Mixing
What exactly is a non-linear orthogonal projection?
$begingroup$
In a Hilbert space of bounded integrable functions, let $P$ be an operator such that
$$P(f(x)) = frac{f(x)+|f(x)|}{2}$$
The complement of $P$ can be written as $Q = I - P$, hence
$$Q(f(x)) = frac{f(x)-|f(x)|}{2}$$
Both $P$ and $Q$ are idempotent. However, in contrary to many textbook examples, $P$ and $Q$ are not linear, nor are they Hermitian. Still
$$langle P(f(x))|Q(f(x))rangle = 0$$
holds. Are we allowed to call $P$ an orthogonal projection and $Q$ its orthogonal complement?
(Sorry for my inaccurate wording, I am a physicist, not a mathematician.)
hilbert-spaces projection
asked Jan 29 at 20:56
user3609959user3609959
212
$endgroup$
add a comment |
$begingroup$
In a Hilbert space of bounded integrable functions, let $P$ be an operator such that
$$P(f(x)) = frac{f(x)+|f(x)|}{2}$$
The complement of $P$ can be written as $Q = I - P$, hence
$$Q(f(x)) = frac{f(x)-|f(x)|}{2}$$
Both $P$ and $Q$ are idempotent. However, in contrary to many textbook examples, $P$ and $Q$ are not linear, nor are they Hermitian. Still
$$langle P(f(x))|Q(f(x))rangle = 0$$
holds. Are we allowed to call $P$ an orthogonal projection and $Q$ its orthogonal complement?
(Sorry for my inaccurate wording, I am a physicist, not a mathematician.)
hilbert-spaces projection
asked Jan 29 at 20:56
user3609959user3609959
212
$endgroup$
2
$begingroup$
No. The term projection in this context is reserved for linear maps
$endgroup$
– Hagen von Eitzen
Jan 29 at 21:13
1
$begingroup$
I would call them like 'nonlinear projections'.
$endgroup$
– Berci
Jan 29 at 21:57
$begingroup$
Moreover, the "space of bounded intregrable functions" isn't (in most cases) a Hilbert space with the usual $L^2$-norm, since it is not complete. Example: $f_n(x) = x^{-1/4} 1_{[1/n,1]}$ converges in $L^2(0,1)$ to $f(x) = x^{-1/4}$.
$endgroup$
– p4sch
Jan 30 at 16:53
$begingroup$
@p4sch Please see my answer below. In fact, any normed vector space will be sufficient to define an orthogonal projection, need not be $L^2$.
$endgroup$
– user3609959
Feb 1 at 9:52
add a comment |
$begingroup$
In a Hilbert space of bounded integrable functions, let $P$ be an operator such that
$$P(f(x)) = frac{f(x)+|f(x)|}{2}$$
The complement of $P$ can be written as $Q = I - P$, hence
$$Q(f(x)) = frac{f(x)-|f(x)|}{2}$$
Both $P$ and $Q$ are idempotent. However, in contrary to many textbook examples, $P$ and $Q$ are not linear, nor are they Hermitian. Still
$$langle P(f(x))|Q(f(x))rangle = 0$$
holds. Are we allowed to call $P$ an orthogonal projection and $Q$ its orthogonal complement?
(Sorry for my inaccurate wording, I am a physicist, not a mathematician.)
hilbert-spaces projection
asked Jan 29 at 20:56
user3609959user3609959
212
$endgroup$
In a Hilbert space of bounded integrable functions, let $P$ be an operator such that
$$P(f(x)) = frac{f(x)+|f(x)|}{2}$$
The complement of $P$ can be written as $Q = I - P$, hence
$$Q(f(x)) = frac{f(x)-|f(x)|}{2}$$
Both $P$ and $Q$ are idempotent. However, in contrary to many textbook examples, $P$ and $Q$ are not linear, nor are they Hermitian. Still
$$langle P(f(x))|Q(f(x))rangle = 0$$
holds. Are we allowed to call $P$ an orthogonal projection and $Q$ its orthogonal complement?
(Sorry for my inaccurate wording, I am a physicist, not a mathematician.)
hilbert-spaces projection
hilbert-spaces projection
asked Jan 29 at 20:56
user3609959user3609959
212
asked Jan 29 at 20:56
user3609959user3609959
212
edited Feb 1 at 9:38
user3609959
asked Jan 29 at 20:56
user3609959user3609959
212
asked Jan 29 at 20:56
user3609959user3609959
212
asked Jan 29 at 20:56
user3609959user3609959
212
212
2
$begingroup$
No. The term projection in this context is reserved for linear maps
$endgroup$
– Hagen von Eitzen
Jan 29 at 21:13
1
$begingroup$
I would call them like 'nonlinear projections'.
