Testing and Non-linear Preconditioning of the Proximal Point Method

Abstract

Employing the ideas of non-linear preconditioning and testing of the classical proximal point method, we formalise common arguments in convergence rate and convergence proofs of optimisation methods to the verification of a simple iteration-wise inequality. When applied to fixed point operators, the latter can be seen as a generalisation of firm non-expansivity or the $\alpha$-averaged property. The main purpose of this work is to provide the abstract background theory for our companion paper “Block-proximal methods with spatially adapted acceleration”. In the present account we demonstrate the effectiveness of the general approach on several classical algorithms, as well as their stochastic variants. Besides, of course, the proximal point method, these method include the gradient descent, forward–backward splitting, Douglas–Rachford splitting, Newton’s method, as well as several methods for saddle-point problems, such as the Alternating Directions Method of Multipliers, and the Chambolle–Pock method.

Appendices

Appendix A: Outer Semicontinuity of Maximal Monotone Operators

We could not find the following result explicitly stated in the literature, although it is hidden in, e.g., the proof of [2, Theorem 1].

Lemma A.1

Let $H: U\rightrightarrows U$ be maximal monotone on a Hilbert space $U$. Then H is is weak-to-strong outer semicontinuous: for any sequence $\{u^i\}_{i \in \mathbb {N}}$, and any $z^i \in H(u^i)$ such that $u^i\mathrel {\rightharpoonup }u$ weakly, and $z^i \rightarrow z$ strongly, we have $z \in H(u)$.

Proof

By monotonicity, for any $u' \in U$ and $z' \in U$ holds $D_i :=\langle u'-u^i,z'-z^i\rangle \ge 0$. Since a weakly convergent sequence is bounded, we have $D_i \ge \langle u'-u^i,z'-z\rangle -C\Vert z-z^i\Vert$ for some $C>0$ independent of i. Taking the limit, we therefore have $\langle u'-u,z'-z\rangle \ge 0$. If we had $z \not \in H(u)$, this would contradict that H is maximal, i.e., its graph not contained in the graph of any monotone operator. $\square$

Appendix B: Three-Point Inequalities

The following three-point formulas are central to handling forward steps with respect to smooth functions.

Lemma B.1

If $J \in \mathrm {cpl}(X)$ has L-Lipschitz gradient. Then

$$\begin{aligned} \langle \nabla J(z)-\nabla J({\widehat{x}}),x-{\widehat{x}}\rangle \ge -\frac{L}{4}\Vert x-z\Vert ^2 \quad ({\widehat{x}}, z, x \in X), \end{aligned}$$

(74)

as well as

$$\begin{aligned} \langle \nabla J(z),x-{\widehat{x}}\rangle \ge J(x)-J({\widehat{x}}) - \frac{L}{2}\Vert x-z\Vert ^2 \quad ({\widehat{x}}, z, x \in X). \end{aligned}$$

(75)

Proof

Regarding the “three-point hypomonotonicity” (74), the L-Lipschitz gradient implies co-coercivity (see [22] or Appendix C)

$$\begin{aligned} \langle \nabla J(z)-\nabla J({\widehat{x}}),z-{\widehat{x}}\rangle \ge L^{-1} \Vert \nabla J(z)-\nabla J({\widehat{x}})\Vert ^2. \end{aligned}$$

Thus using Cauchy’s inequality

$$\begin{aligned} \begin{aligned} \langle \nabla J(z)-\nabla J({\widehat{x}}),x-{\widehat{x}}\rangle&=\langle \nabla J(z)-\nabla J({\widehat{x}}),z-{\widehat{x}}\rangle +\langle \nabla J(z)-\nabla J({\widehat{x}}),x-z\rangle \\&\ge -\frac{L}{4}\Vert x-z\Vert ^2. \end{aligned} \end{aligned}$$

