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Is there a notion of a continuous basis of a Banach space?
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 23:30 UTC (7:30pm US/Eastern)Fourier Transform: Understanding change of basis property with ideas from linear algebraCan Fourier transform be seen as a decomposition over a basis in a space of tempered distributionsFunctions as integrals of basis functionsBasis of $L^2(mathbb R)$ and Fourier transform.Is the kernel in an integral transform considered as some generalized basis?How to define the Fourier transform on arbitrary Hilbert spaces?Is there a notion of basis for Banach spaces?Schauder basis for a separable Banach spaceWhat is the difference between a Hamel basis and a Schauder basis?Hamel basis for subspacesHamel basis and Banach spacesWhy isn't every Hamel basis a Schauder basis?Is there any example of a sequence in Banach space that is not a basic sequence?Importance of a basic sequence in Banach Space TheoryGive an unbounded linear functional on Hilbert spaceIs a linearly independent set whose span is dense a Schauder basis?
$begingroup$
If $X$ is a Banach space, then a Hamel basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as a linear combination of elements of $B$. And a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$.
But my question is, is there a notion of a “continuous basis” of a Banach space? That is, a subset $B$ of $X$ such that every element of $X$ can be written uniquely in terms of some kind of integral involving elements of $B$.
I’m not sure what the integral should look like, but one possibility is this. We define some function $f:mathbbRrightarrow X$, and we let $B$ be the range of $f$. And then for any $xin X$, there exists a unique function $g:mathbbRrightarrowmathbbR$ such that $x = int_-infty^infty g(t)f(t)dt$, where this is a Bochner integral. And if that’s the case we say that $B$ is a continuous basis for $X$. Does any of this make sense?
EDIT: I've realized that my question is related to a whole bunch of other topics, including Fourier transforms, Rigged Hilbert Spaces, and Spectral Theory. See this answer, this answer, this question, this question, and this question.
functional-analysis banach-spaces distribution-theory bochner-spaces schauder-basis
asked Apr 2 at 14:31
Keshav SrinivasanKeshav Srinivasan
2,50821547
$endgroup$
add a comment |
$begingroup$
If $X$ is a Banach space, then a Hamel basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as a linear combination of elements of $B$. And a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$.
But my question is, is there a notion of a “continuous basis” of a Banach space? That is, a subset $B$ of $X$ such that every element of $X$ can be written uniquely in terms of some kind of integral involving elements of $B$.
I’m not sure what the integral should look like, but one possibility is this. We define some function $f:mathbbRrightarrow X$, and we let $B$ be the range of $f$. And then for any $xin X$, there exists a unique function $g:mathbbRrightarrowmathbbR$ such that $x = int_-infty^infty g(t)f(t)dt$, where this is a Bochner integral. And if that’s the case we say that $B$ is a continuous basis for $X$. Does any of this make sense?
EDIT: I've realized that my question is related to a whole bunch of other topics, including Fourier transforms, Rigged Hilbert Spaces, and Spectral Theory. See this answer, this answer, this question, this question, and this question.
functional-analysis banach-spaces distribution-theory bochner-spaces schauder-basis
asked Apr 2 at 14:31
Keshav SrinivasanKeshav Srinivasan
2,50821547
$endgroup$
$begingroup$
Some context might help. Every vector space already has a Hamel basis so it's unclear what you hope to gain.
$endgroup$
– Ben W
Apr 5 at 13:47
$begingroup$
@BenW A Hamel basis involves finite linear combinations. A Schauder basis involves countably infinite linear combinations. I want something that involves “uncountable linear combinations”, i.e. integrals.
$endgroup$
– Keshav Srinivasan
Apr 5 at 14:14
$begingroup$
@BenW I just made an edit that provides more context.
$endgroup$
– Keshav Srinivasan
Apr 8 at 1:22
add a comment |
$begingroup$
If $X$ is a Banach space, then a Hamel basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as a linear combination of elements of $B$. And a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$.
But my question is, is there a notion of a “continuous basis” of a Banach space? That is, a subset $B$ of $X$ such that every element of $X$ can be written uniquely in terms of some kind of integral involving elements of $B$.
I’m not sure what the integral should look like, but one possibility is this. We define some function $f:mathbbRrightarrow X$, and we let $B$ be the range of $f$. And then for any $xin X$, there exists a unique function $g:mathbbRrightarrowmathbbR$ such that $x = int_-infty^infty g(t)f(t)dt$, where this is a Bochner integral. And if that’s the case we say that $B$ is a continuous basis for $X$. Does any of this make sense?
