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Monday, August 20, 2012

The coarea formula for BV functions

Today, we will discuss the coarea formula for BV functions, established by Fleming and Rishel in 1960 (W. Fleming, R. Rishel, An integral formula for total gradient variation, Arch. Math. 11 (1960), 218-222.)

Let us first recall the lower semicontinuity property of the total variation with respect to the strong convergence in L1(Ω):
Let {un}nΩ be a sequence in BV(Ω) strongly converging to some u in L1(Ω) and satisfying |un|BV<. Then uBV(Ω) and |u|BV(Ω)liminfn|un|BV(Ω). The coarea formula is as stated as follows:

Coarea formula: Let u be a given function in BV(Ω). Then for almost every t in R, the level set Et={xΩRN:u(x)>t} of u is a set of finite perimeter in Ω, and
Du=DχEtdt,|u|BV(Ω)=|Du|(Ω)=Ω|DχEt|dt.
The second assertion above is often written in terms of the perimeter of level sets as follows:
|u|BV(Ω)=Per(Et,Ω)dt,
where Per(Et,Ω)=HN1(ΩEt) is the perimeter of the level set Et.

Proof of the coarea formula: We follow the treatment of Variational Analysis in Sobolev and BV spaces by Attouch et al.

Part I: Proof of |Du|(Ω)Ω|DχEt|dt:

Let us assume that DχEt belongs to M(Ω,RN), the set of all RN valued Borel measures which is the set of all the σ additive set functions μ:B(Ω)RN, with μ()=0. For all t in R set
ft={χEtif t0χΩEtif t<0 It is easy to see that u(x)=ft(x)dt for all xΩ. For all ˉϕC1c(Ω,RN) we have Du,ˉϕ=Ωudivˉϕdx=Ωftdivˉϕdtdx=Ω0ftdivˉϕdtdxΩ0ftdivˉϕdtdx=Ω0χΩEtdivˉϕdtdxΩ0χEtdivˉϕdtdx=Ω0(1χEt)divˉϕdtdxΩ0χEtdivˉϕdtdx=Ω0divˉϕdtdxΩ0χEtdivˉϕdtdxΩ0χEtdivˉϕdtdx=Ω0divˉϕdtdxΩχEtdivˉϕdtdx=0ΩχEtdivˉϕdtdx=DχEt,ˉϕdt Hence, Du=DχEtdt, and |Du|(Ω)Ω|DχEt|dxdt.

Part II: (coverse) Proof of Ω|DχEt|dxdt|Du|(Ω):


Step i. We first assume that u belongs to the space A(Ω) of piecewise linear and continuous functions in Ω. By linearity, one can assume that u=ax+b with aRN and bR, so that

Ω|DχEt|=H(ΩEt)=H(Ω{x|ax+b=t})=Ω{x|ax+b=t}dHN1(x)={x|ax+b=t}χΩ(x)dHN1(x)
Thus we get,
Ω|DχEt|dxdt=|a|L(Ω)={x|ax+b=t}χΩ(x)dHN1(x)=Ω|Du|
Hence we have,
Ω|DχEt|dxdt=|Du|(Ω) if u=ax+b.

Step ii. Now we prove that Ω|DχEt|dxdt|Du|(Ω) for all uBV(Ω). We know that the space of piecewise linear and continuous functions A(Ω) is dense in W1,1(Ω) when equipped with the strong topology. We also know that the space CW1,1(Ω) is dense in BV when equipped with the intermediate convergence. Thus, there exists a sequence {un}nNA(Ω) such that unu for the intermediate convergence. For each of the functions un we set En,t:={xΩ:un(x)>t}. Due to the intermediate convergence, we have
Ω|Du|=limnΩ|Dun|the intermediate convergence=limnΩ|DχEn,t|dxdtfrom step iliminfnΩ|DχEn,t|dxdtFatou's lemma
From the intermediate convergence we also have unu in L1(Ω). i.e.
Ω|unu|=Ω|χEn,tχEt| dtdx=(Ω|χEn,tχEt| dx)dt=0.
Thus, for a subsequence Enm,t, and for almost all tR, χEnm,tχEn,t strongly in L1(Ω). The lower semicontinuity of the total variation with respect to the strong convergence in L1(Ω) we get liminfnΩ|DχEnm,t|dxΩ|DχEt|dx. Thus, we get,
Ω|Du|Ω|DχEt|dxdt.
This completes the proof!

