0
$\begingroup$

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.

$\endgroup$

1 Answer 1

4
$\begingroup$

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 $E$.

$\endgroup$
4
  • $\begingroup$ 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. $\endgroup$ May 8, 2016 at 18:01
  • $\begingroup$ That was a mistake on my part, I forgot a $+$ sign. $\endgroup$
    – User8128
    May 8, 2016 at 22:25
  • $\begingroup$ @User8128 Thank you for a detailed explanation. $\endgroup$
    – chris
    May 9, 2016 at 13:58
  • $\begingroup$ @User8128 Thank you for a detailed explanation. $\endgroup$
    – chris
    May 9, 2016 at 14:03

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.