# Prove $\sup_{1 \leq n < \infty} \|\sum_{i=1}^n \alpha_i x_i\| \leq \lim_{n \rightarrow \infty} \sup_{1 \leq k \leq n} \|\sum_{i=1}^k \alpha_i x_i\|$

From book bases in Banach spaces I by ivan singer Proposition

Let $$\{x_n\}$$ be a sequence in a Banach space $$E,$$ such that $$x_n \neq 0 (n = 1,2,...),$$ and let $$Y$$ be the Banach space of sequences of scalars $$(\alpha_n)$$ for which $$\displaystyle \lim_n \sum_{k=1}^n \alpha_k x_k$$ exists in $$E.$$ i.e. $$Y = \{\{\alpha_n\} \subset K | \displaystyle \sum_{i=1}^\infty \alpha_i x_i \text{ }converges \text{ } in \text{ } E\}$$ Then the unit vectors $$\begin{eqnarray} \nonumber e_n = \{\delta_{nj}\}_{j=1}^\infty (n = 1,2,...) = \{\{1,0,0,0,...\}, \{0,1,0,0,...\}, \{0,0,1,0,...\}, ....\} \end{eqnarray}$$ constitute a basis of $$Y.$$

Proof If $$\{\alpha_n\} \in Y,$$ i.e. if $$\displaystyle \sum_{i=1}^\infty \alpha_i x_i$$ converges, then since the tail end of a convergent sequence equals zero $$\|\{\alpha_n\} - \sum_{i=1}^m \alpha_i e_i\| = \|\{\alpha_{m+1}, \alpha_{m+2}, ...\}\|= \displaystyle \sup_{m+1 \leq k < \infty} \|\sum_{i = m+1}^k \alpha_i x_i\| \rightarrow 0 \text{ for } m \rightarrow \infty,$$

whence $$\displaystyle \sum_{i=1}^\infty \alpha_i e_i$$ converges to $$\{\alpha_n\}.$$ On the other hand, to show that the representation is unique, if $$\displaystyle \sum_{i=1}^\infty \alpha_i e_i = 0,$$ then, by the above, $$\begin{eqnarray} \nonumber \|\{\alpha_n\}\| &=& \displaystyle \sup_{1 \leq n < \infty} \|\sum_{i=1}^n \alpha_i x_i\| \leq \lim_{n \rightarrow \infty} \sup_{1 \leq k \leq n} \|\sum_{i=1}^k \alpha_i x_i\| = \lim_{n \rightarrow \infty} \|\{\alpha_1, \alpha_2, ..., \alpha_n\}\| =\\ \nonumber &=& \lim_{n \rightarrow \infty} \|\sum_{i=1}^n \alpha_i e_i\| = \|\lim_{n \rightarrow \infty} \sum_{i=1}^n \alpha_i e_i \| \text{ since norm is a continuous function }\\ \nonumber &=& \|\sum_{i=1}^\infty \alpha_i e_i\| = 0, \end{eqnarray}$$ whence $$\alpha_n = 0 (n = 1,2,...).$$ Thus, every $$\{\alpha_n\} \in Y$$ has a unique expansion $$\displaystyle \sum_{i=1}^\infty \alpha_i e_i,$$ i.e. $$\{e_n\}$$ is a basis of $$Y.$$

So Q1 how to prove that $$\displaystyle \sup_{1 \leq n < \infty} \|\sum_{i=1}^n \alpha_i x_i\| \leq \lim_{n \rightarrow \infty} \sup_{1 \leq k \leq n} \|\sum_{i=1}^k \alpha_i x_i\|$$

Q2 Why $$\displaystyle \sup_{m+1 \leq k < \infty} \|\sum_{i = m+1}^k \alpha_i x_i\| \rightarrow 0 \text{ for } m \rightarrow \infty$$

Thanks Any help will be apreciated

• Q2: the space is complete, so $\sum_{i=1}^n \alpha_i x_i$ is a Cauchy sequence.
– Gary
Commented Jan 28, 2022 at 1:16

Let $$\beta_n$$ be any sequence of numbers. Then $$\sup_n \beta_n = \lim_n \sup_{k \le n} \beta_k$$.
Clearly $$\sup_m \beta_m \ge \sup_{k \le n} \beta_k$$ for all $$k$$ and hence $$\sup_m \beta_m \ge \lim_n \sup_{k \le n} \beta_k$$ (limit of increasing sequence of numbers).
Since $$\beta_m \le \lim_n \sup_{k \le n} \beta_k$$ for all $$m$$, we have $$\sup_m \beta_m \le \lim_n \sup_{k \le n} \beta_k$$.
• i beg your pardon why $\beta_m \leq \lim_n \sup_{k \leq n} \beta_k$ for all $m$
• Because when $n \ge m$ we have $\beta_m \le \sup_{k \le n} \beta_k$. Commented Feb 4, 2022 at 3:41