In the following, let \(\{y^{k+1}\}\) and \(\{u^{k+1}=(v^{k+1},x^{k+1})\}\) be the sequences generated by the PDFP algorithm (1.3), i.e.
\(y^{k+1}={T}_{0}(v^{k},x^{k})\) and \((v^{k+1},x^{k+1})={T}(v^{k},x^{k})\). Let \({u^{*}}= ({v^{*}},{x^{*}} )\) be a fixed point of the operator T.
3.1 Convergence
Lemma 3.1
We have the following estimates:
$$\begin{aligned}& \bigl\Vert v^{k+1}-v^{*}\bigr\Vert ^{2} \leq\bigl\Vert v^{k}-v^{*}\bigr\Vert ^{2}-\bigl\Vert v^{k+1}-v^{k} \bigr\Vert ^{2}+2 \bigl\langle B^{T}\bigl(v^{k+1}-v^{*} \bigr),y^{k+1}-x^{*}\bigr\rangle , \end{aligned}$$
(3.1)
$$\begin{aligned}& \bigl\Vert x^{k+1}-x^{*}\bigr\Vert ^{2} \leq\bigl\Vert x^{k}-x^{*}\bigr\Vert ^{2}-\bigl\Vert x^{k+1}-y^{k+1} \bigr\Vert ^{2}-\bigl\Vert x^{k}-y^{k+1}\bigr\Vert ^{2} \\& \hphantom{\bigl\Vert x^{k+1}-x^{*}\bigr\Vert ^{2} \leq{}}{}+2\bigl\langle x^{k+1}-y^{k+1}, \gamma \nabla{f_{1}}\bigl(x^{k}\bigr)+\lambda B^{T}v^{k} \bigr\rangle \\& \hphantom{\bigl\Vert x^{k+1}-x^{*}\bigr\Vert ^{2} \leq{}}{}-2\bigl\langle x^{k+1}-x^{*}, \gamma\nabla{f_{1}} \bigl(x^{k}\bigr)+\lambda B^{T}v^{k+1} \bigr\rangle +2 \gamma\bigl({f_{3}}\bigl(x^{*}\bigr)-{f_{3}} \bigl(y^{k+1}\bigr)\bigr). \end{aligned}$$
(3.2)
Proof
We first prove (3.1). By Lemma 2.2, we know \(I-\operatorname{prox}_{\frac {\gamma }{\lambda}{{f_{2}}}}\) is firmly nonexpansive, and using (1.3)2 and (2.18) we further have
$$ \bigl\Vert v^{k+1}-v^{*}\bigr\Vert ^{2}\leq\bigl\langle v^{k+1}-v^{*},\bigl(B y^{k+1}+v^{k}\bigr)-\bigl(Bx^{*}+v^{*} \bigr)\bigr\rangle , $$
which implies
$$ \bigl\langle v^{k+1}-v^{*},v^{k+1}-v^{k}\bigr\rangle \leq\bigl\langle v^{k+1}-v^{*},B \bigl(y^{k+1}-x^{*}\bigr)\bigr\rangle =\bigl\langle B^{T} \bigl(v^{k+1}-v^{*} \bigr),y^{k+1}-x^{*}\bigr\rangle .$$
Thus
$$\begin{aligned} \bigl\Vert v^{k+1}-v^{*}\bigr\Vert ^{2} =&\bigl\Vert v^{k}-v^{*}\bigr\Vert ^{2}-\bigl\Vert v^{k+1}-v^{k} \bigr\Vert ^{2}+2\bigl\langle v^{k+1}-v^{*}, v^{k+1}-v^{k}\bigr\rangle \\ \leq& \bigl\Vert v^{k}-v^{*}\bigr\Vert ^{2}-\bigl\Vert v^{k+1}-v^{k}\bigr\Vert ^{2}+2 \bigl\langle B^{T}\bigl(v^{k+1}-v^{*}\bigr),y^{k+1}-x^{*}\bigr\rangle . \end{aligned}$$
Next we prove (3.2). By the optimality condition of (1.3)3 (cf. (2.5)), we have
$$ \bigl(x^{k}-\gamma\nabla{f_{1}}\bigl(x^{k}\bigr)- \lambda B^{T}v^{k+1}\bigr)-x^{k+1} \in\gamma \partial{f_{3}}\bigl(x^{k+1}\bigr). $$
By the property of subdifferentials (cf. (2.3)),
$$ \bigl\langle x^{*}-x^{k+1},\bigl(x^{k}-\gamma \nabla{f_{1}}\bigl(x^{k}\bigr)-\lambda B^{T}v^{k+1} \bigr)-x^{k+1} \bigr\rangle \leq\gamma\bigl({f_{3}} \bigl(x^{*}\bigr)- {f_{3}}\bigl(x^{k+1}\bigr)\bigr), $$
i.e.
