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Please help me with the following problem: Let $\gamma\in (0,1)$ and $a<0<b$, $-a<b$, and $x\geq0$. Solve the following equation $$f(x)=\frac{\gamma}{b-a}\int_{\max(a+x,0)}^{b+x}f(y)dy$$ I try to solve this using Banach contraction theorem. Unfortunately if I put $f_0=1$ I get $f_1$ with two different expression over intervals $[0,-a]$ and $[-a,\infty)$. Then $f_2$ will be define over three intervals $[0,-a], [-a,-2a],[-2a,\infty)$ Besides that one can see that the degree of the polynomials will be increasing in some intervals. Do you have any idea how can I force with this?

copper.hat
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Darek
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    It looks like a contraction to me (using $|f| = \sup_x |f(x)|$), and $f=0$ solves the equation, hence must be the solution. – copper.hat Dec 06 '14 at 23:05
  • Thank you, but there is still problem since : $$|f|\leq\frac{\gamma}{b-a}(b+x-\max(a+x,0))|f|\leq \gamma |f|$$ and everything works great when we seek a solution in uniformly bounded function defined on a half line. Are there still a trivial solution if I want to find any continuous function? – Darek Dec 06 '14 at 23:15
  • Well, the zero solution will always be a solution. Did you mean for the integral to be interpreted as $-\int_{b+x}^{\max(a+x,0)}$ or zero when $\max(a+x,0)>b+x$? – copper.hat Dec 07 '14 at 00:09
  • From the assumption $b>0$, $a<0<b$ we always have $$\max(a+x,0)<b+x$$ since $b>0, x\geq 0$ and $a+x<b+x$ – Darek Dec 07 '14 at 00:18
  • So $f$ is defined on $[a,\infty)$ and the above equation holds for $x \in [0,\infty)$? (That is, the behaviour of $f$ on $[a,0)$ is not specified by the above equation?) – copper.hat Dec 07 '14 at 00:20
  • $f$ is defined on $[0,\infty)$ – Darek Dec 07 '14 at 00:28
  • Right! I was confused. – copper.hat Dec 07 '14 at 00:40

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