Solving inverse problems via hemicontractive maps

Stefan Soltuz
(original), Fixed point theory, JCR/ISI, paper

Abstract

We prove a “collage” theorem for hemicontractive maps and we use it for inverse problems. A numerical example is given.

    Authors

    S.M. Soltuz
    (Tiberiu Popoviciu Institute of Numerical Analysis, Romanian Academy)

    Keywords

    Hemicontractive maps; Inverse problems.

    References

    See the expanding block below.

    Paper coordinates

    Ş.M. Şoltuz, Solving inverse problems via hemicontractive maps, Nonlinear Analysis: Theory, Methods & Applications, 71 (2009), 2387-2390.
    doi: 10.1016/j.na.2009.01.071

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    Solving inverse problems via hemicontractive maps

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    Nonlinear Analysis: Theory, Methods & Applications

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    Elsevier

    DOI

    10.1016/j.na.2009.01.071

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    0362-546X

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    j.na.2009.01.071

    Solving inverse problems via hemicontractive maps

    Ştefan M. ŞoltuzThe Institute of Numerical Analysis, P.O. Box 68-1, 400110, Cluj-Napoca, Romania

    MSC:
    65J22
    47N40
    Keywords:
    Hemicontractive maps
    Inverse problems

    Abstract

    We prove a "collage" theorem for hemicontractive maps and we use it for inverse problems. A numerical example is given. © 2009 Elsevier Ltd. All rights reserved.