(By the way, step ii also proves that DχEtM(Ω,RN) for a.e. tR, something that we used in Part I of the proof.)



Thursday, August 9, 2012

Hausdorff measure and dimension

Let us talk about Hausdorff measure Hs named after Felix Hausdorff, which comes up in image processing often. These measures allow us to measure very small subsets of Rn. The idea is that A is an s dimensional subset of Rn if 0<Hs(A)<. We will follow the notation and treatment of Evans and Gariepy.



Felix Hausdorff
Definitions:
(i) Let ARn, 0s<, 0<δ. Define Hsδ(A)=inf{j=1α(s)(diam(Cj)2)s|Aj=1Cj,diam(Cj)δ}, where {Cj}Rn is the covering of the set A and α(s)=πs2Γ(s2+1). Here Γ(s)=0exxs1dx,(0<s<), is the usual gamma function. (ii) The sdimensional Hausdorff measure of A is denoted by Hs(A) and it is defined as the limit:
Hs(A)=limδ0Hsδ(A)=supδ>0Hsδ(A).

The book: 'Measure theory and fine properties of functions' by Evans and Gariepy includes the normalizing constant α(s) in the definition, whereas some other authors do not include this constant. The constant α(s) is included because if s is an integer, then Hs(A) is equal to the sdimensional surface area for nice sets (graph of Lipschitz function).


Some properties of Hausdorff measure:
1.Hs is a Borel regular measure. Refer to E.G. for the proof.

2. Let us consider a Hausdorff measure H0 of a singleton {a}Rn. First note that α(0)=1. So, H0({a})=H0δ({a})=1. Thus, H0 is a counting measure.

3. H1=L1 on R1, i.e. H1 measure of a line segment is its length.

4. Since Rn is not σfinite, Hs is not a Radon measure for 0s<n.

5. If s>n then Hs0 on Rn.
To see this is easy. Let us consider a unit cube Q in Rn, fix an integer m1. The unit cube can be decomposed into mn cubes with sides 1/m and diameter n1/2/m. Therefore,
Hsn/m(Q)mni=1α(s)(n1/2/m)s=α(s)ns/2mns.If s>n, then mns0 as m i.e. as the size of the covering vanishes. Thus, Hs(Q)=0, so Hs(Rn)=0 for s>n.

6. Scaling: Hs(λA)=λsHs(A) for all λ>0, ARn. (Easy to prove.)

7. Invariance under affine transformation: Hs(L(A))=Hs(A) for each affine isometry L:RnRn,ARn.

8. Hn=Ln on Rn.


Hausdorff-Besicovitch dimension:
The hausdorff dimension of a set ARn is defined to be

Hdim(A)=inf{0s<|Hs(A)=0}.

The following assertions follow from the definition:
Countable sets have Hausdorff dimension 0. The Euclidean space has Hausdorff dimension n.
The circle has Hausdorff dimension 1.
The Cantor set (a zero-dimensional topological space) is a union of two copies of itself, each copy shrunk by a factor 1/3; this fact can be used to prove that its Hausdorff dimension is ln2/ln3.




Thursday, July 26, 2012

Rudin-Osher-Fatemi image denoising model


Today we will finally touch the subject of the celebrated image denoising model proposed by Rudin, Osher and Fatemi in 1992, ( L. Rudin, S. Osher, E. Fatemi, "Nonlinear total variation based noise removal algorithms", Physica D, vol. 60, pp. 259–268, 1992.)

This model is motivated by the following problem in image restoration:
Given a noisy image f:ΩR, where Ω is bounded open subset of R2, we need to obtain a clean image u.

