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A Banach space is a complete normed vector space: A vector space equipped with a norm such that every Cauchy sequence converges.

16 votes

Is the strong operator topology metrizable?

No, it is not sequential (hence non-metrisable) unless $X$ is finite-dimensional. Otherwise, let $(z_n)_{n=1}^\infty$ be a linearly independent sequence that is dense in $X$. For each $k$ we may consi …
Tomasz Kania's user avatar
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16 votes
Accepted

Is the $L^1$-space dual to a Banach space

OP's question was about being isomorphic to a dual space so we need to observe that $L_1$ lacks the Radon–Nikodym property, which is invariant under isomorphisms, and separable dual spaces have this p …
Tomasz Kania's user avatar
  • 11.3k
15 votes
Accepted

$c_0$ is not isometrically isomorphic to $c$

The (multiplicative) Banach–Mazur distance between $c$ and $c_0$ is exactly 3: M. Cambern, On mappings of sequence spaces, Studia Math. 30. (1968), 73-77. Let me take this opportunity to adverti …
Tomasz Kania's user avatar
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13 votes
Accepted

On $C(K)$ spaces embeddable into the Banach space $c_0$

The Szlenk index is the answer. A space $C(K)$, where $K$ is infinite compact Hausdorff space, is embeddable into $c_0$ if and only if $K$ is homeomorphic to an ordinal below $\omega^\omega$ and if …
Tomasz Kania's user avatar
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13 votes

Nonseparable counterexamples in analysis

Sobczyk's theorem: if $Z$ is a subspace of a separable Banach space $X$ that is isomorphic to $c_0$, then $Z$ is complemented in $X$, fails for many non-separable spaces such as $X=\ell_\infty\cong C( …
13 votes
Accepted

Decomposable Banach Spaces

According to the recent preprint by Koszmider, Shelah and Świętek under the generalised continuum hypothesis there is no such bound. In particular, one cannot prove the existence of such a bound worki …
Tomasz Kania's user avatar
  • 11.3k
12 votes

Do non-stable Banach spaces exist?

This is the famous Banach's hyperplane problem that was solved in the negative by W. T. Gowers. There is a whole industry in Banach space theory concerning spaces which have even more peculiar propert …
Tomasz Kania's user avatar
  • 11.3k
12 votes
Accepted

Existence of injective compact operators

No, for cardinality reasons. The range of a compact operator is norm-separable hence has cardinality continuum (if non-zero). It is then enough to take $X$ to have bigger cardinality, for example, $X …
Tomasz Kania's user avatar
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11 votes
Accepted

What is the smallest Lipschitz constant of a Lipschitz retraction of $\ell_\infty([0,1])$ on...

Yes, as we have this theorem of Nigel Kalton: Let $K$ be a compact metric space. Then $C(K)$ is an absolute 2-Lipschitz retract. Please see [1] for details. [1] Kalton, Nigel J. "Extending Lipschitz …
Tomasz Kania's user avatar
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11 votes

B(H) as a direct sum of a closed two sided ideal and a subalgebra

Let me complement Bill's answer. It is basically the same idea. Gramsch and Luft described the lattice of closed ideals of $B(H)$ where is $H$ a non-separable Hilbert space. B. Gramsch, Eine Ide …
Tomasz Kania's user avatar
  • 11.3k
11 votes

Whether Krein-Milman property implies Radon-Nikodym property

These two properties are equivalent for: Dual spaces (R. E. Huff, P. D. Morris, Proc. Amer. Math. Soc. 49 (1975), 104-108, C. Stegall, Trans. Amer. Math. Soc. 206 (1975), 213-223, C. Stegall, Trans. …
Tomasz Kania's user avatar
  • 11.3k
11 votes
Accepted

Subspaces isomorphic with quotients

Every separable Banach space is a quotient of $\ell_1$, so in particular every subspace of $\ell_1$ is a quotient of $\ell_1$.
Tomasz Kania's user avatar
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10 votes

If the cardinality of $B(X)$, the space of operators on $X$, is continuum, must $X$ be separ...

Assume Martin's axiom and the negation of CH. Then $2^{\omega_1}=\mathfrak c$. Let $X=\ell_2(\omega_1)$. Every operator on $X$ is determined by its values on a dense set of cardinality $\omega_1$, hen …
Tomasz Kania's user avatar
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10 votes
Accepted

Does separability of the strong operator topology imply separability of the underlying space?

Let $y$ be a norm-one vector in $X$. Consider the evaluation map $T\mapsto Ty$, which is a (linear) surjection from $B(X)$ onto $X$. This map is SOT-norm continuous. Indeed, suppose that $T_n\to T$ in …
Tomasz Kania's user avatar
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9 votes
Accepted

Radon-Nikodym property for space of signed measures

The spaces you are interested in are abstractly AL-spaces and by Kakutani's representation theorem, they can be represented as $L_1(\mu)$ for some measure. In particular, they have the RNP if and only …
Tomasz Kania's user avatar
  • 11.3k

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