# Banach Space continuous function

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.

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

#### Solutions Collecting From Web of "Banach Space continuous function"

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 $f_2(x)$
$$f_{2}(x)=(T f_1)(x) = \dfrac{1}{3} \displaystyle\int^1_0tx f_1(t)\ dt + e^x – \dfrac{\pi}{3}.$$