February 2017


Here is a lemma that I find myself googling regularly since I always forget it’s exact form.

Lemma 1 Let {A} be a monotone operator, {\lambda>0} and denote by {R_{\lambda A} = (I+\lambda A)^{-1}} the resolvent of {\lambda A}. Then it holds that

\displaystyle  \begin{array}{rcl}  R_{\lambda A^{-1}}(x) = x - \lambda R_{\lambda^{-1}A}(\lambda^{-1}x). \end{array}

Proof: We start with the left hand side {y = R_{\lambda A^{-1}}(x) = (I+\lambda A^{-1})^{-1} x} and deduce

\displaystyle  \begin{array}{rcl}  x &\in& y + \lambda A^{-1}y\\ \iff \frac{x-y}{\lambda} &\in& A^{-1}y\\ \iff y &\in& A(\frac{x-y}{\lambda})\\ \iff x &\in& A(\frac{x-y}{\lambda}) + x-y\\ \iff \frac{x}{\lambda} &\in& \frac{1}{\lambda}A(\frac{x-y}{\lambda}) + \frac{x-y}{\lambda}\\ \iff \frac{x-y}{\lambda} & = &(I + \lambda^{-1}A)^{-1}(\lambda^{-1}x)\\ \iff x - \lambda (I+\lambda^{-1}A)^{-1}(\lambda^{-1}x) & = & y. \end{array}

\Box

I do not know any official name of this, but would call it Moreau’s identity which is the name of the respective statement for proximal operators for convex functions {f} and {g}:

\displaystyle  \begin{array}{rcl}  \mathrm{prox}_{\lambda f^{*}}(x) = x - \lambda\mathrm{prox}_{\lambda^{-1}f}(\lambda^{-1}x). \end{array}

The version for monotone operators is Proposition 23.18 in the first edition of Bauschke and Combette’s book “Convex Analysis and Monotone Operator Theory in Hilbert Spaces”.

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I blogged about the Douglas-Rachford method before and in this post I’d like to dig a bit into the history of the method.

As the name suggests, the method has its roots in a paper by Douglas and Rachford and the paper is

Douglas, Jim, Jr., and Henry H. Rachford Jr., “On the numerical solution of heat conduction problems in two and three space variables.” Transactions of the American mathematical Society 82.2 (1956): 421-439.

At first glance, the title does not suggest that the paper may be related to monotone inclusions and if you read the paper you’ll not find any monotone operator mentioned. So let’s start and look at Douglas and Rachford’s paper.

1. Solving the heat equation numerically

So let us see, what they were after and how this is related to what is known as Douglas-Rachford splitting method today.

Indeed, Douglas and Rachford wanted to solve the instationary heat equation

\displaystyle \begin{array}{rcl} \partial_{t}u &=& \partial_{xx}u + \partial_{yy}u \\ u(x,y,0) &=& f(x,y) \end{array}

with Dirichlet boundary conditions (they also considered three dimensions, but let us skip that here). They considered a rectangular grid and a very simple finite difference approximation of the second derivatives, i.e.

\displaystyle \begin{array}{rcl} \partial_{xx}u(x,y,t)&\approx& (u^{n}_{i+1,j}-2u^{n}_{i,j}+u^{n}_{i-1,j})/h^{2}\\ \partial_{yy}u(x,y,t)&\approx& (u^{n}_{i,j+1}-2u^{n}_{i,j}+u^{n}_{i,j-1})/h^{2} \end{array}

(with modifications at the boundary to accomodate the boundary conditions). To ease notation, we abbreviate the difference quotients as operators (actually, also matrices) that act for a fixed time step

\displaystyle \begin{array}{rcl} (Au^{n})_{i,j} &=& (u^{n}_{i+1,j}-2u^{n}_{i,j}+u^{n}_{i-1,j})/h^{2}\\ (Bu^{n})_{i,j} &=& (u^{n}_{i,j+1}-2u^{n}_{i,j}+u^{n}_{i,j+1})/h^{2}. \end{array}

With this notation, our problem is to solve

\displaystyle \begin{array}{rcl} \partial_{t}u &=& (A+B)u \end{array}

in time.

