Good day to All.

Let $S_{1,n} = \sum_{i=1}^{n}\xi_{i}$, where $(\xi_{i})_{i \in \mathbb{N}}$ be independent RV with values in some Banach space.

On pages 79-80 in this book author provides an example that illustrates the fact that in the infinite dimensional spaces it may be not enough to have assumptions on the distribution of the individual summands $\xi_{i}$ in order to control the deviations of $\lVert {S_{1,n}}\rVert $, where $\lVert \cdot \rVert $ is the norm of corresponding Banach space. Author says:"The following example serves to show the relative independence of restrictions on concentration of the sum and those concerning the distributions of individual summands ".

In his example 3.0.1 let the space $c_0$ (with the norm $\|x\|_{\infty} = \sup_{n \in \mathbb{N}} |x^{(n)}|$) be the space of real-valued convergent to zero sequences and we consider an array of i.i.d. real-valued $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$.

Consider the sequence $\xi_{j} = ( \frac{\xi_{j}^{(i)}}{\ln\ln(10 i)}, i \in \mathbb{N} )$ which are centered, in $c_{0}$ and a.s. bounded, i.e. $\|\xi_j\|_{\infty} = \frac{1}{\ln \ln 10}$. Consider RV's $\eta_{n}^{(i)} = \sum_{j=1}^{n}\xi_{j}^{(i)}$ and notice that they are independent. One easily sees that $\mathbb{P}(\eta_{n}^{i}=n) = 2^{-n}$ and therefore it implies: \begin{align*} \mathbb{P}(\max_{i \leq 2^n}\eta_{n}^{(i)}<n) = \prod_{k=1}^{2^n}(1-\mathbb{P}(\eta_{n}^{(i)}=n)) \leq \exp({-2^{n} 2^{-n})} = e^{-1}, \end{align*} where we used classical $1-x \leq \exp(-x)$ for $x\geq 0$. Thefore, for all $n$ we have: \begin{align*} P\left(\|S_{1,n}\| \geq \frac{n}{\ln \ln (10 * 2^n)}\right) \geq P\left(\max_{i \leq 2^n}\eta_{n}^{(i)} \geq n\right) \geq 1-e^{-1}. \end{align*}

Thus, the norm of sum scales (with constant probability) like $\frac{n}{\ln (n)}$, whereas $\|\xi_{j}\|_{\infty} = \frac{1}{\ln \ln (10)}$.

My question: can't we also consider $l^{\infty} := \{x=(x^{(n)}) : \sup_{n\in \mathbb N}| x^{(n)}| \lt \infty\}$ (which is Banach space with respect to the norm $\|x\|_{\infty} = \sup_{n \in \mathbb N} |x^{(n)}|$) and just the sequences $\xi_{j} = (\xi_{j}^{j}, j \in \mathbb{N})$ (without additional log weight ) and $\xi_{j}^{i} = \frac{1}{2}\delta_{-1}+\frac{1}{2}\delta_{1}$ i.i.d in $i,j$ as in example 3.0.1? Then, it seems the same reasoning as in the book applies for this case.