$$ \bigl\langle x^{k+1}-x^{*}, x^{k+1}-x^{k}\bigr\rangle \leq-\bigl\langle x^{k+1}-x^{*}, \gamma\nabla{f_{1}} \bigl(x^{k}\bigr)+\lambda B^{T}v^{k+1} \bigr\rangle + \gamma \bigl({f_{3}}\bigl(x^{*}\bigr)- {f_{3}} \bigl(x^{k+1}\bigr)\bigr). $$
Therefore,
$$\begin{aligned} \bigl\Vert x^{k+1}-x^{*}\bigr\Vert ^{2} =&\bigl\Vert x^{k}-x^{*}\bigr\Vert ^{2}-\bigl\Vert x^{k+1}-x^{k} \bigr\Vert ^{2}+2\bigl\langle x^{k+1}-x^{*}, x^{k+1}-x^{k}\bigr\rangle \\ \leq& \bigl\Vert x^{k}-x^{*}\bigr\Vert ^{2}-\bigl\Vert x^{k+1}-x^{k}\bigr\Vert ^{2}-2\bigl\langle x^{k+1}-x^{*}, \gamma \nabla{f_{1}}\bigl(x^{k}\bigr)+ \lambda B^{T}v^{k+1} \bigr\rangle \\ &{}+2\gamma\bigl({f_{3}}\bigl(x^{*}\bigr)- {f_{3}}\bigl(x^{k+1}\bigr)\bigr). \end{aligned}$$
(3.3)
On the other hand, by the optimality condition of (1.3)1, it follows that
$$ \bigl(x^{k}-\gamma\nabla{f_{1}}\bigl(x^{k}\bigr)- \lambda B^{T}v^{k}\bigr)-y^{k+1} \in\gamma \partial{f_{3}}\bigl(y^{k+1}\bigr). $$
Thanks to the property of subdifferentials, we have
$$ \bigl\langle x^{k+1}-y^{k+1},\bigl(x^{k}-\gamma \nabla{f_{1}}\bigl(x^{k}\bigr)-\lambda B^{T}v^{k} \bigr)-y^{k+1} \bigr\rangle \leq\gamma\bigl({f_{3}} \bigl(x^{k+1}\bigr)-{f_{3}}\bigl(y^{k+1}\bigr)\bigr). $$
So
$$ \bigl\langle x^{k+1}-y^{k+1}, x^{k}-y^{k+1} \bigr\rangle \leq\bigl\langle x^{k+1}-y^{k+1}, \gamma \nabla{f_{1}}\bigl(x^{k}\bigr)+\lambda B^{T}v^{k} \bigr\rangle +\gamma\bigl({f_{3}}\bigl(x^{k+1} \bigr)-{f_{3}}\bigl(y^{k+1}\bigr)\bigr). $$
Thus
$$\begin{aligned} -\bigl\Vert x^{k+1}-x^{k}\bigr\Vert ^{2} =&- \bigl\Vert x^{k+1}-y^{k+1}\bigr\Vert ^{2}-\bigl\Vert x^{k}-y^{k+1}\bigr\Vert ^{2}+2\bigl\langle x^{k+1}-y^{k+1}, x^{k}-y^{k+1}\bigr\rangle \\ \leq& -\bigl\Vert x^{k+1}-y^{k+1}\bigr\Vert ^{2}- \bigl\Vert x^{k}-y^{k+1}\bigr\Vert ^{2}+2\bigl\langle x^{k+1}-y^{k+1}, \gamma\nabla{f_{1}} \bigl(x^{k}\bigr)+\lambda B^{T}v^{k} \bigr\rangle \\ &{}+2\gamma\bigl({f_{3}}\bigl(x^{k+1}\bigr)-{f_{3}} \bigl(y^{k+1}\bigr)\bigr). \end{aligned}$$