$endgroup$
– Berci
Jan 29 at 21:57
$begingroup$
Moreover, the "space of bounded intregrable functions" isn't (in most cases) a Hilbert space with the usual $L^2$-norm, since it is not complete. Example: $f_n(x) = x^{-1/4} 1_{[1/n,1]}$ converges in $L^2(0,1)$ to $f(x) = x^{-1/4}$.
$endgroup$
– p4sch
Jan 30 at 16:53
$begingroup$
@p4sch Please see my answer below. In fact, any normed vector space will be sufficient to define an orthogonal projection, need not be $L^2$.
$endgroup$
– user3609959
Feb 1 at 9:52
add a comment |
2
$begingroup$
No. The term projection in this context is reserved for linear maps
$endgroup$
– Hagen von Eitzen
Jan 29 at 21:13
1
$begingroup$
I would call them like 'nonlinear projections'.
$endgroup$
– Berci
Jan 29 at 21:57
$begingroup$
Moreover, the "space of bounded intregrable functions" isn't (in most cases) a Hilbert space with the usual $L^2$-norm, since it is not complete. Example: $f_n(x) = x^{-1/4} 1_{[1/n,1]}$ converges in $L^2(0,1)$ to $f(x) = x^{-1/4}$.
$endgroup$
– p4sch
Jan 30 at 16:53
$begingroup$
@p4sch Please see my answer below. In fact, any normed vector space will be sufficient to define an orthogonal projection, need not be $L^2$.
$endgroup$
– user3609959
Feb 1 at 9:52
2
2
$begingroup$
No. The term projection in this context is reserved for linear maps
$endgroup$
– Hagen von Eitzen
Jan 29 at 21:13
$begingroup$
No. The term projection in this context is reserved for linear maps
$endgroup$
– Hagen von Eitzen
Jan 29 at 21:13
1
1
$begingroup$
I would call them like 'nonlinear projections'.
$endgroup$
– Berci
Jan 29 at 21:57
$begingroup$
I would call them like 'nonlinear projections'.
$endgroup$
– Berci
Jan 29 at 21:57
$begingroup$
Moreover, the "space of bounded intregrable functions" isn't (in most cases) a Hilbert space with the usual $L^2$-norm, since it is not complete. Example: $f_n(x) = x^{-1/4} 1_{[1/n,1]}$ converges in $L^2(0,1)$ to $f(x) = x^{-1/4}$.
$endgroup$
– p4sch
Jan 30 at 16:53
$begingroup$
Moreover, the "space of bounded intregrable functions" isn't (in most cases) a Hilbert space with the usual $L^2$-norm, since it is not complete. Example: $f_n(x) = x^{-1/4} 1_{[1/n,1]}$ converges in $L^2(0,1)$ to $f(x) = x^{-1/4}$.
$endgroup$
– p4sch
Jan 30 at 16:53
$begingroup$
@p4sch Please see my answer below. In fact, any normed vector space will be sufficient to define an orthogonal projection, need not be $L^2$.
$endgroup$
– user3609959
Feb 1 at 9:52
$begingroup$
@p4sch Please see my answer below. In fact, any normed vector space will be sufficient to define an orthogonal projection, need not be $L^2$.
$endgroup$
– user3609959
Feb 1 at 9:52
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
After a lot of Googling, I am able to come up with an answer myself. There is a well-known definition of non-linear orthogonal projections. Roughly, the procedure is as follows:
- In a metric space (e.g. a Hilbert space $H$), choose any non-empty subset $M$, which typically might be a manifold, but need not necessarily be a linear subspace.
- For each element $h in H$, define a so-called distance function $$rho(h,M)=inf_{m in M} ||h-m||$$.
- In the domain of all $h$ that have a unique $m$ (i.e. a perpendicular foot point in $M$), we can define a projection $P$ such that $P(h) = m$.
To point it out explicitly, the properties of $P$ totally depend on the norm $||cdot||$ and on the choice of $M$. As a simple example, let $M$ be the unit circle in the complex plain. Then $P(z)=z/|z|$.
If, as a special case, $M$ is a closed linear subspace of $H$, then $P$ turns out to be a linear orthogonal projection.
For more details, see e.g. http://matwbn.icm.edu.pl/ksiazki/apm/apm59/apm5911.pdf.
answered Feb 1 at 9:05
user3609959user3609959
212
$endgroup$
add a comment |
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$begingroup$
After a lot of Googling, I am able to come up with an answer myself. There is a well-known definition of non-linear orthogonal projections. Roughly, the procedure is as follows:
- In a metric space (e.g. a Hilbert space $H$), choose any non-empty subset $M$, which typically might be a manifold, but need not necessarily be a linear subspace.
- For each element $h in H$, define a so-called distance function $$rho(h,M)=inf_{m in M} ||h-m||$$.
- In the domain of all $h$ that have a unique $m$ (i.e. a perpendicular foot point in $M$), we can define a projection $P$ such that $P(h) = m$.