Then they give the following iteration:

\displaystyle Av^{n+1}+Bw^{n} = \frac{v^{n+1}-w^{n}}{\tau} \ \ \ \ \ (1)

 

\displaystyle Bw^{n+1} = Bw^{n} + \frac{w^{n+1}-v^{n+1}}{\tau} \ \ \ \ \ (2)

 

(plus boundary conditions which I’d like to swipe under the rug here). If we eliminate {v^{n+1}} from the first equation using the second we get

\displaystyle (A+B)w^{n+1} = \frac{w^{n+1}-w^{n}}{\tau} + \tau AB(w^{n+1}-w^{n}). \ \ \ \ \ (3)

 

This is a kind of implicit Euler method with an additional small term {\tau AB(w^{n+1}-w^{n})}. From a numerical point of it has one advantage over the implicit Euler method: As equations (1) and (2) show, one does not need to invert {I-\tau(A+B)} in every iteration, but only {I-\tau A} and {I-\tau B}. Remember, this was in 1950s, and solving large linear equations was a much bigger problem than it is today. In this specific case of the heat equation, the operators {A} and {B} are in fact tridiagonal, and hence, solving with {I-\tau A} and {I-\tau B} can be done by Gaussian elimination without any fill-in in linear time (read Thomas algorithm). This is a huge time saver when compared to solving with {I-\tau(A+B)} which has a fairly large bandwidth (no matter how you reorder).

How do they prove convergence of the method? They don’t since they wanted to solve a parabolic PDE. They were after stability of the scheme, and this can be done by analyzing the eigenvalues of the iteration. Since the matrices {A} and {B} are well understood, they were able to write down the eigenfunctions of the operator associated to iteration (3) explicitly and since the finite difference approximation is well understood, they were able to prove approximation properties. Note that the method can also be seen, as a means to calculate the steady state of the heat equation.

We reformulate the iteration (3) further to see how {w^{n+1}} is actually derived from {w^{n}}: We obtain

\displaystyle (-I + \tau(A+B) - \tau^{2}AB)w^{n+1} = (-I-\tau^{2}AB)w^{n} \ \ \ \ \ (4)

 

2. What about monotone inclusions?

What has the previous section to do with solving monotone inclusions? A monotone inclusion is

\displaystyle \begin{array}{rcl} 0\in Tx \end{array}

with a monotone operator, that is, a multivalued mapping {T} from a Hilbert space {X} to (subsets of) itself such that for all {x,y\in X} and {u\in Tx} and {v\in Ty} it holds that

\displaystyle \begin{array}{rcl} \langle u-v,x-y\rangle\geq 0. \end{array}

We are going to restrict ourselves to real Hilbert spaces here. Note that linear operators are monotone if they are positive semi-definite and further note that monotone linear operators need not to be symmetric. A general approach to the solution of monotone inclusions are so-called splitting methods. There one splits {T} additively {T=A+B} as a sum of two other monotone operators. Then one tries to use the so-called resolvents of {A} and {B}, namely

\displaystyle \begin{array}{rcl} R_{A} = (I+A)^{-1},\qquad R_{B} = (I+B)^{-1} \end{array}

to obtain a numerical method. By the way, the resolvent of a monotone operator always exists and is single valued (to be honest, one needs a regularity assumption here, namely one need maximal monotone operators, but we will not deal with this issue here).

The two operators {A = \partial_{xx}} and {B = \partial_{yy}} from the previous section are not monotone, but {-A} and {-B} are, so the equation {-Au - Bu = 0} is a special case of a montone inclusion. To work with monotone operators we rename

\displaystyle \begin{array}{rcl} A \leftarrow -A,\qquad B\leftarrow -B \end{array}

and write the iteration~(4) in terms of monotone operators as

\displaystyle \begin{array}{rcl} (I + \tau(A+B) + \tau^{2}AB)w^{n+1} = (I+\tau^{2}AB)w^{n}, \end{array}

i.e.

