The Poincaré-constant of a domain ${\Omega\subset\mathbb{R}^{n}}$ is the smallest constant ${C}$ such that the estimate

$\displaystyle \|u- \bar u\|_{p}\leq C\|\nabla u\|_{p}$

holds (where ${\bar u = |\Omega|^{-1}\int_{\Omega} u}$ is the mean value of ${u}$).

These constants are known for some classes of domains and some values of ${p}$: E.g. Payne and Weinberger showed in 1960 that for ${p=2}$ and convex ${\Omega}$ the constant is ${\tfrac{\mathrm{diam}(\Omega)}{\pi}}$ and Acosta and Duran showed in 2004 that for ${p=1}$ and convex ${\Omega}$ one gets ${\tfrac{\mathrm{diam}(\Omega)}2}$.

I do not know of any other results on these constants, and while playing around with these kind of things, I derived another one: Since I often work with images ${u:\Omega\rightarrow [0,1]}$ I will assume that ${0\leq u\leq 1}$ in the following.

Using the co-area formula we can express the ${1}$-norm of the gradient via the length of the level sets: Denoting by ${\mathcal{H}^{n-1}}$ the ${n-1}$-dimensional Hausdorff measure (which is, roughly speaking, the length in the case of ${n=2}$) and with ${E_{t}}$ the level set of ${u}$ at level ${t}$, the co-area formula states that

$\displaystyle \int_{\Omega}|\nabla u| = \int_{0}^{1}\mathcal{H}^{n-1}(\partial E_{t}) dt. \ \ \ \ \ (1)$

Now we combine this with the isoperimetric inequality, which is

$\displaystyle \mathcal{H}^{n-1}(\partial E_{t})\geq n |B_{1}|^{\tfrac1n}|E_{t}|^{\tfrac{n-1}{n}}, \ \ \ \ \ (2)$

where ${B_{1}}$ is unit ball and ${|\cdot|}$ denotes the Lebesgue measure. Combining~(1) and~(2) we get

$\displaystyle \int_{\Omega}|\nabla u| \geq n |B_{1}|^{\tfrac1n}\int_{0}^{1}|E_{t}|^{\tfrac{n-1}{n}} dt. \ \ \ \ \ (3)$

Now we use the trivial fact that

$\displaystyle u(x)^{2} = \int_{0}^{u(x)}2t dt = \int_{0}^{1}2t\chi_{\{x\mid u(x)\geq t\}}(x)dt$

combined with Fubini to get

$\displaystyle \int_{\Omega}u^{2} = \int_{0}^{1}2t\int_{\Omega}\chi_{\{x\mid u(x)\geq t\}}(x)dx\, dt = 2\int_{0}^{1}t |E_{t}|dt. \ \ \ \ \ (4)$

Since ${|E_{t}|/|\Omega|\leq 1}$ and ${(n-1)/n<1}$ it holds that ${|E_{t}|\leq |\Omega|^{\tfrac1n}|E_{t}|^{\tfrac{n-1}n}}$. Combining this with~(4) we get

$\displaystyle \int_{\Omega}u^{2}\leq 2|\Omega|^{\tfrac1n}\int_{0}^{1}t |E_{t}|^{\tfrac{n-1}n}dt \leq 2|\Omega|^{\tfrac1n}\int_{0}^{1} |E_{t}|^{\tfrac{n-1}n}dt.$

Finally, we use~(3) to obtain

$\displaystyle \int_{\Omega}u^{2}\leq \frac{2|\Omega|^{\tfrac1n}}{n|B_{1}|^{\tfrac1n}}\int_{\Omega}|\nabla u|$

in other words

$\displaystyle \|u\|_{2}^{2}\leq \frac{2|\Omega|^{\tfrac1n}}{n|B_{1}|^{\tfrac1n}}\|\nabla u\|_{1}.$

This estimate is quite explicit, does not need the subtraction of the mean value, does not need convexity of ${\Omega}$, but also does not obey the scaling ${u\mapsto au}$ (which is of no surprise since we used the condition ${0\leq u\leq 1}$ which also does not obey this scaling).