To prove (75), the Lipschitz gradient implies the smoothness or “descent inequality” (again, [22] or Appendix C)

$$\begin{aligned} J(z)-J(x) \ge \langle \nabla J(z),z-x\rangle - \frac{L}{2}\Vert x-z\Vert ^2. \end{aligned}$$

(76)

By convexity $J({\widehat{x}})-J(z) \ge \langle \nabla J(z),{\widehat{x}}-z\rangle$. Summed, we obtain (75). $\square$

Lemma B.2

If $J \in \mathrm {cpl}(X)$ has L-Lipschitz gradient and is $\gamma$-strongly convex. Then for any $\tau >0$ holds

$$\begin{aligned} \langle \nabla J(z)-\nabla J({\widehat{x}}),x-{\widehat{x}}\rangle \ge \frac{2\gamma -\tau L^2}{2}\Vert x-{\widehat{x}}\Vert ^2 -\frac{1}{2\tau }\Vert x-z\Vert ^2 \quad ({\widehat{x}}, z, x \in X), \end{aligned}$$

(77)

as well as

$$\begin{aligned} \langle \nabla J(z),x-{\widehat{x}}\rangle \ge J(x)-J({\widehat{x}}) + \frac{\gamma -\tau L^2}{2}\Vert x-{\widehat{x}}\Vert ^2 -\frac{1}{2\tau }\Vert x-z\Vert ^2 \quad ({\widehat{x}}, z, x \in X). \end{aligned}$$

(78)

Proof

To prove (78), using strong convexity,the Lipschitz gradient, and Cauchy’s inequality, we have

$$\begin{aligned} \begin{aligned} \langle \nabla J(z),x-{\widehat{x}}\rangle&=\langle \nabla J(x),x-{\widehat{x}}\rangle +\langle \nabla J(z)-\nabla J(x),x-{\widehat{x}}\rangle \\&\ge J(x)-J({\widehat{x}}) + \frac{\gamma }{2}\Vert x-{\widehat{x}}\Vert ^2 -\frac{1}{2\tau }\Vert x-z\Vert ^2 - \frac{\tau L^2}{2}\Vert x-{\widehat{x}}\Vert ^2. \end{aligned} \end{aligned}$$

Regarding (77), using the $\gamma$-strong monotonicity of $\nabla J$, we estimate completely analogously

$$\begin{aligned} \begin{aligned} \langle \nabla J(z)-\nabla J({\widehat{x}}),x-{\widehat{x}}\rangle&=\langle \nabla J(x)-\nabla J({\widehat{x}}),x-{\widehat{x}}\rangle +\langle \nabla J(z)-\nabla J(x),x-{\widehat{x}}\rangle \\&\ge \gamma \Vert x-{\widehat{x}}\Vert ^2 -\frac{1}{2\tau }\Vert x-z\Vert ^2 - \frac{\tau L^2}{2}\Vert x-{\widehat{x}}\Vert ^2. \end{aligned} \end{aligned}$$

$\square$

Since smooth functions with a positive Hessian are locally convex, the above lemmas readily extend to this case, locally. In fact, we have following more precise result:

Lemma B.3

Suppose $J \in C^2(X)$ with $\nabla ^2 J({\widehat{x}}) > 0$ at given ${\widehat{x}}\in X$. Then for any $\tau \in (0, 2]$ and all $z, x, \eta \in X$, we have

$$\begin{aligned} \langle \nabla J(z)-\nabla J({\widehat{x}}),x-{\widehat{x}}\rangle \ge \frac{(1-\delta _{z,\eta })(2-\tau )}{2}\Vert x-{\widehat{x}}\Vert ^2_{\nabla ^2 J(\eta )} -\frac{1+\delta _{z,\eta }}{2\tau } \Vert x-z\Vert ^2_{\nabla ^2 J(\eta )} \end{aligned}$$

