Projection of a $L^2$ function over a set of constant functions
I don't know how to solve the following exercise:
Let $X$ be the subspace of $L^{2}(0,1)$ of a.e. constant functions. What is the projection over $X$ of a function $u in L^2(0,1)$?
First of all, $X$ is convex, and this is trivial to prove. It's also closed. So, by the characterization of projections onto closed and convex subspace I have to find a $v in X$ s.t.
(*) $langle u-v,f rangle=0$ for every $f in X$, i.e. $int_0^1 (u(x)-v(x))f(x)dx=0$, but $u$ and $f$ are constants, so I can't see any way to set that integral equals to $0$.
Maybe I could start by thinking about $Vert u - v Vert_{L^2}^2=int_0^1|u(x)-v(x)|^2dx $ and how to minimize this integral with $v in X$. The best I can do, I guess, is to set $v(x)=v=max {|u(x)|: x in [0,1] }$, but I still can't understand how to find a $u$ such that (*) holds.
EDIT:
I need also to prove that $X$ is closed.To this aim, I take a sequence ${ f_n }$ in $X$ s.t. $f_n rightarrow^{L^2} f$. I want to show that $f$ is a.e. constant. By Riesz-Fisher there exists a subsequence ${f_{n_k} } in X$ such that $f_{n_k} rightarrow f$ a.e. in $X$. How can I conclude?
functional-analysis hilbert-spaces lp-spaces projection
asked Dec 27 '18 at 14:35
VoB
673213
add a comment |
I don't know how to solve the following exercise:
Let $X$ be the subspace of $L^{2}(0,1)$ of a.e. constant functions. What is the projection over $X$ of a function $u in L^2(0,1)$?
First of all, $X$ is convex, and this is trivial to prove. It's also closed. So, by the characterization of projections onto closed and convex subspace I have to find a $v in X$ s.t.
(*) $langle u-v,f rangle=0$ for every $f in X$, i.e. $int_0^1 (u(x)-v(x))f(x)dx=0$, but $u$ and $f$ are constants, so I can't see any way to set that integral equals to $0$.
Maybe I could start by thinking about $Vert u - v Vert_{L^2}^2=int_0^1|u(x)-v(x)|^2dx $ and how to minimize this integral with $v in X$. The best I can do, I guess, is to set $v(x)=v=max {|u(x)|: x in [0,1] }$, but I still can't understand how to find a $u$ such that (*) holds.
EDIT:
I need also to prove that $X$ is closed.To this aim, I take a sequence ${ f_n }$ in $X$ s.t. $f_n rightarrow^{L^2} f$. I want to show that $f$ is a.e. constant. By Riesz-Fisher there exists a subsequence ${f_{n_k} } in X$ such that $f_{n_k} rightarrow f$ a.e. in $X$. How can I conclude?
functional-analysis hilbert-spaces lp-spaces projection
asked Dec 27 '18 at 14:35
VoB
673213
(*) is the characterization of a projection if $X$ is a linear subspace. (Which here luckily is the case.)
– user251257
Dec 27 '18 at 14:40
add a comment |
I don't know how to solve the following exercise:
Let $X$ be the subspace of $L^{2}(0,1)$ of a.e. constant functions. What is the projection over $X$ of a function $u in L^2(0,1)$?
First of all, $X$ is convex, and this is trivial to prove. It's also closed. So, by the characterization of projections onto closed and convex subspace I have to find a $v in X$ s.t.
(*) $langle u-v,f rangle=0$ for every $f in X$, i.e. $int_0^1 (u(x)-v(x))f(x)dx=0$, but $u$ and $f$ are constants, so I can't see any way to set that integral equals to $0$.
Maybe I could start by thinking about $Vert u - v Vert_{L^2}^2=int_0^1|u(x)-v(x)|^2dx $ and how to minimize this integral with $v in X$. The best I can do, I guess, is to set $v(x)=v=max {|u(x)|: x in [0,1] }$, but I still can't understand how to find a $u$ such that (*) holds.
EDIT:
I need also to prove that $X$ is closed.To this aim, I take a sequence ${ f_n }$ in $X$ s.t. $f_n rightarrow^{L^2} f$. I want to show that $f$ is a.e. constant. By Riesz-Fisher there exists a subsequence ${f_{n_k} } in X$ such that $f_{n_k} rightarrow f$ a.e. in $X$. How can I conclude?
functional-analysis hilbert-spaces lp-spaces projection
asked Dec 27 '18 at 14:35
VoB
673213
I don't know how to solve the following exercise:
Let $X$ be the subspace of $L^{2}(0,1)$ of a.e. constant functions. What is the projection over $X$ of a function $u in L^2(0,1)$?
