Bounding the total variation for Banach lattice

Question

Let $E$ be a Banach lattice and $E_+$ denote its positive cone.

For $x_1, \ldots, x_n \in E_+$, is there any way to bound $\sum_1^n\|x_i\|$ with $\|\sum_1^n x_i\|$ without using n?

Similaryly, for a atomless finite measure space $X$, let $f:X \rightarrow E_+$ be an integrable function. Can I bound $\int_X\|f\|d\mu$ with $\|\int_Xfd\mu\|$?

I think this is not usually possible, but the range of the function is only restricted to a positive cone of $E$

Answer 1

No, this seems not to be possible. Take $E = L^2(0,1)$ and consider $$x_i = \chi_{((i-1)/n,n)}.$$ Then, $$\sum_{i=1}^n \|x_i\|_{L^2(0,1)} = \sqrt{n}$$ while $$\|\sum_{i=1}^n x_i\|_{L^2(0,1)} = 1.$$ It is even worse with $L^p(0,1)$, $p > 2$. And with $p = \infty$, you need the constant $n$.