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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”.

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.

Yesterday I uploaded the paper An extended Perona-Malik model based on probabilistic models by Lars Mescheder and myself to the arXiv. Recently I already blogged about this work, so I do not have to add that much. The main theme of the work was that if we have an image that is a blurred and noisy version of some true image, we formulate the reconstruction via Bayesian statistics. As a prior model for images we used a Gaussian scale mixture, i.e. we have a latent variable $z$ (in every pixel $x$) and the joint prior for the image $u$ and the latent variable $z$ is

$p(u,z) \propto \exp\Big(-\sum_x \frac{z(x)}2 |\nabla u(x)|^2 + v(z(x))\Big)$

where $x$ denotes the pixels, $\nabla$ is the discrete gradient of the image and $v$ is some non-negative function defined on the non-negative reals. Besides algorithms to approximate the MAP estimate, Lars proposed a mean-field approximation which does not calculate the most probable image, but iteratively approximates the posterior distribution by distributions which factorize over the variables $u$ and $z$. Using some further approximation (since the resulting algorithm in its plain form is not implementable) one arrives at an algorithm which some keeps more uncertainty in $u$ and, in practice, gives a point estimate for the denoised image that seems more “representative”. Here is an example:

This is the blurred and noisy image (the blur is motions blur and implemented with periodic boundary conditions for simplicity):

The next image is the approximation of the MAP estimate we got and you see the usual drawbacks. Since the MAP neglects all uncertainty in $u$ and maximizes the posterior probability, the image is “too sharp” in the way that smooth transitions (e.g. at the lighthouse) turn into piecewise constant stairs. Also the rocks appear very blocky.

Here is the result from the mean-field approximation. Since uncertainty in $u$ is taken into account, the image does not suffer from staircasing and also has a more natural appeal, most prominently in the part with the rocks.

The paper has some more examples and also shows another relation to Mumford-Shah denoising (loosely speaking, one uses a discrete latent variable $z$ to serve as a binary variable to say if a pixel is an edge or not). Oh, by the way, the algorithms behind the mean-field approximation use some parts of more general duality between random variables that Lars and his co-authors develop in another paper.

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