In dimension ${n=2}$ the estimate takes the simpler form

$\displaystyle \|u\|_{2}^{2}\leq \sqrt{\tfrac{|\Omega|}{\pi}}\|\nabla u\|_{1}.$

In sparse recovery one aims to leverage the sparsity of some vector ${u}$ to get a good reconstruction of this vector from a noisy and indirect measurement ${u^{0}}$. Sparsity can be expressed in two different ways: The analysis way and the synthesis way. For both of them you need a system of other vectors ${v_{k}}$.

• The analysis way: Here you consider the inner products ${\langle u,v_{k}\rangle}$ and the sparsity assumption is, that this sequence is sparse.
• The synthesis way: Here you assume that there is sparse sequence of coefficients ${\psi_{k}}$ such that ${u=\sum_{k} \psi_{k}v_{k}}$.

The first approach is called “analysis” since here one analyzes the vector ${u}$ with the system of vectors. In the second approach, you use the vectors ${v_{k}}$ to synthesize the vector ${u}$.

We will keep things very simple here and only consider a noisy observation (and not an indirect one). The recovery approach in the analysis way is then to find ${u}$ as a solution of

$\displaystyle \tfrac12\|u-u_{0}\| + \alpha R(\langle u,v_{k}\rangle)$

while for the synthesis way you write ${u = \sum_{k} \psi_{k}v_{k} = V\psi}$ (where we collected the vectors ${v_{k}}$ as the columns of the matrix ${V}$) and solve

$\displaystyle \tfrac12\|V\psi-u_{0}\| + \alpha R(\psi)$

The functional ${R}$ should enforce the sparsity of the vector of coefficients in the former case and the sparsity of ${\psi}$ in the latter case. With the matrix ${V}$ the analysis way reads as

$\displaystyle \tfrac12\|u-u_{0}\| + \alpha R(V^{T}u)$

One important observation is, that both approaches coincide if the matrix ${V}$ is orthonormal: Since ${V^{-1} = V^{T}}$ in this case, ${u=V\psi}$ is equivalent to ${\psi = V^{T}u}$, i.e. ${\psi_{k} = \langle u,v_{k}\rangle}$. For other sets of vectors ${v_{k}}$, this is not true and it is not so clear, how the analysis and the synthesis approach are related. In this post I’d like to show one instance where both approaches do lead to results that are not even close to being similar.

1. The “ROF analysis” approach to denoising

If ${u^{0}}$ is an image, the famous Rudin-Osher-Fatemi denoising model fits under the “analysis” framework: The vectors ${v_{k}}$ are all the finite differences of neighboring pixels such that ${V^{T}u = \nabla u}$ with the discrete gradient ${\nabla u}$. As sparsifying functional you take a mixed ${2}$${1}$ norm, i.e. ${R(\nabla u) = \|\nabla u\|_{2,1} = \sum_{ij}\sqrt{(\partial_{1}u_{ij})^{2} + (\partial_{2}u_{ij})^{2}}}$. The denoised ${u}$ is the solution to

$\displaystyle \min_{u} \tfrac12\|u-u^{0}\|^{2} + \alpha\|\nabla u\|_{2,1}$

Thus, the model will lead to some denoised ${u}$ with a sparse gradient. This sparse gradient is partly the reason for the success of the model in that it keeps edges, but is also responsible for the drawback of this approach: The denoised image will be mostly piesewise constant.

2. The “ROF synthesis” approach

As ${V^{T} = \nabla}$, the corresponding synthesis approach would be to solve

$\displaystyle \min_{\psi}\tfrac12\|\nabla^{T}\psi - u^{0}\|^{2} + \alpha \|\psi\|_{2,1}$

and the set ${u = \nabla^{T}\psi}$. In this approach, the denoised image will be a sparse linear combination of particular two-pixel images. If you think about that, it does not really sound like a good idea to approximate an image by a sparse linear combination of two-pixel images. (One technicality: The operator ${\nabla^{T}}$ is not onto – ${\nabla}$ has a non-trivial kernel, consisting of the constant images and hence, all images in the range of ${\nabla^{T}}$ have zero mean. Thus, to get something remotely close to ${u^{0}}$ one should subtract the mean of ${u^{0}}$ before the reconstruction and add it back afterwards.)

