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Sufficient condition to show density of $mathcal{E}$ in $Lambda^2$(Hilbert space)

Analysis operator $T_Phi$ is injective and has a closed rangeClosed linear span of a frame in a Hilbert space $mathcal{H}$ coincide with $mathcal{H}$Operators in Hilbert space ( orthogonal projection)Projection on a Hilbert spaceDoes weak*-convergence imply convergence of the operator norms?Showing $mathcal{H}$ is a hilbert space.Bases in Hilbert spaceDensity of orthogonal projections in $mathcal{B}(mathcal{H})$$fin E mapsto underset{{xin[0,1]}}sup f(x)$ continuousSequences in Hilbert space

1

$begingroup$

I am trying to understand the proof that the space of simple processes is dense in $Lambda^2$. The proof in my lecture notes starts by assuming that
for $phi in Lambda^2$ which is orthogonal to $mathcal{E}$. Then the proof claims that it is sufficient to show that $phiequiv 0 in Lambda^2$.

I cant precisely show why does this imply the density of $mathcal{E}$ in $Lambda^2$.
Clearly since $phi$ is orthogonal to $mathcal{E}$ we have that
$langle phi, Frangle=0$ for every $F in mathcal{E}$. Now if I knew that $mathcal{E}$ was dense in $Lambda^2$, then by the continuity of the inner product I can easily show that $phi$ is orthogonal to $Lambda^2$. Indeed given any element $H in Lambda^2$ , there exists a sequence ${H_n}_{n=1}^infty subseteq mathcal{E}$ such that

$$
lim_{n to infty}vert vert H_n-H vert vert_{Lambda^2}=0
$$

and hence $ lim_{n to infty} vertlangle H_n-H,phi rangle vert le lim_{n to infty}vert vert H_n-Hvertvert_{Lambda^2}^{1/2} vertvert phi vert vert_{Lambda^2}^{1/2}=0$ which would imply
$$
langle H,phi rangle= lim_{n to infty}langle H_n,phi rangle= 0
$$

But all this seems to be useless, since I assume $mathcal{E}$ to be dense where as I want to show the density

functional-analysis hilbert-spaces stochastic-calculus

share|cite|improve this question

edited Mar 11 at 19:55

user3503589

asked Mar 11 at 19:01

user3503589user3503589

1,3051821

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are $Lambda^2$ and $mathcal{E}$ here?
  $endgroup$
  – K.Power
  Mar 11 at 19:35
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $Lambda^2$ is a Hilbert space and $mathcal{E}$ is a subset of $Lambda^2$
  $endgroup$
  – user3503589
  Mar 11 at 19:53
  

  
  
  
  

add a comment | 

1

$begingroup$

I am trying to understand the proof that the space of simple processes is dense in $Lambda^2$. The proof in my lecture notes starts by assuming that
for $phi in Lambda^2$ which is orthogonal to $mathcal{E}$. Then the proof claims that it is sufficient to show that $phiequiv 0 in Lambda^2$.

I cant precisely show why does this imply the density of $mathcal{E}$ in $Lambda^2$.
Clearly since $phi$ is orthogonal to $mathcal{E}$ we have that
$langle phi, Frangle=0$ for every $F in mathcal{E}$. Now if I knew that $mathcal{E}$ was dense in $Lambda^2$, then by the continuity of the inner product I can easily show that $phi$ is orthogonal to $Lambda^2$. Indeed given any element $H in Lambda^2$ , there exists a sequence ${H_n}_{n=1}^infty subseteq mathcal{E}$ such that

$$
lim_{n to infty}vert vert H_n-H vert vert_{Lambda^2}=0
$$

and hence $ lim_{n to infty} vertlangle H_n-H,phi rangle vert le lim_{n to infty}vert vert H_n-Hvertvert_{Lambda^2}^{1/2} vertvert phi vert vert_{Lambda^2}^{1/2}=0$ which would imply
$$
langle H,phi rangle= lim_{n to infty}langle H_n,phi rangle= 0
$$

