Why is $\sum_{i=1}^n |v_i| || Le_i||_W \leq \max_{1 \leq j \leq n} ||Le_j||_W \sum_{i=1}^n ||v||_1$?

Question

Why is $\sum_{i=1}^n |v_i| || Le_i||_W \leq \max_{1 \leq j \leq n} ||Le_j||_W \sum_{i=1}^n ||v||_1$?

Where $v_i$ is a scalar taken by linearity out of $\sum_i Lv_ie_i$, $e_i$s are standard basis of $\mathbb{R}^n$ and $L: V \rightarrow W$ linear transformation. $V$ is finite normed space, $W$ is normed space.

For a basis $\{ f_i \}$ the norm $||\sum v_i f_i||_1=||(v_1,v_2,..., v_n)||_1$ and is an isometry between $(\mathbb{R}^n, ||\cdot||_1)$ and $(V, ||\cdot||_1)$.

I have troubles reading the inequality. I'm not really sure what happens in it. Does it use Cauchy-Schwartz?

Why can one assume that $\sum_{i=1}^n ||Le_i||_w \leq \max_j ||Le_j||_W$?

What about the inequality applied on $\sum_{i=1}^n |v_i|$? Is this a $\ell^p$-space?

Answer 1

\begin{align} \sum_{i=1}^n |v_i| || Le_i||_W &\le \sum_{i=1}^n |v_i| \left(\max_{1 \leq j \leq n} ||Le_j||_W\right) \\ &= \left(\max_{1 \leq j \leq n} ||Le_j||_W\right) \sum_{i=1}^n |v_i| \\ &= \left(\max_{1 \leq j \leq n} ||Le_j||_W\right) ||v||_1 \\ &\le \max_{1 \leq j \leq n} ||Le_j||_W \sum_{i=1}^n ||v||_1 \end{align}