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Suppose ($x_n$) is a normalized, linearly independent, sequence in a reflexive Banach space $X$, and $T$ is an injective, strictly singular, bounded operator on $X$ such that $Tx_n\longrightarrow 0$. Does there exist a subsequence $(y_n)$ of $(x_n)$ such that $T$ restricted to the closed span of $(y_n)$ is compact? When $(x_n)$ is a basic sequence and $\sum||Tx_n||$ converges (so $Tx_n\longrightarrow 0$ fast enough), it is not hard to show that $T$ restricted to closed span of $(x_n)$ is indeed compact (without any need to assume that $T$ is strictly singular or $X$ is reflexive).

An operator is called strictly singular if it is not an isomorphism when restricted to any infinite dimensional subspace. Compact operators are always strictly singular but not the other way around. However, strictly singular are compact in $l_p$ and $c_0$.

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  • $\begingroup$ Maybe I misunderstood the question, it may be not true, for example when $E=\ell^2$, $x_n=n^{-1}\sum_{j=1}^ne_j$ and $T=\sum_{k=1}^{+\infty}T_{2^k}$, where $T_j$ is the diagonal operator, such that $T_j(e_k)=\delta_{jk}$. $\endgroup$
    – Davide Giraudo
    Apr 25, 2012 at 20:59
  • $\begingroup$ I also may have misunderstood the question, but what do you mean when you say "$T(x_n) \to 0$ as fast as I want"? Do you mean that $T(x_n) = 0$ for all $n\geq N$ for some natural number $N$? Or do you mean that for any sequence of positive numbers $\epsilon_n$ which goes to zero, $\|T(x_n)\| \leq \epsilon_n$? Or is there another meaning for this phrase which I am overlooking? $\endgroup$
    – Nicholas Stull
    Apr 25, 2012 at 21:42
  • $\begingroup$ I edited for what I really wanted to ask, hopefully this time is more clear. My initial question, besides being a bit confusing, had an obvious answer, quickly pointed out by Davide. Thank you. $\endgroup$
    – Theo
    Apr 25, 2012 at 22:12
  • $\begingroup$ @Theo, Thank you for the edit. Even though the question is now resolved, it was an interesting problem to think about. $\endgroup$
    – Nicholas Stull
    Apr 26, 2012 at 3:23

1 Answer 1

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The answer is yes, there is such a subsequence.

We do not need to assume strict singularity of $T$, we use boundedness and injectivity of $T$, that $\|T x_n\| \xrightarrow{n\to\infty} 0$ and reflexivity of $X$.

Since $X$ is reflexive, the unit ball is weakly sequentially compact by the Eberlein–Šmulian theorem, so we may pass to a weakly convergent subsequence $x_n \to x$. Since $T$ is bounded, we also have that $Tx_n \to Tx$ weakly, but by hypothesis $Tx_n \to 0$ in norm. Thus $Tx = 0$ and by injectivity of $T$ we must have that $x = 0$.

Since $(x_n)_{n=1}^\infty$ is normalized and $x_n \to 0$ weakly, the Bessaga–Pełczyński selection principle(1) allows us to pass to a basic subsequence of $x_n$.

Passing to yet another subsequence, we finally find a basic sequence $(x_{n_k})_{k=1}^\infty$ such that $\sum_{k=1}^{\infty} \|Tx_{n_k}\| \lt \infty$, and this yields that the restriction of $T$ to the linear span of $(x_{n_k})_{k=1}^\infty$ is compact by what you mentioned in your question.

(1) If $x_n \to 0$ weakly and $\inf_{n\in\mathbb{N}} \|x_n\| \gt 0$ then there is a basic subsequence $(x_{n_k})_{k=1}^\infty$ of $(x_n)_{n=1}^\infty$.

The proof of this is known as the “gliding hump argument”, and was established as Corollary C.1, page 156 in Bessaga–Pełczyński, On bases and unconditional convergence of series in Banach spaces, Studia Math. 17 (1958), 151–164.

See also Albiac–Kalton, Topics in Banach Space Theory, Proposition 1.5.2, as well as Proposition 1.3.10 and Theorem 1.4.4.

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  • $\begingroup$ In another direction it might also be possible to exploit the strict singularity of $T$ and, in turn, drop reflexivity of $X$ from the hypotheses. However, I did not quite see how to find the subsequence of $x_n$ with such an approach. Note that an operator is strictly singular if and only if every infinite-dimensional subspace contains a further infinite-dimensional subspace $Y$ such that $T|_Y$ is compact. $\endgroup$
    – t.b.
    Apr 26, 2012 at 2:33
  • $\begingroup$ Thanks t.b, but it is still not clear to me. If $X$ is reflexive, thus $B(X)$ is weekly compact, thus weakly sequentially compact by Eberlein-Smulian, it follows that any bounded sequence has a weakly convergent subsequence. Therefore, WLOG, we may assume that $x_n\to z$ weakly, so $x_n-z\to 0$ weakly. However, $(x_n-z)$ might be norm null. So what I don't understand why we can assume that $z=0$, so we can extract a basic sequence from $(x_n)$ using Bessaga-Pelczynski. $\endgroup$
    – Theo
    Apr 26, 2012 at 2:54
  • $\begingroup$ @Theo: you're absolutely right. I think I fixed the argument, but it is very late here, so I might have goofed again. Sorry about that. $\endgroup$
    – t.b.
    Apr 26, 2012 at 4:03
  • $\begingroup$ @Theo: out of curiosity: where does this problem come from? It seems to me that the hypotheses are quite redundant; for example strict singularity yields a sequence $x_n$ as in the hypotheses. Also, I did not use reflexivity. $\endgroup$
    – t.b.
    Apr 26, 2012 at 4:17
  • $\begingroup$ Hmmm....If what you say is correct, then you showed that from any normalized, linearly independent sequence you can extract a basic sequence, which as far as I know is not true. I think the mistake is when you first choose a subsequence. As you say Ker $e_1^*$ is a hyperplane, but I don't see why any of the $x_n$ should be in that hyperplane. The closed span of $x_n$ intersects the hyperplane, but all of $x_n$ could be outside. Or am I missing something? Thank you again for spending time on this :) $\endgroup$
    – Theo
    Apr 26, 2012 at 5:10

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