For solving this problem they proposed the following minimization problem:
 minimize Ω|u| subject to Ωu=Ωf and Ω|fu|2=σ2. where σ2 is the estimated global variance of the additive noise. This can be also formulated in terms of soft constraint minimization:
uλ=argminu{|u|BV(Ω)+λfu2L2(Ω)}. where |u|BV(Ω) is the BV-seminorm.
In general, this belongs to the class of minimization problems of the following type
uλ=argminu{J(u)+H(f,u)} where J is a convex nonnegative regularization functional and the fitting functional H is convex nonnegative with respect to u for fixed f. Due to convexity there exists a minimizer uλ such that '0' is in the sub-differential of J(u)+H(f,u) at uλ i.e.
0uJ(uλ)+uH(uλ,f).
(here subdifferential uJ at uλ is the set uJ(uλ):={v:J(u)J(uλ)+v,uuλ}. )

If H(u,f)=λfu2L2(Ω), then uH(u,f)=2λ(uf). Moreover if H(u,f)=λfKu2L2(Ω), where K:L2(Ω)H, H being some Hilbert space, then uH(u,f)=2λK(Kuf), where K denotes an adjoint of K.

Formally, we can write J(u)=div(u|u|), and we can write the Euler-Lagrange differential equation for the Rudin Osher Fatemi model as follows:
0=div(u|u|)+2λ(uf) or equivalently,
u=f+12λdiv(u|u|)
We impose Neumann boundary condition uν=0 in the boundary Ω, this ensures the condition on the mean: Ωu=Ωf.

Dependence on the scaling parameter λ:

The parameter λ determines the denoising level. If λ is small, then it is equivalent to penalizing the fidelity term more and the minimizer uλ is very close to the original image. On the other hand, for large λ the minimizer is a cartoon representation of the image f.



We perform a few experiments on the image of Lenna:
Original image, Lena.

For λ=0.8 and the image in the range [0, 255] we get the following result:
The denoised image uλ for λ=0.8
We see above that for this λ=0.8 the denoising is minimal.


For λ=0.05 and the image in the range [0, 255] we get the following result:

The denoised image uλ for λ=0.05.


Here we see the details are gone, but we still see a lot of structure. Let's see what happens if we take λ=0.01, with image in [0, 255].
The denoised image uλ for λ=0.01.

We here see that we get only a cartoon image of Lena. In other words we get only large-scale structures in the image of Lena.

Properties of the ROF denoising:
Let us look at some properties of the ROF model now.

1. The L2-norm of the minimizer uλ is bounded, and the bound is independent of λ.

This nice property can be proven (Nezzar) by realizing that uλ is a minimizer of the ROF functional, thus for any wBV(Ω) we have:
|uλ|BV(Ω)+λfuλ2L2(Ω)|w|BV(Ω)+λfw2L2(Ω).This is true even if we choose w0, which gives us:|uλ|BV(Ω)+λfuλ2L2(Ω)λf2L2(Ω).As 0|uλ|BV(Ω) we have:fuλ2L2(Ω)f2L2(Ω).By expanding the L2 term and using Cauchy-Schwartz inequality we get Ωf22fuλ+u2λΩf2,Ωu2λΩ2fuλ2uλL2(Ω)fL2(Ω),uλL2(Ω)2fL2(Ω).

2. The average of the minimizer uλ is the same as the average of the original image f.

To prove this it suffices to prove that Ωuλ=Ωf. This follows from integrating the Euler-Lagrange differential equation and Neumann boundary condition.

Now let us discuss properties of the ROF minimization proven by Yves Meyer. Let's define a few things first.

Definition: Let G denote the space as follows G={v=div ˉg:ˉg=(g1,g2); with g1,g2L(R2)}.
The star-norm v of vG is defined as follows:
v=inf{|ˉg|l2L:v=div ˉg}

3. If vL2(R2), then
fv|f|BV(Ω)v.
This property is true for fW1,1(R2). Then we can use the density of W1,1(R2) in BV(R2).

4. Meyer: If f12λ, then the Rudin-Osher-Fatemi decomposition f=uλ+vλ is characterized by the following two conditions:vλ=12λ and uλvλ=12λ|uλ|BV.
Moreover, if f12λ, then uλ=0 and vλ=f.

To prove this assertion, consider that if uλ is the minimizer of the ROF energy then uλ+ϵh is not! for a constant ϵ and a function h. This gives |vh|12λ|h|BV. Secondly, taking h=uλ we obtain equality.

