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I have no idea where to even start, i have never dealt with question like this before, any direction you can give me would be greatly appreciated.

greg
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    Set $tx=a$,then $x\partial {t}=\partial {a}$,do the replacements and then derivate the relation and find $f$ – Haha Nov 03 '13 at 20:28

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Hint: Consider the operator

$$T \colon C([0,1]) \to C([0,1]);\quad T(f) \colon x \mapsto x^2 + \int_0^1 t^2f(tx)\,dt.$$

Show that this operator satisfies the hypotheses of Banach's fixed point theorem.

Daniel Fischer
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  • My partial result is

    |T(f)-T(g)|= int(t^2(f(tx)-g(tx))

    is that ok so far?\

     – greg Nov 03 '13 at 20:30
  • That's a good start. Now estimate $\lvert f(tx) - g(tx)\rvert$. – Daniel Fischer Nov 03 '13 at 20:32
  • Won't this be dependent on the norm i choose? However, i am geussing that |f(tx)-g(tx)| <= norm|f-g|. How am i to know which norm to choose? – greg Nov 03 '13 at 20:34
  • Actually, isnt f(tx) at most 1 and g(tx) at most 1? So, wouldnt that mean that |f(tx)- g(tx)| <= 1 – greg Nov 03 '13 at 20:36
  • Unless explicitly stated otherwise, always use the $\sup$-norm, $\lVert f\rVert = \sup {\lvert f(x)\rvert : x \in [a,b]}$ on $C([a,b])$. – Daniel Fischer Nov 03 '13 at 20:36
  • Thanks, so that means that i should get T(f)-T(g) <= norm(f-g)int(t^2) = norm(f-g)9 – greg Nov 03 '13 at 20:38
  • @greg No, $f(tx)$ can be arbitrarily large, $f$ can be any continuous function. But $\lvert f(tx) - g(tx)\rvert$ is bounded by the norm of $f-g$. – Daniel Fischer Nov 03 '13 at 20:39
  • @greg $\int_0^1 t^2,dt =$? Not $9$. – Daniel Fischer Nov 03 '13 at 20:40
  • OMG LOL! yeah i understood my mistake. its 1/3 so then taking the max t value. i think we should obtain, |T(f)-T(g)| = (1\3)*norm(f-g) which means that it is a contraction and so now we can use the banach fixed point theorem. So that we obtain, f(t) = x^2 + int t^2f(tx)dt and this is enough to prove the existence. Is this correct? – greg Nov 03 '13 at 20:44
  • That's correct. It proves not only existence, also uniqueness. – Daniel Fischer Nov 03 '13 at 20:45
  • Thanks alot, i really appreciate it. I hate it when i make silly mistakes! – greg Nov 03 '13 at 20:46
  • @greg Everybody does (make silly mistakes, and hate making them). – Daniel Fischer Nov 03 '13 at 20:48