(79)

with

$$\begin{aligned} \delta _{z,\eta } :=\inf \left\{ \delta \ge 0 \,\Bigg |\, \begin{array}{r} (1-\delta )\nabla ^2 J(\eta ) \le \nabla ^2 J(\zeta ) \le (1+\delta )\nabla ^2 J(\eta ) \\ \text { for all } \zeta \in {{\mathrm{cl}}}B(\Vert z-{\widehat{x}}\Vert , {\widehat{x}}) \end{array} \right\} . \end{aligned}$$

(80)

If $x \in {{\mathrm{cl}}}B(\Vert z-{\widehat{x}}\Vert , {\widehat{x}})$, then also

$$\begin{aligned} \langle \nabla J(z),x-{\widehat{x}}\rangle \ge J(x)-J({\widehat{x}}) + \frac{(1-\delta _{z,\eta })(1-\tau )-2\delta _{z,\eta }}{2}\Vert x-{\widehat{x}}\Vert _{\nabla ^2 J(\eta )}^2 -\frac{1+\delta _{z,\eta }}{2\tau }\Vert x-z\Vert _{\nabla ^2 J(\eta )}^2. \end{aligned}$$

(81)

Proof

By Taylor expansion, for some $\zeta$ between z and ${\widehat{x}}$, and any $\tau >0$, we have

$$\begin{aligned} \begin{aligned} \langle \nabla J(z)-\nabla J({\widehat{x}}),x-{\widehat{x}}\rangle&=\langle \nabla ^2 J(\zeta )(z-{\widehat{x}}),x-{\widehat{x}}\rangle \\&=\Vert x-{\widehat{x}}\Vert ^2_{\nabla ^2 J(\zeta )} +\langle \nabla ^2 J(\zeta )(z-x),x-{\widehat{x}}\rangle \\&\ge \frac{2-\tau }{2}\Vert x-{\widehat{x}}\Vert ^2_{\nabla ^2 J(\zeta )} -\frac{1}{2\tau } \Vert x-z\Vert ^2_{\nabla ^2 J(\zeta )}. \end{aligned} \end{aligned}$$

(82)

Since $\zeta \in {{\mathrm{cl}}}B(\Vert z-{\widehat{x}}\Vert , {\widehat{x}})$, by the definition of $\delta _{z,\eta }$, we obtain (79).

Similarly, by Taylor expansion, for some $\zeta _0$ between x and ${\widehat{x}}$, we have

$$\begin{aligned} \langle \nabla J(z),x-{\widehat{x}}\rangle - J(x) + J({\widehat{x}}) = \langle \nabla J(z)-\nabla J({\widehat{x}}),x-{\widehat{x}}\rangle -\frac{1}{2}\langle \nabla ^2 J(\zeta _0)(x-{\widehat{x}}),x-{\widehat{x}}\rangle \end{aligned}$$

(83)

Using (82) we obtain

$$\begin{aligned} \begin{aligned} \langle \nabla J(z),x-{\widehat{x}}\rangle \! -\! J(x) + J({\widehat{x}})&\ge \frac{1}{2}\Vert x-{\widehat{x}}\Vert ^2_{(2-\tau )\nabla ^2 J(\zeta ) - \nabla ^2 J(\zeta _0)} -\frac{1}{2\tau } \Vert x-z\Vert ^2_{\nabla ^2 J(\zeta )}. \end{aligned} \end{aligned}$$

Using the assumption $x \in {{\mathrm{cl}}}B(\Vert z-{\widehat{x}}\Vert , {\widehat{x}})$, we have $\zeta _0 \in {{\mathrm{cl}}}B(\Vert z-{\widehat{x}}\Vert , {\widehat{x}})$. Hence we obtain (81) by the definition of $\delta _{z,\eta }$ and $(1-\delta _{z,\eta })(2-\tau )-(1+\delta _{z,\eta })=(1-\delta _{z,\eta })(1-\tau )-2\delta _{z,\eta }$. $\square$