Here are some results:

Note that the results actually fit to what we would have predicted: For larger ${\alpha}$ we get a “sparser gradient” (less edges, less variation) for the analysis approach, and “sparser two-pixel combinations” for the synthesis approach. In fact, for the synthesis approach to give something that is remotely close to the image, one needs quite a lot two-pixel images, i.e. a very low sparsity.

In conclusion, one should not consider the analysis and synthesis approach to be something similar.

Incidentally, the dual of the analysis approach

$\displaystyle \min_{u} \tfrac12\|u-u^{0}\|^{2} + \alpha\|\nabla u\|_{2,1}$

can be written as

$\displaystyle \min_{\|\phi\|_{2,\infty}\leq\alpha}\tfrac12\|\nabla^{T}\phi-u^{0}\|^{2} = \min_{\phi}\tfrac12\|\nabla^{T}\phi-u^{0}\|^{2} + I_{\|\cdot\|_{2,\infty}\leq\alpha}(\phi)$

which looks related to the synthesis approach (one basically replaces the ${2,1}$-norm by its conjugate function). Actually, a similar relation between the dual of the analysis approach and the synthesis approach always holds.

$\displaystyle \min_{x}\max_{y}F(x) + \langle Kx,y\rangle - G(y). \ \ \ \ \ (1)$

(where I omit all the standard assumptions, like convexity, continuity ans such…). Fenchel-Rockafellar duality says that solutions are characterized by the inclusion

$\displaystyle 0 \in\left( \begin{bmatrix} \partial F & 0\\ 0 & \partial G \end{bmatrix} + \begin{bmatrix} 0 & K^{T}\\ -K & 0 \end{bmatrix}\right) \begin{bmatrix} x^{*}\\y^{*} \end{bmatrix}$

Noting that the operators

$\displaystyle A = \begin{bmatrix} \partial F & 0\\ 0 & \partial G \end{bmatrix},\quad B = \begin{bmatrix} 0 & K^{T}\\ -K & 0 \end{bmatrix}$

are both monotone, we may apply any of the splitting methods available, for example the Douglas-Rachford method. In terms of resolvents

$\displaystyle R_{tA}(z) := (I+tA)^{-1}(z)$

$\displaystyle \begin{array}{rcl} z^{k+1} & = & R_{tB}(\bar z^{k})\\ \bar z^{k+1}& = & R_{tA}(2z^{k+1}-\bar z^{k}) + \bar z^{k}-z^{k+1}. \end{array}$

For the saddle point problem, this iteration is (with ${z = (x,y)}$)

$\displaystyle \begin{array}{rcl} x^{k+1} &=& R_{t\partial F}(\bar x^{k})\\ y^{k+1} &=& R_{t\partial G}(\bar y^{k})\\ \begin{bmatrix} \bar x^{k+1}\\ \bar y^{k+1} \end{bmatrix} & = & \begin{bmatrix} I & tK^{T}\\ -tK & I \end{bmatrix}^{-1} \begin{bmatrix} 2x^{k+1}-\bar x^{k}\\ 2y^{k+1}-\bar y^{k} \end{bmatrix} + \begin{bmatrix} \bar x^{k}- x^{k+1}\\ \bar y^{k}-y^{k+1} \end{bmatrix}. \end{array}$

The first two lines involve proximal steps and we assume that they are simple to implement. The last line, however, involves the solution of a large linear system. This can be broken down to a slightly smaller linear system involving the matrix ${(I+t^{2}K^{T}K)}$ as follows: The linear system equals

$\displaystyle \begin{array}{rcl} \bar x^{k+1} & = & x^{k+1} - tK^{T}(y^{k+1}+\bar y^{k+1}-\bar y^{k})\\ \bar y^{k+1} & = & y^{k+1} + tK(x^{k+1} + \bar x^{k+1}-\bar x^{k}). \end{array}$

