+1}-\xi ^{k+1}\Vert . \end{aligned}$$
(41)
Next we estimate the bound of the last term in (41). First,
$$\begin{aligned} \Vert \overline{\xi }^{k+1}-\xi ^{k+1}\Vert ^2= & {} \frac{1}{\tau \sigma }\Vert \overline{y}^{k+1}-y^{k+1}\Vert ^2+\frac{1}{\tau \sigma }\Vert \overline{z}^{k+1}-z^{k+1}\Vert ^2\nonumber \\&\quad +\sigma \Vert \overline{x}^{k+1}-x^{k+1}\Vert ^2, \end{aligned}$$
(42)
and
$$\begin{aligned} \overline{y}^{k+1}-y^{k+1}= & {} y^k-\tau \sigma (\mathcal{A}(\overline{x}^{k+1}-b))-(y^k-\tau \sigma (\mathcal{A}(x^{k+1}-b)))\nonumber \\= & {} \tau \sigma \mathcal{A}(x^{k+1}-\overline{x}^{k+1}) \end{aligned}$$
(43)
$$\begin{aligned} \overline{z}^{k+1}-z^{k+1}= & {} z^k{+}\tau (\Pi _{K^*}(z^k{-}\sigma \overline{x}^{k+1}){-}z^k)-(z^k+\tau (\Pi _{K^*}(z^k-\sigma x^{k+1})-z^k))\nonumber \\= & {} \tau (\Pi _{K^*}(z^k-\sigma \overline{x}^{k+1})-\Pi _{K^*}(z^k-\sigma x^{k+1})). \end{aligned}$$
(44)
Then (42)–(44) imply that
$$\begin{aligned} \Vert \overline{\xi }^{k+1}{-}\xi ^{k+1}\Vert ^2\le & {} \tau \sigma \Vert \mathcal{A}(\overline{x}^{k+1}{-}x^{k+1})\Vert ^2{+}\tau \sigma \Vert \overline{x}^{k+1}{-}x^{k+1}\Vert ^2{+}\sigma \Vert \overline{x}^{k+1}-x^{k+1}\Vert ^2\nonumber \\= & {} \Vert \overline{x}^{k+1}-x^{k+1}\Vert ^2_{(\tau +1)\sigma \mathcal{I}+\tau \sigma \mathcal{A}^*\mathcal{A}}\le \varrho \varepsilon _k^2, \end{aligned}$$
where \(\varrho :=\Vert (\tau +1)\sigma \mathcal{I}+\tau \sigma \mathcal{A}^*\mathcal{A}\Vert /\underline{\lambda }\). Hence
$$\begin{aligned} \Vert \overline{\xi }^{k+1}-\xi ^{k+1}\Vert\le & {} \sqrt{\varrho }\varepsilon _k, \end{aligned}$$
then
$$\begin{aligned} \Vert \xi ^{k+1}\Vert\le & {} \Vert \xi ^{k}\Vert +\sqrt{\varrho }\varepsilon _k. \end{aligned}$$
(45)
From (45), we have
$$\begin{aligned} \Vert \xi ^{k+1}\Vert\le & {} \Vert \xi ^{1}\Vert +\sqrt{\varrho }\mathcal{E}. \end{aligned}$$
(46)
We can see from (46) that the sequences \(\{\xi ^k\}\) and \(\{\overline{\xi }^k\}\) are bounded. From the definition of \(\{\xi ^k\}\), we see that \(\{x^k\}\), \(\{y^k\}\) and \(\{z^k\}\) are also bounded.
