Geometry of Optimal Decomposition for the Couple (ℓ2, X) on Rn

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' $

Department of Mathematics

Geometry of Optimal Decomposition for the

Couple (ℓ

2

, X) on

R

n

Japhet Niyobuhungiro

LiTH-MAT-R–2013/06–SE

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Department of Mathematics

Linkping University

S-581 83 Linkping, Sweden.

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Geometry of Optimal Decomposition for

the Couple

(`

2 , X

)

on

R

n

Japhet Niyobuhungiro

∗†

Abstract

We investigate the geometry of optimal decomposition for the L– func-tional L2,1(t, x;`2, X) = inf x=x0+x1 1 2kx0k 2 `2+tkx1kX , where space`2_{is defined by the standard Euclidean norm}_k_·_k

2and where X is a Banach space onRnand t is a given positive parameter. Our proof is based on some geometrical considerations and Yves Meyer’s duality approach which was considered for the couple L2, BV

in connection with the famous in image processing ROF denoising model. Our goal is also to investigate possibility to extend Meyer’s approach to more general couples than L2, BV.

1 Introduction

The denoising problem is a problem to extract the ideal “clean” image (sig-nal) from its noisy version. The problem of denoising is fundamental in im-age (signal) processing. From a mathematical point of view, it consists of reconstructing approximately an initial, unknown image (signal) f∗ from the observed, noisy image (signal)

fob= f∗+η,

where the term η corresponds to the noise. In a very successful in image pro-cessing Rudin-Osher-Fatemi (ROF) denoising model (see (Rudin et al., 1992; Scherzer et al., 2009; Chan and Shen, 2005)) it is supposed that the initial im-age f∗ and observed image fob are functions defined on a rectangular plane domain D. Moreover it is also supposed that the noise η is a function from

∗_{Department of Mathematics, Linköping University, SE–581 83 Linköping, Sweden. E-mail:}

japhet.niyobuhungiro@liu.se

†_{Department of Applied Mathematics, National University of Rwanda, P.O. Box: 56 Butare}

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L2 = L2(D) and the initial image f∗ belongs to the well-known in analysis space of functions of bounded variation BV, in which the seminorm for the differentiable function f is equal to

kfk_BV = Z Z D ∂ f ∂x(x, y) + ∂ f ∂y(x, y) dxdy.

For the non-differentiable functions, norm in BV is more complicated and we will not discuss it here. The ROF model suggests to use as approximation to the initial image f∗a function ftsuch that

L2,1 t, fob; L2, BV= 1 2kfob− ftk 2 L2 +tkftk_BV,

for some t>0. Such a function ftis called exact minimizer for the functional L2,1 t, fob; L2, BV = inf f ∈BV 1 2kfob− fk 2 L2+tkfk_BV .

Unfortunately even for the known parameter t > 0, the construction of the minimizer ft is not simple. One of the approaches to this problem is based on a very interesting fact that the minimizer ft is also a minimizer for the dual problem (see for example (Scherzer et al., 2009; Meyer, 2002)).

The main idea of this paper is to prove a result which gives a geometrical characterization of such exact minimizer in the case of a finite dimensional couple(`2, X)onRn, where X is defined by an arbitrary norm. Our proof in this situation is rather simple and provides a clear explanation of the general phenomenon.

The paper is organized as follows: In Section 2 we formulate an important result characterizing the exact minimizer under the duality. The proof will be detailed in Section 3. Finally, in Section 4 we will outline some remarks and future considerations.

2 Main result

Let X be a Banach space onRnequipped with normk·k_X. Let x be a fixed el-ement fromRnand consider the following well known in interpolation theory L– functional (see the book by (Bergh and Löfström, 1976))

L2,1(t, x;`2, X) = inf x0∈`2 1 2kx0k 2 `2+tkx−x0k_X ,

where t is a positive parameter. Let x0,optbe an exact minimizer for L2,1(t, x;`2, X) i.e., L2,1(t, x;`2, X) = 1 2 x0,opt 2 `2+t x−x0,opt X.

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Let us define X∗to be the dual space of X equipped with the dual normk·k_X∗

defined by

kyk_X∗=sup{|hy, xi|: x∈X; kxk_X≤1},

where

hy, xi =y · x and kxk_X= (hx, xi)1/2.

