Given a family $(A_{\lambda})_{\lambda\in\Lambda}$ of Banach spaces, let $\bigoplus_{\lambda}A_{\lambda}$ be the set of all $(a_{\lambda})\in\prod_{\lambda}A_{\lambda}$ such that $||(a_{\lambda})||=\sup_{\lambda}||a_{\lambda}||<\infty$. I am trying to show that $\bigoplus_{\lambda}A_{\lambda}$ is complete but I am stuck somewhere.

What I did was to take a Cauchy sequence $(a_{\lambda,n})_{n=1}^{\infty}$ in $\bigoplus_{\lambda}A_{\lambda}$ and saw that for each $\lambda$ I have a Cauchy sequence in $A_{\lambda}$, which converges to some $a_{\lambda}\in A_{\lambda}$. So I have a candidate for the limit in $\bigoplus_{\lambda}A_{\lambda}$ but I'm missing something in trying to show that I have convergence in $\bigoplus_{\lambda}A_{\lambda}$. In fact, I'm not sure whether that candidate is in fact in $\bigoplus_{\lambda}A_{\lambda}$.

  • 2
    $\begingroup$ Hint: Recall, or reread, a proof that $\ell^\infty$ is a Banach space. $\endgroup$ – Nate Eldredge Oct 8 '13 at 22:54
  • $\begingroup$ Indeed that was what I needed. Thanks. $\endgroup$ – cyc Oct 8 '13 at 23:20
  • $\begingroup$ Glad it helped! You could post your solution as an answer, if you like. $\endgroup$ – Nate Eldredge Oct 8 '13 at 23:35
  • $\begingroup$ Just a curious question, why this is called a direct sum, it looks like a product to me. $\endgroup$ – user50618 May 21 '15 at 19:51

For completeness, I post the proof here. The first two facts are valid for all metric spaces.

  1. If a Cauchy sequence has a convergent subsequence, it converges.

  2. Every Cauchy sequence contains a subsequence (denoted $\{x_n\}$ here) such that $$d(x_n,x_{n+1})\le 2^{-n},\quad n\in\mathbb{N} \tag{1}$$ (Note that $(1)$ implies being Cauchy.)

So, it suffices to prove that a sequence $\{x_n\}$ in $\bigoplus_{\lambda}A_{\lambda}$ that satisfies (1) converges. Since the property (1) is inherited by each coordinate, and each $A_\lambda$ is complete, the sequence $\{x_n\}$ converges coordinate-wise. Let $x$ be its coordinate-wise limit. The triangle inequality implies that for every $\lambda$, $$\|x_n^\lambda - x^\lambda\| \le \sum_{k=n}^\infty 2^{-k} = 2^{1-n}$$ Hence $\|x_n - x \| \le 2^{1-n}\to 0$ as desired.


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