\displaystyle \begin{array}{rcl} w^{n+1} = (I+\tau A+\tau B+\tau^{2}AB)^{-1}(I+\tau AB)w^{n}. \end{array}

Using {I+\tau A+\tau B + \tau^{2}A = (I+\tau A)(I+\tau B)} and {(I+\tau^{2}AB) = (I-\tau B) + (I + \tau A)\tau B} we rewrite this in terms of resolvents as

\displaystyle \begin{array}{rcl} w^{n+1} & = &(I+\tau B)^{-1}[(I+\tau A)^{-1}(I-\tau B) + \tau B]w^{n}\\ & =& R_{\tau B}(R_{\tau A}(w^{n}-\tau Bw^{n}) + \tau Bw^{n}). \end{array}

This is not really applicable to a general monotone inclusion since there {A} and {B} may be multi-valued, i.e. the term {Bw^{n}} is not well defined (the iteration may be used as is for splittings where {B} is monotone and single valued, though).

But what to do, when both and {A} and {B} are multivaled? The trick is, to introduce a new variable {w^{n} = R_{\tau B}(u^{n})}. Plugging this in throughout leads to

\displaystyle \begin{array}{rcl} R_{\tau B} u^{n+1} & = & R_{\tau B}(R_{\tau A}(R_{\tau B}u^{n}-\tau B R_{\tau B}u^{n}) + \tau B R_{\tau B}u^{n}). \end{array}

We cancel the outer {R_{\tau B}} and use {\tau B R_{\tau B}u^{n} = u^{n} - R_{\tau B}u^{n}} to get

\displaystyle \begin{array}{rcl} u^{n+1} & = & R_{\tau A}(2R_{\tau B}u^{n} - u^{n}) + u^{n} - R_{\tau B}u^{n} \end{array}

and here we go: This is exactly what is known as Douglas-Rachford method (see the last version of the iteration in my previous post). Note that it is not {u^{n}} that converges to a solution, but {w^{n} = R_{\tau B}u^{n}}, so it is convenient to write the iteration in the two variables

\displaystyle \begin{array}{rcl} w^{n} & = & R_{\tau B}u^{n}\\ u^{n+1} & = & R_{\tau A}(2w^{n} - u^{n}) + u^{n} - w^{n}. \end{array}

The observation, that these splitting method that Douglas and Rachford devised for linear problems has a kind of much wider applicability is due to Lions and Mercier and the paper is

Lions, Pierre-Louis, and Bertrand Mercier. “Splitting algorithms for the sum of two nonlinear operators.” SIAM Journal on Numerical Analysis 16.6 (1979): 964-979.

Other, much older, splitting methods for linear systems, such as the Jacobi method, the Gauss-Seidel method used different properties of the matrices such as the diagonal of the matrix or the upper and lower triangluar parts and as such, do not generalize easily to the case of operators on a Hilbert space.

Assume that we are given a probability distribution {p} on {{\mathbb R}^{n}} and for simplicity we further assume that {p} is continuous, i.e. {p(x)} somehow indicates, how likely {x\in{\mathbb R}^{n}} is. To be more precise, the probability, that {x\in A} for some (measurable) set {A} is, is {\int_{A}dp}. We do I start like this? There are cases, in which one known some probability distribution, then obtains some {x} and wants to know, if this {x} was particularly representative for this distribution. One example of this situation is a statistical inverse problem where the model is {y = Ax + \epsilon} with a linear (known) map {A} and some error {\epsilon}. From some observed {y}, one wants to know as much as possible about the underlying {x}. Assuming that the noise {\epsilon} has a known distribution {p(\epsilon)} and that one has prior information on {x} (i.e. likely {x} are ones where the prior distribution {p(x)} is large) one can model the following probability distributions:

\displaystyle  \begin{array}{rcl}  p(y\mid x) = p(y-Ax) = p(\epsilon) &&\text{prob. to see}\ y,\ \text{provided}\ x\\ p(x) && \text{prior distribution of}\ x\\ p(x\mid y) = \frac{p(y\mid x)p(x)}{p(y)} && \text{prob. of}\ x,\ \text{provided}\ y; \text{posterior}. \end{array}

The latter distribution, the posterior, is what one is interested in, i.e. given that one has observed {y}, what is the probability that some {x} (which we may generate in some way) is the true solution? So, in principle, all information we have is contained in the posterior, but how to get on grip on it?

