4
$\begingroup$

Let $X$ be an infinite-dimensional Banach space and $\beta$ denote Istrățescu's spreading measure of noncompactness, i.e. $$\beta(M) = \sup \{ \varepsilon > 0 \colon \exists_{(x_n)^{\mathbb N} \in X^{\mathbb N}} \forall_{m \ne n} \colon \|x_n-x_m\| > \varepsilon \}$$ for any $M \subset X$.

If we denote the (closed) unit ball of $X$ by $B_X$, it is clear that we have $1 \le \beta(B_X) \le 2$ (by Riesz's lemma and the triangle inequality). I believe one has $\beta(B_X) = \sqrt 2$ whenever $X$ is a Hilbert space and $\beta(B_X) = 2$ if $X$ is one of the sequence spaces $c_0$, $\ell^1$, and $\ell^\infty$ or one of the function spaces $L^1[0,1]$ or $L^\infty[0,1]$, so that intuitively, the less round a space, the larger the value $\beta(B_X)$.

Q: There is a large zoo of notions of non-roundness, like the (alternative) Daugavet property, the notion of (almost) CL-spaces, lushness, $n(X) = 1$ (where $n$ denotes the numerical index). Is it known how the property $\beta(B_X) = 2$ fits in here? Is it known if this property is inherited by duals or pre-duals? (biduals obviously inherit it). Is it known if such spaces can be reflexive?

It is known, e.g., that spaces with the Daugavet property cannot be reflexive, and neither can those with $n(X) = 1$, at least in the real case; see also

Kadets, Vladimir; Martín, Miguel; Payá, Rafael. Recent progress and open questions on the numerical index of Banach spaces. RACSAM. Rev. R. Acad. Cienc. Exactas Fís. Nat. Ser. A Mat. 100 (2006), no. 1-2, 155--182. MR2267407

Consequently, neither can lush or (almost) CL spaces in the real case.

Edit: As pointed out by Bill Johnson, it is shown in

Elton, J.; Odell, E. The unit ball of every infinite-dimensional normed linear space contains a $(1+\varepsilon )$-separated sequence. Colloq. Math. 44 (1981), no. 1, 105--109. MR0633103

that we always have $\beta(B_X) > 1$. For non-reflexive spaces, even more is known: It was shown in

Kryczka, Andrzej; Prus, Stanisław. Separated sequences in nonreflexive Banach spaces. Proc. Amer. Math. Soc. 129 (2001), no. 1, 155--163. MR1695123

that $\beta(B_X) \ge \sqrt[5]4 \approx 1.3195$ holds for any non-reflexive space $X$. They also constructed a non-reflexive space $X$ with $\beta(B_X) \approx 1.7086$, thereby providing an upper bound for how much further this lower bound could be improved.

Edit: The following computations show $\beta(B_X) = 2$ for the spaces that were given as examples earlier.

  • $X = c_0$ or $X = \ell^\infty$: Consider the sequence $x_1 = (1, 0, \dotsc)$, $x_2 = (-1, 1, 0, \dotsc)$, $x_3 = (-1, -1, 1, \dotsc)$; then $x_m$ and $x_n$ differ by 2 in the coordinate $\min(m,n)$ for $m \ne n$.

  • $X = L^\infty[0,1]$: Analogously, consider the sequence $(f_n)$ given by $$f_n = \begin{cases} -1 & \text{on $[0, 1-2^{1-n})$}\\ 1 & \text{on $[1-2^{1-n},1-2^{-n})$}\\ 0 & \text{on $[1-2^{-n},1]$} \end{cases}$$ for $n \ge 1$. Then $f_n$ and $f_m$ differ by 2 on the interval $[1-2^{1-k},1-2^{-k}]$ with $k = \min(m,n)$ whenever $m \ne n$.

  • $X = C[0,1]$: Replacing the constant 1-intervals with hat-functions in the previous example produces a sequence of continuous functions that show $\beta(B_X) = 2$. This sequence works in $L^\infty[0,1]$, too, naturally.

  • $X = \ell^1$: Here, it suffices to consider the canonical sequence of unit vectors $x_1 = (1, 0, \dotsc)$, $x_2 = (0, 1, 0, \dotsc)$, $x_3 = (0, 0, 1, 0, \dotsc)$ and so forth (so that $x_n$ and $x_m$ differ by 1 in the $k$-th coordinate for $k \in \{ m, n \}$ whenever $n \ne m$).

  • $X = L^1[0,1]$: Analogously, consider the sequence $(f_n)$ given by $$f_n = \begin{cases} 0 & \text{on $[0, 1-2^{1-n})$}\\ 2^n & \text{on $[1-2^{1-n},1-2^{-n})$}\\ 0 & \text{on $[1-2^{-n},1]$} \end{cases}$$ for $n \ge 1$. Then $f_n$ and $f_m$ differ by $2^k$ on the interval $[1-2^{1-k},1-2^{-k}]$ of length $2^{-k}$ for $k \in \{ m, n \}$ whenever $m \ne n$.

