I want to demonstrate that this transformation is a contraction: $$Tx(t)=\frac{1}{2}\left (x^2(t) + 1 -t^2 \right )$$ In the closed set $S$: $$S=\{x\in C([-1,1; \mathbb{R} ]):\Vert x\Vert_0 \leq 1\}$$ I'm supposed to find a $\lambda$, s.t. $\vert \lambda \vert \leq 1$
So, I got this: $$\Vert Tx_n - x_n\Vert = \Vert x_{n+1} - x_{n}\Vert = \Vert Tx_n - Tx_{n-1}\Vert$$
$$\Vert Tx_n - Tx_{n-1}\Vert = \left \Vert \frac{1}{2}\left (x_{n}^2(t) + 1 -t^2 \right ) - \frac{1}{2}\left (x_{n-1}^2(t) + 1 -t^2 \right ) \right \Vert$$ $$=\frac{1}{2}\left \Vert \left (x_{n}(t) + x_{n-1}(t)\right )\left (x_{n}(t) - x_{n-1}(t)\right )\right \Vert$$ $$\leq\frac{1}{2}\left \Vert x_{n}(t) \Vert + \Vert x_{n-1}(t) \Vert \Vert x_{n}(t) - x_{n-1}(t)\right \Vert$$
In the next step I have some doubts, since my $S$ set is defining that my $\Vert x\Vert \leq 1$, could I state that if I sum two of those elements I shall get always $\Vert x_n + x_m \Vert \leq 2 $
By considering the later I get,
$$\leq\frac{2}{2}\left \Vert x_{n}(t) \Vert \Vert x_{n}(t) - x_{n-1}(t)\right \Vert$$
Or,
$$\leq\frac{2}{2}\left \Vert x_{n-1}(t) \Vert \Vert x_{n}(t) - x_{n-1}(t)\right \Vert$$
By the contraction mapping theorem, $T$ is a contraction because for $\vert \lambda \vert \leq 1$:
$$\Vert x_{n+1} - x_{n}\Vert = \lambda^n\Vert x_{1} - x_{0}\Vert$$
So in my expression,
$$\leq\frac{2}{2}\left \Vert x_{n-1}(t) \Vert \Vert x_{n}(t) - x_{n-1}(t)\right \Vert = \left \Vert x_{n-1}(t) \Vert \Vert x_{n}(t) - x_{n-1}(t)\right \Vert$$
$$\Vert x_{n+1} - x_{n}\Vert \leq \left \Vert x_{n-1}(t) \Vert \Vert x_{n}(t) - x_{n-1}(t)\right \Vert$$
Is $ \Vert x_{n-1}(t)\Vert$ my first $\lambda_1$ (I'm saying this because it is always less than 1)?