But all this seems to be useless, since I assume $mathcal{E}$ to be dense where as I want to show the density

functional-analysis hilbert-spaces stochastic-calculus

share|cite|improve this question

edited Mar 11 at 19:55

user3503589

asked Mar 11 at 19:01

user3503589user3503589

1,3051821

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  What are $Lambda^2$ and $mathcal{E}$ here?
  $endgroup$
  – K.Power
  Mar 11 at 19:35
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $Lambda^2$ is a Hilbert space and $mathcal{E}$ is a subset of $Lambda^2$
  $endgroup$
  – user3503589
  Mar 11 at 19:53
  

  
  
  
  

add a comment | 

$begingroup$

I am trying to understand the proof that the space of simple processes is dense in $Lambda^2$. The proof in my lecture notes starts by assuming that
for $phi in Lambda^2$ which is orthogonal to $mathcal{E}$. Then the proof claims that it is sufficient to show that $phiequiv 0 in Lambda^2$.

I cant precisely show why does this imply the density of $mathcal{E}$ in $Lambda^2$.
Clearly since $phi$ is orthogonal to $mathcal{E}$ we have that
$langle phi, Frangle=0$ for every $F in mathcal{E}$. Now if I knew that $mathcal{E}$ was dense in $Lambda^2$, then by the continuity of the inner product I can easily show that $phi$ is orthogonal to $Lambda^2$. Indeed given any element $H in Lambda^2$ , there exists a sequence ${H_n}_{n=1}^infty subseteq mathcal{E}$ such that

$$
lim_{n to infty}vert vert H_n-H vert vert_{Lambda^2}=0
$$

and hence $ lim_{n to infty} vertlangle H_n-H,phi rangle vert le lim_{n to infty}vert vert H_n-Hvertvert_{Lambda^2}^{1/2} vertvert phi vert vert_{Lambda^2}^{1/2}=0$ which would imply
$$
langle H,phi rangle= lim_{n to infty}langle H_n,phi rangle= 0
$$

But all this seems to be useless, since I assume $mathcal{E}$ to be dense where as I want to show the density

functional-analysis hilbert-spaces stochastic-calculus

share|cite|improve this question

edited Mar 11 at 19:55

user3503589

asked Mar 11 at 19:01

user3503589user3503589

1,3051821

$endgroup$

share|cite|improve this question

edited Mar 11 at 19:55

user3503589

asked Mar 11 at 19:01

user3503589user3503589

1,3051821

share|cite|improve this question

edited Mar 11 at 19:55

user3503589

asked Mar 11 at 19:01

user3503589user3503589

1,3051821

asked Mar 11 at 19:01

user3503589user3503589

1,3051821

asked Mar 11 at 19:01

user3503589user3503589

1,3051821

1 Answer
1

active

oldest

votes

2

$begingroup$

If $H$ is a hilbert space and $X$ a subspace we have that $X$ is dense in $H$ if and only if $hperp Ximplies h=0$. The forward direction is fairly trivial utilising the continuity of the innerproduct.

For the converse let us assume that $hperp Ximplies h=0$. We first note that $H=bar Xoplus bar X^perp$ (it is a basic theorem that should be covered in any first course on Hilbert spaces. Specifically closed subspaces are complemented by their orthogonal complement). Now suppose $xin bar X^perp$. This implies $xin X^perp$, which by assumption means $x=0$. Thus $bar X^perp={0}$, meaning $H=bar X$.

All that is left is to note that the space of simple functions is a subspace of any "ordinary" function space.

share|cite|improve this answer

answered Mar 11 at 19:58

K.PowerK.Power

3,370926

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $X^{perp}=bar{X^{perp}}$ because $X^{perp}$ is a closed linear subspace right(even if $X$ is not a linear)?
  $endgroup$
  – user3503589
  Mar 11 at 21:40
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, via continuity of the inner-product. We don't even need that for this proof though. All we need is that $Asubset Bimplies B^perpsubset A^perp$
  $endgroup$
  – K.Power
  Mar 11 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much .
  $endgroup$
  – user3503589
  Mar 11 at 21:48
  

  
  
  
  

add a comment | 

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2

$begingroup$

If $H$ is a hilbert space and $X$ a subspace we have that $X$ is dense in $H$ if and only if $hperp Ximplies h=0$. The forward direction is fairly trivial utilising the continuity of the innerproduct.

