Banach Fixed Point Theorem Proof

Question

(Banach Fixed Point Theorem) Let $X$ be a Banach space, $D$ ⊆ $X$ closed and $T$: $D$ → $D$ a contraction, which means that $T$ is Lipschitz continuous with Lipschitz constant $L < 1$:
$$||T(u) − T(v)||_X ≤ L||u − v||_X \hspace{1cm} ∀u, v ∈ D.$$ Then, we have:

1. $T$ has a unique fixed-point $\bar{u} ∈ D$.
2. Let $u_0$ ∈ $D$ be arbitrary and $u_{k+1}$ := $T(u_k), \hspace{1cm}$ $k = 0, 1, . . . ⇒ u_k$ → $u$
3. $||\bar{u} − u_k|| ≤ L||\bar{u} − u_{k−1}||$ $\hspace{1cm}$ $(k ≥ 1)$

$\textbf{To Prove}$,
a) $||\bar{u}-u_k|| \leq \frac {L^k}{1-L}||T(u_0) - u_0||\hspace{1cm} (k ≥ 1)$

b)$||\bar{u}-u_k|| \leq \frac {L}{1-L}||u_k - u_{k-1}||\hspace{1cm} (k ≥ 1)$.

So far I have proved by induction that,
$||u_{k+1} - u_k||≤ L^k||Tu_0 - u_0||$

Now, by considering $m<n$, I proved that,
$||u_n - u_m|| ≤ \frac{L^m}{1-L}||Tu_0-u_0||$
So can I prove a) by saying let, n= ∞ and m = k?,

also how to I go from here to b)