We can also derive the following alternate result:

Lemma B.4

Suppose $J \in C^2(X)$ with $\nabla ^2 J({\widehat{x}}) > 0$ at given ${\widehat{x}}\in X$. Then for all $z, x, \eta \in X$ we have

$$\begin{aligned} \langle \nabla J(z)-\nabla J({\widehat{x}}),x-{\widehat{x}}\rangle \ge \frac{1-\delta _{z,\eta }}{2}\Vert x-{\widehat{x}}\Vert ^2_{\nabla ^2 J(\eta )} + \frac{1-\delta _{z,\eta }}{2}\Vert z-{\widehat{x}}\Vert ^2_{\nabla ^2 J(\eta )} - \frac{1}{2}\Vert x-z\Vert ^2_{\nabla ^2 J(\eta )} \end{aligned}$$

(84)

for $\delta _{z,\eta }$ given by (80). If $x \in {{\mathrm{cl}}}B(\Vert z-{\widehat{x}}\Vert , {\widehat{x}})$, then also

$$\begin{aligned} \begin{aligned} \langle \nabla J(z),x-{\widehat{x}}\rangle \ge&-\delta _{z,\eta }\Vert x-{\widehat{x}}\Vert ^2_{\nabla ^2 J(\eta )} + \frac{1-\delta _{z,\eta }}{2}\Vert z-{\widehat{x}}\Vert ^2_{\nabla ^2 J(\eta )} - \frac{1}{2}\Vert x-z\Vert ^2_{\nabla ^2 J(\eta )} \\&+ J(x)-J({\widehat{x}}). \end{aligned} \end{aligned}$$

(85)

Proof

By Taylor expansion, for some $\zeta$ between z and ${\widehat{x}}$, we have

$$\begin{aligned} \begin{aligned} \langle \nabla J(z)-\nabla J({\widehat{x}}),x-{\widehat{x}}\rangle&=\langle \nabla ^2 J(\zeta )(z-{\widehat{x}}),x-{\widehat{x}}\rangle \\&= \langle \nabla ^2 J(\eta )(z-{\widehat{x}}),x-{\widehat{x}}\rangle \\&\quad +\,\langle [\nabla ^2 J(\zeta )-\nabla ^2 J(\eta )](z-{\widehat{x}}),x-{\widehat{x}}\rangle \\&\ge \langle \nabla ^2 J(\eta )(z-{\widehat{x}}),x-{\widehat{x}}\rangle \\&\quad -\, \frac{\delta _{z,\eta }}{2}\Vert x-{\widehat{x}}\Vert _{\nabla ^2 J(\eta )} - \frac{\delta _{z,\eta }}{2}\Vert z-{\widehat{x}}\Vert _{\nabla ^2 J(\eta )}. \end{aligned} \end{aligned}$$

(86)

In the last step we have used Cauchy’s inequality, and the definition of $\delta _{z,\eta }$ following $\zeta \in {{\mathrm{cl}}}B(\Vert z-{\widehat{x}}\Vert , {\widehat{x}})$. The standard three-point or Pythagoras’ identity states

$$\begin{aligned} \langle \nabla ^2 J(\eta )(z-{\widehat{x}}),x-{\widehat{x}}\rangle = \frac{1}{2}\Vert z-{\widehat{x}}\Vert ^2_{\nabla ^2 J(\eta )} + \frac{1}{2}\Vert x-{\widehat{x}}\Vert ^2_{\nabla ^2 J(\eta )} - \frac{1}{2}\Vert x-z\Vert ^2_{\nabla ^2 J(\eta )}. \end{aligned}$$

Applying this in (86), we obtain (84).

To prove (85), we use (83), the definition of $\delta _{z,\eta }$, and (84). $\square$

Appendix C: Projected Gradients and Smoothness

The next lemma generalises well-known properties (see, e.g., [22]) of smooth convex functions to projected gradients, when we take P as projection operator. With P a random projection, taking the expectation in (89), we in particular obtain a connection to the Expected Separable Over-approximation property in the stochastic coordinate descent literature [34].

