# Schröder's equation and functional analysis

1. Feb 3, 2004

### phoenixthoth

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 $$f\circ g=rf$$.

motivation: if there is an invertible f, then the nth iterate of g is given by $$g^{n}=f^{-1}\left( r^{n}f\right)$$. 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 $$\left\{ f\in V:f\circ g=rf\right\}$$. S(r,g) is a subspace of V.

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

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, $$x+\frac{b}{a-1}\in S\left( a,ax+b\right)$$ , 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?

2. Feb 3, 2004

### phoenixthoth

pseudoinverse

i was also thinking of using some kind of pseudoinverse.

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

two things:
1. how would i define $$k^{T}\left( r,g\right)$$ 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 $$k^{T}\left( r,g\right) \circ k\left( r,g\right)$$ is invertible?

3. Feb 4, 2004

### phoenixthoth

i'm suspecting that $$\left\| k\left( r,g\right) \right\| =\min \left\{ 1+\left| r\right| ,\left\| g\right\| +\left| r\right| \right\}$$ but i'm not sure how to prove it.

in general, $$\left\| k\left( r,g\right) \right\| =\sup_{f}\frac{\left\| k\left( r,g\right) \left( f\right) \right\| }{\left\| f\right\| }$$.

it should be something that depends on g and r.

Last edited: Feb 4, 2004
4. Feb 5, 2004

### phoenixthoth

if $$V$$ is the set of all bounded continuous functions mapping $$A$$ to $$R$$, can someone exhibit an orthonormal basis $$B$$ 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 $$\left\| k\left( r,g\right) \right\| :=\sup_{f\in V}\frac{\left\| k\left( r,g\right) \left( f\right) \right\| }{\left\| f\right\| }=\sum_{\phi \in B}\left\| k\left( r,g\right) \left( \phi \right) \right\|$$?

5. Feb 6, 2004

### phoenixthoth

possible method of attack

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

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

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

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