Replacing the term \(-\|x^{k+1}-x^{k}\|^{2}\) in (3.3) with the right side term of the above inequality, we immediately obtain (3.2). □
Lemma 3.2
We have
$$\begin{aligned} \bigl\Vert u^{k+1} -u^{*}\bigr\Vert _{\lambda}^{2} \leq& \bigl\Vert u^{k}-u^{*}\bigr\Vert _{\lambda}^{2}- \lambda\bigl\Vert v^{k+1}-v^{k}\bigr\Vert ^{2}_{M}-\bigl\Vert x^{k+1}-y^{k+1}+ \lambda B^{T}\bigl(v^{k+1}-v^{k}\bigr)\bigr\Vert ^{2} \\ &{} -\bigl\Vert \bigl(x^{k}-y^{k+1}\bigr)-\bigl(\gamma \nabla{f_{1}}\bigl(x^{k}\bigr)-\gamma\nabla{f_{1}} \bigl(x^{*}\bigr)\bigr)\bigr\Vert ^{2} \\ &{}-\gamma(2\beta- {\gamma} )\bigl\Vert \nabla{f_{1}}\bigl(x^{k}\bigr)-\nabla{f_{1}} \bigl(x^{*}\bigr)\bigr\Vert ^{2}. \end{aligned}$$
(3.4)
Proof
Summing the two inequalities (3.1) and (3.2) and re-arranging the terms, we have
$$\begin{aligned}& \lambda\bigl\Vert v^{k+1}-v^{*}\bigr\Vert ^{2}+\bigl\Vert x^{k+1}-x^{*}\bigr\Vert ^{2} \\& \quad \leq \lambda\bigl\Vert v^{k}-v^{*}\bigr\Vert ^{2}+ \bigl\Vert x^{k}-x^{*}\bigr\Vert ^{2}-\lambda\bigl\Vert v^{k+1}-v^{k}\bigr\Vert ^{2}-\bigl\Vert x^{k+1}-y^{k+1}\bigr\Vert ^{2}-\bigl\Vert x^{k}-y^{k+1}\bigr\Vert ^{2} \\& \qquad {} +2 \bigl\langle \lambda B^{T} \bigl(v^{k+1}-v^{*} \bigr),y^{k+1}-x^{*}\bigr\rangle +2\bigl\langle x^{k+1}-y^{k+1}, \gamma\nabla{f_{1}}\bigl(x^{k}\bigr)+\lambda B^{T}v^{k} \bigr\rangle \\& \qquad {} -2\bigl\langle x^{k+1}-x^{*}, \gamma\nabla{f_{1}} \bigl(x^{k}\bigr)+\lambda B^{T}v^{k+1}\bigr\rangle +2 \gamma\bigl({f_{3}}\bigl(x^{*}\bigr)-{f_{3}} \bigl(y^{k+1}\bigr)\bigr) \\& \quad = \lambda\bigl\Vert v^{k}-v^{*}\bigr\Vert ^{2}+\bigl\Vert x^{k}-x^{*}\bigr\Vert ^{2}-\lambda\bigl\Vert v^{k+1}-v^{k}\bigr\Vert ^{2}-\bigl\Vert x^{k+1}-y^{k+1}\bigr\Vert ^{2}-\bigl\Vert x^{k}-y^{k+1}\bigr\Vert ^{2} \\& \qquad {} +2 \bigl\langle \lambda B^{T} \bigl(v^{k+1}-v^{k} \bigr),y^{k+1}-x^{k+1}\bigr\rangle +2\bigl\langle x^{k} -y^{k+1},\gamma\nabla{f_{1}}\bigl(x^{k}\bigr)-\gamma \nabla{f_{1}}\bigl(x^{*}\bigr)\bigr\rangle \\& \qquad {} -2\bigl\langle x^{k}-x^{*},\gamma\nabla{f_{1}} \bigl(x^{k}\bigr)-\gamma\nabla {f_{1}}\bigl(x^{*}\bigr)\bigr\rangle \\& \qquad {} +2\bigl(\bigl\langle y^{k+1}-x^{*},-\gamma\nabla{f_{1}} \bigl(x^{*}\bigr)-\lambda B^{T} v^{*} \bigr\rangle +\gamma \bigl({f_{3}}\bigl(x^{*}\bigr)-{f_{3}}\bigl(y^{k+1} \bigr)\bigr)\bigr) \\& \quad = \lambda\bigl\Vert v^{k}-v^{*}\bigr\Vert ^{2}+\bigl\Vert x^{k}-x^{*}\bigr\Vert ^{2}-\lambda\bigl\Vert v^{k+1}-v^{k}\bigr\Vert _{M}^{2}-\bigl\Vert x^{k+1}-y^{k+1}+\lambda B^{T} \bigl(v^{k+1}-v^{k}\bigr)\bigr\Vert ^{2} \\& \qquad {} -\bigl\Vert \bigl(x^{k}-y^{k+1}\bigr)-\bigl(\gamma \nabla{f_{1}}\bigl(x^{k}\bigr)-\gamma\nabla{f_{1}} \bigl(x^{*}\bigr)\bigr)\bigr\Vert ^{2}+\bigl\Vert \gamma \nabla{f_{1}}\bigl(x^{k}\bigr)-\gamma\nabla{f_{1}} \bigl(x^{*}\bigr)\bigr\Vert ^{2} \\& \qquad {} -2\bigl\langle x^{k}-x^{*},\gamma\nabla{f_{1}} \bigl(x^{k}\bigr)-\gamma\nabla {f_{1}}\bigl(x^{*}\bigr)\bigr\rangle \\& \qquad {} +2\bigl(\bigl\langle y^{k+1}-x^{*},-\gamma\nabla{f_{1}} \bigl(x^{*}\bigr)-\lambda B^{T} v^{*} \bigr\rangle +\gamma \bigl({f_{3}}\bigl(x^{*}\bigr)-{f_{3}}\bigl(y^{k+1} \bigr)\bigr)\bigr), \end{aligned}$$
(3.5)
where \(\|\cdot\|_{M}\) is given in (2.14) and (2.15). Meanwhile, by the optimality condition of (2.19), we have
$$ -\gamma\nabla{f_{1}}\bigl(x^{*}\bigr)-\lambda B^{T}v^{*}\in \gamma\partial{f_{3}}\bigl(x^{*}\bigr), $$
which implies
$$ \bigl\langle y^{k+1}-x^{*},-\gamma\nabla{f_{1}}\bigl(x^{*}\bigr)- \lambda B^{T}v^{*}\bigr\rangle +\gamma\bigl({f_{3}} \bigl(x^{*}\bigr)- {f_{3}}\bigl(y^{k+1}\bigr)\bigr) \leq0. $$
(3.6)
On the other hand, it follows from (2.8) that
$$ -\bigl\langle x^{k}- x^{*},\nabla{f_{1}}\bigl(x^{k} \bigr)-\nabla{f_{1}}\bigl(x^{*}\bigr)\bigr\rangle \leq -{\beta}\bigl\Vert \nabla{f_{1}}\bigl(x^{k}\bigr)- \nabla{f_{1}} \bigl(x^{*}\bigr)\bigr\Vert ^{2}. $$
(3.7)
Recalling (2.16), we immediately obtain (3.4) in terms of (3.5)-(3.7). □
Lemma 3.3
Let
\(0<\lambda<1/{\lambda_{\mathrm{max}}(BB^{T})}\)
and
\(0<\gamma<2\beta\). Then the sequence
\(\{\|u^{k}-u^{*}\|_{\lambda}\}\)
is non-increasing and
\(\lim_{k\to+\infty} \|u^{k+1}-u^{k}\|_{\lambda}=0\).