Plugging ${\bar y^{k+1}}$ from the second equation into the first gives

$\displaystyle \bar x^{k+1} = x^{k+1} - tK^{T}(2y^{k+1}-\bar y^{k}) - tK^{T}K(x^{k+1}-\bar x^{k+1}-\bar x^{k})$

Denoting ${d^{k+1}= x^{k+1}+\bar x^{k+1}-\bar x^{k}}$ this can be written as

$\displaystyle (I+t^{2}K^{T}K)d^{k+1} = (2x^{k+1}-\bar x^{k}) - tK^{T}(2y^{k+1}-\bar y^{k}).$

and the second equation is just

$\displaystyle \bar y^{k+1} = y^{k+1} + tKd^{k+1}.$

This gives the overall iteration

$\displaystyle \begin{array}{rcl} x^{k+1} &=& R_{t\partial F}(\bar x^{k})\\ y^{k+1} &=& R_{t\partial G}(\bar y^{k})\\ d^{k+1} &=& (I+t^{2}K^{T}K)^{-1}(2x^{k+1}-\bar x^{k} - tK(2y^{k+1}-\bar y^{k}))\\ \bar x^{k+1}&=& \bar x^{k}-x^{k+1}+d^{k+1}\\ \bar y^{k+1}&=& y^{k+1}+tKd^{k+1} \end{array}$

This is nothing else than using the Schur complement or factoring as

$\displaystyle \begin{bmatrix} I & tK^{T}\\ -tK & I \end{bmatrix} = \begin{bmatrix} 0 & 0\\ 0 & I \end{bmatrix} + \begin{bmatrix} I\\tK \end{bmatrix} (I + t^{2}K^{T}K)^{-1} \begin{bmatrix} I & -tK^{T} \end{bmatrix}$

and has been applied to imaging problems by O’Connor and Vandenberghe in “Primal-Dual Decomposition by Operator Splitting and Applications to Image Deblurring” (doi). For many problems in imaging, the involved inversion may be fairly easy to perform (if ${K}$ is the image gradient, for example, we only need to solve an equation with an operator like ${(I - t^{2}\Delta)}$ and appropriate boundary conditions). However, there are problems where this inversion is a problem.

I’d like to show the following trick to circumvent the matrix inversion, which I learned from Bredies and Sun’s “Accelerated Douglas-Rachford methods for the solution of convex-concave saddle-point problems”: Here is a slightly different saddle point problem

$\displaystyle \min_{x}\max_{y,x_{p}}F(x) + \langle Kx,y\rangle + \langle Hx,x_{p}\rangle- G(y) - I_{\{0\}}(x_{p}). \ \ \ \ \ (2)$

We added a new dual variable ${x_{p}}$, which is forced to be zero by the additional indicator functional ${I_{\{0\}}}$. Hence, the additional bilinear term ${\langle Hx,x_{p}\rangle}$ is also zero, and we see that ${(x,y)}$ is a solution of (1) if and only if ${(x,y,0)}$ is a solution of (2). In other words: The problem just looks differently, but is, in essence, the same as before.

Now let us write down the Douglas-Rachford iteration for (2). We write this problem as

$\displaystyle \min_{x}\max_{\tilde y} F(x) + \langle \tilde Kx,\tilde y\rangle -\tilde G(\tilde y)$

with

$\displaystyle \tilde y = \begin{bmatrix} y\\x_{p} \end{bmatrix}, \quad \tilde K = \begin{bmatrix} K\\H \end{bmatrix}, \quad \tilde G(\tilde y) = \tilde G(y,x_{p}) = G(y) + I_{\{0\}}(x_{p}).$