Thursday, May 31, 2012

More on intermediate convergence


According to the vectorial version of the Riesz-Alexandroff representation theorem, the dual norm |u|BV=Du=sup{Ωu div ϕ:ϕC1c(Ω,RN) and ϕ1} is also the total mass |Du|(Ω)=Ω|Du| of the total variation |Du| of the measure Du.

Moreover, from classical integration theory, the integral Ωϕd(Du) is well defined for all Du integrable functions ϕ from Ω into RN, e.g. for functions in Cb(Ω,RN). For the same reasons, Ωϕd|Du| is well defined for all |Du| integrable functions ϕ, in particular for functions in Cb(Ω).

Thus, we can say that |Dun||Du| narrowly in M(Ω) if Ωϕd|Dun|Ωϕd|Du|, for all ϕCb(Ω). If we choose ϕ1 on Ω then DunDu i.e. |un|BV|u|BV.

Hence, unu in L1(Ω) and the narrow convergence of measures, |Dun||Du| in M(Ω) implies the intermediate convergence, unu (i.e. unu in L1(Ω) and |un|BV|u|BV).


We have the following proposition regarding the intermediate convergence.

Proposition
The following three assertions are equivalent
(i) unu in the sense of intermediate convergence.
(ii) unu in BV(Ω) and |un|BV|u|BV.
(iii) unu strongly in L1(Ω) and |Dun||Du| narrowly in M(Ω).

Proof
The equivalence (i) to (ii) is a consequence of a proposition from the previous post. We have just seen above that (i) and (iii) are equivalent. We only need to prove that (ii) implies (iii).

Recall that unu weakly in BV(Ω) means that
(a) unu strongly in L1(Ω) and
(b) the measures DunDu weakly in M(Ω,RN).

Also, recall that for non-negative Borel measures μn and μ in M(Ω) the following statements are equivalent:
(c) μn(Ω)μ(Ω) and μ(U)liminfnμn(U) for all open subsets UΩ.
(d) μnμ narrowly in M(Ω).

Let, μn=|Dun| and μ=|Du|. We have been given that |Dun|(Ω)|Du|(Ω) i.e. we have μn(Ω)μ(Ω). For showing the narrow convergence μnμ we just need to prove the lower-semicontinuity property on UΩ,
i.e. we need to show that for μ(U)liminfnμn(U) for all open subsets UΩ,
i.e. we need to show that for |Du|(U)liminfn|Dun|(U) for all open subsets UΩ.

To this effect we use the proposition from the previous blog which says that if unBV(Ω) with |un|BV(Ω) bounded, with unu in L1(Ω), then we have the lower semicontinuity property i.e. |Du|(Ω)liminfn|Dun|(Ω), and uBV(Ω).

Let UΩ be any open subset. We have unBV(Ω), thus as UΩ, we have unBV(U). Thus, we have supn|Dun|(U)supn|Dun|(Ω)<.

The weak convergence unu in L1(Ω) implies the strong convergence unu in L1(Ω) and thus the strong convergence unu in L1(U) follows as UΩ.

 Now that we have the conditions in the proposition: unBV(U), supn|Dun|(U)<, and unu in L1(U) we have the lower semicontinuity property: |Du|(U)liminfn|Dun|(U) for all open subsets UΩ

As we have |Dun|(Ω)|Du|(Ω) and |Du|(U)liminfn|Dun|(U) for all open subsets UΩ, from the equivalence of (c) and (d) we get the narrow convergence: |Dun||Dun| narrowly in M(Ω)..

In conclusion, intermediate convergence implies narrow convergence of |Dun|.


Friday, February 24, 2012

Narrow convergence of measures and intermediate convergence of BV functions




Recall By the Riesz-Alexandroff representation theorem the dual of the space C0(Ω,RN) (and thus of Cc(Ω,RN) can be isometrically identified with M(Ω,RN). i.e. any Borel measure μ is a bounded linear functional (continuous linear form if you like that) on C0(Ω,RN) or Cc(Ω,RN). Thus, the weak convergence of a sequence of Borel measures μn is defined as follows.

Weak convergence of Borel measures: We say that a sequence of Borel measures μn converges weakly to μ i.e. μnμ if μn,ϕμ,ϕ for all ϕC0(Ω,RN).

