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.