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I'm reading the book Topics in Banach Space Theory by Albiac F. Kalton N. J. I got stuck at the proof of Pitt's theorem. enter image description here

In the second paragraphs authors tries to prove ad absurdum that for weakly nuul sequence $\lim\limits_{n\to\infty}\Vert T(x_n)\Vert=0$. They say that without loss of generality one may suppose that $\{x_n\}_{n=1}^\infty$ is a weakly null sequence with $\Vert x_n\Vert=1$ and $\Vert T(x_n)\Vert>\delta$ for all $n\in\mathbb{N}$. I think they normalized original sequence $\{x_n\}_{n=1}^\infty$ and claims that it is also weakly null.

Why is weakly null sequence remains weakly null after normalization?

Another place I got stuck is the place where authors claims that passing to subsequence in $\{T(x_n)\}_{n=1}^\infty$ gives subsequence equivalent to the natural basis of $\ell_p$. And they also assume that after passing to subsequence $\{x_n\}_{n=1}^\infty$ remains to be equivalent to natural basis of $\ell_p$.

Why is $\{x_n\}_{n=1}^\infty$ remains to be equivalent to natural basis of $\ell_p$?

Thank you.

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  • $\begingroup$ Is there anything wrong with my argument to the OP first question? If $x_n\to 0$ weakly, then for any $x^*\in X^*,$ we have $$x^*\left( \frac{x_n}{\|x_n\|} \right) = \frac{1}{\|x_n\|} x^*(x_n) \to 0.$$ $\endgroup$
    – Idonknow
    Aug 2 '18 at 7:26
  • $\begingroup$ @Idonknow how do you deduce that last step? $\endgroup$
    – Andrei Kh
    Apr 9 '19 at 4:24
  • $\begingroup$ @Tomasz Kania Why is it important that the weak topolgy on $B_X$ is $w$-metrizable? We could have concluded the last sentence in that paragraph from Eberlein-Smulian, right? $\endgroup$
    – Markus
    May 10 '20 at 20:04
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  1. We assume that $\{x_n\}$ doesn't converge strongly to $0$. We show, that each subsequence of $\{Tx_n\}$ has a convergent subsequence. So here, we can find a constant $C>0$ and $A$ infinite such that $\lVert x_n\rVert\geq C$ for all $n\in A$. Let $y_n:=\frac 1{\lVert x_n\rVert}x_n$ for $n\in A$. Then $\{y_n\}_{n\in A}$ converges weakly to $0$, as for $f\in (\ell^r)^*$ and $n\in A$, we have $$|f(y_n)|=|f(x_n)|\frac 1{\lVert y_n\rVert}\leq \frac{|f(x_n)|}C.$$

  2. By definition of equivalence $\{u_n\}$ and $\{v_n\}$ are equivalent if for all sequence $\{a_n\}$ of scalars, $\sum_{n=1}^{+\infty}a_nu_n$ is convergent if and only if so is $\sum_{n=1}^{+\infty}a_nv_n$. So $\{x_n\}_{n\in A}$ is equivalent to $\{e_n\}_{n\in A}$ (not to the whole sequence). But it's enough to conclude, as we would have boundedness from an infinite subspace of $\ell^r$ to $\ell^p$.

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I think that for your first question the answer is:

If $\|x_{n}\|\rightarrow 0$, then $x_{n}\rightarrow 0$ which implies that $T(x_n)\rightarrow 0$. So you can suppose that $\|x_{n}\|>\delta$ for some $\delta>0$. By hypothesis you have $$\langle y,x_{n}\rangle\rightarrow 0,\ \forall\ y\in X^{\star}$$

hence $$\Big|\Big\langle y,\frac{x_{n}}{\|x_{n}\|}\Big\rangle\Big|\leq \delta|\langle y,x_{n}\rangle|,\ \forall\ y\in X^{\star} $$

From the last inequality you can conclude that $\frac{x_{n}}{\|x_{n}\|}$ is an weakly null sequence.

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    $\begingroup$ It's true after having taken a subsequence of $\{x_n\}$ (otherwise take $x_{2k}=e_k$, $x_{2k+1}=0$. But that the idea. $\endgroup$ Oct 21 '12 at 21:50
  • $\begingroup$ @DavideGiraudo, yes you are right, but im assuming that Norbert is aware of that. $\endgroup$
    – Tomás
    Oct 21 '12 at 21:56
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    $\begingroup$ Of course I understand necessity of subsequences. $\endgroup$
    – Norbert
    Oct 21 '12 at 22:05
  • $\begingroup$ That's what I assumed, but I needed to be clarified, for example if an other user reads the question. $\endgroup$ Oct 22 '12 at 11:40

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