# prove the map is a contraction

Let $L$ be a fixed positive real number. Let $K:[0,T] \times \mathbb{R} \rightarrow \mathbb{R}$ be continuous and satisfy the lipschitz condition

$|K(s,x)-K(s,y)| \leq L|x-y|$

for all $s \in [0,T]$ and $x,y \in \mathbb{R}$. Consider the space $E$ of all continuous real valued functions on $[0,T]$ equipped with the norm

$||g||= \max_{0 \leq t \leq T} e^{-Lt} |g(t)|$.

Prove that the map $F:E \rightarrow E$ defined by

$F(g)(t)=\sin(t) + \int_0^T K(s,g(s)) ds$

is a contraction in $(E,||\cdot||)$.

Help me out please. There is so much packed in the question I don't even know where to start. Thanks.

A slight correction, you'll need to define $$(Fg)(t) = \sin(t) +\int^t_{0} K(s,g(s)) ds;$$ that is, the upper bound needs to be $t$, not $T$.
Take $g,h \in E$. We see \begin{align*} \lvert (Fg)(t)-(Fh)(t) \rvert &= \left \lvert \int^t_{0} [K(s,g(s)) - K(s,h(s))] ds\right \rvert \\ &\le \int^t_0 \lvert K(s,g(s)) - K(s,h(s)) \rvert ds \\ &\le \int^t_0 L\lvert g(s) - h(s) \rvert ds\\ &=L \int^t_0 e^{Ls} e^{-Ls} \lvert g(s) - h(s) \rvert ds. \end{align*} However, for each $s$, we have $e^{-Ls} \lvert g(s) - h(s) \rvert \le \max_{0\le s\le T} e^{-Ls} \lvert g(s) - h(s) \rvert = \| g- h \|.$ Thus \begin{align*} \lvert (Fg)(t)-(Fh)(t) \rvert &\le L \| g - h\| \int^t_0 e^{Ls} ds \\ &= (e^{Lt} - 1) \| g- h\|. \end{align*} Then $$e^{-Lt} \lvert (Fg)(t)-(Fh)(t) \rvert \le (1 - e^{-Lt}) \| g - h\| \le (1-e^{-LT})\|g - h\|.$$ Passing to the maximum, we see $$\| Fg - Fh \| \le (1-e^{-LT}) \|g - h\|.$$ But $0 \le 1-e^{-LT} < 1$ so this shows that $F$ is a contraction on $E$.
• In the Q we have $F(g)(t)=\sin t +...$, not $F(g)(t)=\sin x \cdot \; ...$, which does look like another mistake, but I have not examined it. May 8, 2016 at 18:01
• That was a mistake on my part, I forgot a $+$ sign. May 8, 2016 at 22:25