Is the span of all invertible functions equal to the span of all functions?

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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




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    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














up vote
4
down vote

favorite
2












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




share|cite|improve this question

















  • 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












up vote
4
down vote

favorite
2









up vote
4
down vote

favorite
2






2





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




share|cite|improve this question













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





share|cite|improve this question












share|cite|improve this question




share|cite|improve this question








edited Jul 24 at 0:15









Anonymous

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asked Jul 23 at 23:54









Robert

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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












  • 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















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