    1. Introduction

    Let X X XXX be a real Banach space, T : X → X T : X → X T:X rarr XT: X \rightarrow XT:X→X be an operator. The map J : X → 2 X ∗ J : X → 2 X ∗ J:X rarr2^(X^(**))J: X \rightarrow 2^{X^{*}}J:X→2X∗ given by J x := { f ∈ X ∗ : ⟨ x , f ⟩ = ‖ x ‖ 2 , ‖ f ‖ = ‖ x ‖ } , ∀ x ∈ X J x := f ∈ X ∗ : ⟨ x , f ⟩ = ‖ x ‖ 2 , ‖ f ‖ = ‖ x ‖ , ∀ x ∈ X Jx:={f inX^(**):(:x,f:)=:}{:||x||^(2),||f||=||x||},AA x in XJ x:=\left\{f \in X^{*}:\langle x, f\rangle=\right. \left.\|x\|^{2},\|f\|=\|x\|\right\}, \forall x \in XJx:={f∈X∗:⟨x,f⟩=‖x‖2,‖f‖=‖x‖},∀x∈X, is called the normalized duality mapping. It is easy to see that
    (1) ⟨ y , j ( x ) ⟩ ≤ ‖ x ‖ ‖ y ‖ , ∀ x , y ∈ X , ∀ j ( x ) ∈ J ( x ) . (1) ⟨ y , j ( x ) ⟩ ≤ ‖ x ‖ ‖ y ‖ , ∀ x , y ∈ X , ∀ j ( x ) ∈ J ( x ) . {:(1)(:y","j(x):) <= ||x||||y||","quad AA x","y in X","AA j(x)in J(x).:}\begin{equation*} \langle y, j(x)\rangle \leq\|x\|\|y\|, \quad \forall x, y \in X, \forall j(x) \in J(x) . \tag{1} \end{equation*}(1)⟨y,j(x)⟩≤‖x‖‖y‖,∀x,y∈X,∀j(x)∈J(x).
    Definition 1. A map T : X → X T : X → X T:X rarr XT: X \rightarrow XT:X→X is called hemicontractive if there exist k ∈ ( 0 , 1 ) k ∈ ( 0 , 1 ) k in(0,1)k \in(0,1)k∈(0,1) and q ∈ X q ∈ X q in Xq \in Xq∈X with q = T q q = T q q=Tqq=T qq=Tq such that for every x ∈ X x ∈ X x in Xx \in Xx∈X there exists j ( x − q ) ∈ J ( x − q ) j ( x − q ) ∈ J ( x − q ) j(x-q)in J(x-q)j(x-q) \in J(x-q)j(x−q)∈J(x−q) satisfying
    (2) ⟨ T x − T q , j ( x − q ) ⟩ ≤ k ‖ x − q ‖ 2 , ∀ x ∈ X . (2) ⟨ T x − T q , j ( x − q ) ⟩ ≤ k ‖ x − q ‖ 2 , ∀ x ∈ X . {:(2)(:Tx-Tq","j(x-q):) <= k||x-q||^(2)","quad AA x in X.:}\begin{equation*} \langle T x-T q, j(x-q)\rangle \leq k\|x-q\|^{2}, \quad \forall x \in X . \tag{2} \end{equation*}(2)⟨Tx−Tq,j(x−q)⟩≤k‖x−q‖2,∀x∈X.
    Remark 2. The fixed point q q qqq in Definition 1 is uniquely determined and, sometimes, will be denoted by x T ∗ x T ∗ x_(T)^(**)x_{T}^{*}xT∗.
    Indeed, if p = T p p = T p p=Tpp=T pp=Tp is another fixed point of the hemicontractive mapping T T TTT, then
    ‖ p − q ‖ 2 = ⟨ p − q , j ( p − q ) ⟩ = ⟨ T p − T q , j ( p − q ) ⟩ ≤ k ‖ p − q ‖ 2 , ‖ p − q ‖ 2 = ⟨ p − q , j ( p − q ) ⟩ = ⟨ T p − T q , j ( p − q ) ⟩ ≤ k ‖ p − q ‖ 2 , {:[||p-q||^(2)=(:p-q","j(p-q):)],[=(:Tp-Tq","j(p-q):) <= k||p-q||^(2)","]:}\begin{aligned} \|p-q\|^{2} & =\langle p-q, j(p-q)\rangle \\ & =\langle T p-T q, j(p-q)\rangle \leq k\|p-q\|^{2}, \end{aligned}‖p−q‖2=⟨p−q,j(p−q)⟩=⟨Tp−Tq,j(p−q)⟩≤k‖p−q‖2,
    implying ‖ p − q ‖ = 0 ‖ p − q ‖ = 0 ||p-q||=0\|p-q\|=0‖p−q‖=0, i.e., p = q p = q p=qp=qp=q.
    It is well known that T T TTT is a contraction if there exists k ∈ ( 0 , 1 ) k ∈ ( 0 , 1 ) k in(0,1)k \in(0,1)k∈(0,1) such that ‖ T x − T y ‖ ≤ k ‖ x − y ‖ , ∀ x , y ∈ X ‖ T x − T y ‖ ≤ k ‖ x − y ‖ , ∀ x , y ∈ X ||Tx-Ty|| <= k||x-y||,AA x,y in X\|T x-T y\| \leq k\|x-y\|, \forall x, y \in X‖Tx−Ty‖≤k‖x−y‖,∀x,y∈X.
    Remark 3. The class of contractions is a subclass of hemicontractions.
    Let T T TTT be a k k kkk-contraction of the Banach space X X XXX. Then T T TTT has a unique fixed point q q qqq and
    ⟨ T x − T q , j ( x − q ) ⟩ ≤ ‖ T x − T q ‖ ‖ x − q ‖ ≤ k ‖ x − q ‖ 2 , ⟨ T x − T q , j ( x − q ) ⟩ ≤ ‖ T x − T q ‖ ‖ x − q ‖ ≤ k ‖ x − q ‖ 2 , {:[(:Tx-Tq","j(x-q):) <= ||Tx-Tq||||x-q||],[ <= k||x-q||^(2)","]:}\begin{aligned} \langle T x-T q, j(x-q)\rangle & \leq\|T x-T q\|\|x-q\| \\ & \leq k\|x-q\|^{2}, \end{aligned}⟨Tx−Tq,j(x−q)⟩≤‖Tx−Tq‖‖x−q‖≤k‖x−q‖2,
