Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

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
Know someone interested in this topic? Share this thread via Reddit, Google+, Twitter, or Facebook

Have something to add?

Similar Discussions: Schröder's equation and functional analysis
  1. Functional analysis (Replies: 5)

  2. Functional analysis (Replies: 1)