Since the sequence \(\{(x^k,y^k,z^k)\}\) is bounded, it has a subsequence \(\{(x^{k_i},y^{k_i},z^{k_i})\}\) which converges to an accumulation point \((x^{\infty },y^{\infty },z^{\infty })\). Now we show that \((x^{\infty },y^{\infty },z^{\infty })\) is a KKT point. By Proposition 3.3, we have that
$$\begin{aligned}&\sum _{k=1}^{\infty }\left( \Vert \overline{x}^{k+1}_e\Vert ^2_{2\Sigma _f+(2-\tau )\sigma \mathcal{A}^*\mathcal{A}-\sigma \mathcal{I}}+\frac{2-\tau }{\sigma }\Vert \Pi _{K^*}(z^k-\sigma \overline{x}^{k+1})-z^k\Vert ^2\right) \\&\qquad \le \sum _{k=1}^{\infty }\left( (\phi _k(\overline{x},\overline{y},\overline{z})-\phi _{k+1}(\overline{x},\overline{y},\overline{z}))+(\phi _{k+1}(\overline{x},\overline{y},\overline{z})-\overline{\phi }_{k+1}(\overline{x},\overline{y},\overline{z}))\right) \\&\qquad \le \phi _1(\overline{x},\overline{y},\overline{z})+\sum _{k=1}^{\infty }\Vert \xi ^{k+1}-\overline{\xi }^{k+1}\Vert (\Vert \xi ^{k+1}\Vert +\Vert \overline{\xi }^{k+1}\Vert )\\&\qquad \le \phi _1(\overline{x},\overline{y},\overline{z})+\sqrt{\varrho }\mathcal{E}(\Vert \xi ^{k+1}\Vert +\Vert \overline{\xi }^{k+1}\Vert )<\infty , \end{aligned}$$
where we have used the inequality that \(\phi _{k+1}(\overline{x},\overline{y},\overline{z})-\overline{\phi }_{k+1}(\overline{x},\overline{y},\overline{z})=\Vert \xi ^{k+1}\Vert ^2-\Vert \overline{\xi }^{k+1}\Vert ^2\le \Vert \xi ^{k+1}-\overline{\xi }^{k+1}\Vert (\Vert \xi ^{k+1}\Vert +\Vert \overline{\xi }^{k+1}\Vert )\).
From the summability of the sequences \(\{\Vert \overline{x}^{k+1}_e\Vert ^2_{2\Sigma _f+(2-\tau )\sigma \mathcal{A}^*\mathcal{A}-\sigma \mathcal{I}}\}\) and \(\{\Vert \Pi _{K^*}(z^k-\sigma \overline{x}^{k+1})-z^k\Vert ^2\}\), we have that
$$\begin{aligned} \lim _{k\rightarrow \infty }\left( \Vert \overline{x}^{k+1}_e\Vert ^2_{2\Sigma _f+(2-\tau )\sigma \mathcal{A}^*\mathcal{A}-\sigma \mathcal{I}}+\frac{2-\tau }{\sigma }\Vert \Pi _{K^*}(z^k-\sigma \overline{x}^{k+1})-z^k\Vert ^2\right) =0.\qquad \end{aligned}$$
(47)
By (47) and (22), we obtain that
$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert \overline{x}^{k+1}_e\Vert =0=\lim _{k\rightarrow \infty }\Vert \Pi _{K^*}(z^k-\sigma \overline{x}^{k+1})-z^k\Vert . \end{aligned}$$
(48)
From (48), we get
$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert \overline{x}^{k+1}-\overline{x}^{k}\Vert= & {} 0. \end{aligned}$$
