Riesz, Hilbert and Hamel bases
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I was surprised to read both at PlanetMath and in Wikipedia (apparently copied from PlanetMath) that
If $H$ is a finite-dimensional [Hilbert] space, then every basis of $H$ is a Riesz basis.
I thought that in the context of Hilbert spaces the unqualified term "basis" is usually taken to mean "Hilbert basis" (i.e. orthonormal basis), not "Hamel basis" (i.e. basis in the linear-algebra sense), but under that interpretation the restriction to finite-dimensional spaces makes no sense, since every Hilbert basis is a Riesz basis (with constants $c=C=1$). Am I right in thinking that
- they mean a Hamel basis, and
- it would be preferable to disambiguate that statement by inserting "Hamel"?
terminology hilbert-spaces hamel-basis
edited Dec 28 '18 at 18:28
José Carlos Santos
176k24135244
asked Oct 28 '12 at 15:33
jorikijoriki
171k10191352
$endgroup$
add a comment |
$begingroup$
I was surprised to read both at PlanetMath and in Wikipedia (apparently copied from PlanetMath) that
If $H$ is a finite-dimensional [Hilbert] space, then every basis of $H$ is a Riesz basis.
I thought that in the context of Hilbert spaces the unqualified term "basis" is usually taken to mean "Hilbert basis" (i.e. orthonormal basis), not "Hamel basis" (i.e. basis in the linear-algebra sense), but under that interpretation the restriction to finite-dimensional spaces makes no sense, since every Hilbert basis is a Riesz basis (with constants $c=C=1$). Am I right in thinking that
- they mean a Hamel basis, and
- it would be preferable to disambiguate that statement by inserting "Hamel"?
terminology hilbert-spaces hamel-basis
edited Dec 28 '18 at 18:28
José Carlos Santos
176k24135244
asked Oct 28 '12 at 15:33
jorikijoriki
171k10191352
$endgroup$
$begingroup$
On the other hand, in the context of Banach spaces the unqualified "basis" is usually a "Schauder basis" (in my experience). Now what is the reader to assume if the Banach space is also a Hilbert space...
$endgroup$
– user53153
Dec 19 '12 at 0:32
add a comment |
$begingroup$
I was surprised to read both at PlanetMath and in Wikipedia (apparently copied from PlanetMath) that
If $H$ is a finite-dimensional [Hilbert] space, then every basis of $H$ is a Riesz basis.
I thought that in the context of Hilbert spaces the unqualified term "basis" is usually taken to mean "Hilbert basis" (i.e. orthonormal basis), not "Hamel basis" (i.e. basis in the linear-algebra sense), but under that interpretation the restriction to finite-dimensional spaces makes no sense, since every Hilbert basis is a Riesz basis (with constants $c=C=1$). Am I right in thinking that
- they mean a Hamel basis, and
- it would be preferable to disambiguate that statement by inserting "Hamel"?
terminology hilbert-spaces hamel-basis
edited Dec 28 '18 at 18:28
José Carlos Santos
176k24135244
asked Oct 28 '12 at 15:33
jorikijoriki
171k10191352
$endgroup$
I was surprised to read both at PlanetMath and in Wikipedia (apparently copied from PlanetMath) that
If $H$ is a finite-dimensional [Hilbert] space, then every basis of $H$ is a Riesz basis.
I thought that in the context of Hilbert spaces the unqualified term "basis" is usually taken to mean "Hilbert basis" (i.e. orthonormal basis), not "Hamel basis" (i.e. basis in the linear-algebra sense), but under that interpretation the restriction to finite-dimensional spaces makes no sense, since every Hilbert basis is a Riesz basis (with constants $c=C=1$). Am I right in thinking that
- they mean a Hamel basis, and
- it would be preferable to disambiguate that statement by inserting "Hamel"?
terminology hilbert-spaces hamel-basis
terminology hilbert-spaces hamel-basis
edited Dec 28 '18 at 18:28
José Carlos Santos
176k24135244
asked Oct 28 '12 at 15:33
jorikijoriki
171k10191352
edited Dec 28 '18 at 18:28
José Carlos Santos
176k24135244
asked Oct 28 '12 at 15:33
jorikijoriki
171k10191352
edited Dec 28 '18 at 18:28
José Carlos Santos
176k24135244
edited Dec 28 '18 at 18:28
José Carlos Santos
176k24135244
edited Dec 28 '18 at 18:28
José Carlos Santos
176k24135244
176k24135244
asked Oct 28 '12 at 15:33
jorikijoriki
171k10191352
asked Oct 28 '12 at 15:33
jorikijoriki
171k10191352
asked Oct 28 '12 at 15:33
jorikijoriki
171k10191352
171k10191352
$begingroup$
On the other hand, in the context of Banach spaces the unqualified "basis" is usually a "Schauder basis" (in my experience). Now what is the reader to assume if the Banach space is also a Hilbert space...