First of all, $X$ is convex, and this is trivial to prove. It's also closed. So, by the characterization of projections onto closed and convex subspace I have to find a $v in X$ s.t.
(*) $langle u-v,f rangle=0$ for every $f in X$, i.e. $int_0^1 (u(x)-v(x))f(x)dx=0$, but $u$ and $f$ are constants, so I can't see any way to set that integral equals to $0$.
Maybe I could start by thinking about $Vert u - v Vert_{L^2}^2=int_0^1|u(x)-v(x)|^2dx $ and how to minimize this integral with $v in X$. The best I can do, I guess, is to set $v(x)=v=max {|u(x)|: x in [0,1] }$, but I still can't understand how to find a $u$ such that (*) holds.
EDIT:
I need also to prove that $X$ is closed.To this aim, I take a sequence ${ f_n }$ in $X$ s.t. $f_n rightarrow^{L^2} f$. I want to show that $f$ is a.e. constant. By Riesz-Fisher there exists a subsequence ${f_{n_k} } in X$ such that $f_{n_k} rightarrow f$ a.e. in $X$. How can I conclude?
functional-analysis hilbert-spaces lp-spaces projection
functional-analysis hilbert-spaces lp-spaces projection
asked Dec 27 '18 at 14:35
VoB
673213
asked Dec 27 '18 at 14:35
VoB
673213
edited Dec 27 '18 at 14:49
asked Dec 27 '18 at 14:35
VoB
673213
asked Dec 27 '18 at 14:35
VoB
673213
asked Dec 27 '18 at 14:35
VoB
673213
673213
(*) is the characterization of a projection if $X$ is a linear subspace. (Which here luckily is the case.)
– user251257
Dec 27 '18 at 14:40
add a comment |
(*) is the characterization of a projection if $X$ is a linear subspace. (Which here luckily is the case.)
– user251257
Dec 27 '18 at 14:40
(*) is the characterization of a projection if $X$ is a linear subspace. (Which here luckily is the case.)
– user251257
Dec 27 '18 at 14:40
(*) is the characterization of a projection if $X$ is a linear subspace. (Which here luckily is the case.)
– user251257
Dec 27 '18 at 14:40
add a comment |
1 Answer
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The space $X$ is spanned by the constant function $1$, whose norm is $1$. Therefore, the orthogonal projection of $f$ on $X$ is $langle u,1rangle1$. That is, it's the constant function $int_0^1u$.
answered Dec 27 '18 at 14:38
José Carlos Santos
151k22123224
Okay, now that makes sense. Actually, I didn't prove it's closed, and now I see that it's not so easy. I edited my original post
– VoB
Dec 27 '18 at 14:49
In any Banach space, a subspace spanned by a single vector (or, more generally, any finite-dimensional subspace) is closed.
– José Carlos Santos
Dec 27 '18 at 14:54
Yes, I know this. But I want to show it by using the classical definition of closed set.
– VoB
Dec 27 '18 at 14:57
Use the fact that any convergent sequence of elements of $mathbb{R}v$ is of the form $(lambda_nv)_{ninmathbb N}$, where each $lambda_n$ is a real number. But then the sequence $(lambda_n)_{ninmathbb N}$ converges to some $lambdainmathbb R$ and therefore $lim_{ntoinfty}lambda_nv=lambda vinmathbb{R}v$.
– José Carlos Santos
Dec 27 '18 at 15:01
With $mathbb{R}v$ you mean real multiples of a constant $v$, right? So you say: assume that ${ f_n } in X$ converges. By definitions, each $f_n$ is of the form $f_n=lambda_n v$, where $v$ is a constant. But then, since the sequence converges, $lambda_n rightarrow lambda$, and then $f_n=lambda_n v rightarrow lambda v in mathbb{R}$, so $X$ is closed
– VoB
Dec 27 '18 at 15:17
|
show 3 more comments
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1 Answer
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The space $X$ is spanned by the constant function $1$, whose norm is $1$. Therefore, the orthogonal projection of $f$ on $X$ is $langle u,1rangle1$. That is, it's the constant function $int_0^1u$.
answered Dec 27 '18 at 14:38
José Carlos Santos
151k22123224
Okay, now that makes sense. Actually, I didn't prove it's closed, and now I see that it's not so easy. I edited my original post
– VoB
Dec 27 '18 at 14:49
In any Banach space, a subspace spanned by a single vector (or, more generally, any finite-dimensional subspace) is closed.