Proof
If \(0<\lambda< 1/{\lambda_{\mathrm{max}}(BB^{T})}\) and \(0<\gamma<2\beta\), it follows from (3.4) that \(\|u^{k+1}-u^{*}\|_{\lambda}\leq\| u^{k}-u^{*}\|_{\lambda}\), i.e. the sequence \(\{\|u^{k}-u^{*}\| _{\lambda}\}\) is non-increasing. Moreover, summing the inequalities (3.4) from \(k=0\) to \(k=+\infty\), we get
$$\begin{aligned}& \lim_{k\to+\infty} \bigl\| v^{k+1}-v^{k} \bigr\| _{M}=0, \end{aligned}$$
(3.8)
$$\begin{aligned}& \lim_{k\to+\infty} \bigl\| x^{k+1}-y^{k+1}+\lambda B^{T}\bigl(v^{k+1}-v^{k}\bigr)\bigr\| =0, \end{aligned}$$
(3.9)
$$\begin{aligned}& \lim_{k\to+\infty} \bigl\| \bigl(x^{k}-y^{k+1}\bigr)- \bigl(\gamma\nabla{f_{1}}\bigl(x^{k}\bigr)-\gamma \nabla{f_{1}}\bigl(x^{*}\bigr)\bigr)\bigr\| =0, \end{aligned}$$
(3.10)
$$\begin{aligned}& \lim_{k\to+\infty} \bigl\| \nabla{f_{1}}\bigl(x^{k} \bigr)-\nabla{f_{1}}\bigl(x^{*}\bigr)\bigr\| =0. \end{aligned}$$
(3.11)
The combination of (3.10) and (3.11) gives
$$ \lim_{k\to+\infty} \bigl\Vert x^{k}-y^{k+1}\bigr\Vert =0. $$
(3.12)
Noting that \(0<\lambda<1/{\lambda_{\mathrm{max}}(BB^{T})}\), we know M is symmetric and positive definite, so (3.8) is equivalent to
$$ \lim_{k\to+\infty} \bigl\Vert v^{k+1}-v^{k}\bigr\Vert =0. $$
(3.13)
Hence, we have from the above inequality and (3.9) that
$$ \lim_{k\to+\infty} \bigl\Vert x^{k+1}-y^{k+1}\bigr\Vert =0. $$
(3.14)
The combination of (3.12) and (3.14) then gives rise to
$$ \lim_{k\to+\infty} \bigl\Vert x^{k+1}-x^{k}\bigr\Vert =0. $$
(3.15)
According to (3.13), (3.15), and (2.16), we have \(\lim_{k\to+\infty} \|u^{k+1}-u^{k}\|_{\lambda}=0\). □
As a direct consequence of Lemma 3.3 and Lemma 2.3, we obtain the convergence of the PDFP as follows.
Theorem 3.1
Let
\(0<\lambda<1/{\lambda_{\mathrm{max}}(BB^{T})}\)
and
\(0<\gamma<2\beta\). Then the sequence
\(\{u^{k}\}\)
is bounded and converges to a fixed point of
T, and both
\(\{x^{k}\}\)
and
\(\{y^{k}\}\)
converge to a solution of (1.1).
Proof
By Lemma 2.2, both \(\operatorname{prox}_{{\gamma }{f_{3}}}\) and \(I-\operatorname{prox}_{\frac{\gamma}{\lambda}{f_{2}}}\) are firmly nonexpansive, thus the operator T defined by (2.9)-(2.12) is continuous. From Lemma 3.3, we know that the sequence \(\{\|u^{k}-u^{*}\| _{\lambda}\}\) is non-increasing and \(\lim_{k\to+\infty} \| u^{k+1}-u^{k}\|_{\lambda}=0\). By using Lemma 2.3, we know that the sequence \(\{ u^{k}\}\) is bounded and converges to a fixed point of T. By using Theorem 2.1 and (3.14), we can conclude that both \(\{x^{k}\}\) and \(\{y^{k}\} \) converge to a solution of (1.1). □
Remark 3.1
For the special case \(f_{3}=0\), the PDFP reduces naturally to the \(\mathrm{PDFP}^{2}\mathrm{O}\) (4.2) proposed in [15], where the conditions for the parameters are \(0<\lambda\leq1/\lambda _{\mathrm{max}}(BB^{T})\), \(0<\gamma<2\beta\). In the proof of Lemma 3.3, we utilize the positive definitiveness of M to obtain (3.13) from (3.8). So the condition for the parameter λ is slightly more restricted as \(0<\lambda< 1/\lambda_{\mathrm{max}}(BB^{T})\) in Lemma 3.3 and Theorem 3.1. When \(f_{3}=0\), the conditions in the proof of Lemma 3.3 can also be relaxed to \(0<\lambda\leq1/\lambda_{\mathrm{max}}(BB^{T})\). As a matter of fact, it is easy to check by the definition of \(y^{k+1}\) (see (1.3)1) and the optimality condition of (2.19) that
$$ \bigl\Vert \bigl(x^{k}-y^{k+1}\bigr)-\bigl( \gamma\nabla{f_{1}}\bigl(x^{k}\bigr)-\gamma \nabla{f_{1}}\bigl(x^{*}\bigr)\bigr)\bigr\Vert ^{2}=\bigl\Vert \lambda B^{T}\bigl(v^{k}-v^{*}\bigr)\bigr\Vert ^{2}. $$
(3.16)
Observing that \(\|v^{k+1}-v^{k}\|^{2}=\|v^{k+1}-v^{k}\|_{M}^{2}+\lambda\| B^{T}(v^{k+1}-v^{k})\|^{2}\), we have by (3.16), (3.10) and (3.8) that \(\lim_{k\to+\infty} \|v^{k+1}-v^{k}\|=0\). Therefore we can derive the convergence whenever M is semi-positive definite for \(f_{3}=0\).