$endgroup$
– user53153
Dec 19 '12 at 0:32
add a comment |
$begingroup$
On the other hand, in the context of Banach spaces the unqualified "basis" is usually a "Schauder basis" (in my experience). Now what is the reader to assume if the Banach space is also a Hilbert space...
$endgroup$
– user53153
Dec 19 '12 at 0:32
$begingroup$
On the other hand, in the context of Banach spaces the unqualified "basis" is usually a "Schauder basis" (in my experience). Now what is the reader to assume if the Banach space is also a Hilbert space...
$endgroup$
– user53153
Dec 19 '12 at 0:32
$begingroup$
On the other hand, in the context of Banach spaces the unqualified "basis" is usually a "Schauder basis" (in my experience). Now what is the reader to assume if the Banach space is also a Hilbert space...
$endgroup$
– user53153
Dec 19 '12 at 0:32
add a comment |
1 Answer
1
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oldest
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Yes, they certainly mean a "Hamel basis" for the reasons you give. I'm not too sure how widespread the use of "basis" = "orthonormal basis" in Hilbert spaces really is, but the literature is vast. I've certainly seen this (ab)use and I would interpret the term "basis" in a Hilbert space the same way as you did and would be taken aback if it should turn out to mean anything else than "orthonormal basis". However, it seems to me (without thorough checking) that throughout functional analysis literature I know people tend to add the qualifier "orthonormal" even if there's no other kind of basis under consideration in the entire book or article.
To avoid confusion of the kind you mention, it would certainly be better to disambiguate the statement.
On a final note, I slightly prefer the term "algebraic basis" over "Hamel basis" because some people (e.g. mathworld here) reserve the term for a basis of the $mathbb{Q}$-vector space $mathbb{R}$ --- Hamel used such a basis to construct discontinuous solutions of Cauchy's functional equation.
answered Oct 28 '12 at 16:01
a ghost from the pasta ghost from the past
311
$endgroup$
$begingroup$
The last link should be correct, but the server of the Goettinger Digitalisierungszentrum seems to be down at the moment.
$endgroup$
– a ghost from the past
Oct 28 '12 at 16:02
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Yes, they certainly mean a "Hamel basis" for the reasons you give. I'm not too sure how widespread the use of "basis" = "orthonormal basis" in Hilbert spaces really is, but the literature is vast. I've certainly seen this (ab)use and I would interpret the term "basis" in a Hilbert space the same way as you did and would be taken aback if it should turn out to mean anything else than "orthonormal basis". However, it seems to me (without thorough checking) that throughout functional analysis literature I know people tend to add the qualifier "orthonormal" even if there's no other kind of basis under consideration in the entire book or article.
To avoid confusion of the kind you mention, it would certainly be better to disambiguate the statement.
On a final note, I slightly prefer the term "algebraic basis" over "Hamel basis" because some people (e.g. mathworld here) reserve the term for a basis of the $mathbb{Q}$-vector space $mathbb{R}$ --- Hamel used such a basis to construct discontinuous solutions of Cauchy's functional equation.
answered Oct 28 '12 at 16:01
a ghost from the pasta ghost from the past
311
$endgroup$
$begingroup$
The last link should be correct, but the server of the Goettinger Digitalisierungszentrum seems to be down at the moment.
$endgroup$
– a ghost from the past
Oct 28 '12 at 16:02
add a comment |
$begingroup$
Yes, they certainly mean a "Hamel basis" for the reasons you give. I'm not too sure how widespread the use of "basis" = "orthonormal basis" in Hilbert spaces really is, but the literature is vast. I've certainly seen this (ab)use and I would interpret the term "basis" in a Hilbert space the same way as you did and would be taken aback if it should turn out to mean anything else than "orthonormal basis". However, it seems to me (without thorough checking) that throughout functional analysis literature I know people tend to add the qualifier "orthonormal" even if there's no other kind of basis under consideration in the entire book or article.