Here, μ,ϕμ(ϕ)ΩϕdμNi=1Ωϕidμi.

Narrow convergence: A sequence {μn} in M(Ω,RN) narrowly converges to μ (μnμ narrowly) in M(Ω,RN) if and only if ΩfdμnΩfdμ, for all fCb(Ω,RN).

Narrow convergence is stronger than weak convergence.
If Ω=(0,1) and μn=δ1n then μn0 weakly. This is because for all ϕC0(Ω), we have, μn(ϕ)=ϕ(1n)ϕ(0)=0=μ(ϕ).

On the other hand, if ϕ1 on Ω, (note that ϕC0(Ω) but ϕCb(Ω)), then we see that Ωdμn=μn(Ω)=1 for all n. Essentially what happened with μn=δ1n is that it converges weakly to μ=0, but μn(Ω), the size of Ω measured in μn does not converge to the size of Ω in measure μ(=0), that is μnμ but μn(Ω)μ(Ω).

We have the following compactness result for the narrow topology.

PROKHOROV's sequential compactness theorem
If for a bounded subset H of M+(Ω) it is true that for all ϵ there exists a compact subset KϵΩ such that sup{μ(Ω Kϵ):μH}ϵ, then H is sequentially compact for the narrow topology.

The following equivalence proposition says that the convergence μn(Ω)μ(Ω) is precisely what we need to ensure narrow convergence of measures.

Proposition
Let μn and μ be in M+(Ω), non-negative Borel measures, then the following assertions are equivalent:
(i) μnμ narrowly.
(ii)μn(Ω)μ(Ω) and μμ weakly.

Proof
Note that as narrow convergence is stronger than weak convergence, it is clear that if μnμ narrowly then μμ. Also, μn(Ω)μ(Ω) follows by taking f1.

So, we need to only prove that if μμ weakly and μn(Ω)μ(Ω) then μnμ narrowly. That is, we need to show that for any function f in Cb(Ω,RN) we have |ΩfdμnΩfdμ|0. To do this we add and subtract Ωfϕdμn and Ωfϕdμ where ϕ is continuous function that is compactly supported in Ω. The result follows as f is bounded i.e. f is a finite number and ΩfϕdμnΩfϕdμ due to compact support of ϕ in Ω.

More equivalence proposition
Let μn and μ be in M+(Ω), non-negative Borel measures, then the following assertions are equivalent:
(i) μnμ narrowly.
(ii) μn(Ω)μ(Ω) and μ(U)liminfnμn(U) for all open subsets UΩ.
(iii) μn(Ω)μ(Ω) and μ(F)liminfnμn(F) for all closed subsets FΩ.
(iv) μn(B)μ(B) for all Borel subsets BΩ.

Ok, so far so good. But images are not necessarily continuous functions. They might have jump discontinuities, i.e. edges. For this the following result due to Marle (Measure et probabilities) is helpful:

Proposition
Let μn and μ be non-negative Borel measures M+(Ω), with μnμ narrowly and let f be a μn measurable bounded function for all n, such that it is discontinuous only on a set of null μ measurable set then
ΩfdμnΩfdμ

Now, that we have seen a notion of convergence for sequence of measures that is stronger than weak convergence, let's look at the a notion of convergence for sequence of functions in BV space that is stronger than weak convergence.

Intermediate convergence
Let {un} be a sequence in BV(Ω) and uBV(Ω). we say that unu intermediately (or in the sense of intermediate convergence) if and only if
unu in L1(Ω),
|un|BV|u|BV

Recall, from the previous post that for {un}BV(Ω) with bounded TV mass and unu strongly in L1(Ω), then we have unu in BV, but |u|BVliminfn|un|BV i.e. some of the mass is lost.

Ok, now I am hungry, so more about this tomorrow.


Tuesday, February 21, 2012

There is something about the BV space

I have been putting it off for a while. But now that I can write Latex in the blogger I can talk about BV spaces.
Solutions of some mathematical problems which have discontinuities along one-codimensional manifolds where the first distributional derivatives are measures prompt us to consider the BV space. I will refer to Attouch et al's book: Variational Analysis in Sobolev and BV spaces.

