Let $A \in \mathbb R^{n,n} $ invertible, $b \in \mathbb R^n$ and suppose to solve $Ax=b$ with $LR$ decomposition with pivoting. Why does $\Vert L \Vert_\infty \le n$ always hold?
$$\Vert L\Vert_\infty=\max_{1\le i\le n}\sum_{j=1}^n \vert l_{ij}\vert$$ I do not see how to continue here. Help is highly needed and appreciated!
Assuming your pivoting scheme picks the largest element every time (which is more or less the case when numerical stability is important), then w.l.o.g. you can assume that $|a_{jj}^{(j-1)}| \geq |a_{ij}^{(j-1)}|, \forall i$. The notation followed here is the one mentioned in Wikipedia's article.
By the definition of $l_{ij}$, you can see that
$$ \begin{align} l_{ij} &= -\frac{a^{(j-1)}_{ij}}{a^{(j-1)}_{jj}} \Rightarrow \\ |l_{ij}| &= \frac{|a_{ij}^{(j-1)}|}{|a_{jj}^{(j-1)}|} \leq 1 \end{align} $$
Therefore, $\sum_{j=1}^n |l_{ij}| \leq \sum_{j=1}^n 1 = n, \forall i$ which implies that $|| L ||_{\infty} \leq n$.