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Let $T$ be the following operator on $C[0,1]$: $$(Tu)(t) = u(0) + \lambda\int_0^t u(\tau)d\tau$$ where $\lambda \in (-1,1) \subset \mathbb{R}$. Then I need to show $T$ is a contraction. So I need

$$||Tu - Tv|| \leq c||u-v||,$$

equivalently,

$$\max_{0\leq t\leq 1}\left|u(0) - v(0) + \lambda\int_0^t (u-v)(\tau) d\tau\right| \leq c \max_{0\leq t \leq 1} |(u-v)(t)|,$$

but I don't know what to do with the left-hand side, and I don't know if there's a convenient constant $c$ to pick ahead of time or if the right $c$ will fall out of the proof.

Nick
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1 Answers1

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You can show that $c = 1+|\lambda|$ and that this cannot be improved. The last statement first: if $u = 1$ for $\lambda \geq 0$ ($u = -1$ for $\lambda <0$) and $v = 0$ the the RHS is $c$ and the LHS is $1+\lambda$, so $c\geq 1+\lambda$. To show the converse, recall that $|a +b|\leq |a| + |b|$ and that $|\int f \mathrm dt |\leq \int |f|\mathrm d\tau$.

SBF
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  • As a follow-up, and I'm sorry if this is a silly question, but does that mean that the mapping is not a contraction unless $\lambda$ < 0? – Nick Jul 24 '14 at 14:28
  • I should have put $c = 1+ |\lambda|$ there: if $\lambda<0$ then take $u = -1$. This mapping is never a contraction. – SBF Jul 24 '14 at 15:29