Definition of BV space: We say that a function u:ΩR is a function of bounded variations i.e. uBV if and only if it belongs to L1(Ω) and its gradient Du in the distributional sense is in M(Ω,RN).

The following statements are equivalent.
(i) uBV(Ω)
(ii) uL1(Ω) and uxiM(Ω) for all i=1,2,...,N.
(iii) uL1(Ω) and |u|BV<, where |u|BV:=sup{Du,ϕ:ϕCc(Ω,RN) and ϕ1}.
(here Du,ϕ:=Ni=1Ωϕiuxi.)
(iv) uL1(Ω) and |u|BV=sup{Ωu div ϕ:ϕC1c(Ω,RN) and ϕ1}<.

The implication (ii)(iii) follows as Cc(Ω,RN) is dense in C0(Ω,RN). The implication (ii)(iii) follows as Cc(Ω,RN) is dense in C0(Ω,RN) and Cc(Ω,RN).

Notations and facts:
ΩRN.
C0(Ω,RN) is the space of all continuous functions that vanish at infinity. i.e. for a function ϕC0(Ω,RN) the following is true: Given any ϵ>0 there exists a compact set KϵΩ such that supxΩKϵ|ϕ(x)|ϵ.
The functions ϕ in C0(Ω,RN) are equipped with the uniform norm
ϕ:=supxΩ{|ϕ(x)|}.
Cc(Ω,RN) is the subspace of C0(Ω,RN) with a compact support in Ω.
Cb(Ω,RN) is the subset of all bounded continuous functions from Ω into RN.

Density results:
Cc(Ω,RN) is dense in C0(Ω,RN).
The space of infinitely differentiable and compactly supported functions Cc(Ω,RN) is dense is C0(Ω,RN) and Cc(Ω,RN).

Duality results:
M(Ω,RN) denotes the space of all RN valued Borel measures.
M(Ω,RN) is isomorphic to the product space MN(Ω).
By the Riesz-Alexandroff representation theorem the dual of the space C0(Ω,RN) (and thus of Cc(Ω,RN)) can be isometrically identified with M(Ω,RN). i.e. any Borel measure μ is a bounded linear functional (continuous linear form if you like that) on C0(Ω,RN) or Cc(Ω,RN) and the dual norms C0(Ω,RN) and Cc(Ω,RN) are equal to the total mass ||(Ω) :
|μ|(Ω)Ω|μ|=sup{μ,ϕ:ϕC0(Ω,RN) and ϕ1}=sup{μ,ϕ:ϕCc(Ω,RN) and ϕ1}. Here, μ,ϕμ(ϕ)ΩϕdμNi=1Ωϕidμi.

The relation between BV and the Sobolev space W1,1:

According to Radon-Nykodym theorem there exists uL1(Ω,RN) and measure Dsu that is singular with respect to LN|Ω, the N dimensional Lebesgue measure restricted to Ω, such that: Du=uLN|Ω+Dsu. Thus, W1,1 is a subspace of the BV space and for functions in W1,1 we can write Du=u.

Norm on the BV space:

The BV space is equipped with the norm uBV=uL1+|u|BV.
Equipped with this norm the space of BV functions is complete. The completeness of the BV space follows from the completeness of L1 and lower semicontinuity property which is stated below.

Proposition about the lower semicontinuity property of BV seminorm:
If {un} is a sequence in BV(Ω) with supn|un|BV< that converges strongly unu in L1(Ω) then |u|BVliminfn|un|BV and uBV(Ω).
Proof:
We use the following definition of BV seminorm here: uL1(Ω) and |u|BV=sup{Ωu div ϕ:ϕC1c(Ω,RN) and ϕ1}<.
Let ϕC1c(Ω,RN) with ϕ1.
We first observe that
limnΩun div ϕ=Ωu div ϕ.(1)
This follows as unu in L1(Ω) strongly, and that ϕC1c(Ω,RN). Indeed, |Ωun div ϕu div ϕ|Ω|unu|| div ϕ|0, as n. Note that Ωun div ϕsup{Ωun div ϕ:ϕC1c(Ω,RN) and ϕ1}, i.e. Ωun div ϕ|un|BV. Taking liminf of both sides, we get limnΩun div ϕliminfn|un|BV.(2) From (1) and (2) we get
Ωu div ϕliminfn|un|BV.(3)
Taking supremum on the left side of (3) over all ϕC1c(Ω,RN) and ϕ1, we get |u|BVliminfn|un|BV. Moreover, if |un|BV< for all n we get |u|BV<, i.e. uBV.