What we know are the distribution {p(y\mid x)} and {p(x)} but note that {p(y)} is not know and is usually not simple to compute. Another expression for {p(y)} is {\int p(y\mid x)p(x) dy} since the right hand side in the definition of the posterior has to be a probability distribution. So to calculate {p(y)} we have to evaluate one integral, but note that this is in an integral over {{\mathbb R}^{n}} and in the context of statistical inverse problems, this {n} is usually not in the ten-thousands, but easily in the millions. So we see, that the calculation of {p(y)} is usually out of reach (unless all distributions are very simple). So where are we: If we get some {x} we can calculate {p(y\mid x)} and {p(x)}, but what would it say if {p(y\mid x)p(x) =0.01}, for example? Not much, because we do not know the the normalization constant. While {0.01} seems pretty small and hence {x} to be somehow unlikely, it may be that {p(y) = 0.011} and then { p(x\mid y) = 0.01/0.011 \approx 0.91} which would make {x} to be rather likely. On the other hand we would really would like some more information about {p(x\mid y)} to judge about that {x} since {p(x\mid y)} may have a large variance and a value of {0.01} may be among the largest possible values, again rendering {x} quite likely, but subject to large variance. To recap:

  1. We have a probability distribution {p} but we can only generate values {Cp(x)} for some unknown {C>0}.
  2. The domain of definition of {p} has a huge dimension.
  3. We want to know as much as possible about {p}, e.g. its expected value, its variance or its mode…

Note that some questions about {p} can be answered without knowing the normalization constant, e.g. the mode {x^{*}}, i.e. the {x} which maximizes {p}. That may one prime reason while MAP (maximum-a-posterior) estimators are so widely used… One approach, to get more information out of {p} is to use sampling.

Before we come to methods that can sample from unnormalized distributions, we describe what we actually mean by sampling, and give the main building blocks.

1. Sampling from simple distribution

Sampling from a distribution {p} means a method to generate values {x} such that the values are distributed according to {p}. Expressed in formula: For every integrable function {f} and sequence {x_{k}} of generated samples, we want that

\displaystyle  \begin{array}{rcl}  \lim_{n\rightarrow\infty}\frac1n\sum_{k=1}^{n}f(x_{k}) = \int f(x)dp(x). \end{array}

In the following we want to describe a few methods to generate samples. All the methods will build on the ability to sample from some simple basic distributions, and these are

  1. The uniform distribution on {[0,1]} denoted by {\mathcal{U}(0,1)}. On a computer this is possible to a good approximation by pseudorandom number generators like the Mersenne twister (used, e.g., in MATLAB).
  2. The standard normal distribution denoted by {\mathcal{N}(0,1)}. If you generate pseudorandom normally distributed numbers, then, under the hood, the machine generates uniformly distributed numbers and cleverly transforms these to be normally distributed, e.g. with the Box-Muller method.

Note that simple scaling allows to sample from {\mathcal{U}(a,b)} and {\mathcal{N}(\mu,\sigma^{2})}.

If we sample {x} from some distribution {p} we will denote this by

\displaystyle  \begin{array}{rcl}  x\sim p \end{array}

so {x\sim \mathcal{U}(0,1)} means that {x} is drawn from a uniform distribution over {[0,1]}.

2. Rejection sampling

For rejection sampling from {p} (having access to {Cp} only) we need a so-called proposal distribution {q} from which we can sample (i.e. a uniform one or a normal one) and we need to know some {M>0} such that

\displaystyle  \begin{array}{rcl}  Mq(x) \geq Cp(x) \end{array}

for all {x}. The sampling scheme goes as follows: Draw proposals from the proposal distribution and accept them based on some rule that will ensure the we will end up with samples distributed according to {p}. In more detail:

  1. generate {x\sim q}, i.e. sample {x} from the proposal distribution,
  2. calculate the acceptance rate

    \displaystyle  \begin{array}{rcl}  \alpha = \frac{Cp(x)}{Mq(x)}, \end{array}

  3. generate {t\sim \mathcal{U}(0,1)},
  4. accept {x} if {t\leq \alpha}, otherwise reject it.

Proposition 1 The samples generated by rejection sampling are distributed according to {p}.

Proof: Let us denote by {A} the event that a sample {x}, drawn from {q} has been accepted. Further, we denote by {\pi(x\mid A)} the distribution of {x}, provided it has been accepted. So we aim to show that {\pi(x\mid A) = p(x)}.