Remark 3.2
For the special case \(f_{1}=0\), the problem (1.1) only corresponds to two proper lower semi-continuous convex functions. The convergence condition \(0<\gamma<2\beta\) in the PDFP becomes \(0<\gamma<+\infty\). Although γ is an arbitrary positive number in theory, the range of γ will affect the convergence speed and it is also a difficult problem to choose a best value in practice.
3.2 Linear convergence rate for special cases
In the following, we will show the convergence rate results with some additional assumptions on the basic problem (1.1). In particular, for \(f_{3}=0\), the algorithm reduces to the \(\mathrm{PDFP}^{2}\mathrm{O}\) proposed in [15]. The conditions for a linear convergence given there as Condition 3.1 in [15] is as follows: for \(0<\lambda\leq1/\lambda_{\mathrm{max}}(BB^{T})\) and \(0<\gamma<{2}\beta\), there exist \(\eta_{1}, \eta_{2}\in[0,1)\) such that
$$\begin{aligned} &\bigl\Vert I-\lambda BB^{T}\bigr\Vert _{2}\leq\eta_{1}^{2}, \\ &\bigl\Vert g(x)-g(y)\bigr\Vert \le\eta_{2} \Vert x-y\Vert \quad \mbox{for all } x, y\in\mathbb {R}^{n}, \end{aligned}$$
(3.17)
where \(g(x)\) is given in (2.13). It is easy to see that a strongly convex function \(f_{1}\) satisfies the condition (3.17). For a general \(f_{3}\), we need stronger conditions on the functions.
Theorem 3.2
Suppose that (3.17) holds and
\(f_{2}^{*}\)
is strongly convex. Then we have
$$ \bigl\Vert u^{k+1}-u^{*}\bigr\Vert _{(1+\lambda\delta/\gamma)\lambda} \leq\eta\bigl\Vert u^{k}-u^{*}\bigr\Vert _{(1+\lambda\delta/\gamma)\lambda}, $$
where
\(0<\eta<1\)
is the convergence rate (indicated in the proof) and
\(\delta>0\)
is a parameter describing the strongly monotone property of
\(\partial f_{2}^{*}\) (cf. (2.4)).
