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gsa
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Take a look at the papers "Über Normtopologien in linearen Räumen" (Link to the article) as well as "Über vollständige Normtopologien in linearen Räumen" (Link to the article) by D. Laugwitz.

He proves the following. Let $E$ be a vector space, denote by $a(E)$ the cardinality of a Hamel basis of $E$ and denote by $n(E)$ the number of mutually non-equivalent norms on $E$. Then

$$n(E) = 2^{a(E)}, \quad\text{if}~\mathfrak{c}\leq a(E),$$ $$ \mathfrak{c}\leq 2^{a(E)} \leq n(E) \leq 2^{\mathfrak{c}}, \quad\text{if}~\aleph_0\leq a(E)<\mathfrak{c},$$

and (of course)

$$ n(E) = 1 \quad\text{if}~1\leq a(E) \leq \aleph_0.$$$$ n(E) = 1 \quad\text{if}~1\leq a(E) < \aleph_0.$$

Here $\aleph_0$ denotes the cardinality of countable sets and $\mathfrak{c}$ the cardinality of the continuum. In the second article you can find the answer for complete norms.

Take a look at the papers "Über Normtopologien in linearen Räumen" (Link to the article) as well as "Über vollständige Normtopologien in linearen Räumen" (Link to the article) by D. Laugwitz.

He proves the following. Let $E$ be a vector space, denote by $a(E)$ the cardinality of a Hamel basis of $E$ and denote by $n(E)$ the number of mutually non-equivalent norms on $E$. Then

$$n(E) = 2^{a(E)}, \quad\text{if}~\mathfrak{c}\leq a(E),$$ $$ \mathfrak{c}\leq 2^{a(E)} \leq n(E) \leq 2^{\mathfrak{c}}, \quad\text{if}~\aleph_0\leq a(E)<\mathfrak{c},$$

and (of course)

$$ n(E) = 1 \quad\text{if}~1\leq a(E) \leq \aleph_0.$$

Here $\aleph_0$ denotes the cardinality of countable sets and $\mathfrak{c}$ the cardinality of the continuum. In the second article you can find the answer for complete norms.

Take a look at the papers "Über Normtopologien in linearen Räumen" (Link to the article) as well as "Über vollständige Normtopologien in linearen Räumen" (Link to the article) by D. Laugwitz.

He proves the following. Let $E$ be a vector space, denote by $a(E)$ the cardinality of a Hamel basis of $E$ and denote by $n(E)$ the number of mutually non-equivalent norms on $E$. Then

$$n(E) = 2^{a(E)}, \quad\text{if}~\mathfrak{c}\leq a(E),$$ $$ \mathfrak{c}\leq 2^{a(E)} \leq n(E) \leq 2^{\mathfrak{c}}, \quad\text{if}~\aleph_0\leq a(E)<\mathfrak{c},$$

and (of course)

$$ n(E) = 1 \quad\text{if}~1\leq a(E) < \aleph_0.$$

Here $\aleph_0$ denotes the cardinality of countable sets and $\mathfrak{c}$ the cardinality of the continuum. In the second article you can find the answer for complete norms.

fixed typo
Source Link
gsa
gsa
  • 655
  • 6
  • 12

I don't know the answer for infinite dimensional vector spaces. But every finite dimensional vector space is complete, so it might also be interesting to consider Banach spaces. In this case the answer is given byTake a look at the paper of Laugwitz:papers "Über Normtopologien in linearen Räumen" (Link to the article) as well as "Über vollständige Normtopologien in linearen Räumen" (Link to the article) by D. GivenLaugwitz.

He proves the following. Let $E$ be a Banachvector space, denote by $a(E)$ the cardinality of a Hamel basis of $E$, Laugwitz shows how to construct sets and denote by $n(E)$ the number of mutually non-equivalent complete norms on $E$ with cardinality as large as. Then

$$n(E) = 2^{a(E)}, \quad\text{if}~\mathfrak{c}\leq a(E),$$ $$ \mathfrak{c}\leq 2^{a(E)} \leq n(E) \leq 2^{\mathfrak{c}}, \quad\text{if}~\aleph_0\leq a(E)<\mathfrak{c},$$

and (of course)

$$ n(E) = 1 \quad\text{if}~1\leq a(E) \leq \aleph_0.$$

Here $\aleph_0$ denotes the power setcardinality of a Hamel basiscountable sets and he also shows that this cannot be improved$\mathfrak{c}$ the cardinality of the continuum. In the second article you can find the answer for complete norms.

I don't know the answer for infinite dimensional vector spaces. But every finite dimensional vector space is complete, so it might also be interesting to consider Banach spaces. In this case the answer is given by the paper of Laugwitz: "Über vollständige Normtopologien in linearen Räumen" (Link to the article). Given a Banach space $E$, Laugwitz shows how to construct sets of mutually non-equivalent complete norms on $E$ with cardinality as large as the power set of a Hamel basis and he also shows that this cannot be improved.

Take a look at the papers "Über Normtopologien in linearen Räumen" (Link to the article) as well as "Über vollständige Normtopologien in linearen Räumen" (Link to the article) by D. Laugwitz.

He proves the following. Let $E$ be a vector space, denote by $a(E)$ the cardinality of a Hamel basis of $E$ and denote by $n(E)$ the number of mutually non-equivalent norms on $E$. Then

$$n(E) = 2^{a(E)}, \quad\text{if}~\mathfrak{c}\leq a(E),$$ $$ \mathfrak{c}\leq 2^{a(E)} \leq n(E) \leq 2^{\mathfrak{c}}, \quad\text{if}~\aleph_0\leq a(E)<\mathfrak{c},$$

and (of course)

$$ n(E) = 1 \quad\text{if}~1\leq a(E) \leq \aleph_0.$$

Here $\aleph_0$ denotes the cardinality of countable sets and $\mathfrak{c}$ the cardinality of the continuum. In the second article you can find the answer for complete norms.

Source Link
gsa
gsa
  • 655
  • 6
  • 12

I don't know the answer for infinite dimensional vector spaces. But every finite dimensional vector space is complete, so it might also be interesting to consider Banach spaces. In this case the answer is given by the paper of Laugwitz: "Über vollständige Normtopologien in linearen Räumen" (Link to the article). Given a Banach space $E$, Laugwitz shows how to construct sets of mutually non-equivalent complete norms on $E$ with cardinality as large as the power set of a Hamel basis and he also shows that this cannot be improved.