To do so, we employ Bayes’ theorem and write

\displaystyle  \begin{array}{rcl}  \pi(x\mid A) = \frac{\pi(A\mid x)\pi(x)}{\pi(A)}. \end{array}

For the three probabilities on the right hand side we know:

  • {\pi(x)} is the distribution of the sample, i.e the proposal distribution, {\pi(x) = q(x)}.
  • The probability {\pi(A\mid x)} is the probability of acceptance, provided that {x} has been sampled. Looking at the acceptance step (step 2), we see that this probability is exactly {\alpha}, i.e.

    \displaystyle  \begin{array}{rcl}  \pi(A\mid x) & = & \alpha = \frac{Cp(x)}{Mq(x)}. \end{array}

  • The probability of the event {A} is the probability of acceptance, and this is given by the integral of the joint distribution {\pi(x,A)} over the values {x}. Since the joint distribution fulfills {\pi(x,A) = \pi(A\mid x)\pi(x)} and {\pi(x) = q(x)} we get

    \displaystyle  \begin{array}{rcl}  \pi(A) & = & \int \pi(x,A) dx = \int \pi(A\mid x)q(x)dx\\ & = & \int \frac{Cp(x)}{Mq(x)}q(x) dx\\ & = & \int \frac{C}{M}p(x) dx = \frac{M}{C} \int p(x)dx = \frac{M}{C} \end{array}

    since {p} is a probability distribution.

Together we get

\displaystyle  \begin{array}{rcl}  \pi(x\mid A) = \frac{\frac{Cp(x)}{Mq(x)}\, q(x)}{\frac{C}{M}} = p(x). \end{array}

\Box

The crucial steps to apply rejection sampling is to find a good proposal distribution with small constant {M} (and also, to calculate {M} can be tricky). As an example, consider that you want to sample from the tail of a standard normal distribution, e.g. you want to obtain “normally distributed random numbers larger that {2}”. Rejection sampling is pretty straightforward: You choose {q} as standard normal, get samples {x} from that and only accept if {x\geq 2}. In this case you have {M=1}, but {C} is small, namely {C\approx 0.023} which means that less than 1 out of 43 samples are accepted…

3. (Non-adaptive) Metropolis sampling

Here comes another method. This method is from the class of Markov-Chain methods, i.e. we will generate a sequence {x_{1},x_{2},\dots} such that they form a Markov chain with a given transition probability. This means, we have a distribution {\kappa} such that {x_{k+1}\sim \kappa(\cdot,x_{k})}. Again, our goal is that the sequence is distributed according to {p}. A central result in the theory of Markov-chain methods is, that this is the case when the so-called (detailed) balance equation is fulfilled , i.e. if

\displaystyle  \begin{array}{rcl}  p(x)\kappa(y,x) = p(y)\kappa(x,y). \end{array}

Remember, that we only have access to {Cp(x)} and we don’t know {C}. Here is the method which is called Metropolis-Hastings sampling: To begin, choose a symmetric proposal distribution {q}, (i.e. {q(x,y)} is the distribution for “go from {y} to {x}” and fulfills {q(x,y) = q(y,x)}), initialize with some {x_{0}} and set {k=1}. Then:

  1. generate {x^{*}\sim q(\cdot,x_{k-1})}, i.e. make a trial step from {x_{k-1}},
  2. set {\alpha = \min(1,\tfrac{p(x^{*})}{p(x_{k-1})})},
  3. generate {t\sim\mathcal{U}(0,1)}
  4. accept {x^{*}} with probability {\alpha}, i.e. set {x_{k} = x^{*}} if {t\leq \alpha}, increase {k} and repeat, otherwise do reject, i.e. do not increase {k} and repeat.

Note that in step 3 we don’t really need {p(x^{*})} and {p(x_{k+1})} but only {Cp(x^{*})} and {Cp(x_{k-1})} since the unknown {C}‘s cancel out.

Proposition 2 The samples generated from Metropolis-Hastings sampling are distributed according to {p}.

