Linear subsets of nonlinear sets in topological vector spaces (Q2864923)

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scientific article; zbMATH DE number 6233822
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Linear subsets of nonlinear sets in topological vector spaces
scientific article; zbMATH DE number 6233822

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    Linear subsets of nonlinear sets in topological vector spaces (English)
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    27 November 2013
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    lineability
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    spaceability
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    algebrability
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    real analysis
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    complex analysis
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    special functions
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    operator theory
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    Baire category theorem
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    hypercyclic manifolds
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    zeros of polynomials
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    In the last few decades, increasing attention has been given to the following type of ``phenomenon'': We are given a linear space \(X\) and a (possibly strange) property \(\mathcal P\) that some elements of \(X\) might satisfy. Next, we consider the subset \(\mathcal S(\mathcal P) \subset X\), \(\mathcal S(\mathcal P) \equiv \{x \in X \;| \;x\) satisfies \(\mathcal P\} \cup \{0\}\). The motivation for this fascinating survey is the fact that in many cases, for many \(X\) and many \(\mathcal P\), the set \(\mathcal S(\mathcal P)\) contains large linear subspaces. If this occurs, we say that \(\mathcal P\) is a lineable property.NEWLINENEWLINEAs an example, let \(X = \mathcal C[0,1]\) and let \(\mathcal P\) be the property that a function is continuous but nowhere differentiable. It is known that this is a lineable property. It is also spaceable; that is, the set \(\mathcal S(\mathcal P)\) contains an infinite-dimensional closed subspace (cf. [\textit{V. Fonf} et al., C. R. Acad. Bulg. Sci. 52, No. 11--12, 13--16 (1999; Zbl 0945.26010)] and [\textit{L. Rodríguez-Piazza}, Proc. Am. Math. Soc. 123, No. 12, 3649--3654 (1995; Zbl 0844.46007)]). In fact, even more is true: \(\mathcal S(\mathcal P)\) is algebrable, essentially meaning that there is an infinitely generated algebra \(\mathcal A \subset \mathcal S(\mathcal P)\) (cf. [\textit{F. Bayart} and \textit{L. Quarta}, Isr. J. Math. 158, 285--296 (2007; Zbl 1138.46017)]). It seems difficult to guess what properties are lineable, spaceable, algebrable, or one of several other natural conditions. For instance, with the same \(X = \mathcal C[0,1]\) but now \(\mathcal P\) being the much nicer property that the function is everywhere differentiable, it turns out that \(\mathcal S(\mathcal P)\) is nothing more than lineable (noting, e.g., that each polynomial is in \(\mathcal S(\mathcal P)\)).NEWLINENEWLINENearly half of this article deals with such situations, where \(X\) is a space of continuous, or differentiable, or measurable, or \dots functions. Another section focuses on the relation between lineability and hypercyclicity. Recall that if \(X\) is a separable Banach or Fréchet space, an operator \(T:X \to X\) is called hypercyclic on \(X\) if there is a vector \(x_0 \in X\) whose orbit under \(T\), \(\mathrm{orb}(T,x_0) = \{x_0, T(x_0), T^2(x_0),\dots\}\) is dense in \(X.\) In this case, we say that \(x_0\) is a hypercyclic vector for \(T\). One basic example, due to \textit{S. Rolewicz} [Stud. Math. 32, 17--22 (1969; Zbl 0174.44203)], is the weighted backward shift \(2B:\ell_2 \to \ell_2\), \(2B(x_1,x_2,\dots) = 2(x_2,x_3,\dots)\). It was observed by \textit{P. S. Bourdon} [Proc. Am. Math. Soc. 118, No. 3, 845--847 (1993; Zbl 0809.47005)] that if \(T\) is hypercyclic on \(X\), then \(\{x \in X \;| \;x\) is hypercyclic for \(T \} \cup \{0\}\) contains an infinite-dimensional vector space. In fact, in the particular example of the weighted backward shift, this set does not contain an infinite-dimensional closed subspace. In other words, we have lineability but not spaceability for such a hypercyclic operator. On the other hand, when \(X = \mathcal H(\mathbb C),\) the space of entire functions, and \(D\) is the (somewhat) analogous differentiation operator, \(D(f) = f^\prime,\) then \textit{G. R. MacLane} [J. Anal. Math. 2, 72--87 (1952; Zbl 0049.05603)] showed that there are indeed functions \(f_0 \in \mathcal H(\mathbb C)\) whose orbit under \(D\) is dense in \(\mathcal H(\mathbb C)\). Moreover, \textit{S. Shkarin} [Isr. J. Math. 180, 271--283 (2010; Zbl 1218.47017)] proved that there is a closed subspace of such functions \(f_0\) (assuming, of course, that the \(0\)-function is included).NEWLINENEWLINEThe article concludes with a discussion of the existence of linear subspaces of \(P^{-1}(0),\) where \(P:X \to {\mathbb {K}}\) is a \({\mathbb {K}} = \mathbb R\)- or \(\mathbb C\)-valued polynomial on a finite- or infinite-dimensional space \(X\) such that \(P(0) = 0\), followed by an extensive bibliography.NEWLINENEWLINEOne must be careful to avoid making sweeping generalizations about lineability properties. For instance, building on an example of \textit{C. J. Read} [``Banach spaces with no proximinal subspaces of codimension 2'', \url{arXiv:1307.7958}, Isr. J. Math. 223, 493--504 (2018; Zbl 1397.46010)], very recently \textit{M. Rmoutil} [Private communication, J. Funct. Anal. 272, No.~3, 918--928 (2017; Zbl 1359.46008)] has solved an important lineability problem due to \textit{P. Bandyopadhyay} and \textit{G. Godefroy} [J. Convex Anal. 13, No. 3--4, 489--497 (2006; Zbl 1118.46024)] involving the Bishop-Phelps theorem: There is a renorming of \(c_0\) such that the set of norm-attaining continuous linear functionals in its dual does not contain a 2-dimensional subspace. In another direction, in very many cases, the set \(\mathcal S(\mathcal P)\) contains a dense \(G_\delta\) set, and so it is natural to wonder whether the Baire category theorem is somehow related to lineability. As the authors note, this does not appear to be the case. For instance, using \(X = \mathcal C[0,1]\), where only \(\mathbb R\)-valued functions are considered [\textit{V. I. Gurariy} and \textit{L. Quarta}, J. Math. Anal. Appl. 294, No. 1, 62--72 (2004; Zbl 1053.46014)], let \(\mathcal P = \{f \in X \;| \;f\) attains its maximum at exactly one point \(\} \cup \{0\}\). They showed that, although \(\mathcal S(\mathcal P)\) is a dense \(G_\delta\) in \(\mathcal C[0,1],\) it does not even contain a 2-dimensional subspace.NEWLINENEWLINETo summarize, this survey is likely to be, and deserves to be, studied for some considerable time.
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