Writing down the Douglas-Rachford iteration gives

$\displaystyle \begin{array}{rcl} x^{k+1} &=& R_{t\partial F}(\bar x^{k})\\ \tilde y^{k+1} &=& R_{t\partial \tilde G}(\bar{ \tilde y}^{k})\\ \begin{bmatrix} \bar x^{k+1}\\ \bar {\tilde y}^{k+1} \end{bmatrix} & = & \begin{bmatrix} I & t\tilde K^{T}\\ -t\tilde K & I \end{bmatrix}^{-1} \begin{bmatrix} 2x^{k+1}-\bar x^{k}\\ 2\tilde y^{k+1}-\bar {\tilde y}^{k} \end{bmatrix} + \begin{bmatrix} \bar x^{k}- x^{k+1}\\ \bar {\tilde y}^{k}-\tilde y^{k+1} \end{bmatrix}. \end{array}$

Switching back to variables without a tilde, we get, using ${R_{tI_{\{0\}}}(x) = 0}$,

$\displaystyle \begin{array}{rcl} x^{k+1} &=& R_{t\partial F}(\bar x^{k})\\ y^{k+1} &=& R_{t\partial \tilde G}(\bar{ y}^{k})\\ x_{p}^{k+1} &=& 0\\ \begin{bmatrix} \bar x^{k+1}\\ \bar {y}^{k+1}\\ \bar x_{p}^{k+1} \end{bmatrix} & = & \begin{bmatrix} I & tK^{T} & tH^{T}\\ -t K & I & 0\\ -t H & 0 & I \end{bmatrix}^{-1} \begin{bmatrix} 2x^{k+1}-\bar x^{k}\\ 2 y^{k+1}-\bar {y}^{k}\\ 2x_{p}^{k+1}-\bar x_{p}^{k} \end{bmatrix} + \begin{bmatrix} \bar x^{k}- x^{k+1}\\ \bar {y}^{k}-y^{k+1}\\ \bar x_{p}^{k}-x_{p}^{k+1} \end{bmatrix}. \end{array}$

First not that ${x_{p}^{k+1}=0}$ throughout the iteration and from the last line of the linear system we get that

$\displaystyle \begin{array}{rcl} -tH\bar x^{k+1} + \bar x_{p}^{k+1} = -\bar x_{p}^{k} -tH(\bar x^{k}-x^{k+1}) + \bar x_{p}^{k} \end{array}$

which implies that

$\displaystyle \bar x_{p}^{k+1} = tH\bar x^{k+1}.$

Thus, both variables ${x_{p}^{k}}$ and ${\bar x_{p}^{k}}$ disappear in the iteration. Now we rewrite the remaining first two lines of the linear system as

$\displaystyle \begin{array}{rcl} \bar x^{k+1} + tK^{T}\bar y^{k+1} + t^{2}H^{T}H\bar x^{k+1} &=& x^{k+1} + tK^{T}(\bar y^{k}-y^{k+1}) + t^{2}H^{T}H\bar x^{k}\\ \bar y^{k+1}-tK\bar x^{k+1} &=& y^{k+1} + tK(x^{k+1}-\bar x^{k}). \end{array}$

Again denoting ${d^{k+1}=x^{k+1}+\bar x^{k+1}-\bar x^{k}}$, solving the second equation for ${\bar y^{k+1}}$ and plugging the result in the first gives

$\displaystyle (I+t^{2}H^{T}H)\bar x^{k+1} +tK^{T}(y^{k+1}+tKd^{k+1}) = x^{k+1}+tK(\bar y^{k}-y^{k+1}) + t^{2}H^{T}H\bar x^{k}.$

To eliminate ${\bar x^{k+1}}$ we add ${(I+t^{2}H^{T}H)(x^{k+1}-\bar x^{k})}$ on both sides and get

$\displaystyle (I+t^{2}(H^{T}H+K^{T}K))d^{k+1} = 2x^{k+1}-\bar x^{k} -tK(y^{k+1}+\bar y^{k+1}-\bar y^{k}) + t^{2}H^{T}Hx^{k+1}.$

In total we obtain the following iteration:

$\displaystyle \begin{array}{rcl} x^{k+1} &=& R_{t\partial F}(\bar x^{k})\\ y^{k+1} &=& R_{t\partial G}(\bar y^{k})\\ d^{k+1} &=& (I+t^{2}(H^{T}H + K^{T}K))^{-1}(2x^{k+1}-\bar x^{k} - tK(2y^{k+1}-\bar y^{k}) + t^{2}H^{T}Hx^{k+1})\\ \bar x^{k+1}&=& \bar x^{k}-x^{k+1}+d^{k+1}\\ \bar y^{k+1}&=& y^{k+1}+tKd^{k+1} \end{array}$

and note that only the third line changed.