To avoid confusion of the kind you mention, it would certainly be better to disambiguate the statement.
On a final note, I slightly prefer the term "algebraic basis" over "Hamel basis" because some people (e.g. mathworld here) reserve the term for a basis of the $mathbb{Q}$-vector space $mathbb{R}$ --- Hamel used such a basis to construct discontinuous solutions of Cauchy's functional equation.
answered Oct 28 '12 at 16:01
a ghost from the pasta ghost from the past
311
$endgroup$
$begingroup$
The last link should be correct, but the server of the Goettinger Digitalisierungszentrum seems to be down at the moment.
$endgroup$
– a ghost from the past
Oct 28 '12 at 16:02
add a comment |
$begingroup$
Yes, they certainly mean a "Hamel basis" for the reasons you give. I'm not too sure how widespread the use of "basis" = "orthonormal basis" in Hilbert spaces really is, but the literature is vast. I've certainly seen this (ab)use and I would interpret the term "basis" in a Hilbert space the same way as you did and would be taken aback if it should turn out to mean anything else than "orthonormal basis". However, it seems to me (without thorough checking) that throughout functional analysis literature I know people tend to add the qualifier "orthonormal" even if there's no other kind of basis under consideration in the entire book or article.
To avoid confusion of the kind you mention, it would certainly be better to disambiguate the statement.
On a final note, I slightly prefer the term "algebraic basis" over "Hamel basis" because some people (e.g. mathworld here) reserve the term for a basis of the $mathbb{Q}$-vector space $mathbb{R}$ --- Hamel used such a basis to construct discontinuous solutions of Cauchy's functional equation.
answered Oct 28 '12 at 16:01
a ghost from the pasta ghost from the past
311
$endgroup$
Yes, they certainly mean a "Hamel basis" for the reasons you give. I'm not too sure how widespread the use of "basis" = "orthonormal basis" in Hilbert spaces really is, but the literature is vast. I've certainly seen this (ab)use and I would interpret the term "basis" in a Hilbert space the same way as you did and would be taken aback if it should turn out to mean anything else than "orthonormal basis". However, it seems to me (without thorough checking) that throughout functional analysis literature I know people tend to add the qualifier "orthonormal" even if there's no other kind of basis under consideration in the entire book or article.
To avoid confusion of the kind you mention, it would certainly be better to disambiguate the statement.
On a final note, I slightly prefer the term "algebraic basis" over "Hamel basis" because some people (e.g. mathworld here) reserve the term for a basis of the $mathbb{Q}$-vector space $mathbb{R}$ --- Hamel used such a basis to construct discontinuous solutions of Cauchy's functional equation.
answered Oct 28 '12 at 16:01
a ghost from the pasta ghost from the past
311
answered Oct 28 '12 at 16:01
a ghost from the pasta ghost from the past
311
answered Oct 28 '12 at 16:01
a ghost from the pasta ghost from the past
311
answered Oct 28 '12 at 16:01
a ghost from the pasta ghost from the past
311
311
$begingroup$
The last link should be correct, but the server of the Goettinger Digitalisierungszentrum seems to be down at the moment.
$endgroup$
– a ghost from the past
Oct 28 '12 at 16:02
add a comment |
$begingroup$
The last link should be correct, but the server of the Goettinger Digitalisierungszentrum seems to be down at the moment.
$endgroup$
– a ghost from the past
Oct 28 '12 at 16:02
$begingroup$
The last link should be correct, but the server of the Goettinger Digitalisierungszentrum seems to be down at the moment.
$endgroup$
– a ghost from the past
Oct 28 '12 at 16:02
$begingroup$
The last link should be correct, but the server of the Goettinger Digitalisierungszentrum seems to be down at the moment.
$endgroup$
– a ghost from the past
Oct 28 '12 at 16:02
add a comment |
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$begingroup$
On the other hand, in the context of Banach spaces the unqualified "basis" is usually a "Schauder basis" (in my experience). Now what is the reader to assume if the Banach space is also a Hilbert space...
$endgroup$
– user53153
Dec 19 '12 at 0:32