To point it out explicitly, the properties of $P$ totally depend on the norm $||cdot||$ and on the choice of $M$. As a simple example, let $M$ be the unit circle in the complex plain. Then $P(z)=z/|z|$.
If, as a special case, $M$ is a closed linear subspace of $H$, then $P$ turns out to be a linear orthogonal projection.
For more details, see e.g. http://matwbn.icm.edu.pl/ksiazki/apm/apm59/apm5911.pdf.
answered Feb 1 at 9:05
user3609959user3609959
212
$endgroup$
add a comment |
$begingroup$
After a lot of Googling, I am able to come up with an answer myself. There is a well-known definition of non-linear orthogonal projections. Roughly, the procedure is as follows:
- In a metric space (e.g. a Hilbert space $H$), choose any non-empty subset $M$, which typically might be a manifold, but need not necessarily be a linear subspace.
- For each element $h in H$, define a so-called distance function $$rho(h,M)=inf_{m in M} ||h-m||$$.
- In the domain of all $h$ that have a unique $m$ (i.e. a perpendicular foot point in $M$), we can define a projection $P$ such that $P(h) = m$.
To point it out explicitly, the properties of $P$ totally depend on the norm $||cdot||$ and on the choice of $M$. As a simple example, let $M$ be the unit circle in the complex plain. Then $P(z)=z/|z|$.
If, as a special case, $M$ is a closed linear subspace of $H$, then $P$ turns out to be a linear orthogonal projection.
For more details, see e.g. http://matwbn.icm.edu.pl/ksiazki/apm/apm59/apm5911.pdf.
answered Feb 1 at 9:05
user3609959user3609959
212
$endgroup$
add a comment |
$begingroup$
After a lot of Googling, I am able to come up with an answer myself. There is a well-known definition of non-linear orthogonal projections. Roughly, the procedure is as follows:
- In a metric space (e.g. a Hilbert space $H$), choose any non-empty subset $M$, which typically might be a manifold, but need not necessarily be a linear subspace.
- For each element $h in H$, define a so-called distance function $$rho(h,M)=inf_{m in M} ||h-m||$$.
- In the domain of all $h$ that have a unique $m$ (i.e. a perpendicular foot point in $M$), we can define a projection $P$ such that $P(h) = m$.
To point it out explicitly, the properties of $P$ totally depend on the norm $||cdot||$ and on the choice of $M$. As a simple example, let $M$ be the unit circle in the complex plain. Then $P(z)=z/|z|$.
If, as a special case, $M$ is a closed linear subspace of $H$, then $P$ turns out to be a linear orthogonal projection.
For more details, see e.g. http://matwbn.icm.edu.pl/ksiazki/apm/apm59/apm5911.pdf.
answered Feb 1 at 9:05
user3609959user3609959
212
$endgroup$
After a lot of Googling, I am able to come up with an answer myself. There is a well-known definition of non-linear orthogonal projections. Roughly, the procedure is as follows:
- In a metric space (e.g. a Hilbert space $H$), choose any non-empty subset $M$, which typically might be a manifold, but need not necessarily be a linear subspace.
- For each element $h in H$, define a so-called distance function $$rho(h,M)=inf_{m in M} ||h-m||$$.
- In the domain of all $h$ that have a unique $m$ (i.e. a perpendicular foot point in $M$), we can define a projection $P$ such that $P(h) = m$.
To point it out explicitly, the properties of $P$ totally depend on the norm $||cdot||$ and on the choice of $M$. As a simple example, let $M$ be the unit circle in the complex plain. Then $P(z)=z/|z|$.
If, as a special case, $M$ is a closed linear subspace of $H$, then $P$ turns out to be a linear orthogonal projection.
For more details, see e.g. http://matwbn.icm.edu.pl/ksiazki/apm/apm59/apm5911.pdf.
answered Feb 1 at 9:05
user3609959user3609959
212
edited Feb 3 at 10:25
answered Feb 1 at 9:05
user3609959user3609959
212
answered Feb 1 at 9:05
user3609959user3609959
212
answered Feb 1 at 9:05
user3609959user3609959
212
212
add a comment |
add a comment |
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2
$begingroup$
No. The term projection in this context is reserved for linear maps
$endgroup$
– Hagen von Eitzen
Jan 29 at 21:13
1
$begingroup$
I would call them like 'nonlinear projections'.
$endgroup$
– Berci
Jan 29 at 21:57
$begingroup$
Moreover, the "space of bounded intregrable functions" isn't (in most cases) a Hilbert space with the usual $L^2$-norm, since it is not complete. Example: $f_n(x) = x^{-1/4} 1_{[1/n,1]}$ converges in $L^2(0,1)$ to $f(x) = x^{-1/4}$.
$endgroup$
– p4sch
Jan 30 at 16:53
$begingroup$
@p4sch Please see my answer below. In fact, any normed vector space will be sufficient to define an orthogonal projection, need not be $L^2$.
$endgroup$
– user3609959
Feb 1 at 9:52