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

  • 2
    $\begingroup$ Q2: the space is complete, so $\sum_{i=1}^n \alpha_i x_i$ is a Cauchy sequence. $\endgroup$
    – Gary
    Commented Jan 28, 2022 at 1:16

1 Answer 1


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 $\sup_m \beta_m \le \lim_n \sup_{k \le n} \beta_k$.

For the second question, since the sum converges it is Cauchy.

  • $\begingroup$ i beg your pardon why $\beta_m \leq \lim_n \sup_{k \leq n} \beta_k$ for all $m$ $\endgroup$
    – Roba
    Commented Feb 3, 2022 at 22:54
  • 1
    $\begingroup$ Because when $n \ge m$ we have $\beta_m \le \sup_{k \le n} \beta_k$. $\endgroup$
    – copper.hat
    Commented Feb 4, 2022 at 3:41

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