Proof: We first calculate the transition probability of the method. The probability to go from {x_{k-1}} to the proposed {x^{*}} is “{q(x^{*},x_{k-1})} multiplied by the acceptance rate {\alpha}” and the probability to stay in {x_{k-1}} is {1-\alpha}. In formula: We write the acceptance ratio as

\displaystyle  \begin{array}{rcl}  \alpha(x_{k-1},x^{*}) = \min(1,\frac{p(x^{*})}{p(x_{k-1})}) = \begin{cases} 1 & p(x^{*})\geq p(x_{k-1})\\ \frac{p(x^{*})}{p(x_{k-1})} & p(x_{k-1})\geq p(x^{*}) \end{cases} \end{array}

and can write the transition probability as

\displaystyle  \begin{array}{rcl}  \kappa(y,x) = \alpha(x,y)q(y,x) + (1-\alpha^{*}(x))\delta_{x}(y) \end{array}

with

\displaystyle  \begin{array}{rcl}  \alpha^{*}(x) = \int \alpha(x,y)q(y,x) dy. \end{array}

We aim to show the detailed balance equation {p(x)\kappa(y,x) = p(y)\kappa(x,y)}. First by a simple distinction of cases, we get

\displaystyle  \begin{array}{rcl}  \frac{\alpha(x,y)}{\alpha(y,x)} = \frac{p(y)}{p(x)}. \end{array}

This gives {p(x)\alpha(x,y) = p(y)\alpha(y,x)} and since {q} is symmetric, we get

\displaystyle  \begin{array}{rcl}  p(x)\alpha(x,y)q(x,y) = p(y)\alpha(y,x)q(x,y). \end{array}

Then we note that

\displaystyle  \begin{array}{rcl}  (1-\alpha^{*}(x))\delta_{x}(y) p(x) = (1-\alpha^{*}(y))\delta_{y}(x)p(y) \end{array}

(which can be seen by integration both sides against some {f(x,y)}). Putting things together, we get

\displaystyle  \begin{array}{rcl}  p(x)\kappa(y,x) & = & p(x) \alpha(x,y)q(y,x) + p(x)(1-\alpha^{*}(x))\delta_{x}(y)\\ & = & p(y)\alpha(y,x)q(x,y) + p(y)(1-\alpha^{*}(y)\delta_{y}(x)\\ & = & p(y)\kappa(x,y) \end{array}

as desired. \Box

Note that Metropolis-Hastings sampling is fairly easy to implement as soon as it is simple to get samples from the proposal distribution. A quick way is to use the standard normal distribution as proposal distribution. In this case, Metropolis-Hastings performs a “restricted random walk”, i.e. if we would accept every step, the sequence {x_{k}} would be a random walk. However, we allow for the possibility to stop the walk in the cases where it would lead us to a region of lower probability — in the cases where it leads to a point of higher probability, we follow the random walk. Although this approach is simple to implement, it may take a lot of time for the chain to get something reasonable. The reasons are, that the random walk steps may be rejected quite often, that it is a long way to get to where the “most probability is” from the initialization or that the distribution {p} has several modes, separated by valleys and that the random walk has difficulties to get from one mode to the other (again due to frequent rejection).

Also note that one can use Gaussian distributions with any variance as proposal distributions and that the variance acts like some stepsize. As often with stepsizes, there is a trade-off: smaller stepsizes mean, that the sampler will need more time to explore the distribution while larger stepsizes would allow faster exploration but also lead to more frequent rejection.

Consider a convex optimization problem of the form

\displaystyle \begin{array}{rcl} \min_{x}F(x) + G(Ax) \end{array}

with convex {F} and {G} and matrix {A}. (We formulate everything quite loosely, skipping over details like continuity and such, as they are irrelevant for the subject matter). Optimization problems of this type have a specific type of dual problem, namely the Fenchel-Rockafellar dual, which is

\displaystyle \begin{array}{rcl} \max_{y}-F^{*}(-A^{T}y) - G^{*}(y) \end{array}

and under certain regularity conditions it holds that the optimal value of the dual equals the the objective value of the primal and, moreover, that a pair {(x^{*},y^{*})} is both primal and dual optimal if and only if the primal dual gap is zero, i.e. if and only if

\displaystyle \begin{array}{rcl} F(x^{*})+G(Ax^{*}) + F^{*}(-A^{T}y^{*})+G^{*}(y^{*}) = 0. \end{array}

Hence, it is quite handy to use the primal dual gap as a stopping criteria for iterative methods to solve these problems. So, if one runs an algorithm which produces primal iterates {x^{k}} and dual iterates {y^{k}} one can monitor

\displaystyle \begin{array}{rcl} \mathcal{G}(x^{k},y^{k}) = F(x^{k})+G(Ax^{k}) + F^{*}(-A^{T}y^{k})+G^{*}(y^{k}). \end{array}

and stop if the value falls below a desired tolerance.

