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I have two half-questions that tie into one another.

Suppose $T$ is an operator on $C([0, 1])$ defined by $$(Tu)(t) = \int_0^t (u(x))^2dx.$$ Show that T is not a contraction on the closed unit ball in $C([0, 1])$, but that it is one on the closed ball of radius $\frac{1}{4}$ in $C([0, 1])$.

Alternatively, for $|\lambda|<1$, $\lambda\in\mathbb{R}$, what makes the operator $T:C([0,1])\to C([0,1])$ defined by $$(Tx)(t) = x(0)+\lambda\int_0^t x(\tau)d\tau,$$ a contraction?

I don't have a handle on contractions just yet, so any help/hints would be greatly appreciated.

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  • $\begingroup$ for the second one which is a linear operator, a contraction for what norm ? if it is for $||.||_{\infty}$ and with $\lambda > 0$ you get that $||x||_{\infty} < 1 \implies ||T x ||_{\infty} < 1+\lambda $ and that the maximum $ ||Tx|| = 1+\lambda$ is attained for $x(t) = 1$ $\endgroup$ – reuns Jan 26 '16 at 17:32
  • $\begingroup$ @user1952009 I wasn't given any norm. Just what I wrote. Is it alright to assume $|\cdot|_\infty$? $\endgroup$ – Desperate Fluffy Jan 26 '16 at 17:44
  • $\begingroup$ an operator can be a contraction for a certain norm but not for another one... $\endgroup$ – reuns Jan 26 '16 at 17:46
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Regarding the first one, $(Tu)(t)={\displaystyle \int_0^t (u(x))^2 dx}$. To show that it is not a contraction, it is easy to construct a counter example where $u(x) = 1$ and $v(x)=0$ and then compare $\lVert Tu-Tv \rVert$ with $\lVert u-v \rVert$. Remember that the norm on $\mathcal{C}([a,b])$ is defined by $\lVert f \rVert = max_{[a,b]} \lvert f(x) \rvert$ and, by definition, $T$ is a contraction if $\lVert Tu-Tv \rVert \leq \alpha \lVert u-v \rVert$, where $\alpha < 1$.

To show that it is a contraction in a closed ball of radius $\frac{1}{4}$, use the fact that $u^2 - v^2 = (u+v)(u-v)$ and work with that when you check the norm for $Tu-Tv$, having in mind that now the norm of $u$ and $v$ are limited to $\frac{1}{4}$. Hope that helps.

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