I understand that in order to prove the existence and uniqueness of a solution to a first order non-linear ODE, we first convert it into an integral equation:
$y_{n+1} := Ty_{n} = y_{0} + \int_{0}^{t}f(s, y_{n})ds,$
and then show that the resulting operator is a contraction, provided that the right-hand side function is Lipschitz:
$|Ty_{n+1} - Ty_{n}| \le L \tau |y_{n+1} - y_{n}|$,
where $L$ is the Lipschitz constant and $t \in [0,\tau]$. The operator is a contraction if $L \tau<1$.
If $L \tau>1$, we define the norm $|| y ||_{\lambda} :=\sup_{0 \le t \le \tau} e^{-\lambda t}||y||$ with $\lambda>L$, which can be shown to be equivalent to $| \cdot |$. Then
$||Ty_{n+1} - Ty_{n}||_{\lambda} \leq \sup_{0 \le t \le \tau} e^{-\lambda t} ||y_{n+1}-y_{n}||_{\lambda} \frac{L}{\lambda}(e^{\lambda t}-1) \leq \frac{L}{\lambda}||y_{n+1}-y_{n}||_{\lambda} < ||y_{n+1}-y_{n}||_{\lambda},$
i.e. the Lipschitz condition is enough for $T$ to be a contraction; we no longer need that $L \tau <1$.
If we look at the distance between Picard iterates in the $|\cdot|$ norm, we have
$|y_{n+1}-y_{n}|=|Ty_{n}-Ty_{n-1}| \leq (L\tau)^{n}|y_{n-1}-y_{n-2}| $
and
$|y_{j}-y_{i}| \leq \sum_{k=i}^{j-1}|y_{k+1}-y_{k}| \leq \sum_{k=i}^{j-1} (L \tau)^{k} |y_{k+1}-y_{k}| $.
The sequence $y_{n}$ is Cauchy w.r.t the norm $|\cdot |$ iff $L \tau < 1$, because otherwise the geometric series above diverges.
Then how is it possible that even if $L \tau > 1$, we can still show the existence and uniqueness of solution, which is given by the limit of Picard's iterates as $n \rightarrow \infty$? Doesn't the fact that $y_{n}$ is not Cauchy in $| \cdot|$ for $L \tau >1 $ imply that it is also not Cauchy in $|| \cdot ||$, since the norms are equivalent?
I am missing something here, and I would greatly appreciate if someone can tell me where in my reasoning I'm going wrong.