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I am solving the following problem and its parts.

Let (C[0,1],$d_\infty$) be the metric space of continuous functions on [0,1] where the distance function is defined by

$d_\infty(f,g)=\sup_{x∈[0,1]}|f(x)−g(x)|. $

Let $T : (C[0, 1], d_\infty)\to (C[0, 1],d_\infty$) be defined by

$(Tf)(x)=\int_0^xf(t)dt$

Prove that:

T is not a contraction, i.e. there does not exist 0 < K < 1 such that $d_\infty(T f, T g)\leq K · d_\infty(f, g)$ holds for any $f,g ∈ C[0,1]$.

Ive tried an example but I haven't really got anywhere. My work is as follows:

$$f'(t)=\int_0^xtdt=\frac{x^2}{2}=F(x)-F(0)$$ so we pick $t$ & $t^2$ such that $d$($t$,$t^2$)=sup[$t$-$t^2$] with max value = $\frac{1}{4}$ and $d$(T$t$,T$t^2$)=sup[$\frac{t^2}{2}-\frac{t^3}{3}$] with max value =.167

2 Answers2

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Take $f \equiv 1$ and $g\equiv 0$. Then, $$d_\infty(Tf,Tg) = 1 = d_\infty(f,g).$$

Gary Moon
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Let $E:=\mathcal{C}^0([a,b],d_\infty)$ for any $-\infty<a<b<+\infty$. Then $(E,d_\infty)$ is a Banach space, so assuming by contradiction that $T$ is a contraction, there exists an unique fixed point $f$ for $T$ by Banach's fixed point theorem.

Take $\alpha>0$ and let $g_\alpha\in E$ be defined by $g_\alpha(x):=\mathrm{e}^{\alpha x}$. Then, as $Tf=f$, we get: $$d_\infty(f-Tg_\alpha)=d_\infty(Tf-Tg_\alpha)<d_\infty(f-g_\alpha).$$ By continuity in $\alpha>0$ of the LHS and the RHS above, and by boundedness of $f$ over $[0,1]$, we have $$d_\infty(f-Tg_\alpha)=\sup_{x\in[0,1]}\left|f(x)-\frac{\mathrm{e}^{\alpha x}}{\alpha}\right|\underset{\alpha\to0^+}{\longrightarrow}+\infty$$ while $$d_\infty(f-g_\alpha)=\sup_{x\in[0,1]}\left|f(x)-\mathrm{e}^{\alpha x}\right|\underset{\alpha\to0^+}{\longrightarrow}\sup_{x\in[0,1]}\left|f(x)-1\right|<+\infty,$$ a contradiction.

The conclusion is therefore that $T$ can not be a contraction on $E$.

EDIT: In fact, $T$ has an unique fixed point which is the trivial function $f=0_E$, see there: Why $f(x)=e^x$ is a fixed point of $T$? The above argument can be adapted to situations where the fixed point, if it exists, is not trivial.

Nicolas
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