Mixing
Is the span of all invertible functions equal to the span of all functions?
Clash Royale CLAN TAG#URR8PPP
up vote
4
down vote
favorite
In Koopman analysis, any non-linear dynamical system can be exactly represented as an infinte dimensional linear operator acting on a Hilbert space of all functions. I am looking to prove that the same property holds (or doesn't hold) when you look at only invertible functions.
I think this comes down to the question of showing equality between the span of invertible functions and the span of all functions - but I am not really familiar with how to do this, or if this is even a well posed question as I have presented it
vector-spaces hilbert-spaces nonlinear-system
edited Jul 24 at 0:15
Anonymous
4,8033940
asked Jul 23 at 23:54
Robert
9110
add a comment |Â
up vote
4
down vote
favorite
In Koopman analysis, any non-linear dynamical system can be exactly represented as an infinte dimensional linear operator acting on a Hilbert space of all functions. I am looking to prove that the same property holds (or doesn't hold) when you look at only invertible functions.
I think this comes down to the question of showing equality between the span of invertible functions and the span of all functions - but I am not really familiar with how to do this, or if this is even a well posed question as I have presented it
vector-spaces hilbert-spaces nonlinear-system
edited Jul 24 at 0:15
Anonymous
4,8033940
asked Jul 23 at 23:54
Robert
9110
2
I don't know anything about Koopman theory, but: 1) by "Hilbert space if all functions", you mean "Hilbert space of all square-integrable functions"? 2) By "invertible", you mean "vanishing nowhere"? 3) The "space" of invertible functions is not a vector space.
– PhoemueX
Jul 24 at 4:05
1) I don't think I mean just square-integrable functions although I think this set of functions does span the same space. 2) By invertible I mean that there exists a $g$ such that if $f(x) = y$, $g(y) = x$ which probably means vanishing nowhere and that $f$ is bijective. 3) I don't think this is a problem since I am not trying to span the space of invertible functions. The goal is to span the (infinite dimensional) space of all functions using an (infinite dimensional) basis set of invertible functions.
– Robert
Jul 24 at 18:30
add a comment |Â
up vote
4
down vote
favorite
up vote
4
down vote
favorite
In Koopman analysis, any non-linear dynamical system can be exactly represented as an infinte dimensional linear operator acting on a Hilbert space of all functions. I am looking to prove that the same property holds (or doesn't hold) when you look at only invertible functions.
I think this comes down to the question of showing equality between the span of invertible functions and the span of all functions - but I am not really familiar with how to do this, or if this is even a well posed question as I have presented it
vector-spaces hilbert-spaces nonlinear-system
edited Jul 24 at 0:15
Anonymous
4,8033940
asked Jul 23 at 23:54
Robert
9110
In Koopman analysis, any non-linear dynamical system can be exactly represented as an infinte dimensional linear operator acting on a Hilbert space of all functions. I am looking to prove that the same property holds (or doesn't hold) when you look at only invertible functions.
I think this comes down to the question of showing equality between the span of invertible functions and the span of all functions - but I am not really familiar with how to do this, or if this is even a well posed question as I have presented it
vector-spaces hilbert-spaces nonlinear-system
edited Jul 24 at 0:15
Anonymous
4,8033940
asked Jul 23 at 23:54
Robert
9110
edited Jul 24 at 0:15
Anonymous
4,8033940
edited Jul 24 at 0:15
Anonymous
4,8033940
edited Jul 24 at 0:15
Anonymous
4,8033940
4,8033940
asked Jul 23 at 23:54
Robert
9110
asked Jul 23 at 23:54
Robert
9110
asked Jul 23 at 23:54
Robert
9110
9110
2
I don't know anything about Koopman theory, but: 1) by "Hilbert space if all functions", you mean "Hilbert space of all square-integrable functions"? 2) By "invertible", you mean "vanishing nowhere"? 3) The "space" of invertible functions is not a vector space.
– PhoemueX
Jul 24 at 4:05
1) I don't think I mean just square-integrable functions although I think this set of functions does span the same space. 2) By invertible I mean that there exists a $g$ such that if $f(x) = y$, $g(y) = x$ which probably means vanishing nowhere and that $f$ is bijective. 3) I don't think this is a problem since I am not trying to span the space of invertible functions. The goal is to span the (infinite dimensional) space of all functions using an (infinite dimensional) basis set of invertible functions.
– Robert
Jul 24 at 18:30
add a comment |Â
2
I don't know anything about Koopman theory, but: 1) by "Hilbert space if all functions", you mean "Hilbert space of all square-integrable functions"? 2) By "invertible", you mean "vanishing nowhere"? 3) The "space" of invertible functions is not a vector space.
– PhoemueX
Jul 24 at 4:05
1) I don't think I mean just square-integrable functions although I think this set of functions does span the same space. 2) By invertible I mean that there exists a $g$ such that if $f(x) = y$, $g(y) = x$ which probably means vanishing nowhere and that $f$ is bijective. 3) I don't think this is a problem since I am not trying to span the space of invertible functions. The goal is to span the (infinite dimensional) space of all functions using an (infinite dimensional) basis set of invertible functions.
– Robert
Jul 24 at 18:30
2
2
I don't know anything about Koopman theory, but: 1) by "Hilbert space if all functions", you mean "Hilbert space of all square-integrable functions"? 2) By "invertible", you mean "vanishing nowhere"? 3) The "space" of invertible functions is not a vector space.
– PhoemueX
Jul 24 at 4:05
I don't know anything about Koopman theory, but: 1) by "Hilbert space if all functions", you mean "Hilbert space of all square-integrable functions"? 2) By "invertible", you mean "vanishing nowhere"? 3) The "space" of invertible functions is not a vector space.
– PhoemueX
Jul 24 at 4:05
1) I don't think I mean just square-integrable functions although I think this set of functions does span the same space. 2) By invertible I mean that there exists a $g$ such that if $f(x) = y$, $g(y) = x$ which probably means vanishing nowhere and that $f$ is bijective. 3) I don't think this is a problem since I am not trying to span the space of invertible functions. The goal is to span the (infinite dimensional) space of all functions using an (infinite dimensional) basis set of invertible functions.
– Robert
Jul 24 at 18:30
1) I don't think I mean just square-integrable functions although I think this set of functions does span the same space. 2) By invertible I mean that there exists a $g$ such that if $f(x) = y$, $g(y) = x$ which probably means vanishing nowhere and that $f$ is bijective. 3) I don't think this is a problem since I am not trying to span the space of invertible functions. The goal is to span the (infinite dimensional) space of all functions using an (infinite dimensional) basis set of invertible functions.
– Robert
Jul 24 at 18:30
add a comment |Â
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
active
oldest
votes
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2860884%2fis-the-span-of-all-invertible-functions-equal-to-the-span-of-all-functions%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
2
I don't know anything about Koopman theory, but: 1) by "Hilbert space if all functions", you mean "Hilbert space of all square-integrable functions"? 2) By "invertible", you mean "vanishing nowhere"? 3) The "space" of invertible functions is not a vector space.
– PhoemueX
Jul 24 at 4:05
1) I don't think I mean just square-integrable functions although I think this set of functions does span the same space. 2) By invertible I mean that there exists a $g$ such that if $f(x) = y$, $g(y) = x$ which probably means vanishing nowhere and that $f$ is bijective. 3) I don't think this is a problem since I am not trying to span the space of invertible functions. The goal is to span the (infinite dimensional) space of all functions using an (infinite dimensional) basis set of invertible functions.
– Robert
Jul 24 at 18:30