    for j ( x − q ) ∈ J ( x − q ) j ( x − q ) ∈ J ( x − q ) j(x-q)in J(x-q)j(x-q) \in J(x-q)j(x−q)∈J(x−q).
    Remark 4. The above inclusion is proper.
    Indeed, note that T ( x , y ) = ( − y , x ) T ( x , y ) = ( − y , x ) T(x,y)=(-y,x)T(x, y)=(-y, x)T(x,y)=(−y,x) is not a contraction while it is hemicontractive with q = ( 0 , 0 ) q = ( 0 , 0 ) q=(0,0)q=(0,0)q=(0,0) and k = 0.5 k = 0.5 k=0.5k=0.5k=0.5,
    ⟨ T ( x , y ) , ( x , y ) ⟩ = ⟨ ( − y , x ) , ( x , y ) ⟩ = 0 ≤ ( 1 / 2 ) ( x 2 + y 2 ) = 0.5 ‖ ( x , y ) ‖ 2 ⟨ T ( x , y ) , ( x , y ) ⟩ = ⟨ ( − y , x ) , ( x , y ) ⟩ = 0 ≤ ( 1 / 2 ) x 2 + y 2 = 0.5 ‖ ( x , y ) ‖ 2 {:[(:T(x","y)","(x","y):)=(:(-y","x)","(x","y):)=0],[ <= (1//2)(x^(2)+y^(2))=0.5||(x","y)||^(2)]:}\begin{aligned} \langle T(x, y),(x, y)\rangle & =\langle(-y, x),(x, y)\rangle=0 \\ & \leq(1 / 2)\left(x^{2}+y^{2}\right)=0.5\|(x, y)\|^{2} \end{aligned}⟨T(x,y),(x,y)⟩=⟨(−y,x),(x,y)⟩=0≤(1/2)(x2+y2)=0.5‖(x,y)‖2
    Recently, Kunze et al. (see [1-3]) have considered a class of inverse problems for ordinary differential equations and provided a mathematical basis for solving them within the framework of Banach spaces and contractions. We shall consider the same framework of Banach spaces and the larger class of hemicontractive maps.
    Notation 5. Denote by HemiLip := { T , T : X → X , T := { T , T : X → X , T :={T,T:X rarr X,T:=\{T, T: X \rightarrow X, T:={T,T:X→X,T a hemicontractive map with constant k ∈ ( 0 , 1 ) k ∈ ( 0 , 1 ) k in(0,1)k \in(0,1)k∈(0,1), Lipschitzian with constant L ≥ 1 L ≥ 1 L >= 1L \geq 1L≥1 and T ( X ) T ( X ) T(X)T(X)T(X) bounded } } }\}}.
    A typical inverse problem is the following:
    Problem 6. For given ε > 0 ε > 0 epsi > 0\varepsilon>0ε>0 and a "target" x ¯ x ¯ bar(x)\bar{x}x¯, find T ε ∈ T ε ∈ T_(epsi)inT_{\varepsilon} \inTε∈ HemiLip such that ‖ x ¯ − x T ε ∗ ‖ < ε x ¯ − x T ε ∗ < ε ||( bar(x))-x_(T_(epsi))^(**)|| < epsi\left\|\bar{x}-x_{T_{\varepsilon}}^{*}\right\|<\varepsilon‖x¯−xTε∗‖<ε, where x T ε ∗ = T ε ( x T ε ∗ ) x T ε ∗ = T ε x T ε ∗ x_(T_(epsi))^(**)=T_(epsi)(x_(T_(epsi))^(**))x_{T_{\varepsilon}}^{*}=T_{\varepsilon}\left(x_{T_{\varepsilon}}^{*}\right)xTε∗=Tε(xTε∗) is the unique fixed point of the hemicontractive mapping T ε T ε T_(epsi)T_{\varepsilon}Tε.
    According to [1], randomly selecting various maps in HemiLip, finding their fixed points and computing the distance from our target is an extremely tedious procedure. Consider now the following problem which we shall fit in our framework and which is very useful for practitioners.
    Problem 7. Let x ¯ ∈ X x ¯ ∈ X bar(x)in X\bar{x} \in Xx¯∈X be a target and let δ > 0 δ > 0 delta > 0\delta>0δ>0 be given. Find T δ ∈ T δ ∈ T_(delta)inT_{\delta} \inTδ∈ HemiLip such that ‖ x ¯ − T δ x ¯ ‖ < δ x ¯ − T δ x ¯ < δ ||( bar(x))-T_(delta)( bar(x))|| < delta\left\|\bar{x}-T_{\delta} \bar{x}\right\|<\delta‖x¯−Tδx¯‖<δ.
    In other words, instead of searching for hemicontractive maps whose fixed points lie close to target x ¯ x ¯ bar(x)\bar{x}x¯, we search for hemicontractive maps that send x ¯ x ¯ bar(x)\bar{x}x¯ close to itself.