Proof
Use Moreau’s identity (cf. (2.7)) to get
$$(I-\operatorname{prox}_{\frac{\gamma}{\lambda}{f_{2}}}) \bigl(By^{k+1}+v^{k} \bigr) =\frac{\gamma}{\lambda}\operatorname{prox}_{\frac{\lambda}{\gamma }{f_{2}^{*}}} \biggl( \frac{\lambda}{\gamma}By^{k+1}+\frac{\lambda}{\gamma}v^{k}\biggr). $$
So (1.3)2 is equivalent to
$$ \frac{\lambda}{\gamma}v^{k+1} =\operatorname{prox}_{\frac{\lambda}{\gamma}{f_{2}^{*}}} \biggl(\frac{\lambda}{\gamma}By^{k+1}+\frac{\lambda}{\gamma}v^{k} \biggr). $$
(3.18)
According to the optimality condition of (3.18),
$$ \frac{\lambda}{\gamma}By^{k+1}+\frac{\lambda}{\gamma}v^{k}- \frac {\lambda}{\gamma}v^{k+1} \in\frac{\lambda}{\gamma}\partial {f_{2}^{*}}\biggl(\frac{\lambda}{\gamma}v^{k+1}\biggr). $$
(3.19)
Similarly, according to the optimality condition of (2.18),
$$ \frac{\lambda}{\gamma}Bx^{*} \in\frac{\lambda}{\gamma}\partial {f_{2}^{*}}\biggl(\frac{\lambda}{\gamma}v^{*}\biggr). $$
(3.20)
Observing that \(\partial f_{2}^{*}\) is δ-strongly monotone, we have by (3.19) and (3.20)
$$ \bigl\langle v^{k+1}-v^{*},\bigl(B y^{k+1}+v^{k}-v^{k+1} \bigr)-Bx^{*}\bigr\rangle \geq\frac {\lambda}{\gamma} \delta\bigl\Vert v^{k+1}-v^{*}\bigr\Vert ^{2}, $$
i.e.
$$ \bigl\langle v^{k+1}-v^{*},v^{k+1}-v^{k}\bigr\rangle \leq\bigl\langle B^{T} \bigl(v^{k+1}-v^{*}\bigr),y^{k+1}-x^{*} \bigr\rangle -\frac{\lambda}{\gamma}\delta\bigl\Vert v^{k+1}-v^{*}\bigr\Vert ^{2}. $$
Thus
$$\begin{aligned} \bigl\Vert v^{k+1}-v^{*}\bigr\Vert ^{2} =& \bigl\Vert v^{k}-v^{*}\bigr\Vert ^{2}-\bigl\Vert v^{k+1}-v^{k}\bigr\Vert ^{2}+2\bigl\langle v^{k+1}-v^{*}, v^{k+1}-v^{k}\bigr\rangle \\ \leq&\bigl\Vert v^{k}-v^{*}\bigr\Vert ^{2}-\bigl\Vert v^{k+1}-v^{k}\bigr\Vert ^{2}+2 \bigl\langle B^{T}\bigl(v^{k+1}-v^{*}\bigr),y^{k+1}-x^{*}\bigr\rangle \\ &{}- \frac{\lambda}{\gamma}\delta\bigl\Vert v^{k+1}-v^{*}\bigr\Vert ^{2}. \end{aligned}$$
(3.21)
Summing the two inequalities (3.21) and (3.2), and then using the same argument for driving (3.5), we arrive at
$$\begin{aligned} \biggl(1+\frac{\lambda}{\gamma} \delta\biggr)\lambda\bigl\Vert v^{k+1}-v^{*}\bigr\Vert ^{2}+ \bigl\Vert x^{k+1}-x^{*}\bigr\Vert ^{2} \leq& \lambda\bigl\Vert v^{k}-v^{*}\bigr\Vert ^{2}+ \bigl\Vert g \bigl(x^{k}\bigr)-g\bigl(x^{*}\bigr)\bigr\Vert ^{2} \\ \leq& \lambda\bigl\Vert v^{k}-v^{*}\bigr\Vert ^{2}+ \eta_{2}^{2}\bigl\Vert x^{k}-x^{*}\bigr\Vert ^{2}, \end{aligned}$$
(3.22)
where we have also used the condition (3.17) and the inequality (3.6).
Let \(\eta_{3}=1/\sqrt{1+\lambda\delta/\gamma}\) and \(\eta=\max\{ \eta_{2},\eta_{3}\}\). It is clear that \(0<\eta<1\). Hence, according to the notation (2.16), the estimate (3.22) can be rewritten as required. □
We note that a linear convergence rate for strongly convex \(f_{2}^{*}\) and \(f_{3}\) are obtained in [19]. They introduced two preconditioned operators for accelerating the algorithm, while a clear relation between the convergence rate and the preconditioned operators is still missing. Meanwhile, introducing preconditioned operators could be beneficial in practice, and we can also introduce a preconditioned operator to deal with \(\nabla f_{1}\) in our scheme. Since the analysis is rather similar to the current one, we will omit it in this paper.