Note that in the case whenk·k_X is a seminorm, the dual normk·k_X∗ could

be equal to+∞ for some elements. Let us consider a closed ball of radius t in

the normk·k_X∗ of X∗centered at 0 and denote it by tBX∗:

tBX∗={y∈Rn:kyk_X∗≤t}.

Theorem 2.1 (Stability of a minimizer under duality). Let x0,opt be an exact minimizer for L2,1(t, x;`2, X). Then x0,optis the nearest element of tBX∗to the point

x in the metric of`2: inf kx0kX∗≤t kx−x0k_`2 = x−x0,opt `2.

The situation is illustrated by Figure 1 below, where x1,opt =x−x0,opt.

Figure 1: Illustration of Theorem 2.1.

Remark 1. Theorem2.1 can be formulated in terms of the E-functional on the couple

`2, X∗. More precisely, let x0,optbe an exact minimizer for L2,1(t, x;`2, X). Then Et, x;`2, X∗= inf kx0kX∗≤t kx−x0k_`2 = x−x0,opt _`₂. (1)

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The Theorem 2.1 provides ways to compute the exact minimizers x0,opt for L2,1(t, x;`2, X) using E t, x;`2, X∗. For illustration, let us consider a simple example.

Example 1. Consider the couple `2,`1 and consider the problem of calculating the exact minimizer for the functional

L2,1(t, x;`2,`1) = inf x0∈`2 1 2kx0k 2 `2+tkx−x0k_`1 . In this case, X = `1and the ball tBX∗ is defined by

tBX∗={y∈Rn:kyk_`_∞ ≤t}.

From Theorem 2.1 ( see also Remark 1 ), the solution x0,optto this problem is an exact minimizer for the E-functional

E(t, x;`2,`∞) = inf {x0: kx0k`∞≤t}

kx−x0k`2.

It can be calculated analytically as follows

x0,opt i=    t, if xi >t xi, if −t≤xi≤t −t, if xi < −t , for i=1, 2, . . . , n.

Remark 2. The paper by (Nilsson and Peetre, 1986) provides ways to find the exact

minimizer for the couple(L1, L2)of the functional K(t, x; L1, L2) = inf

x0∈L1

(kx0k_L1+tkx−x0k_L2).

3 Proof of the main result

In this section we will proceed to prove Theorem 2.1. The proof will be based on three lemmas that we will prove first. Assume that x0,opt is an exact mini-mizer for L(t, x;`2, X). This means that

L2,1(t, x;`2, X) = 1 2 x0,opt 2 `2+t x−x0,opt _X. (2)

Moreover, we will assume that both x0,opt 6= 0 and x−x0,opt 6= 0. The more general case will be investigated in our fiture study . Let us put the following notation α= 1 2 x0,opt 2 `2 and β=t x−x0,opt _X. (3)

Consider the closed balls B1=nu∈Rn_:_k_u_k `2 ≤ √ 2αo and B2= u∈Rn _:_k_x₋_u_k X≤ β t . (4)

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Figure 2: Illustration of lemma 1.

We note that B1 and B2 are nonempty convex subsets of Rn. Let T1 be the tangent hyperplane to B1 passing through the point x0,opt as illustrated in Figure 2.

It is clear that since T1is orthogonal to the vector x0,opt. Therefore we have u, x0,opt

=x0,opt, x0,opt

∀u∈T1.

The hyperplane T1 divides Rn into two closed halfspaces. We will denote these hyperplanes by H1and H2which contain the origin 0 and x respectively.

Lemma 1. Let H1and H2be defined as above and let B1and B2be defined as in (4). Then B1is contained in H1and B2is contained in H2.

Proof. As T1 is a tangent hyperplane to B1 therefore B1 ⊂ H1 is obviously satisfied. So we need only to prove that B2 ⊂ H2 as illustrated in Figure 2. Suppose the contrary, that there exists v∈H1\H2such that v∈B2. Note also that from definition of B2and (3) follows that x0,opt ∈bd(B2), where bd(B2)

is the boundary of B2. Consider the segmentv, x0,opt. From convexity of B2, this segment must be contained in B2and therefore since B1is a ball of`2and T1 is tangent to B1 through x0,opt, then we can find on this segment a point which belongs to the interior of B1. We will denote this element by w and this situation is illustrated in Figure 3.

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Figure 3: The impossible situation where B2contains an element of H1\H2. So, in this case we have:

kwk_`2 <

√

2α and tkx−wk_X ≤β. (5)

From (5) and (3), it follows that 1 2kwk 2 `2+tkx−wk_X < 1 2 x0,opt 2 `2+t x−x0,opt X.

