I am reading Peter Lax's "Functional analysis".

Let $y_n$ be a bounded sequence of vectors in a Banach reflexive space, $X, Y$ their closed linear span. Take a countable set $m_j$ of applications belonging to the dual $Y^*$ of $Y$ ($Y^*$ is separable).

The author then says that we can apply the "classical diagonal process" to obtain a subsequence $z_n$ of vectors such that $\lim_n m_j(z_n)$ exists for every $j$.

I do not understand how he builds this subsequence.

  • $\begingroup$ How is $X$ defined ? $\endgroup$
    – Jean Marie
    Jul 20, 2016 at 15:16
  • $\begingroup$ X is a reflexive Banach space $\endgroup$ Jul 20, 2016 at 15:29

1 Answer 1


The key ingredient is the observation that if a sequence is convergent, then any subsequence is also convergent.

  1. First, you construct recursively a sequence of subsequences of $y_n$ as follows.
  2. To begin, put $z_{n,0}:=y_n$.
  3. Given $(z_{n,j})_n$, you can find (by compactness) a subsequence $(z_{n,j+1})_n$ such that $\lim_{n}m_j(z_{n,j+1})$ exists.
  4. Note that then all $\lim_n m_{j'}(z_{n,j+1})$ exist, for $j'\leq j$ (because in this case $(z_{n,j+1})_n$ is a subsequence of $(z_{n,j'+1})_n$).
  5. Finally, put $z_n:=z_{n,n}$ (i.e. $z_n$ is the diagonal of $z_{n,j}$, hence the name). To see that $\lim_n m_j(z_n)$ exists, note that except for finitely many elements, $z_n$ is a subsequence of $z_{n,j}$.

You don't need $X$ to be reflexive, or even a Banach space for this argument. All that really matters is that all the functions $m_j$ map the sequence $y_n$ into compact metric spaces (so that we can use compactness in step 3.) -- the topology on $X$ (or its linear structure) is completely irrelevant beyond this, guaranteed by the fact that bounded functionals by definition map bounded sequences to bounded subsets of the real line.

  • $\begingroup$ Thank you very much Tomasz, now it is clear. It has been really instructive. $\endgroup$ Jul 20, 2016 at 18:35
  • $\begingroup$ The only thing it would like to understand better is passage 4. Could you be more explicit please? $\endgroup$ Jul 20, 2016 at 19:14
  • $\begingroup$ @GiovanniSiclari: There you go. $\endgroup$
    – tomasz
    Jul 21, 2016 at 9:25
  • $\begingroup$ Thank you. This morning I have finally understood it. $\endgroup$ Jul 21, 2016 at 9:27

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