Lemma C.1

Let $J \in \mathrm {cpl}(X)$, and $P \in \mathcal {L}(X; X)$ be self-adjoint and positive semi-definite on a Hilbert space X. Suppose P has a pseudo-inverse $P^\dag$ satisfying $P P^\dag P = P$. Consider the properties:

• (i) P-relative Lipschitz continuity of $\nabla J$ with factor L:

  $$\begin{aligned} \Vert \nabla J(x)-\nabla J(y)\Vert _P \le L \Vert x-y\Vert _{P^\dag } \quad (x, y \in X). \end{aligned}$$

  (87)

• (ii) The P-relative property

  $$\begin{aligned} \langle \nabla J(x+Ph) - \nabla J(x),Ph\rangle \le L\Vert h\Vert _P^2 \quad (x, h \in X). \end{aligned}$$

  (88)

• (iii) P-relative smoothness of J with factor L:

  $$\begin{aligned} J(x+Ph) \le J(x) + \langle \nabla J(x),Ph\rangle +\frac{L}{2}\Vert h\Vert _P^2 \quad (x, h \in X). \end{aligned}$$

  (89)

• (iv) The P-relative property

  $$\begin{aligned} J(y) \le J(x) + \langle \nabla J(y),y-x\rangle -\frac{1}{2L}\Vert \nabla J(x)-\nabla J(y)\Vert _P^2 \quad (x, h \in X). \end{aligned}$$

  (90)

• (v) P-relative co-coercivity of $\nabla J$ with factor $L^{-1}$:

  $$\begin{aligned} L^{-1} \Vert \nabla J(x)-\nabla J(y)\Vert _P^2 \le \langle \nabla J(x)-\nabla J(y),x-y\rangle \quad (x, y \in X). \end{aligned}$$

  (91)

We have (i) $\implies$ (ii) $\iff$ (iii) $\implies$ (iv) $\implies$ (v). If P is invertible, all are equivalent.

Proof

(i) $\implies$ (ii): Take $y=x+Ph$ and multiply (87) by $\Vert h\Vert _P$. Then use Cauchy–Schwarz.

(ii) $\implies$ (iii): Using the mean value theorem and (88), we compute (89):

$$\begin{aligned} \begin{aligned}&J(x+Ph) - J(x) - \langle \nabla J(x),Ph\rangle =\int _0^1 \langle \nabla J(x+tPh),Ph\rangle \,dt - \langle \nabla J(x),Ph\rangle \\&\quad =\int _0^1 \langle \nabla J(x+tPh)-\nabla J(x),Ph\rangle \,dt \le \int _0^1 t \,dt \cdot L\Vert h\Vert _P^2 = \frac{L}{2} \Vert h\Vert _P^2. \end{aligned} \end{aligned}$$

(iii) $\implies$ (ii): Add together (89) for $x=x'$ and $x=x'+Ph$.

(iii) $\implies$ (iv): Adding $-\langle \nabla J(y),x+Ph\rangle$ on both sides of (89), we get

$$\begin{aligned} J(x+Ph) - \langle \nabla J(y),x+Ph\rangle \le J(x) - \langle \nabla J(y),x\rangle + \langle \nabla J(x)-\nabla J(y),Ph\rangle +\frac{L}{2}\Vert h\Vert _P^2. \end{aligned}$$

The left hand side is minimised with respect to x by taking $x=y-Ph$. Taking on the right-hand side $h=L^{-1}(\nabla J(y)-\nabla J(x))$ therefore gives (90).

(iv) $\implies$ (v): Summing the estimate (90) with the same estimate with x and y exchanged, we obtain (91).

(v) $\implies$ (i) when P is invertible: Cauchy–Schwarz. $\square$