    2. Main results

    Theorem 8 (Collage Theorem for Hemicontractive Maps). Let X X XXX be a real Banach space and T T TTT a hemicontractive map with contraction factor k ∈ ( 0 , 1 ) k ∈ ( 0 , 1 ) k in(0,1)k \in(0,1)k∈(0,1) and fixed point x ∗ ∈ X x ∗ ∈ X x^(**)in Xx^{*} \in Xx∗∈X. Then for any x ∈ X x ∈ X x in Xx \in Xx∈X,
    ‖ x − x ∗ ‖ ≤ 1 1 − k ‖ x − T x ‖ x − x ∗ ≤ 1 1 − k ‖ x − T x ‖ ||x-x^(**)|| <= (1)/(1-k)||x-Tx||\left\|x-x^{*}\right\| \leq \frac{1}{1-k}\|x-T x\|‖x−x∗‖≤11−k‖x−Tx‖
    Proof. The hemicontractive condition assures that the fixed point x ∗ x ∗ x^(**)x^{*}x∗ exists and it is unique. If x = x ∗ x = x ∗ x=x^(**)x=x^{*}x=x∗, the above inequality holds. If x ≠ x ∗ , ∀ x ∈ X x ≠ x ∗ , ∀ x ∈ X x!=x^(**),AA x in Xx \neq x^{*}, \forall x \in Xx≠x∗,∀x∈X, then using (1) and (2) one obtains
    ‖ x − x ∗ ‖ 2 = ⟨ x − x ∗ , j ( x − x ∗ ) ⟩ = ⟨ T x − T x ∗ , j ( x − x ∗ ) ⟩ + ⟨ x − T x , j ( x − x ∗ ) ⟩ ≤ k ‖ x − x ∗ ‖ 2 + ⟨ x − T x , j ( x − x ∗ ) ⟩ ≤ k ‖ x − x ∗ ‖ 2 + ‖ x − T x ‖ ‖ x − x ∗ ‖ x − x ∗ 2 = x − x ∗ , j x − x ∗ = T x − T x ∗ , j x − x ∗ + x − T x , j x − x ∗ ≤ k x − x ∗ 2 + x − T x , j x − x ∗ ≤ k x − x ∗ 2 + ‖ x − T x ‖ x − x ∗ {:[||x-x^(**)||^(2)=(:x-x^(**),j(x-x^(**)):)],[=(:Tx-Tx^(**),j(x-x^(**)):)+(:x-Tx,j(x-x^(**)):)],[ <= k||x-x^(**)||^(2)+(:x-Tx,j(x-x^(**)):)],[ <= k||x-x^(**)||^(2)+||x-Tx||||x-x^(**)||]:}\begin{aligned} \left\|x-x^{*}\right\|^{2} & =\left\langle x-x^{*}, j\left(x-x^{*}\right)\right\rangle \\ & =\left\langle T x-T x^{*}, j\left(x-x^{*}\right)\right\rangle+\left\langle x-T x, j\left(x-x^{*}\right)\right\rangle \\ & \leq k\left\|x-x^{*}\right\|^{2}+\left\langle x-T x, j\left(x-x^{*}\right)\right\rangle \\ & \leq k\left\|x-x^{*}\right\|^{2}+\|x-T x\|\left\|x-x^{*}\right\| \end{aligned}‖x−x∗‖2=⟨x−x∗,j(x−x∗)⟩=⟨Tx−Tx∗,j(x−x∗)⟩+⟨x−Tx,j(x−x∗)⟩≤k‖x−x∗‖2+⟨x−Tx,j(x−x∗)⟩≤k‖x−x∗‖2+‖x−Tx‖‖x−x∗‖
    From which one gets the conclusion.
    The above "Collage Theorem" allows us to reformulate the inverse Problem 6 in the particular and more convenient Problem 7.
    Theorem 9. If Problem 7 has a solution, then Problem 6 has a solution too.
    Proof. Let ε > 0 ε > 0 epsi > 0\varepsilon>0ε>0 and x ¯ ∈ X x ¯ ∈ X bar(x)in X\bar{x} \in Xx¯∈X be given. For δ := ( 1 − k ) ε δ := ( 1 − k ) ε delta:=(1-k)epsi\delta:=(1-k) \varepsilonδ:=(1−k)ε, let T δ ∈ T δ ∈ T_(delta)inT_{\delta} \inTδ∈ HemiLip be such that ‖ x ¯ − T δ x ¯ ‖ < δ x ¯ − T δ x ¯ < δ ||( bar(x))-T_(delta)( bar(x))|| < delta\left\|\bar{x}-T_{\delta} \bar{x}\right\|<\delta‖x¯−Tδx¯‖<δ. If x T δ ∗ x T δ ∗ x_(T_(delta))^(**)x_{T_{\delta}}^{*}xTδ∗ is the unique fixed point of the hemicontractive mapping T δ T δ T_(delta)T_{\delta}Tδ, then, by Theorem 8,