Since the above works for any matrix ${H}$, we have a lot of freedom. Let us see, that it is even possible to avoid any inversion whatsoever: We would like to choose ${H}$ in a way that ${I+t^{2}(H^{T}H + K^{T}K) = \lambda I}$ for some positive ${\lambda}$. This is equivalent to

$\displaystyle H^{T}H = \tfrac{\lambda-1}{t^{2}}I - K^{T}K.$

As soon as the right hand side is positive definite, Cholesky decomposition shows that such an ${H}$ exists, and this happens if ${\lambda\geq 1+t^{2}\|K\|^{2}}$. Further note, that we do need ${H}$ in any way, but only ${H^{T}H}$, and we can perform the iteration without ever solving any linear system since the third row reads as

$\displaystyle d^{k+1} = \tfrac{1}{\lambda}\left(2x^{k+1}-\bar x^{k} - tK(2y^{k+1}-\bar y^{k}) + ((\lambda-1)I - t^{2}K^{T}K)x^{k+1})\right).$

This is a short note to self: Let $A$ be a symmetric positive semidefinite matrix with one-dimensional kernel spanned by $v$. How to solve $Ax=b$ (if you know that $b$ is in the range of $A$)? Just typing

x = A\b

should give a warning in a reasonable software (but also should produce some correct result, if it returns anything at all).

If you don’t want that warning and also want to get the solution that is orthogonal to the kernel you should do

x = (A+v*v')\b.

Note that $A + vv^T$ has full rank (and $v$ is still an eigenvector, but now for the eigenvalue $\|v\|^2$).

Surely, the solution of $Ax=b$ which is orthogonal to the kernel of $A$  also solves this $(A+vv^T)x = b$ since $(A+vv^T)x = Ax + vv^Tx = Ax = b$. Conversely, if $x$ solves $(A + vv^T)x = b$, then taking the inner product with $v$ gives $(Ax)^Tv + (v^Tx)^2 = b^Tv$ and since $b^Tv = 0$ and $(Ax)^T v = x^TAv = 0$ it follows that $v^T x = 0$ which shows that both $Ax=b$ and that $x$ is orthogonal to the kernel.

Also, if you want the solution which is orthogonal to some $z$ (and not to the kernel of $A$) you can solve $(A + zz^T)x=b$. By taking the inner product with $v$, you get that $v^T z\, x^T z=0$ and you get $x\bot z$ as soon as $v^Tz\neq 0$.

I have an open PhD position starting this fall. Here is the job ad:

The workgroup of Prof. Lorenz at the Institute of Analysis and Algebra at the TU Braunschweig is searching a Scientific Assistant (75\% TV-L EG 13). The position is available from 01.10.2017 and is initially limited to three years.

We are looking for candidates
– with a degree (Masters or Diploma) in mathematics above average,
– with a focus on at least one of the areas optimization, numerical mathematics or functional analysis,
– good knowledge of German and
– strong interest in applied mathematics. We suppose that the candidate brings a high commitment for scientific research.

The responsibilities include
– participation in teaching and
– independent research in one of the areas of the work group.

Equally qualified severely challenged persons will be given preference. The TU Braunschweig especially encourages women to apply for the position. Please send your application including CV, copies of certificates and letters of recommendation (if any) in electronic form via e-mail to Dirk Lorenz.

Contact: Prof. Dirk Lorenz, Tel. 0531 391 7423, d.lorenz@tu-braunschweig.de

An here is the job ad as a pdf.

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”.

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

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

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.