Which is a contradiction of (2), because x0,optis the exact minimizer for L2,1(t, x;`2, X). Therefore T1separates B1and B2.

Lemma 2. The exact minimizer x0,optsatisfies x0,opt, x−x0,opt =t x−x0,opt _X. (6)

Proof. Put x1,opt∈X such that

x0,opt=x−x1,opt. (7)

Then the expression (3) can be rewritten as L2,1(t, x;`2, X) = 1 2 x−x1,opt 2 `2+t x1,opt _X, Let us consider arbitrary ε∈ (−1, 1)and put

xε

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It is clear that L2,1(t, x;`2, X) ≤ 1 2kx−x ε 1k 2 `2+tkxε₁k_X. This is equivalent to 1 2kx−x ε 1k 2 `2+tkxε₁k_X ≥ 1 2 x−x1,opt 2 `2+t x1,opt X. Now putting the expression for xε

1, we get 1 2 x−x1,opt+εx1,opt 2 `2+t x1,opt−εx1,opt _X ≥ 1 2 x−x1,opt 2 `2+t x1,opt _X. Expanding we get 1 2 x−x1,opt 2 `2+εx−x1,opt, x1,opt +ε 2 2 x1,opt 2 `2+t(1−ε) x1,opt X ≥ 1 2 x−x1,opt 2 `2+t x1,opt X. Equivalently, ε x−x1,opt, x1,opt −t x1,opt X +ε 2 2 x1,opt 2 `2 ≥0. (8)

Let us consider two cases: • Case 1: ε≥0.

If ε≥0, then dividing the two sides of (8) by ε yields x−x1,opt, x1,opt −t x1,opt X+ ε 2 x1,opt 2 `2 ≥0.

Taking limit when ε tends to zero from the right, we obtain x−x1,opt, x1,opt −t x1,opt _X≥0. (9) • Case 2: ε≤0.

If ε≤0, then a similar reasoning as in case 1 yields x−x1,opt, x1,opt

−t x1,opt

X≤0. (10) Combining (9) and (10), we get

x−x1,opt, x1,opt −t x1,opt X =0. Therefore in view of (7), we obtain that

x0,opt, x−x0,opt

=t x−x0,opt

X, which completes the proof of (6).

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Remark 3. By definition of β (see(3)) we see that the equality in Lemma 2 can be rewritten as

x0,opt, x−x0,opt

=β. (11)

Lemma 3. The exact minimizer x0,optsatisfies

x0,opt _X∗=t.

Proof. In view of (11), it is enough to prove that

β t x0,opt X∗ =x0,opt, x−x0,opt .

By definition of the normk·k_X∗, we have

x0,opt X∗ = sup kzk_X≤1 x0,opt, z ,

which, by homogeneity property of the norm, is equivalent to

β t x0,opt X∗ = sup kzk_X≤β t x0,opt, z . (12) Since x−x0,opt X= β

t (see (3)), we have in particular that

β t x0,opt _X∗ ≥x0,opt, x−x0,opt . (13)

We need to show that

β t x0,opt X∗ ≤x0,opt, x−x0,opt .

Assume, by contradiction that

β t x0,opt X∗ >x0,opt, x−x0,opt .

Using (12), this can be rewritten as: sup kzk_X≤β t x0,opt, z >x0,opt, x−x0,opt . (14)

Denoting by Px_0,optv the vector projection of v onto the vector x0,opt and by putting z=x−u, the expression (14) becomes

sup u∈B2 Px0,opt(x−u) _`2 > Px0,opt x−x0,opt _`2. (15)

From Lemma 1, we know that T1separates B1and B2and x−x0,opt _X = β_t. In Figure 4, T₁00is parallel to T1.

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Figure 4: Orthogonal Projections.

The situation in (15) is impossible because we must have sup u∈B2 Px0,opt(x−u) _`2 ≤ Px0,opt −x+x0,opt _`2.

Therefore we must have sup u∈B2 Px0,opt(x−u) _`₂ ≤ Px0,opt x−x0,opt _`₂, or equivalently sup kzk_X≤β t x0,opt, z ≤x0,opt, x−x0,opt . (16)

From (12), the expression (16) can be rewritten as

β t x0,opt _X∗ ≤x0,opt, x−x0,opt . (17)

Combining (13) and (17), we get

β t x0,opt X∗ =x0,opt, x−x0,opt .