In addition to \(\Vert \overline{x}^{k+1}-x^{k+1}\Vert \le \varepsilon _k/\sqrt{\underline{\lambda }}\), we can get
$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert x^{k+1}-x^{k}\Vert =0=\lim _{k\rightarrow \infty }\Vert x^{k+1}-\overline{x}\Vert , \end{aligned}$$
(49)
and
$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert \Pi _{K^*}(z^k-\sigma x^{k+1})-z^k\Vert= & {} 0. \end{aligned}$$
(50)
Based on (11) and (49), we obtain that
$$\begin{aligned} \lim _{k\rightarrow \infty }\Vert \mathcal{A}(x^{k+1})-b\Vert =\lim _{k\rightarrow \infty }\Vert y^{k+1}-y^{k}\Vert =0. \end{aligned}$$
(51)
From (48), (49), (50), and (51), we see that by taking limits on both sides of (34) along the subsequence \(\{x^{k_i},y^{k_i},z^{k_i}\}\), we obtain that
$$\begin{aligned} \nabla f(x^{\infty }) - \mathcal{A}^*y^{\infty } - z^{\infty }=0. \end{aligned}$$
Moreover, from (51) and (50), we also obtain that
$$\begin{aligned} \mathcal{A}(x^{\infty })-b=0,\ x^{\infty }-\frac{1}{\sigma }\Pi _{K}(\sigma x^{\infty }-z^{\infty })=0. \end{aligned}$$
So \((x^{\infty },y^{\infty },z^{\infty })\) is a KKT point. Without loss of generality, we can assume \((x^{\infty },y^{\infty },z^{\infty })=(\overline{x},\overline{y},\overline{z})\). From (45), we have for any \(k\ge k_i\),
$$\begin{aligned} \Vert \xi ^{k+1}\Vert\le & {} \Vert \xi ^{k_i}\Vert +\sum _{j=k_i}^{k}\sqrt{\varrho }\varepsilon _j. \end{aligned}$$
Since \(\lim _{k_i\rightarrow \infty }\Vert \xi ^{k_i}\Vert =0\) and \(\{\varepsilon _k\}\) is summable, we have that \(\lim _{k\rightarrow \infty }\Vert \xi ^k\Vert =0\). Finally, making use of the definition of \(\xi ^k\), we get
$$\begin{aligned} \lim _{k\rightarrow \infty }x^{k}=x^{\infty }=\overline{x},\ \lim _{k\rightarrow \infty }y^{k}=y^{\infty }=\overline{y},\ \text {and}\ \lim _{k\rightarrow \infty }z^{k}=x^{\infty }=\overline{z}. \end{aligned}$$
That is, we have proved the convergence of the sequence \(\{x^k,y^k,z^k\}\) to an optimal solution. \(\square \)
Finally, we present the proof of Lemma 3.1.
Proof
(Proof of Lemma 3.1) In combination of (32) and (11), we have
$$\begin{aligned} \nabla f(x^{k+1})= & {} \mathcal{A}^*[y^{k+1}-(1-\tau )\sigma (\mathcal{A}(x^{k+1})-b)]\nonumber \\&\quad -\sigma (x^{k+1}-x^{k})+\Pi _{K^*}(z^{k}-\sigma x^k)+d^k. \end{aligned}$$
(52)
By the convexity of \(f(\cdot )\), (52), (11) and (12), we conclude that
