Diametral dimensions of Fréchet spaces (Q2835996)
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scientific article; zbMATH DE number 6658335
| Language | Label | Description | Also known as |
|---|---|---|---|
| English | Diametral dimensions of Fréchet spaces |
scientific article; zbMATH DE number 6658335 |
Statements
Diametral dimensions of Fréchet spaces (English)
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30 November 2016
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Fréchet spaces
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Kolmogorov widths
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diametral dimension
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topological invariants
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0.7759968
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0.7177468
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0.67697555
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\textit{B. S. Mityagin} [Russ. Math. Surv. 16, No. 4, 59--128 (1961); translation from Usp. Mat. Nauk 16, No. 4(100), 63--132 (1961; Zbl 0104.08601)] utilized Kolmogorov diameters to introduce a topological invariant of locally convex spaces \(X\) called the diametral dimension \(\Delta(X)\). He also implicitly studied another invariant \(\Delta_b(X)\), which always contains \(\Delta(X)\). These two diametral dimensions coincide for each non-Montel Fréchet space and they coincide with the Banach space \(c_0\) of null sequences. \textit{A. T. Terz{i}oğlu} investigated in [Turk. J. Math. 37, No. 5, 847--851 (2013; Zbl 1287.46003)] the equality \(\Delta(X)=\Delta_b(X)\) for Fréchet-Schwartz spaces.NEWLINENEWLINEIn the article under review, the authors introduce two new, larger variants of the diametral dimensions, replacing null sequences by bounded sequences in the definitions, and they show that if \(X\) is a Fréchet-Schwartz space, then \(\Delta^{\infty}(X)=\Delta_b^{\infty}(X)\). They then prove that the equality \(\Delta(X)=\Delta_b(X)\) holds for hilbertizable Fréchet-Schwartz spaces, in particular for nuclear Fréchet spaces. The proof is based on a result about the Kolmogorov diameters of the composition of two compact operators on Hilbert spaces.NEWLINENEWLINEIn the last section, the authors extend several results of {Terz{i}oğlu} about prominent bounded sets. They prove that if a Fréchet space \(X\) has the topological invariant \((\overline{\Omega})\) of Vogt and Wagner, then \(X\) has a prominent bounded set. The converse result holds for regular Köthe echelon spaces. However, the equivalence does not hold in general. In fact, \(X=H(\mathbb{D}) \times H(\mathbb{C})\) has a prominent bounded set, but it does not satisfy \((\overline{\Omega})\).
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