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On the Banach space $(C([-1,1]), ||\cdot||_\infty ) $ consider the operator given by

$(Tf)(x)= \dfrac{1}{3} \displaystyle\int^1_0txf(t)\ dt + e^x – \dfrac{\pi}{3} $

1) prove that the mapping is a continuous function for all $ f \in (C([-1,1]) $ I.e. that T maps $ (C([-1,1]),||\cdot||_{\infty}) $ to itself.

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2) Show that T is a contraction mapping on $(C([-1,1]),||\cdot||_{\infty})$

3) Lt $f_0 (x) =1 $ Calculate $f_1$ and $ f_2$ where $f_n:= Tf_{n-1}$

This is a past exam question I’ve come across and don’t know how to solve it

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A related problem.

1) Continuity, note that $|x|\leq 1$ and $|t|\leq 1$,

$$ |(Tf)(x)-(Tg)(x) |\leq \dfrac{1}{3} \displaystyle\int^1_0 |tx||f(t)-g(t)|\ dt \leq \int^1_0 |f(t)-g(t)|\ dt $$

$$ \implies \sup|(Tf)(x)-(Tg)(x) | \leq \frac{1}{3} \int^{1}_{0} \sup|f(t)-g(t)|\ dt $$

$$ ||Tf-Tg||_{\infty} \leq \frac{1}{3} ||f-g||_{\infty}<\frac{\epsilon}{3}=\delta . $$

2) The operator is a contraction mapping, since

$$ ||Tf-Tg||_{\infty} \leq \frac{1}{3} ||f-g||_{\infty}. $$

3) Define

$$ f_{n+1}(x)=(T f_n)(x) = \dfrac{1}{3} \displaystyle\int^1_0tx f_n(t)\ dt + e^x – \dfrac{\pi}{3} \longrightarrow (*) $$

$$ \implies f_{1}(x)=(T f_0)(x) = \dfrac{1}{3} \displaystyle\int^1_0tx f_0(t)\ dt + e^x – \dfrac{\pi}{3} $$

$$ \implies f_{1}(x) = \dfrac{1}{3} \displaystyle\int^1_0tx \ dt + e^x – \dfrac{\pi}{3}. $$

To find $f_2$, subs $f_1$ in $(*)$ and carry on the calculations. This technique is known as the Picard iteration. A related problem.

**Added:** Here is $f_2(x)$

$$ f_{2}(x)=(T f_1)(x) = \dfrac{1}{3} \displaystyle\int^1_0tx f_1(t)\ dt + e^x – \dfrac{\pi}{3}. $$

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