$$\begin{aligned}&Q(\overline{x},\overline{s},y,z;w^{k+1}) = [f(x^{k+1})-\langle y, \, \mathcal{A}(x^{k+1})-b\rangle -\langle z, \, x^{k+1}-s^{k+1}\rangle ]-[f(\overline{x})\nonumber \\&\quad -\langle y^{k+1}, \, \mathcal{A}(\overline{x})-b\rangle -\langle z^{k+1}, \, \overline{x}-\overline{s}\rangle ]\nonumber \\&\le \langle \nabla f(x^{k+1}), \, x^{k+1}-\overline{x}\rangle -\frac{1}{2}\Vert x^{k+1}-\overline{x}\Vert ^2_{\Sigma _f}\nonumber \\&\quad -\langle y, \, \mathcal{A}(x^{k+1})-b\rangle -\langle z, \, x^{k+1}-s^{k+1}\rangle \nonumber \\&= \langle \mathcal{A}^*[y^{k+1}-(1-\tau )\sigma (\mathcal{A}(x^{k+1})-b)]-\sigma (x^{k+1}-x^{k})+\Pi _{K^*}(z^{k}-\sigma x^k)\nonumber \\&\quad +d^k, x^{k+1}-\overline{x}\rangle -\frac{1}{2}\Vert x^{k+1}-\overline{x}\Vert ^2_{\Sigma _f}-\langle y, \, \mathcal{A}(x^{k+1})-b\rangle -\langle z, \, x^{k+1}-s^{k+1}\rangle \nonumber \\&= \langle y^{k+1}-y, \, \mathcal{A}(x^{k+1})-b\rangle -(1-\tau )\sigma \Vert \mathcal{A}(x^{k+1})-b\Vert ^2\nonumber \\&\quad +\sigma \langle x^{k}-x^{k+1}, \, x^{k+1}-\overline{x}\rangle \nonumber \\&\quad +\langle \Pi _{K^*}(z^{k}-\sigma x^k)-\Pi _{K^*}(z^{k}-\sigma x^{k+1}), \, x^{k+1}-\overline{x}\rangle \nonumber \\&\quad +\langle \Pi _{K^*}(z^{k}-\sigma x^{k+1}), x^{k+1}-\overline{x}\rangle -\langle z, \, x^{k+1}-s^{k+1}\rangle +\langle d^k, \, x^{k+1}-\overline{x}\rangle \nonumber \\&\quad -\frac{1}{2}\Vert x^{k+1}-\overline{x}\Vert ^2_{\Sigma _f}\nonumber \\&= \frac{1}{\tau \sigma }\langle y^{k}-y^{k+1}, \, y^{k+1}-y\rangle -(1-\tau )\sigma \Vert \mathcal{A}(x^{k+1})-b\Vert ^2\nonumber \\&\quad +\sigma \langle x^{k}-x^{k+1}, \, x^{k+1}-\overline{x}\rangle \nonumber \\&\quad +\langle \Pi _{K^*}(z^{k}-\sigma x^k)-\Pi _{K^*}(z^{k}-\sigma x^{k+1}), \, x^{k+1}-\overline{x}\rangle -\langle z, \, x^{k+1}-s^{k+1}\rangle \nonumber \\&\quad + \langle \Pi _{K^*}(z^{k}-\sigma x^{k+1}), \, s^{k+1}-\overline{x}-\frac{1}{\tau \sigma }(z^{k+1}-z^k)\rangle +\langle d^k, \, x^{k+1}-\overline{x}\rangle \nonumber \\&\quad -\frac{1}{2}\Vert x^{k+1}-\overline{x}\Vert ^2_{\Sigma _f} \end{aligned}$$
By taking advantage of the fact that \(\langle \Pi _{K^*}(z^{k}-\sigma x^{k+1}), \, s^{k+1}-\overline{x}\rangle \le 0\), we have