There is some problem with this approach which appears if the method produces infeasible iterates in the sense that one of the four terms in {\mathcal{G}} is actually {+\infty}. This may be the case if {F} or {G} are not everywhere finite or, loosely speaking, have linear growth in some directions (since then the respective conjugate will not be finite everywhere). In the rest of the post, I’ll sketch a general method that can often solve this particular problem.

For the sake of simplicity, consider the following primal dual algorithm

\displaystyle \begin{array}{rcl} x^{k+1} & = &\mathrm{prox}_{\tau F}(x^{k}-A^{T}y^{k})\\ y^{k+1} & = &\mathrm{prox}_{\sigma G^{*}}(y^{k}+\sigma A(2x^{k+1}-x^{k})) \end{array}

(also know as primal dual hybrid gradient method or Chambolle-Pock’s algorithm). It converges as soon as {\sigma\tau\leq \|A\|^{-2}}.

While the structure of the algorithm ensures that {F(x^{k})} and {G^{*}(y^{k})} are always finite (since always {\mathrm{prox}_{F}(x)\in\mathrm{dom}(F)}), is may be that {F^{*}(-A^{T}y^{k})} or {G(Ax^{k})} are indeed infinite, rendering the primal dual gap useless.

Let us assume that the problematic term is {F^{*}(-A^{T}y^{k})}. Here is a way out in the case where one can deduce some a-priori bounds on {x^{*}}, i.e. a bounded and convex set {C} with {x^{*}\in C}. In fact, this is often the case (e.g. one may know a-priori that there exist lower bounds {l_{i}} and upper bounds {u_{i}}, i.e. it holds that {l_{i}\leq x^{*}_{i}\leq u_{i}}). Then, adding these constraints to the problem will not change the solution.

Let us see, how this changes the primal dual gap: We set {\tilde F(x) = F(x) + I_{C}(x)} where {C} is the set which models the bound constraints. Since {C} is a bounded convex set and {F} is finite on {C}, it is clear that

\displaystyle \begin{array}{rcl} \tilde F^{*}(\xi) = \sup_{x\in C}\,\langle \xi,x\rangle - F(x) \end{array}

is finite for every {\xi}. This leads to a finite duality gap. However, one should also adapt the prox operator. But this is also simple in the case where the constraint {C} and the function {F} are separable, i.e. {C} encodes bound constraints as above (in other words {C = [l_{1},u_{1}]\times\cdots\times [l_{n},u_{n}]}) and

\displaystyle \begin{array}{rcl} F(x) = \sum_{i} f_{i}(x_{i}). \end{array}

Here it holds that

\displaystyle \begin{array}{rcl} \mathrm{prox}_{\sigma \tilde F}(x)_{i} = \mathrm{prox}_{\sigma f_{i} + I_{[l_{i},u_{i}]}}(x_{i}) \end{array}

and it is simple to see that

\displaystyle \begin{array}{rcl} \mathrm{prox}_{\sigma f_{i} + I_{[l_{i},u_{i}]}}(x_{i}) = \mathrm{proj}_{[l_{i},u_{i}]}\mathrm{prox}_{\tau f_{i}}(x_{i}), \end{array}

i.e., only uses the proximal operator of {F} and project onto the constraints. For general {C}, this step may be more complicated.

One example, where this makes sense is {L^{1}-TV} denoising which can be written as

\displaystyle \begin{array}{rcl} \min_{u}\|u-u^{0}\|_{1} + \lambda TV(u). \end{array}

Here we have

\displaystyle \begin{array}{rcl} F(u) = \|u-u^{0}\|_{1},\quad A = \nabla,\quad G(\phi) = I_{|\phi_{ij}|\leq 1}(\phi). \end{array}

The guy that causes problems here is {F^{*}} which is an indicator functional and indeed {A^{T}\phi^{k}} will usually be dual infeasible. But since {u} is an image with a know range of gray values one can simple add the constraints {0\leq u\leq 1} to the problem and obtains a finite dual while still keeping a simple proximal operator. It is quite instructive to compute {\tilde F} in this case.