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Schröder's equation and functional analysis

  1. Feb 3, 2004 #1
    schröder's equation is a functional equation. let's assume A is a subset of the real numbers and g maps A to itself. the goal is to find a nonzero (invertible, if possible) function f and a real number r such that [tex]f\circ g=rf[/tex].

    motivation: if there is an invertible f, then the nth iterate of g is given by [tex]g^{n}=f^{-1}\left( r^{n}f\right) [/tex]. this can lead to understanding the dynamics of g better.

    let V be the set of functions from A to R, the set of real numbers.

    let S(r,g) denote [tex]\left\{ f\in V:f\circ g=rf\right\} [/tex]. S(r,g) is a subspace of V.

    define a map k(r,g) from V to V by the following:
    [tex]k\left( r,g\right) \left( f\right) =f\circ g-rf[/tex].

    k(r,g) is a linear operator on V with kernel S(r,g).

    putting in this context, are there any theorems (with more assumptions perhaps) that
    1. give one at least a sense of when the kernel of a map is not {0}
    2. can find a basis for the kernel of a map explicitly
    3. can approximate solutions to k(r,g)(f)=0?

    any thoughts would be helpful. if you want to get real specific, assume A=[0,1] and g is either 2^(x-1) or x^2.

    as an example, [tex]x+\frac{b}{a-1}\in S\left( a,ax+b\right) [/tex] , abusing the notation a little to let ax+b represent the function g such that g(x)=ax+b.

    i know that S(r,g)=fix(k(r-1,g)), where fix(h) is the set of fixed points of h. this opens me up to all the fixed point theorems in functional analysis except that i already know it has a fixed point, namely the zero function. what will be interesting is uniqueness fixed point theorems to show that 0 is the only fixed point. however, under one or two more assumptions on g and A, for example, i can turn my attention to bounded functions and use the sup norm and perhaps talk about contractions. all i can do in that case, and here it's critical that g maps A to A, is show that the norm of k(r,g) is at most 1+|r| which doesn't prove it's a contraction. it would be extremely helpful if anyone were able to tell me what the norm of this operator k(r,g) is. in other words, the sup of ||k(r,g)(f)|| where ||f||=1. then i can tell for which r k is a contraction.

    also, is it true that if an operator is NOT a contraction then it doesn't have a unique fixed point or could it not be a contraction and still have a unique fixed point?

    any thoughts would be helpful.
  2. jcsd
  3. Feb 3, 2004 #2

    i was also thinking of using some kind of pseudoinverse.

    if i can somehow define a transpose of k(r,g), [tex]k^{T}\left( r,g\right) [/tex], then one form of its pseudoinverse [tex]k^{+}[/tex]would be maybe
    [tex]k^{+}\left( r,g\right) =\left( k^{T}\left( r,g\right) \circ k\left( r,g\right) \right) ^{-1}k^{T}\left( r,g\right) [/tex] and so the solution to [tex]k\left( r,g\right) \left( f\right) =0[/tex] could be approximated by [tex]k^{+}\left( r,g\right) \left( 0\right) [/tex].

    two things:
    1. how would i define [tex]k^{T}\left( r,g\right) [/tex] in that my space is not equipped, as far as i know, an inner product that gives rise to the sup norm and
    2. ensuring that [tex]k^{T}\left( r,g\right) \circ k\left( r,g\right) [/tex] is invertible?
  4. Feb 4, 2004 #3
    i'm suspecting that [tex]\left\| k\left( r,g\right) \right\| =\min \left\{ 1+\left| r\right| ,\left\| g\right\| +\left| r\right| \right\} [/tex] but i'm not sure how to prove it.

    in general, [tex]\left\| k\left( r,g\right) \right\| =\sup_{f}\frac{\left\| k\left( r,g\right) \left( f\right) \right\| }{\left\| f\right\| }[/tex].

    it should be something that depends on g and r.
    Last edited: Feb 4, 2004
  5. Feb 5, 2004 #4
    if [tex]V[/tex] is the set of all bounded continuous functions mapping [tex]A[/tex] to [tex]R[/tex], can someone exhibit an orthonormal basis [tex]B[/tex] for V? or at least confirm my suspicion that the set of power functions (restricted to A--let's say A is the unit interval even) is dense in V? if so, i can use something like grahm-schmit to find an orthonormal basis for V, right?

    is it true that [tex]\left\| k\left( r,g\right) \right\| :=\sup_{f\in V}\frac{\left\| k\left( r,g\right) \left( f\right) \right\| }{\left\| f\right\| }=\sum_{\phi \in B}\left\| k\left( r,g\right) \left( \phi \right) \right\| [/tex]?
  6. Feb 6, 2004 #5
    possible method of attack

    the projection theorem: let [tex]X[/tex] be a Hilbert space, [tex]K[/tex] a closed convex subset, and [tex]x\in X[/tex]. there there is a unique [tex]x^{\prime }\in X[/tex] such that [tex]\left\| x-x^{\prime }\right\| =\inf_{y\in K}\left\| x-y\right\| [/tex].

    i want to take X to be V, the set of continuous real-valued functions from [tex][0,1][/tex] to [tex]R[/tex]. i'm not sure if this is a hilbert space. i want to take [tex]K[/tex] to be [tex]S(r,g)=ker(k(r,g))[/tex]; i'm fairly sure that's closed but i'm not sure it's convex.

    this unique [tex]x^{\prime}[/tex] is often denoted [tex]P_{K}x[/tex] and referred to as the projection of [tex]x[/tex] onto [tex]K[/tex].

    then what i want to do is take a seed function and apply this projection to it to get a function in the kernel and be able to specify when this projection is nonzero. i could at least maybe squeeze an existence/uniqueness theorem out of this though i'm not sure how a calculation would be done.
    Last edited: Feb 6, 2004
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