For the converse let us assume that $hperp Ximplies h=0$. We first note that $H=bar Xoplus bar X^perp$ (it is a basic theorem that should be covered in any first course on Hilbert spaces. Specifically closed subspaces are complemented by their orthogonal complement). Now suppose $xin bar X^perp$. This implies $xin X^perp$, which by assumption means $x=0$. Thus $bar X^perp={0}$, meaning $H=bar X$.

All that is left is to note that the space of simple functions is a subspace of any "ordinary" function space.

share|cite|improve this answer

answered Mar 11 at 19:58

K.PowerK.Power

3,370926

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $X^{perp}=bar{X^{perp}}$ because $X^{perp}$ is a closed linear subspace right(even if $X$ is not a linear)?
  $endgroup$
  – user3503589
  Mar 11 at 21:40
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, via continuity of the inner-product. We don't even need that for this proof though. All we need is that $Asubset Bimplies B^perpsubset A^perp$
  $endgroup$
  – K.Power
  Mar 11 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much .
  $endgroup$
  – user3503589
  Mar 11 at 21:48
  

  
  
  
  

add a comment | 

2

$begingroup$

If $H$ is a hilbert space and $X$ a subspace we have that $X$ is dense in $H$ if and only if $hperp Ximplies h=0$. The forward direction is fairly trivial utilising the continuity of the innerproduct.

For the converse let us assume that $hperp Ximplies h=0$. We first note that $H=bar Xoplus bar X^perp$ (it is a basic theorem that should be covered in any first course on Hilbert spaces. Specifically closed subspaces are complemented by their orthogonal complement). Now suppose $xin bar X^perp$. This implies $xin X^perp$, which by assumption means $x=0$. Thus $bar X^perp={0}$, meaning $H=bar X$.

All that is left is to note that the space of simple functions is a subspace of any "ordinary" function space.

share|cite|improve this answer

answered Mar 11 at 19:58

K.PowerK.Power

3,370926

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  $X^{perp}=bar{X^{perp}}$ because $X^{perp}$ is a closed linear subspace right(even if $X$ is not a linear)?
  $endgroup$
  – user3503589
  Mar 11 at 21:40
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, via continuity of the inner-product. We don't even need that for this proof though. All we need is that $Asubset Bimplies B^perpsubset A^perp$
  $endgroup$
  – K.Power
  Mar 11 at 21:44
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much .
  $endgroup$
  – user3503589
  Mar 11 at 21:48
  

  
  
  
  

add a comment | 

$begingroup$

If $H$ is a hilbert space and $X$ a subspace we have that $X$ is dense in $H$ if and only if $hperp Ximplies h=0$. The forward direction is fairly trivial utilising the continuity of the innerproduct.

For the converse let us assume that $hperp Ximplies h=0$. We first note that $H=bar Xoplus bar X^perp$ (it is a basic theorem that should be covered in any first course on Hilbert spaces. Specifically closed subspaces are complemented by their orthogonal complement). Now suppose $xin bar X^perp$. This implies $xin X^perp$, which by assumption means $x=0$. Thus $bar X^perp={0}$, meaning $H=bar X$.

All that is left is to note that the space of simple functions is a subspace of any "ordinary" function space.

share|cite|improve this answer

answered Mar 11 at 19:58

K.PowerK.Power

3,370926

$endgroup$

share|cite|improve this answer

answered Mar 11 at 19:58

K.PowerK.Power

3,370926

answered Mar 11 at 19:58

K.PowerK.Power

3,370926

answered Mar 11 at 19:58

K.PowerK.Power

3,370926