Using the fact that x0,opt, x−x0,opt

= β (see (11) in Remark 3), we obtain

that x0,opt

_X∗ =t.

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Now we are ready to prove Theorem 2.1.

Proof of Theorem 2.1. To prove Theorem 2.1, we consider the closed balls

B3={u∈Rn :kuk_X∗ ≤t} and B4= n u∈Rn :kx−uk_`2 ≤ x−x0,opt `2 o . In Figure 5 we illustrate these balls. Moreover we consider a hyperplane T2 tangent to B4through the point x0,opt (and of course orthogonal to the vector x−x0,opt because B4 is a ball of`2), then we have two closed halfspaces H3 and H4 which contain the origin 0 and the point x respectively. From this consideration, the ball B4is obviously contained in the halfspace H4.

Figure 5: Closed balls B3and B4ink·kX∗andk·k_`2 respectively.

It will be enough to prove that B3∩B4=x0,opt ,

as illustrated in Figure 5. Suppose by contradiction that there exists another element v∈ H4\H3such that

kx−vk_`2 ≤

x−x0,opt

_`₂ and kvk_X∗≤t, (18)

This situation is illustrated in Figure 6, where T₂00is parallel to T2. Sincekx−vk_`2 ≤

x−x0,opt

`2, i.e., v ∈ B4, by considering orthogonal

pro-jections of the vectors x0,opt and v on the direction of x−x0,opt, we obtain that Px−x0,opt x0,opt _`2 < Px−x0,optv _`2, which is equivalent to x−x0,opt, x0,opt <x−x0,opt, v .

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Figure 6: The situation that B3contains an element v∈H4\H3is impossible. It follows from Lemma 2, that

t x−x0,opt

X<x−x0,opt, v . Using Cauchy-Schwartz inequality, we get

t x−x0,opt X<x−x0,opt, v ≤ x−x0,opt XkvkX∗.

Which implies that

kvk_X∗>t,

which contradicts (18). Therefore B3∩H4=x0,opt .

Since B4is strictly convex, it follows that B3∩B4=x0,opt .

Whence we conclude that the unique nearest element (in the metric of `2_{) of} B3to B4is x0,opt. More precisely:

inf kx0kX∗≤t kx−x0k`2 = x−x0,opt _`₂. This completes the proof of Theorem 2.1.

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4 Final remarks and discussions

1. In (Meyer, 2002), Yves Meyer mentioned that his result (see (Meyer, 2002), Theorem 3 on page 32) could be derived from Theorem 11 on page 192 of (Aubin and Ekeland, 1984). Our result can also be derived from the same theorem. However we have given an independent proof based on the geometry of optimal decomposition that we are interested in.

2. As we see from the proof of Lemma 2, Hilbert structure is very important in our approach. Indeed we used equality

1 2 x−x1,opt+εx1,opt 2 `2 = 1 2 x−x1,opt 2 `2 +εx−x1,opt, x1,opt +ε 2 2 x1,opt 2 `2,

which is not true in general. In fact this approach could not work in the same way for the couple (`p, X) for p 6= 2. Therefore we need a different approach to find an extension of Theorem 2.1 for the couple

(`p, X), 1< p< +∞. We postpone this for our future investigation.

References

J. P. Aubin and I. Ekeland. Applied nonlinear analysis. John Wiley and Sons, 1984.

J. Bergh and J. Löfström. Interpolation Spaces. An Introduction. Springer, Berlin, 1976.

T. F. Chan and J. Shen. Image Processing and Analysis. Variational, PDE, Wavelet, and Stochastic Methods. Siam, 2005.

Y. Meyer. Oscillating Patterns in Image Processing and Nonlinear Evolution Equa-tions. University Lecture Series, Vol. 22, AMS Providence., 2002.

P. Nilsson and J. Peetre. On the K−functional between L1and L2 and some other K− functionals. Journal of Approximation Theory, Academic Press, Bel-gium, 1986.

L. I. Rudin, S. Osher, and E. Fatemi. Nonlinear total variation based noise removal algorithms. Physica, North-Holland, D(60):259–268, 1992.

O. Scherzer, M. Grasmair, H. Grossauer, M. Haltmeier, and F. Lenzen. Varia-tional Methods in Imaging. Springer, New York, 2009.