EDIT: I've realized that my question is related to a whole bunch of other topics, including Fourier transforms, Rigged Hilbert Spaces, and Spectral Theory. See this answer, this answer, this question, this question, and this question.
functional-analysis banach-spaces distribution-theory bochner-spaces schauder-basis
asked Apr 2 at 14:31
Keshav SrinivasanKeshav Srinivasan
2,50821547
$endgroup$
If $X$ is a Banach space, then a Hamel basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as a linear combination of elements of $B$. And a Schauder basis of $X$ is a subset $B$ of $X$ such that every element of $X$ can be written uniquely as an infinite linear combination of elements of $B$.
But my question is, is there a notion of a “continuous basis” of a Banach space? That is, a subset $B$ of $X$ such that every element of $X$ can be written uniquely in terms of some kind of integral involving elements of $B$.
I’m not sure what the integral should look like, but one possibility is this. We define some function $f:mathbbRrightarrow X$, and we let $B$ be the range of $f$. And then for any $xin X$, there exists a unique function $g:mathbbRrightarrowmathbbR$ such that $x = int_-infty^infty g(t)f(t)dt$, where this is a Bochner integral. And if that’s the case we say that $B$ is a continuous basis for $X$. Does any of this make sense?
EDIT: I've realized that my question is related to a whole bunch of other topics, including Fourier transforms, Rigged Hilbert Spaces, and Spectral Theory. See this answer, this answer, this question, this question, and this question.
functional-analysis banach-spaces distribution-theory bochner-spaces schauder-basis
functional-analysis banach-spaces distribution-theory bochner-spaces schauder-basis
asked Apr 2 at 14:31
Keshav SrinivasanKeshav Srinivasan
2,50821547
asked Apr 2 at 14:31
Keshav SrinivasanKeshav Srinivasan
2,50821547
edited Apr 8 at 1:22
Keshav Srinivasan
asked Apr 2 at 14:31
Keshav SrinivasanKeshav Srinivasan
2,50821547
asked Apr 2 at 14:31
Keshav SrinivasanKeshav Srinivasan
2,50821547
asked Apr 2 at 14:31
Keshav SrinivasanKeshav Srinivasan
2,50821547
2,50821547
$begingroup$
Some context might help. Every vector space already has a Hamel basis so it's unclear what you hope to gain.
$endgroup$
– Ben W
Apr 5 at 13:47
$begingroup$
@BenW A Hamel basis involves finite linear combinations. A Schauder basis involves countably infinite linear combinations. I want something that involves “uncountable linear combinations”, i.e. integrals.
$endgroup$
– Keshav Srinivasan
Apr 5 at 14:14
$begingroup$
@BenW I just made an edit that provides more context.
$endgroup$
– Keshav Srinivasan
Apr 8 at 1:22
add a comment |
$begingroup$
Some context might help. Every vector space already has a Hamel basis so it's unclear what you hope to gain.
$endgroup$
– Ben W
Apr 5 at 13:47
$begingroup$
@BenW A Hamel basis involves finite linear combinations. A Schauder basis involves countably infinite linear combinations. I want something that involves “uncountable linear combinations”, i.e. integrals.
$endgroup$
– Keshav Srinivasan
Apr 5 at 14:14
$begingroup$
@BenW I just made an edit that provides more context.
$endgroup$
– Keshav Srinivasan
Apr 8 at 1:22
$begingroup$
Some context might help. Every vector space already has a Hamel basis so it's unclear what you hope to gain.
$endgroup$
– Ben W
Apr 5 at 13:47
$begingroup$
Some context might help. Every vector space already has a Hamel basis so it's unclear what you hope to gain.
$endgroup$
– Ben W
Apr 5 at 13:47
$begingroup$
@BenW A Hamel basis involves finite linear combinations. A Schauder basis involves countably infinite linear combinations. I want something that involves “uncountable linear combinations”, i.e. integrals.
$endgroup$
– Keshav Srinivasan
Apr 5 at 14:14
$begingroup$
@BenW A Hamel basis involves finite linear combinations. A Schauder basis involves countably infinite linear combinations. I want something that involves “uncountable linear combinations”, i.e. integrals.
$endgroup$
– Keshav Srinivasan
Apr 5 at 14:14
$begingroup$
@BenW I just made an edit that provides more context.
$endgroup$
– Keshav Srinivasan
Apr 8 at 1:22
$begingroup$
@BenW I just made an edit that provides more context.
$endgroup$
– Keshav Srinivasan
Apr 8 at 1:22
add a comment |
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$begingroup$
Some context might help. Every vector space already has a Hamel basis so it's unclear what you hope to gain.
$endgroup$
– Ben W
Apr 5 at 13:47
$begingroup$
@BenW A Hamel basis involves finite linear combinations. A Schauder basis involves countably infinite linear combinations. I want something that involves “uncountable linear combinations”, i.e. integrals.
$endgroup$
– Keshav Srinivasan
Apr 5 at 14:14
$begingroup$
@BenW I just made an edit that provides more context.
$endgroup$
– Keshav Srinivasan
Apr 8 at 1:22