    ‖ x ¯ − x T δ ∗ ‖ ≤ 1 1 − k ‖ x ¯ − T δ x ¯ ‖ ≤ 1 1 − k δ = ε . x ¯ − x T δ ∗ ≤ 1 1 − k x ¯ − T δ x ¯ ≤ 1 1 − k δ = ε . ||( bar(x))-x_(T_(delta))^(**)|| <= (1)/(1-k)||( bar(x))-T_(delta)( bar(x))|| <= (1)/(1-k)delta=epsi.\left\|\bar{x}-x_{T_{\delta}}^{*}\right\| \leq \frac{1}{1-k}\left\|\bar{x}-T_{\delta} \bar{x}\right\| \leq \frac{1}{1-k} \delta=\varepsilon .‖x¯−xTδ∗‖≤11−k‖x¯−Tδx¯‖≤11−kδ=ε.
    Note that shrinking the distance between two operators, one of them from HemiLip, reduces the distance between their fixed points.
    Proposition 10. Let X X XXX be a real Banach space and T 1 ∈ T 1 ∈ T_(1)inT_{1} \inT1∈ HemiLip with contraction factor k 1 ∈ ( 0 , 1 ) k 1 ∈ ( 0 , 1 ) k_(1)in(0,1)k_{1} \in(0,1)k1∈(0,1) and T 2 : X → X T 2 : X → X T_(2):X rarr XT_{2}: X \rightarrow XT2:X→X a map such that x 1 ∗ , x 2 ∗ ∈ X x 1 ∗ , x 2 ∗ ∈ X x_(1)^(**),x_(2)^(**)in Xx_{1}^{*}, x_{2}^{*} \in Xx1∗,x2∗∈X are distinct fixed points for T 1 T 1 T_(1)T_{1}T1 and T 2 T 2 T_(2)T_{2}T2. Then,
    ‖ x 1 ∗ − x 2 ∗ ‖ ≤ 1 1 − k 1 sup x ∈ X ‖ T 1 x − T 2 x ‖ x 1 ∗ − x 2 ∗ ≤ 1 1 − k 1 sup x ∈ X   T 1 x − T 2 x ||x_(1)^(**)-x_(2)^(**)|| <= (1)/(1-k_(1))s u p_(x in X)||T_(1)x-T_(2)x||\left\|x_{1}^{*}-x_{2}^{*}\right\| \leq \frac{1}{1-k_{1}} \sup _{x \in X}\left\|T_{1} x-T_{2} x\right\|‖x1∗−x2∗‖≤11−k1supx∈X‖T1x−T2x‖
    Proof. Using (2) one obtains
    ‖ x 1 ∗ − x 2 ∗ ‖ 2 = ⟨ x 1 ∗ − x 2 ∗ , j ( x 1 ∗ − x 2 ∗ ) ⟩ = ⟨ T 1 x 1 ∗ − T 2 x 2 ∗ , j ( x 1 ∗ − x 2 ∗ ) ⟩ = ⟨ T 1 x 1 ∗ − T 1 x 2 ∗ , j ( x 1 ∗ − x 2 ∗ ) ⟩ + ⟨ T 1 x 2 ∗ − T 2 x 2 ∗ , j ( x 1 ∗ − x 2 ∗ ) ⟩ ≤ k 1 ‖ x 1 ∗ − x 2 ∗ ‖ 2 + ‖ x 1 ∗ − x 2 ∗ ‖ ‖ T 1 x 2 ∗ − T 2 x 2 ∗ ‖ ≤ k 1 ‖ x 1 ∗ − x 2 ∗ ‖ 2 + ‖ x 1 ∗ − x 2 ∗ ‖ ( sup x ∈ X ‖ T 1 x − T 2 x ‖ ) , x 1 ∗ − x 2 ∗ 2 = x 1 ∗ − x 2 ∗ , j x 1 ∗ − x 2 ∗ = T 1 x 1 ∗ − T 2 x 2 ∗ , j x 1 ∗ − x 2 ∗ = T 1 x 1 ∗ − T 1 x 2 ∗ , j x 1 ∗ − x 2 ∗ + T 1 x 2 ∗ − T 2 x 2 ∗ , j x 1 ∗ − x 2 ∗ ≤ k 1 x 1 ∗ − x 2 ∗ 2 + x 1 ∗ − x 2 ∗ T 1 x 2 ∗ − T 2 x 2 ∗ ≤ k 1 x 1 ∗ − x 2 ∗ 2 + x 1 ∗ − x 2 ∗ sup x ∈ X   T 1 x − T 2 x , {:[||x_(1)^(**)-x_(2)^(**)||^(2)=(:x_(1)^(**)-x_(2)^(**),j(x_(1)^(**)-x_(2)^(**)):)=(:T_(1)x_(1)^(**)-T_(2)x_(2)^(**),j(x_(1)^(**)-x_(2)^(**)):)],[=(:T_(1)x_(1)^(**)-T_(1)x_(2)^(**),j(x_(1)^(**)-x_(2)^(**)):)+(:T_(1)x_(2)^(**)-T_(2)x_(2)^(**),j(x_(1)^(**)-x_(2)^(**)):)],[ <= k_(1)||x_(1)^(**)-x_(2)^(**)||^(2)+||x_(1)^(**)-x_(2)^(**)||||T_(1)x_(2)^(**)-T_(2)x_(2)^(**)||],[ <= k_(1)||x_(1)^(**)-x_(2)^(**)||^(2)+||x_(1)^(**)-x_(2)^(**)||(s u p_(x in X)||T_(1)x-T_(2)x||)","]:}\begin{aligned} \left\|x_{1}^{*}-x_{2}^{*}\right\|^{2} & =\left\langle x_{1}^{*}-x_{2}^{*}, j\left(x_{1}^{*}-x_{2}^{*}\right)\right\rangle=\left\langle T_{1} x_{1}^{*}-T_{2} x_{2}^{*}, j\left(x_{1}^{*}-x_{2}^{*}\right)\right\rangle \\ & =\left\langle