$\endgroup$
  • 1
    $\begingroup$ I have no idea, but out of curiosity: What are examples of $\beta(B_X) = 1$? I suspect that their unit balls are not round, either. $\endgroup$ – Dirk Dec 7 '16 at 14:10
  • 1
    $\begingroup$ @Dirk I'm not actually sure if such spaces exist. I've now added examples that show $\beta(B_X) = 2$ also for $\ell^1$ and $L^1[0,1]$. Maybe $\sqrt 2$ is actually the minimal value for $\beta(B_X)$. $\endgroup$ – anonymous Dec 7 '16 at 15:07
  • $\begingroup$ Oh, interesting. Another notion of "roundness" is the modulus of convexity. $\endgroup$ – Dirk Dec 7 '16 at 15:26
  • $\begingroup$ @Dirk It is shown in the article by Elton and Odell which Bill Johnson has just mentioned (available e.g. from eudml.org/doc/266720) that $\beta(B_X)$ is always strictly greater than 1. $\endgroup$ – anonymous Dec 7 '16 at 15:46
  • $\begingroup$ One has $\beta(B_{\ell^p}) = 2^{1/p}$ for $1 \le p < \infty$. Thus, $\lim_{p \to \infty} \beta(B_{\ell^p}) = 1$, but $\beta(B_{\ell^\infty}) = 2$. $\endgroup$ – anonymous Dec 7 '16 at 16:01
2
$\begingroup$

Edit

Tsirelson space admits spreading models 1-equivalent to the unit vector basis of $\ell_1$. See the paper

Odell, E.; Schlumprecht, Th. A problem on spreading models. J. Funct. Anal. 153 (1998), no. 2, 249--261. MR1614578

So $\beta =2$ and yet reflexivity holds.

Ignore my first answer below.

If you allow renormings of the space when computing $\beta$ then $\beta(S_X)=2$ implies that $X$ is non-reflexive. This follows (I think) from a theorem of Odell and Schlumprecht.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ How does it follow from [OS]? $\endgroup$ – Bill Johnson Dec 7 '16 at 21:34
  • $\begingroup$ What they actually prove is: if under all renorming $X$ admits a spreading model $(e_i)$ with $\|e_1\pm e_2\|=2$, then $\ell_1$ embeds in $X$. So I should have said if $\beta_{|\cdot|}(S_X)=2$ for all renorming $|\cdot|$ of $X$. Even so this is still not quite enough, will need unconditionality and $\beta=2$ need to be attained. Not sure at the moment if one can overcome these. $\endgroup$ – Bunyamin Sari Dec 7 '16 at 22:46
  • $\begingroup$ Interesting. So if I understand correctly, not even the stronger version, where the supremum in the definition of $\beta(B_X) = 2$ is attained (which is what one actually has in $\ell^1$, $\ell^\infty$, etc.) suffices to guarantee non-reflexivity. $\endgroup$ – anonymous Dec 8 '16 at 13:59
  • $\begingroup$ @anonymous. See my second comment after my answer and correct the typo (non reflexivity instead of reflexivity). $\endgroup$ – Bill Johnson Dec 8 '16 at 16:11
  • 1
    $\begingroup$ As these example show, the question should be really what if $\beta=2$ under all renormings. I still think this may imply non-reflexivity and if you are interested you may have to dig out the proof from [OS] paper i mentioned in the first answer. For instance, Corollary 3.5 of the paper seems to imply the unconditionality concern I had before isn't an issue. $\endgroup$ – Bunyamin Sari Dec 8 '16 at 17:05
1
$\begingroup$

Elton, J.; Odell, E. The unit ball of every infinite-dimensional normed linear space contains a (1+ε)-separated sequence. Colloq. Math. 44 (1981), no. 1, 105–109.

| cite | improve this answer | |
$\endgroup$
  • 1
    $\begingroup$ Tomek, look at $\sum \ell_1^n)_p$ as $p\to \infty$. $\endgroup$ – Bill Johnson Dec 7 '16 at 19:50
  • $\begingroup$ As much as that's an interesting paper and I'm thankful for the reference: I don't think this answers any of my questions. $\endgroup$ – anonymous Dec 7 '16 at 20:45
  • 2
    $\begingroup$ Well, $\beta = 2$ clearly does not imply reflexivity. Reform $\ell_2$ with the maximum of the $\ell_2$ norm and $\max_{j\not= k} |x(j)|+|x(k)|$ to make the unit vector basis elements a distance $2$ apart. $\endgroup$ – Bill Johnson Dec 7 '16 at 21:40

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.