The weak convergence of unu in BV :
We say that a sequence un in BV(Ω) converges weakly to some u in BV(Ω) (i.e. unu in BV(Ω)) if and only if the following two convergences hold:
(i) unu strongly in L1(Ω)
(ii) DunDu weakly in M(Ω,(R)N).

Proposition: If {un} is a sequence in BV(Ω) with supn|un|BV< that converges strongly unu in L1(Ω) then unu in BV.
Proof:
As unu in L1 we only have to show the weak convergence DunDu.

For ϕCc(Ω,RN) we have that Dun,ϕ=Ωun div ϕΩu div ϕ=Du,ϕ.

As result of the density of Cc(Ω,RN) in C0(Ω,RN) and the boundedness of |un|BV we conclude that DunDu.


APPENDIX: Basic definitions

Let X be some set, and 2X symbolically represent its power set, the collection of all subsets of X.

Measure: A mapping μ:2X[0,] is called a measure on X if
(i) μ()=0 and
(ii) μ(A)k=1μ(Ak), whenever Ak=1Ak.

μ measurable set: A set AX is μ measurable if for each set BX,
μ(B)=μ(BA)+μ(BA).


σ-algebra: A subset A2X is called a σ-algebra if it satisfies the following three properties:
(i) A contains the null set and the set X, i.e. ,XA.
(ii) A is closed under complementation: AA implies XAA.
(iii) A is closed under countable unions: If AkA for k=1,2,, then so is the union, k=1Ak.

Borel σ-algebra of Rn is the smallest σ-algebra of Rn containing the open subsets of Rn.

Regular measure: A measure μ on X is regular if for each set AX there exists a μ measurable set B such that AB and μ(A)=μ(B).

Borel set: A Borel set is any set in a topological space that can be formed from open sets (or, equivalently, from closed sets) through the operations of countable union, countable intersection, and relative complement.

Borel measure: A measure μ on Rn is called Borel measure if every Borel set is μ measurable.

Borel regular measure: A measure μ on Rn is called Borel regular measure if μ is Borel and for each ARn there exists a Borel set B such that AB and μ(A)=μ(B).

Radon measure: A measure μ on Rn is called Radon measure if μ is Borel regular and μ(K)< for each compact set KRn.

Tuesday, January 24, 2012

Segmentation of Boo's book

Segmentation is one of the most difficult problems in image processing, where the meaning of the problem depends on the context. For example, for the image of Boo, I want to segment only the book that he is reading in the following image.

Boo: the cutest dog in the world! 

Here is the segmentation result. I will talk about the code some other day.

The book that Boo reads

Red Rose in 3D

Today, we will see how to produce 3D images from normal images.

There are many ways to create  3D images. One of the ancient ways is to shift the colors in the images and used differently colored pair of glasses.

Today I bought a two dollar pair of 3D glasses from Indiglo's childrens' section.
The left glass is red and the right is blue, which means that we have to shift the red channel to the right by a few pixels depending on how far you are sitting from your computer. Can you generate this 5 lines of code in Matlab?

Here is the image that works for my glasses, enjoy:
(You will not see the rose in 3D using the 3D movie glasses where the 3D is obtained using differently polarized light, which I can not do on a computer screen.)

Red rose in 3D
Three 3 is most effective if we have two objects, out of which one is more close to you. In these cases one segments the objects which are closer and apply 3d shift in these objects alone.

Here is an example:

Stars in 3D

Stars in 2D




Sunday, January 22, 2012

Gamma correction: seeing in the darkness!

Here is an example where we can see what is hidden in the background of a picture. The gamma correction in its basic form is: u=fγ.

The following is an illustration where I used a picture I took in Washington DC, the picture of the World War II memorial, with a γ=0.5 I got the trees in the background.

Original picture of the world war II memorial

Gamma corrected image with γ=0.5