$$\begin{aligned} Q(\overline{x},\overline{s},y,z;w^{k+1})\le & {} \frac{1}{\tau \sigma }\langle y^{k}-y^{k+1}, \, y^{k+1}-y\rangle -(1-\tau )\sigma \Vert \mathcal{A}(x^{k+1})-b\Vert ^2\nonumber \\&+\sigma \langle x^{k}-x^{k+1}, \, x^{k+1}-\overline{x}\rangle +\langle \Pi _{K^*}(z^{k}-\sigma x^k)\nonumber \\&-\Pi _{K^*}(z^{k}-\sigma x^{k+1}), \, x^{k+1}-\overline{x}\rangle -\langle z, \, x^{k+1}-s^{k+1}\rangle \nonumber \\&- \frac{1}{\tau \sigma }\langle \Pi _{K^*}(z^{k}-\sigma x^{k+1}), \, z^{k+1}-z^k\rangle +\langle d^k, \, x^{k+1}-\overline{x}\rangle \nonumber \\&-\frac{1}{2}\Vert x^{k+1}-\overline{x}\Vert ^2_{\Sigma _f}\nonumber \\\le & {} \frac{1}{\tau \sigma }\langle y^{k}-y^{k+1}, \, y^{k+1}-y\rangle -(1-\tau )\sigma \Vert \mathcal{A}(x^{k+1})-b\Vert ^2\nonumber \\&+\sigma \langle x^{k}-x^{k+1}, \, x^{k+1}-\overline{x}\rangle \nonumber \\&+\langle \Pi _{K^*}(z^{k}-\sigma x^k)-\Pi _{K^*}(z^{k}-\sigma x^{k+1}), \, x^{k+1}-\overline{x}\rangle \nonumber \\&-\frac{1}{\tau \sigma }\langle \Pi _{K^*}(z^{k}-\sigma x^{k+1})-z, \, z^{k+1}-z^k\rangle \nonumber \\&+\langle d^k, \, x^{k+1}-\overline{x}\rangle -\frac{1}{2}\Vert x^{k+1}-\overline{x}\Vert ^2_{\Sigma _f}\nonumber \\= & {} \frac{1}{\tau \sigma }\langle y^{k}-y^{k+1}, \, y^{k+1}-y\rangle -(1-\tau )\sigma \Vert \mathcal{A}(x^{k+1})-b\Vert ^2\nonumber \\&+\sigma \langle x^{k}-x^{k+1}, \, x^{k+1}-\overline{x}\rangle \nonumber \\&+\langle \Pi _{K^*}(z^{k}-\sigma x^k)-\Pi _{K^*}(z^{k}-\sigma x^{k+1}), \, x^{k+1}-\overline{x}\rangle \nonumber \\&-\frac{1}{\tau \sigma }\langle z^{k+1}+(1-\tau )(\Pi _{K^*}(z^{k}\nonumber \\&-\sigma x^{k+1})-z^k)-z, \, z^{k+1}-z^k\rangle +\langle d^k, \, x^{k+1}-\overline{x}\rangle \nonumber \\&-\frac{1}{2}\Vert x^{k+1}-\overline{x}\Vert ^2_{\Sigma _f}\nonumber \\= & {} \frac{1}{\tau \sigma }\langle y^{k}-y^{k+1}, \, y^{k+1}-y\rangle -(1-\tau )\sigma \Vert \mathcal{A}(x^{k+1})-b\Vert ^2\nonumber \\&+\sigma \langle x^{k}-x^{k+1}, \, x^{k+1}-\overline{x}\rangle \nonumber \\&+\langle \Pi _{K^*}(z^{k}-\sigma x^k)-\Pi _{K^*}(z^{k}\nonumber \\&-\sigma x^{k+1}), \, x^{k+1}-\overline{x}\rangle -\frac{1}{\tau \sigma }\langle z^{k+1}-z, \, z^{k+1}-z^k\rangle \nonumber \\&-\frac{1-\tau }{\tau \sigma }\langle \Pi _{K^*}(z^{k}\nonumber \\&-\sigma x^{k+1}){-}z^k, \, \tau (\Pi _{K^*}(z^{k}{-}\sigma x^{k+1})-z^k)\rangle +\langle d^k, \, x^{k+1}-\overline{x}\rangle \nonumber \\&-\frac{1}{2}\Vert x^{k+1}-\overline{x}\Vert ^2_{\Sigma _f} \end{aligned}$$
Making use of the fact that \(\langle \Pi _{K^*}(z^{k}-\sigma x^k)-\Pi _{K^*}(z^{k}-\sigma x^{k+1}), \, x^{k+1}-\overline{x}\rangle \le \sigma \Vert x^{k+1}-x^{k}\Vert \Vert x^{k+1}-\overline{x}\Vert \le \frac{\sigma }{2}(\Vert x^{k+1}-x^{k}\Vert ^2+\Vert x^{k+1}-\overline{x}\Vert ^2)\), we have
