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[SOLVED] two topology questions

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Poirot

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Feb 15, 2012
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1) Define a function S on the set of polynomials X by S(a+bx+.....)=max{|a|,|b|,.......}. define the distance (metric) on X by d(p,q)=S(p-q)

let f: X -> X Prove that if f(p(x))=1+1/2$\int_{0}\,^{x} p(t)\,dt$, then f is a contraction mapping with no fixed points.

2) Let the function space F = C ([0,1]; [0, 1]) be endowed with the sup metric d. For every f in F define $g(x)=sinx.|x^0.5-f(x)|$. Show that there is exactly one f in F such that f=g. You make take it that F is complete.

I know what I need to do for 1) but I can't seem to simplify it. For 2) I don't know.
 

Opalg

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Feb 7, 2012
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1) Define a function S on the set of polynomials X by S(a+bx+.....)=max{|a|,|b|,.......}. define the distance (metric) on X by d(p,q)=S(p-q)

let f: X -> X Prove that if f(p(x))=1+1/2$\int_{0}\,^{x} p(t)\,dt$, then f is a contraction mapping with no fixed points.
Given two polynomials $p(x) = a_0+a_1x+a_2x^2+\ldots$ and $q(x) = b_0+b_1x+b_2x^2+\ldots$ (only finitely many nonzero coefficients in each case), find explicit expressions for $f(p(x))$ and $f(q(x))$ by integrating, form the difference $f(p(x)) - f(q(x))$, write down $d(f(p(x)) , f(q(x)))$ and show that it is less than a suitable multiple of $d(p(x),q(x)).$

If $p(x)$ is a fixed point then $\displaystyle p(x) = 1 + \frac12\int_0^xp(t)\,dt$. Differentiate, to get $p'(x) = \frac12p(x)$. Solve that differential equation and you will find that the solution is not a polynomial. So there is no fixed point in $X$.

2) Let the function space F = C ([0,1]; [0,1]) be endowed with the sup metric d. For every f in F define $g(x)=\sin x.|x^{0.5}-f(x)|$. Show that there is exactly one f in F such that f=g. You make take it that F is complete.

For 2) I don't know.
You have to show that the map $f \to g$ is a contraction map. The result will then follow from the contraction mapping theorem.

Given $f_1,\; f_2 \in F$, with images $g_1,\;g_2$, $$d(g_1,g_2) = \sup_{0\leqslant x\leqslant1}\bigl\{\sin x \bigl(|x^{0.5}-f_1(x)| - |x^{0.5}-f_2(x)|\bigr)\bigr\}.$$ You need to show that is less than some suitable multiple of $\displaystyle\sup_{0\leqslant x\leqslant1}|f_1(x) - f_2(x)|.$
 
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Poirot

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Feb 15, 2012
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Very helpful thanks. For 2), I will be honest. I put ' I don't know' because I have tried showing it's a contraction mapping but I don't see how to get to desired form and so concluded that it wasn't right. I used traingle inequality and |sinx|<=1, but since 0<x<1, sqrt(x)>x and I can't manipulate it.
 

Opalg

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Feb 7, 2012
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Very helpful thanks. For 2), I will be honest. I put ' I don't know' because I have tried showing it's a contraction mapping but I don't see how to get to desired form and so concluded that it wasn't right. I used traingle inequality and |sinx|<=1, but since 0<x<1, sqrt(x)>x and I can't manipulate it.
Just for the record, you need to use the triangle inequality in the form $\bigl||z|-|w|\bigr| \leqslant |z-w|$. It follows that $\bigl||x^{0.5}-f_1(x)| - |x^{0.5}-f_2(x)|\bigr| \leqslant |(x^{0.5}-f_1(x)) - (x^{0.5}-f_2(x))| = |f_2(x) - f_1(x)|$. Then $d(g_1,g_2) \leqslant \displaystyle\sup_{0\leqslant x\leqslant1}\sin x|f_1(x) - f_2(x)| \leqslant \sin (1) d(f_1,f_2).$ Since $\sin(1)<1$, the mapping $f\to g$ is a contraction map.
 
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Poirot

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Feb 15, 2012
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I have done 1) now. Just wondering about the DE that is obtained p'=0.5p. Clearly 0 satifies this, but 0 is not a fixed point so where, when applying the fundamental theorem of calculus, was it assumed that p is non- zero
 

Opalg

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Feb 7, 2012
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I have done 1) now. Just wondering about the DE that is obtained p'=0.5p. Clearly 0 satifies this, but 0 is not a fixed point so where, when applying the fundamental theorem of calculus, was it assumed that p is non- zero
I neglected to mention the initial condition for the differential equation. If $\displaystyle p(x) = 1 + \tfrac12\int_0^xp(t)\,dt$ then putting $x=0$ you see that $p(0) = 1.$ So $p$ cannot be the zero polynomial.
 
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Poirot

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Feb 15, 2012
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I'm realising these topolgy questions are actually ok as long as one is familiar with the definitions (which I wasn't with contraction mapping), and is not intimidated.
 

Opalg

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Feb 7, 2012
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I'm realising these topolgy questions are actually ok as long as one is familiar with the definitions (which I wasn't with contraction mapping), and is not intimidated.
Knowing the definitions is always a powerful help. (Muscle) (Smile)