A Telescopic Bregmanian Proximal Gradient Method Without the Global Lipschitz Continuity Assumption

Abstract

The problem of minimization of the sum of two convex functions has various theoretical and real-world applications. One of the popular methods for solving this problem is the proximal gradient method (proximal forward–backward algorithm). A very common assumption in the use of this method is that the gradient of the smooth term is globally Lipschitz continuous. However, this assumption is not always satisfied in practice, thus casting a limitation on the method. In this paper, we discuss, in a wide class of finite- and infinite-dimensional spaces, a new variant of the proximal gradient method, which does not impose the above-mentioned global Lipschitz continuity assumption. A key contribution of the method is the dependence of the iterative steps on a certain telescopic decomposition of the constraint set into subsets. Moreover, we use a Bregman divergence in the proximal forward–backward operation. Under certain practical conditions, a non-asymptotic rate of convergence (that is, in the function values) is established, as well as the weak convergence of the whole sequence to a minimizer. We also obtain a few auxiliary results of independent interest.

Appendix: Some Proofs

Here, we provide the proofs of some claims mentioned earlier.

Proof of some claims mentioned in Remark 5.2

We first show that in the backtracking step size rule, if $S_j\ne C$ for some j, then $L_k\le \eta L(f',S_k\cap U)$ for each $k\in \mathbb {N}$. The case $k=1$ holds by our assumption on $L_1$ since $S_j\ne C$ for some $j\in \mathbb {N}$. Let $k\ge 2$ and suppose, by induction, that the claim holds for all natural numbers between 1 to $k-1$. If, to the contrary, we have $L_k>\eta L(f',S_k\cap U)$, then $\eta ^{i_k-1}L_{k-1}>L(f',S_k\cap U)$ because $L_k/\eta =\eta ^{i_k-1}L_{k-1}$. Hence, by using (16) with $L:=\eta ^{i_k-1}L_{k-1}$, we conclude that (14) holds with L instead of $L_k$, a contradiction to the minimality of $i_k$ unless $i_k=0$. But when $i_k=0$, we have $L_k=L_{k-1}$, hence, using the induction hypothesis and the fact that $L(f',S_{k-1}\cap U)\le L(f',S_{k}\cap U)$ (this latter fact follows immediately from the equality $L(f',S_k\cap U):=\sup \{\Vert f'(x)-f'(y)\Vert /\Vert x-y\Vert : x,y\in S_k\cap U, x\ne y\}$ and the assumption $S_{k-1}\subseteq S_k$), we have $L_k=L_{k-1}\le \eta L(f',S_{k-1}\cap U)\le \eta L(f',S_k\cap U)$, a contradiction to the assumption on $L_k$.

Now, we show that if we are in the backtracking step size rule, and we also have $S_k=C$ for all $k\in \mathbb {N}$ and $L_1>\eta L(f',S_1\cap U)$, then $L_{k+1}=L_k$ for each $k\in \mathbb {N}$. Indeed, since $S_1=C$ and $L_{k+1}\ge L_k$ for each $k\in \mathbb {N}$ (as shown in Remark 5.2), we have $L_k>\eta L(f',C\cap U)$ for all $k\in \mathbb {N}$. If we do not have $L_{k+1}=L_k$ for each $k\in \mathbb {N}$, then $i_{k+1}>0$ for some $k\in \mathbb {N}$ and for this k we have $\eta ^{i_{k+1}-1}L_{k}=L_{k+1}/\eta >L(f',C\cap U)$. Thus, (16) with $L:=\eta ^{i_{k+1}-1}L_{k}$ implies that (14) holds with L instead of $L_k$, a contradiction to the minimality of $i_{k+1}$. $\square$

Proof of Lemma 8.1

Since $(x_k)_{k=1}^{\infty }$ is a bounded sequence in a reflexive Banach space, a well-known classical result implies that $(x_k)_{k=1}^{\infty }$ has at least one weak cluster point $q\in X$, which, by our assumption, is in U. Suppose to the contrary that there are at least two different weak cluster points $q_1:=w$-$\lim _{k\rightarrow \infty , k\in N_1}x_k$ and $q_2:=w$-$\lim _{k\rightarrow \infty , k\in N_2}x_k$ in X, where $N_1$ and $N_2$ are two infinite subsets of $\mathbb {N}$. By our assumption, $q_1,q_2\in U$, and hence, since b satisfies the limiting difference property, we have

$$\begin{aligned} B(q_2,q_1)=\lim _{k\rightarrow \infty , k\in N_1}(B(q_2,x_k)-B(q_1,x_k)) \end{aligned}$$

(40a)

and

$$\begin{aligned} B(q_1,q_2)=\lim _{k\rightarrow \infty , k\in N_2}(B(q_1,x_k)-B(q_2,x_k)). \end{aligned}$$

(40b)

Since we assume that $L_1:=\lim _{k\rightarrow \infty }B(q_1,x_k)$ and $L_2:=\lim _{k\rightarrow \infty }B(q_2,x_k)$ exist and are finite, we conclude from (40) that $B(q_2,q_1)=L_2-L_1=-(L_1-L_2)=-B(q_1,q_2)$. The assumptions on b imply that B is nonnegative (see, for example, [22, Proposition 4.13(III)]), and hence $0\le B(q_2,q_1)=-B(q_1,q_2)\le 0$. Thus, $B(q_1,q_2)=B(q_2,q_1)=0$. Since b is strictly convex on U, for all $z_1\in \text {dom}(b)$ and $z_2\in U$, we have $B(z_1,z_2)=0$ if and only if $z_1=z_2$ (see, for example, [22, Proposition 4.13(III)]). Hence $q_1=q_2$, a contradiction to the initial assumption. Thus, all the weak cluster points of $(x_k)_{k=1}^{\infty }$ coincide.

We claim that $(x_k)_{k=1}^{\infty }$ converges weakly to the unique cluster point q. Indeed, otherwise there are a weak neighborhood V of q and a subsequence $(x_{k_j})_{j=1}^\infty$ of $(x_k)_{k=1}^\infty$ which is located outside V. But this subsequence is a bounded sequence in a reflexive Banach space (since $(x_k)_{k=1}^{\infty }$ is bounded), and hence, it has a subsequence which converges weakly, as proved above, to q. Thus, infinitely many elements of $(x_{k_j})_{j=1}^{\infty }$ are in V, a contradiction. $\square$