T_{1} x_{1}^{*}-T_{1} x_{2}^{*}, j\left(x_{1}^{*}-x_{2}^{*}\right)\right\rangle+\left\langle T_{1} x_{2}^{*}-T_{2} x_{2}^{*}, j\left(x_{1}^{*}-x_{2}^{*}\right)\right\rangle \\ & \leq k_{1}\left\|x_{1}^{*}-x_{2}^{*}\right\|^{2}+\left\|x_{1}^{*}-x_{2}^{*}\right\|\left\|T_{1} x_{2}^{*}-T_{2} x_{2}^{*}\right\| \\ & \leq k_{1}\left\|x_{1}^{*}-x_{2}^{*}\right\|^{2}+\left\|x_{1}^{*}-x_{2}^{*}\right\|\left(\sup _{x \in X}\left\|T_{1} x-T_{2} x\right\|\right), \end{aligned}‖x1∗−x2∗‖2=⟨x1∗−x2∗,j(x1∗−x2∗)⟩=⟨T1x1∗−T2x2∗,j(x1∗−x2∗)⟩=⟨T1x1∗−T1x2∗,j(x1∗−x2∗)⟩+⟨T1x2∗−T2x2∗,j(x1∗−x2∗)⟩≤k1‖x1∗−x2∗‖2+‖x1∗−x2∗‖‖T1x2∗−T2x2∗‖≤k1‖x1∗−x2∗‖2+‖x1∗−x2∗‖(supx∈X‖T1x−T2x‖),
    from which we get the conclusion.
    Theorem 11. Let X X XXX be a real Banach space, T : X → X , x ¯ = T x ¯ T : X → X , x ¯ = T x ¯ T:X rarr X, bar(x)=T bar(x)T: X \rightarrow X, \bar{x}=T \bar{x}T:X→X,x¯=Tx¯ and suppose there exists T 1 ∈ T 1 ∈ T_(1)inT_{1} \inT1∈ HemiLip such that sup x ∈ X ‖ T 1 x − T x ‖ ≤ ε sup x ∈ X   T 1 x − T x ≤ ε s u p_(x in X)||T_(1)x-Tx|| <= epsi\sup _{x \in X}\left\|T_{1} x-T x\right\| \leq \varepsilonsupx∈X‖T1x−Tx‖≤ε. Then
    ‖ x ¯ − T 1 x ¯ ‖ ≤ 1 + L 1 − k ε x ¯ − T 1 x ¯ ≤ 1 + L 1 − k ε ||( bar(x))-T_(1)( bar(x))|| <= (1+L)/(1-k)epsi\left\|\bar{x}-T_{1} \bar{x}\right\| \leq \frac{1+L}{1-k} \varepsilon‖x¯−T1x¯‖≤1+L1−kε
    Proof. Let x ∗ = T 1 x ∗ x ∗ = T 1 x ∗ x^(**)=T_(1)x^(**)x^{*}=T_{1} x^{*}x∗=T1x∗, and by use of Proposition 10 we obtain
    ‖ x ¯ − x ∗ ‖ ≤ 1 1 − k ( sup x ∈ X ‖ T 1 x − T x ‖ ) x ¯ − x ∗ ≤ 1 1 − k sup x ∈ X   T 1 x − T x ||( bar(x))-x^(**)|| <= (1)/(1-k)(s u p_(x in X)||T_(1)x-Tx||)\left\|\bar{x}-x^{*}\right\| \leq \frac{1}{1-k}\left(\sup _{x \in X}\left\|T_{1} x-T x\right\|\right)‖x¯−x∗‖≤11−k(supx∈X‖T1x−Tx‖)
    Thus,
    ‖ x ¯ − T 1 x ¯ ‖ ≤ ‖ x ¯ − x ∗ ‖ + ‖ x ∗ − T 1 x ¯ ‖ ≤ ‖ x − x 1 ‖ + ‖ T 1 x ∗ − T 1 x ¯ ‖ ≤ ( 1 + L ) ‖ x ¯ − x ∗ ‖ ≤ 1 + L 1 − k ( sup x ∈ X ‖ T 1 x − T x ‖ ) ≤ 1 + L 1 − k ε . x ¯ − T 1 x ¯ ≤ x ¯ − x ∗ + x ∗ − T 1 x ¯ ≤ x − x 1 + T 1 x ∗ − T 1 x ¯ ≤ ( 1 + L ) x ¯ − x ∗ ≤ 1 + L 1 − k sup x ∈ X   T 1 x − T x ≤ 1 + L 1 − k ε . {:[||( bar(x))-T_(1)( bar(x))|| <= ||( bar(x))-x^(**)||+||x^(**)-T_(1)( bar(x))||],[ <= ||x-x_(1)||+||T_(1)x^(**)-T_(1)( bar(x))||],[ <= (1+L)||( bar(x))-x^(**)||],[ <= (1+L)/(1-k)(s u p_(x in X)||T_(1)x-Tx||) <= (1+L)/(1-k)epsi.]:}\begin{aligned} \left\|\bar{x}-T_{1} \bar{x}\right\| & \leq\left\|\bar{x}-x^{*}\right\|+\left\|x^{*}-T_{1} \bar{x}\right\| \\ & \leq\left\|x-x_{1}\right\|+\left\|T_{1} x^{*}-T_{1} \bar{x}\right\| \\ & \leq(1+L)\left\|\bar{x}-x^{*}\right\| \\ & \leq \frac{1+L}{1-k}\left(\sup _{x \in X}\left\|T_{1} x-T x\right\|\right) \leq \frac{1+L}{1-k} \varepsilon . \end{aligned}‖x¯−T1x¯‖≤‖x¯−x∗‖+‖x∗−T1x¯‖≤‖x−x1‖+‖T1x∗−T1x¯‖≤(1+L)‖x¯−x∗‖≤1+L1−k(supx∈X‖T1x−Tx‖)≤1+L1−kε.

