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Prove that there is a unique bounded continuous real-valued function f on $X=\{ t \in \mathbb{R} : t\geq 0 \}$ , equipped with the sup metric, such that $$ f(s) = 1+ \int_{0}^s e^{-t^2} f(st)dt \quad for\quad s\in X $$

I define a new function $ T :C([0, \infty) \rightarrow C([0, \infty) ) $ by $$ Tf(s)= 1+ \int_{0}^s e^{-t^2} f(st)dt$$ where $C([0, \infty)$ is the space of continuous functions on $ [0, \infty).$

If $T$ is a contraction then by Banach Fixed Point thm, it has a fixed point and it is the unique solution of the above integral equation.

First question is that how do I show that $T$ is a well defined function, meaning that it takes continuous function to continuous function.

Second question is that how do I show $T$ is a contraction ? To answer the second question, I tried to consider $$ |Tf(s) - Th(s)| \leq s d(f,h) $$ Taking the $\sup $ of both sides over $s$, RHS blows off. Does it mean that all I have done is useless?

Esat Koç
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  • The operator $T$ you have defined is not linear, as if $c\in\mathbb R$, then $$T(cf(s)) = 1 + \int_0^s e^{-t^2}cf(st),\mathsf d t$$ while $$cTf(s) = c\left( 1 + \int_0^s e^{-t^2}f(st),\mathsf d t\right),$$ which in general are not equal. – Math1000 Dec 24 '16 at 20:56
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    $T$ is a contraction since for all $s\in[0,\infty)$ we have

    $\left\lvert (T(f))(s)-(T(g))(s)\right\rvert\le\left\lvert\left\lvert f-g\right\rvert\right\rvert_{C^{0}\left([0,\infty)\right)}\int_{0}^{\infty}e^{-t^{2}}dt\le\sqrt{\pi}d\left(f,g\right)$

     – user71352 Dec 24 '16 at 21:15
  • Is $\sqrt{\pi} < 1 $? – Esat Koç Dec 24 '16 at 22:16
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    @Quantes Sorry, forgot the constant had to be less than $1$. More careful estimation gives $\int_{0}^{\infty}e^{-t^{2}}dt=\frac{\sqrt{\pi}}{2}$ which should be less than $1$. – user71352 Dec 24 '16 at 22:20
  • @Math1000 I think linearity is not an issue here. – Esat Koç Dec 25 '16 at 07:23

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