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Show that T is a contraction on a metric space

  1. Apr 3, 2014 #1
    1. The problem statement, all variables and given/known data

    Consider the metric space [itex](R^{n}, d_{∞})[/itex], where if [itex]\underline{x}=(x_{1}, x_{2}, x_{3},...,x_{n})[/itex] and [itex]\underline{y}=(y_{1}, y_{2}, y_{3},...,y_{n})[/itex] we define

    [itex]d_{∞}(\underline{x},\underline{y}) = max_{i=1,2,3...,n} |x_{i} - y_{i}|[/itex]

    Assume that [itex](R^{n}, d_{∞})[/itex] is complete.

    Let [itex]T: R^{n} → R^{n}[/itex] be the mapping given by [itex]T\underline{x}=C\underline{x} + \underline{b}[/itex]. If C has the following property

    [itex]∑_{j}|C_{ij}| < 1[/itex], for [itex]i=1,2,3,....,n[/itex]

    show that [itex]T: R^{n} → R^{n}[/itex] is a contraction on [itex](R^{n}, d_{∞})[/itex]


    2. Relevant equations

    [itex]∑_{j}|C_{ij}| < 1[/itex], for [itex]i=1,2,3,....,n[/itex]

    Therefore the sum of every row of the matrix C is less than 1.

    T is a contraction on [itex](R^{n}, d_{∞})[/itex] if
    [itex]d_{∞}(T(\underline{x}),T(\underline{y}) )≤ Kd_{∞}(\underline{x}, \underline{y}), 0≤K<1[/itex]

    3. The attempt at a solution

    For d(T(y), T(x)) I get

    [itex]max_{i=1,2,3,...,n} |C\underline{x} + \underline{b} - C\underline{y} - \underline{b}|[/itex]

    [itex]=max_{i=1,2,3,...,n} |∑_{j} c_{ij}x{j} - c_{ij}y_{j}|[/itex]

    The best that I can get from this is that [itex]∑_{j} c_{ij}x{j}[/itex] is less than the maximum value of x. But I don't think that's particularly useful and I'm not sure what else to do.
     
  2. jcsd
  3. Apr 3, 2014 #2

    HallsofIvy

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    Staff Emeritus
    Science Advisor

    You have already said that "Therefore the sum of every row of the matrix C is less than 1." So [itex]\sum_j c_{ij}< 1[/itex]. What does that tell you about [itex]\sum_j c_{ij}x_j[/itex]?
     
  4. Apr 4, 2014 #3
    Thanks for the reply, but I have the solution now. The key step was (for anyone else who might be having trouble with a question similar to this) to recognize that

    [itex]|∑_{j} c_{ij}x_{j} - c_{ij}y_{j}| ≤ ∑_{j} |c_{ij}x_{j} - c_{ij}y_{j}|[/itex]

    and then to note how [itex]|x_{j} - y_{j}| ≤ max_{i=1,2,3,...,n} |x_{i} - y_{i}|[/itex]

    Then everything is straight forward from there.
     
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