– José Carlos Santos
Dec 27 '18 at 14:54
Yes, I know this. But I want to show it by using the classical definition of closed set.
– VoB
Dec 27 '18 at 14:57
Use the fact that any convergent sequence of elements of $mathbb{R}v$ is of the form $(lambda_nv)_{ninmathbb N}$, where each $lambda_n$ is a real number. But then the sequence $(lambda_n)_{ninmathbb N}$ converges to some $lambdainmathbb R$ and therefore $lim_{ntoinfty}lambda_nv=lambda vinmathbb{R}v$.
– José Carlos Santos
Dec 27 '18 at 15:01
With $mathbb{R}v$ you mean real multiples of a constant $v$, right? So you say: assume that ${ f_n } in X$ converges. By definitions, each $f_n$ is of the form $f_n=lambda_n v$, where $v$ is a constant. But then, since the sequence converges, $lambda_n rightarrow lambda$, and then $f_n=lambda_n v rightarrow lambda v in mathbb{R}$, so $X$ is closed
– VoB
Dec 27 '18 at 15:17
|
show 3 more comments
The space $X$ is spanned by the constant function $1$, whose norm is $1$. Therefore, the orthogonal projection of $f$ on $X$ is $langle u,1rangle1$. That is, it's the constant function $int_0^1u$.
answered Dec 27 '18 at 14:38
José Carlos Santos
151k22123224
Okay, now that makes sense. Actually, I didn't prove it's closed, and now I see that it's not so easy. I edited my original post
– VoB
Dec 27 '18 at 14:49
In any Banach space, a subspace spanned by a single vector (or, more generally, any finite-dimensional subspace) is closed.
– José Carlos Santos
Dec 27 '18 at 14:54
Yes, I know this. But I want to show it by using the classical definition of closed set.
– VoB
Dec 27 '18 at 14:57
Use the fact that any convergent sequence of elements of $mathbb{R}v$ is of the form $(lambda_nv)_{ninmathbb N}$, where each $lambda_n$ is a real number. But then the sequence $(lambda_n)_{ninmathbb N}$ converges to some $lambdainmathbb R$ and therefore $lim_{ntoinfty}lambda_nv=lambda vinmathbb{R}v$.
– José Carlos Santos
Dec 27 '18 at 15:01
With $mathbb{R}v$ you mean real multiples of a constant $v$, right? So you say: assume that ${ f_n } in X$ converges. By definitions, each $f_n$ is of the form $f_n=lambda_n v$, where $v$ is a constant. But then, since the sequence converges, $lambda_n rightarrow lambda$, and then $f_n=lambda_n v rightarrow lambda v in mathbb{R}$, so $X$ is closed
– VoB
Dec 27 '18 at 15:17
|
show 3 more comments
The space $X$ is spanned by the constant function $1$, whose norm is $1$. Therefore, the orthogonal projection of $f$ on $X$ is $langle u,1rangle1$. That is, it's the constant function $int_0^1u$.
answered Dec 27 '18 at 14:38
José Carlos Santos
151k22123224
The space $X$ is spanned by the constant function $1$, whose norm is $1$. Therefore, the orthogonal projection of $f$ on $X$ is $langle u,1rangle1$. That is, it's the constant function $int_0^1u$.
answered Dec 27 '18 at 14:38
José Carlos Santos
151k22123224
answered Dec 27 '18 at 14:38
José Carlos Santos
151k22123224
answered Dec 27 '18 at 14:38
José Carlos Santos
151k22123224
answered Dec 27 '18 at 14:38
José Carlos Santos
151k22123224
151k22123224
Okay, now that makes sense. Actually, I didn't prove it's closed, and now I see that it's not so easy. I edited my original post
– VoB
Dec 27 '18 at 14:49
In any Banach space, a subspace spanned by a single vector (or, more generally, any finite-dimensional subspace) is closed.
– José Carlos Santos
Dec 27 '18 at 14:54
Yes, I know this. But I want to show it by using the classical definition of closed set.
– VoB
Dec 27 '18 at 14:57
Use the fact that any convergent sequence of elements of $mathbb{R}v$ is of the form $(lambda_nv)_{ninmathbb N}$, where each $lambda_n$ is a real number. But then the sequence $(lambda_n)_{ninmathbb N}$ converges to some $lambdainmathbb R$ and therefore $lim_{ntoinfty}lambda_nv=lambda vinmathbb{R}v$.