    3. Application

    Example 12. Let A ∈ ( 0 , 1 ) , B , C , D ∈ R A ∈ ( 0 , 1 ) , B , C , D ∈ R A in(0,1),B,C,D inRA \in(0,1), B, C, D \in \mathbb{R}A∈(0,1),B,C,D∈R be fixed numbers and F : [ 0 , 3 ] × [ 0 , 3 ] → R 2 F : [ 0 , 3 ] × [ 0 , 3 ] → R 2 F:[0,3]xx[0,3]rarrR^(2)F:[0,3] \times[0,3] \rightarrow \mathbb{R}^{2}F:[0,3]×[0,3]→R2 be given by F ( x , y ) = ( A x + B x y − C y , A y − B x 2 + C x ) F ( x , y ) = A x + B x y − C y , A y − B x 2 + C x F(x,y)=(Ax+Bxy-Cy,Ay-Bx^(2)+Cx)F(x, y)= \left(A x+B x y-C y, A y-B x^{2}+C x\right)F(x,y)=(Ax+Bxy−Cy,Ay−Bx2+Cx). Then F F FFF is Lipschitzian and hemicontractive with bounded range.
    Proof. It is obvious that F F FFF is Lipschitzian and has bounded range. In order to prove that it is hemicontractive, note that
    ⟨ F ( x , y ) , ( x , y ) ⟩ = ⟨ ( A x + B x y − C y , A y − B x 2 + C x ) , ( x , y ) ⟩ = A x 2 + B x 2 y − C x y + A y 2 − B x 2 y + C x y = A ‖ ( x , y ) ‖ 2 . ◻ ⟨ F ( x , y ) , ( x , y ) ⟩ = A x + B x y − C y , A y − B x 2 + C x , ( x , y ) = A x 2 + B x 2 y − C x y + A y 2 − B x 2 y + C x y = A ‖ ( x , y ) ‖ 2 . ◻ {:[(:F(x","y)","(x","y):)=(:(Ax+Bxy-Cy,Ay-Bx^(2)+Cx),(x,y):)],[=Ax^(2)+Bx^(2)y-Cxy+Ay^(2)-Bx^(2)y+Cxy],[=A||(x","y)||^(2).quad◻]:}\begin{aligned} \langle F(x, y),(x, y)\rangle & =\left\langle\left(A x+B x y-C y, A y-B x^{2}+C x\right),(x, y)\right\rangle \\ & =A x^{2}+B x^{2} y-C x y+A y^{2}-B x^{2} y+C x y \\ & =A\|(x, y)\|^{2} . \quad \square \end{aligned}⟨F(x,y),(x,y)⟩=⟨(Ax+Bxy−Cy,Ay−Bx2+Cx),(x,y)⟩=Ax2+Bx2y−Cxy+Ay2−Bx2y+Cxy=A‖(x,y)‖2.◻
    Set C = 0 C = 0 C=0C=0C=0, to obtain our T ε T ε T_(epsi)T_{\varepsilon}Tε function:
    Example 13. Let A ∈ ( 0 , 1 ) , B ∈ R A ∈ ( 0 , 1 ) , B ∈ R A in(0,1),B inRA \in(0,1), B \in \mathbb{R}A∈(0,1),B∈R be fixed numbers and H : [ 0 , 3 ] × [ 0 , 3 ] → R 2 H : [ 0 , 3 ] × [ 0 , 3 ] → R 2 H:[0,3]xx[0,3]rarrR^(2)H:[0,3] \times[0,3] \rightarrow \mathbb{R}^{2}H:[0,3]×[0,3]→R2 be given by H ( x , y ) = ( A x + B x y , A y − B x 2 ) H ( x , y ) = A x + B x y , A y − B x 2 H(x,y)=(Ax+Bxy,Ay-Bx^(2))H(x, y)= \left(A x+B x y, A y-B x^{2}\right)H(x,y)=(Ax+Bxy,Ay−Bx2). Then H H HHH is Lipschitzian and hemicontractive with bounded range.
    Remark 14. Let h ¯ h ¯ bar(h)\bar{h}h¯ be the "target", in order to find T ε T ε T_(epsi)T_{\varepsilon}Tε. As for fitting, we shall look for an appropriate T δ T δ T_(delta)T_{\delta}Tδ. Then by using Matlab (i.e. fminsearch) for min ‖ h ¯ − T δ h ¯ ‖ min h ¯ − T δ h ¯ min||( bar(h))-T_(delta)( bar(h))||\min \left\|\bar{h}-T_{\delta} \bar{h}\right\|min‖h¯−Tδh¯‖, by Theorem 11 we find the parameters which minimize the problem. Set in Example 13, A = 0.3 , B = 6 , T ε := H A = 0.3 , B = 6 , T ε := H A=0.3,B=6,T_(epsi):=HA=0.3, B=6, T_{\varepsilon}:=HA=0.3,B=6,Tε:=H and let h ¯ = ( x ¯ , y ¯ ) = H ( ( x ¯ , y ¯ ) ) h ¯ = ( x ¯ , y ¯ ) = H ( ( x ¯ , y ¯ ) ) bar(h)=( bar(x), bar(y))=H(( bar(x), bar(y)))\bar{h}=(\bar{x}, \bar{y})=H((\bar{x}, \bar{y}))h¯=(x¯,y¯)=H((x¯,y¯)) on ( [ − 1 , 3 ] × [ − 1 , 3 ] ) ( [ − 1 , 3 ] × [ − 1 , 3 ] ) ([-1,3]xx[-1,3])([-1,3] \times[-1,3])([−1,3]×[−1,3]) be the target generated by T ε T ε T_(epsi)T_{\varepsilon}Tε. Use the above algorithm with T δ := F T δ := F T_(delta):=FT_{\delta}:=FTδ:=F, to obtain the H H HHH map, i.e. ( A , B , C ) = ( 0.3000 , 6.0000 , 0.0000 ) ( A , B , C ) = ( 0.3000 , 6.0000 , 0.0000 ) (A,B,C)=(0.3000,6.0000,0.0000)(A, B, C)=(0.3000,6.0000,0.0000)(A,B,C)=(0.3000,6.0000,0.0000) starting from each point between ( 0.3000 , 2.0000 , 2.0000 0.3000 , 2.0000 , 2.0000 0.3000,2.0000,2.00000.3000,2.0000,2.00000.3000,2.0000,2.0000 ) and ( 0.9000 , 8.0000 , 6.0000 0.9000 , 8.0000 , 6.0000 0.9000,8.0000,6.00000.9000,8.0000,6.00000.9000,8.0000,6.0000 ) .
