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Question

Let $I=[0,a]$ and define the norm $||f||_{\lambda}=\sup_I |e^{-\lambda x}f(x)|$ for $f\in C(I)$. Let $\phi:\;\mathbb{R}^2\to\mathbb{R}$ satify $|\phi(x,u)-\phi(y,v)|\leq\rho |u-v|$ for all $x,y,u,v\in\mathbb{R}$ and some $\rho >0$. Define $\tau:\;f\mapsto \int_0^x \phi(t,f(t))\;dt$

I need to find a $\lambda$ such that $\tau$ is a contraction under the norm $||\cdot||_{\lambda}$

Thoughts

I am not too sure how to do this; my first line of thought was:

$$\begin{aligned}||\tau (f)-\tau (g)||_{\lambda} &=\sup_I\Big| e^{-\lambda x} \int_0^x \phi(t,f(t))-\phi(t,g(t))\;dt\Big| \\ &\leq \sup_I e^{-\lambda x} \int_0^x |\phi(t,f(t))-\phi(t,g(t))|\;dt\\ &\leq\sup_I \rho e^{-\lambda x} \int_0^x |f(t)-g(t)|\;dt \end{aligned}$$

But I can't see how to get $\cdots \leq \alpha\sup_I |e^{-\lambda x}(f(x)-g(x))|$ for some $\alpha<1 $ and some $\lambda$ from this. Any help would be appreciated.

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    $\begingroup$ Your assumption $|\phi(x,u)-\phi(y,v)|\leq\rho |u-v|$ looks strange; it implies $\phi$ does not depend on the first argument. $\endgroup$ – user147263 Nov 25 '14 at 5:13
  • $\begingroup$ No, it just says that the modulus of the difference doesn't depend on the first component $\endgroup$ – Joe May 22 at 21:37
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In general you can prove that if $w : I := [a,b] \longmapsto \mathbb{R}$ is strictly positive and continuos, it defines a norm equivalent to $\|f\|_{\infty}$ on $C^{0}(I)$, defined as $\|f\|_{w} := \sup\limits_{t \in I} \left\lbrace w(t) |f(t)|\right\rbrace$

Defining $w(t) := e^{-2L |t-t_{0}|}$ we define the norm $\|f\|_{w} := \sup\limits_{t \in I} \left\lbrace e^{-2L |t-t_{0}|} |f(t)|\right\rbrace$

Let's prove that with this choice, $T$ is a contraction on $I = [t_{0}-\delta,t_{0}+\delta]$ :

$$e^{-2L|t-t_{0}|} |\int_{t_{0}}^{t}[f(s,v(s))-f(s,u(s))]ds| \leq e^{-2L|t-t_{0}|} |\int_{t_{0}}^{t}|f(s,v(s))-f(s,u(s))|ds| \leq$$

$$e^{-2L|t-t_{0}|} |\int_{t_{0}}^{t} L |v(s)-u(s)|ds| \leq Le^{-2L|t-t_{0}|} |\int_{t_{0}}^{t} \|v-u\|_{w}e^{2L|s-t_{0}|} ds|$$

$$ = \frac{\|v-u\|_{w}}{2}(1-e^{-2L|t-t_{0}|}) \leq \frac{1}{2}\|v-u\|_{w}$$

So, taking the supremum for $t \in I$ we get exactly

$$\|T(v)(t)-T(u)(t)\|_{w} \leq \frac{1}{2}\|v-u\|_{w}$$

In other words $T$ is a contraction $\hspace{0.2cm} \Box.$

In you case $I$ becomes $[0,a]$, $T$ becomes $\tau$, $L$ becomes $\rho$ and $\lambda$ is the constant chosen in the definition of the weighted norm.

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Well the supremum can split under multiplication, therefore if $f$ and $g$ were bounded in $I$, then you could call the supremum for the difference between $f$ and $g$ out of the integral, because the supremum of that difference remains constant in $I$ and the supremum and after done that you could join both and constrain the difference of $I$ on the delta and choose the delta you need.

Maybe not the cleanest but I do not see something wrong, but still, a person who can make mistakes, If I made some, please feel free to do all the corrections needed...

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