– José Carlos Santos
Dec 27 '18 at 15:01
With $mathbb{R}v$ you mean real multiples of a constant $v$, right? So you say: assume that ${ f_n } in X$ converges. By definitions, each $f_n$ is of the form $f_n=lambda_n v$, where $v$ is a constant. But then, since the sequence converges, $lambda_n rightarrow lambda$, and then $f_n=lambda_n v rightarrow lambda v in mathbb{R}$, so $X$ is closed
– VoB
Dec 27 '18 at 15:17
|
show 3 more comments
Okay, now that makes sense. Actually, I didn't prove it's closed, and now I see that it's not so easy. I edited my original post
– VoB
Dec 27 '18 at 14:49
In any Banach space, a subspace spanned by a single vector (or, more generally, any finite-dimensional subspace) is closed.
– José Carlos Santos
Dec 27 '18 at 14:54
Yes, I know this. But I want to show it by using the classical definition of closed set.
– VoB
Dec 27 '18 at 14:57
Use the fact that any convergent sequence of elements of $mathbb{R}v$ is of the form $(lambda_nv)_{ninmathbb N}$, where each $lambda_n$ is a real number. But then the sequence $(lambda_n)_{ninmathbb N}$ converges to some $lambdainmathbb R$ and therefore $lim_{ntoinfty}lambda_nv=lambda vinmathbb{R}v$.
– José Carlos Santos
Dec 27 '18 at 15:01
With $mathbb{R}v$ you mean real multiples of a constant $v$, right? So you say: assume that ${ f_n } in X$ converges. By definitions, each $f_n$ is of the form $f_n=lambda_n v$, where $v$ is a constant. But then, since the sequence converges, $lambda_n rightarrow lambda$, and then $f_n=lambda_n v rightarrow lambda v in mathbb{R}$, so $X$ is closed
– VoB
Dec 27 '18 at 15:17
Okay, now that makes sense. Actually, I didn't prove it's closed, and now I see that it's not so easy. I edited my original post
– VoB
Dec 27 '18 at 14:49
Okay, now that makes sense. Actually, I didn't prove it's closed, and now I see that it's not so easy. I edited my original post
– VoB
Dec 27 '18 at 14:49
In any Banach space, a subspace spanned by a single vector (or, more generally, any finite-dimensional subspace) is closed.
– José Carlos Santos
Dec 27 '18 at 14:54
In any Banach space, a subspace spanned by a single vector (or, more generally, any finite-dimensional subspace) is closed.
– José Carlos Santos
Dec 27 '18 at 14:54
Yes, I know this. But I want to show it by using the classical definition of closed set.
– VoB
Dec 27 '18 at 14:57
Yes, I know this. But I want to show it by using the classical definition of closed set.
– VoB
Dec 27 '18 at 14:57
Use the fact that any convergent sequence of elements of $mathbb{R}v$ is of the form $(lambda_nv)_{ninmathbb N}$, where each $lambda_n$ is a real number. But then the sequence $(lambda_n)_{ninmathbb N}$ converges to some $lambdainmathbb R$ and therefore $lim_{ntoinfty}lambda_nv=lambda vinmathbb{R}v$.
– José Carlos Santos
Dec 27 '18 at 15:01
Use the fact that any convergent sequence of elements of $mathbb{R}v$ is of the form $(lambda_nv)_{ninmathbb N}$, where each $lambda_n$ is a real number. But then the sequence $(lambda_n)_{ninmathbb N}$ converges to some $lambdainmathbb R$ and therefore $lim_{ntoinfty}lambda_nv=lambda vinmathbb{R}v$.
– José Carlos Santos
Dec 27 '18 at 15:01
With $mathbb{R}v$ you mean real multiples of a constant $v$, right? So you say: assume that ${ f_n } in X$ converges. By definitions, each $f_n$ is of the form $f_n=lambda_n v$, where $v$ is a constant. But then, since the sequence converges, $lambda_n rightarrow lambda$, and then $f_n=lambda_n v rightarrow lambda v in mathbb{R}$, so $X$ is closed
– VoB
Dec 27 '18 at 15:17
With $mathbb{R}v$ you mean real multiples of a constant $v$, right? So you say: assume that ${ f_n } in X$ converges. By definitions, each $f_n$ is of the form $f_n=lambda_n v$, where $v$ is a constant. But then, since the sequence converges, $lambda_n rightarrow lambda$, and then $f_n=lambda_n v rightarrow lambda v in mathbb{R}$, so $X$ is closed
– VoB
Dec 27 '18 at 15:17
|
show 3 more comments
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(*) is the characterization of a projection if $X$ is a linear subspace. (Which here luckily is the case.)
– user251257
Dec 27 '18 at 14:40