    Remark 15. In Example 13, set A = 0 , B = 0.5 , T ε := H A = 0 , B = 0.5 , T ε := H A=0,B=0.5,T_(epsi):=HA=0, B=0.5, T_{\varepsilon}:=HA=0,B=0.5,Tε:=H and let h ¯ = ( x ¯ , y ¯ ) = H ( ( x ¯ , y ¯ ) ) h ¯ = ( x ¯ , y ¯ ) = H ( ( x ¯ , y ¯ ) ) bar(h)=( bar(x), bar(y))=H(( bar(x), bar(y)))\bar{h}=(\bar{x}, \bar{y})=H((\bar{x}, \bar{y}))h¯=(x¯,y¯)=H((x¯,y¯)) on ( [ − 1 , 3 ] × [ − 1 , 3 ] ) ( [ − 1 , 3 ] × [ − 1 , 3 ] ) ([-1,3]xx[-1,3])([-1,3] \times[-1,3])([−1,3]×[−1,3]) be the target generated by T ε T ε T_(epsi)T_{\varepsilon}Tε. Use again the above algorithm with T δ := F T δ := F T_(delta):=FT_{\delta}:=FTδ:=F, to obtain the H H HHH map, i.e. ( A , B , C ) = ( 0.00 , 0.50 , 0.00 ) ( A , B , C ) = ( 0.00 , 0.50 , 0.00 ) (A,B,C)=(0.00,0.50,0.00)(A, B, C)=(0.00,0.50,0.00)(A,B,C)=(0.00,0.50,0.00) starting from each point between ( 0.0000 , 0.3000 , 0.0000 ) ( 0.0000 , 0.3000 , 0.0000 ) (0.0000,0.3000,0.0000)(0.0000,0.3000,0.0000)(0.0000,0.3000,0.0000) and ( 1.0000 , 3.0000 , 1.0000 ) ( 1.0000 , 3.0000 , 1.0000 ) (1.0000,3.0000,1.0000)(1.0000,3.0000,1.0000)(1.0000,3.0000,1.0000).
    Remark 16. In Example 13, set A = 0 , B = 1 , T ε := H A = 0 , B = 1 , T ε := H A=0,B=1,T_(epsi):=HA=0, B=1, T_{\varepsilon}:=HA=0,B=1,Tε:=H and let h ¯ = ( x ¯ , y ¯ ) = H ( ( x ¯ , y ¯ ) ) h ¯ = ( x ¯ , y ¯ ) = H ( ( x ¯ , y ¯ ) ) bar(h)=( bar(x), bar(y))=H(( bar(x), bar(y)))\bar{h}=(\bar{x}, \bar{y})=H((\bar{x}, \bar{y}))h¯=(x¯,y¯)=H((x¯,y¯)) on ( [ − 1 , 3 ] × [ − 1 , 3 ] ) ( [ − 1 , 3 ] × [ − 1 , 3 ] ) ([-1,3]xx[-1,3])([-1,3] \times[-1,3])([−1,3]×[−1,3]) be the target generated by T ε T ε T_(epsi)T_{\varepsilon}Tε. Use fminsearch with T δ := F T δ := F T_(delta):=FT_{\delta}:=FTδ:=F, to obtain the H H HHH map, i.e. ( A , B , C ) = ( 0.00 , 1.00 , 0.00 ) ( A , B , C ) = ( 0.00 , 1.00 , 0.00 ) (A,B,C)=(0.00,1.00,0.00)(A, B, C)=(0.00,1.00,0.00)(A,B,C)=(0.00,1.00,0.00) starting from each point between ( 0.2000 , 0.3000 , 0.2000 ) ( 0.2000 , 0.3000 , 0.2000 ) (0.2000,0.3000,0.2000)(0.2000,0.3000,0.2000)(0.2000,0.3000,0.2000) and ( 0.7000 , 0.5000 , 0.5000 ) ( 0.7000 , 0.5000 , 0.5000 ) (0.7000,0.5000,0.5000)(0.7000,0.5000,0.5000)(0.7000,0.5000,0.5000).
    Remark 17. In Example 13, set A = 1 , B = 0 , T ε := H A = 1 , B = 0 , T ε := H A=1,B=0,T_(epsi):=HA=1, B=0, T_{\varepsilon}:=HA=1,B=0,Tε:=H and let h ¯ = ( x ¯ , y ¯ ) = H ( ( x ¯ , y ¯ ) ) h ¯ = ( x ¯ , y ¯ ) = H ( ( x ¯ , y ¯ ) ) bar(h)=( bar(x), bar(y))=H(( bar(x), bar(y)))\bar{h}=(\bar{x}, \bar{y})=H((\bar{x}, \bar{y}))h¯=(x¯,y¯)=H((x¯,y¯)) on ( [ − 1 , 3 ] × [ − 1 , 3 ] ) ( [ − 1 , 3 ] × [ − 1 , 3 ] ) ([-1,3]xx[-1,3])([-1,3] \times[-1,3])([−1,3]×[−1,3]) be the target generated by T ε T ε T_(epsi)T_{\varepsilon}Tε. Note that T ε T ε T_(epsi)T_{\varepsilon}Tε is not strongly pseudocontractive. Use fminsearch with T δ := F T δ := F T_(delta):=FT_{\delta}:=FTδ:=F, to obtain the H H HHH map, i.e. ( A , B , C ) = ( 1.00 , 0.00 , 0.00 ) ( A , B , C ) = ( 1.00 , 0.00 , 0.00 ) (A,B,C)=(1.00,0.00,0.00)(A, B, C)=(1.00,0.00,0.00)(A,B,C)=(1.00,0.00,0.00) starting from each point between ( 0.2000 , 0.2000 , 0.2000 ) ( 0.2000 , 0.2000 , 0.2000 ) (0.2000,0.2000,0.2000)(0.2000,0.2000,0.2000)(0.2000,0.2000,0.2000) and ( 0.7000 , 0.7000 , 0.7000 ) ( 0.7000 , 0.7000 , 0.7000 ) (0.7000,0.7000,0.7000)(0.7000,0.7000,0.7000)(0.7000,0.7000,0.7000).

    References

    [1] H.E. Kunze, E.R. Vrscay, Solving inverse problems for ordinary differential equations using the Picard contraction mapping, Inverse Problems 15 (1999) 745-770.
    [2] H.E. Kunze, S. Gomes, Solving an inverse problem for Urison-type integral equations using Banach's fixed point theorem, Inverse Problems 19 (2003) 411-418.
    [3] H.E. Kunze, J.E. Hicken, E.R. Vrscay, Inverse problems for ODEs using contraction maps and suboptimality for the 'collage method', Inverse Problems 20 (2004) 977-991.
    1. E-mail address: smsoltuz@gmail.com.
    2009

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