$$\begin{aligned} Q(\overline{x},\overline{s},y,z;w^{k+1})\le & {} \frac{1}{2\tau \sigma }\Vert y^k-y\Vert ^2-\frac{1}{2\tau \sigma }\Vert y^{k+1}-y\Vert ^2\nonumber \\&-(1-\frac{\tau }{2}) \sigma \Vert \mathcal{A}(x^{k+1})-b\Vert ^2+\frac{\sigma }{2}\Vert x^k-\overline{x}\Vert ^2\nonumber \\&-\frac{1}{2}\Vert x^{k+1}-\overline{x}\Vert ^2_{\Sigma _f}+\frac{1}{2\tau \sigma }\Vert z^k-z\Vert ^2-\frac{1}{2\tau \sigma }\Vert z^{k+1}-z\Vert ^2\nonumber \\&-\frac{1-\frac{1}{2}\tau }{\sigma }\Vert \Pi _{K^*}(z^{k}-\sigma x^{k+1})-z^k\Vert ^2+\langle d^k, \, x^{k+1}-\overline{x}\rangle \nonumber \\\le & {} \frac{1}{2\tau \sigma }\Vert y^k-y\Vert ^2-\frac{1}{2\tau \sigma }\Vert y^{k+1}-y\Vert ^2+\frac{1}{2\tau \sigma }\Vert z^k-z\Vert ^2\nonumber \\&-\frac{1}{2\tau \sigma }\Vert z^{k+1}-z\Vert ^2\nonumber \\&+\frac{\sigma }{2}\Vert x^k-\overline{x}\Vert ^2-\frac{\sigma }{2}\Vert x^{k+1}-\overline{x}\Vert ^2-\frac{1}{2}\Vert x^{k+1}\nonumber \\&-\overline{x}\Vert ^2_{\Sigma _f+(2-\tau )\sigma \mathcal{A}^*\mathcal{A}-\sigma \mathcal{I}} +\langle d^k, \, x^{k+1}-\overline{x}\rangle . \end{aligned}$$
(53)
Summing on both sides of the inequality (53), we could get the conclusion (25) as below
$$\begin{aligned}&\sum _{i=1}^{k}Q(\overline{x},\overline{s},y,z;w^{i+1})\nonumber \\&\le \sum _{i=1}^{k} \Big \{\frac{1}{2\tau \sigma }(\Vert y^i-y\Vert ^2-\Vert y^{i+1}-y\Vert ^2)+\frac{1}{2\tau \sigma }(\Vert z^i-z\Vert ^2-\Vert z^{i+1}-z\Vert ^2)\nonumber \\&\quad +\frac{\sigma }{2}(\Vert x^i-\overline{x}\Vert ^2-\Vert x^{i+1}-\overline{x}\Vert ^2)-\frac{1}{2}\Vert x^{i+1}-\overline{x}\Vert ^2_{\Sigma _f+(2-\tau )\sigma \mathcal{A}^*\mathcal{A}-\sigma \mathcal{I}}\Big \} \nonumber \\&\quad +\sum _{i=1}^{k}\langle d^i, \, x^{i+1}-\overline{x}\rangle \nonumber \\&\le \frac{1}{2\tau \sigma }(\Vert y^1-y\Vert ^2-\Vert y^{k+1}-y\Vert ^2) + \frac{1}{2\tau \sigma }(\Vert z^1-z\Vert ^2-\Vert z^{k+1}-z\Vert ^2) \nonumber \\&\quad +\frac{\sigma }{2}(\Vert x^1-\overline{x}\Vert ^2 - \Vert x^{k+1}-\overline{x}\Vert ^2)-\sum _{i=1}^{k}\frac{1}{2}\Vert x^{i+1}-\overline{x}\Vert ^2_{\Sigma _f+(2-\tau )\sigma \mathcal{A}^*\mathcal{A}-\sigma \mathcal{I}} \nonumber \\&\quad + \sum _{i=1}^{k}\langle d^i, \, x^{i+1}-\overline{x}\rangle . \end{aligned}$$
(54)
The result (27) follows immediately from (54) and the condition (26). \(\square \)
