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Riesz Representation Theorem for Functional on Hilbert space

1

$begingroup$

Let $H$ be a Hilbert space and $f in H^*$. Then there is unique $y in H$ such that $$ f(x)= langle x,y rangle$$ for all $xin H. $

In the proof of this, first we use projection theorem and express $H$ as direct sum of null space of functional and it's orthogonal complement.

Then to find the value of $y$ we take non zero element say $z$ from orthogonal complement of null space of functional and we can show that the required candidate of $y$ is some scalar multiple of $z$. In particular $y=fracf(z)z^2.$

My doubt is here $y$ is dependent on choice of $z$ so if we choose another element say $z^1$ then $y$ also varies according to this, so there are more than one value for fix $f(x)$.

Can anyone explain it.

Thank you.

functional-analysis hilbert-spaces riesz-representation-theorem

share|cite|improve this question

asked Mar 31 at 14:55

user462999user462999

265

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You should split your reasoning into two steps. First existence and then uniqueness. You say that $y$ depends on the choice of $z$. But for existence this is fine, it is not allowed to depend on $x$, but this is fine as $z$ is just in the orthogonal complement of the null space of $f$ (ie does not depend on a particular $x$). What you should do, is to verify that this choice of $y$ does what we want, namely $f(x)=langle x, y rangle$.
  $endgroup$
  – Severin Schraven
  Mar 31 at 15:31
  
  

  
  
  
  

add a comment | 

1

$begingroup$

Let $H$ be a Hilbert space and $f in H^*$. Then there is unique $y in H$ such that $$ f(x)= langle x,y rangle$$ for all $xin H. $

In the proof of this, first we use projection theorem and express $H$ as direct sum of null space of functional and it's orthogonal complement.

Then to find the value of $y$ we take non zero element say $z$ from orthogonal complement of null space of functional and we can show that the required candidate of $y$ is some scalar multiple of $z$. In particular $y=fracf(z)z^2.$

My doubt is here $y$ is dependent on choice of $z$ so if we choose another element say $z^1$ then $y$ also varies according to this, so there are more than one value for fix $f(x)$.

Can anyone explain it.

Thank you.

functional-analysis hilbert-spaces riesz-representation-theorem

share|cite|improve this question

asked Mar 31 at 14:55

user462999user462999

265

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  You should split your reasoning into two steps. First existence and then uniqueness. You say that $y$ depends on the choice of $z$. But for existence this is fine, it is not allowed to depend on $x$, but this is fine as $z$ is just in the orthogonal complement of the null space of $f$ (ie does not depend on a particular $x$). What you should do, is to verify that this choice of $y$ does what we want, namely $f(x)=langle x, y rangle$.
  $endgroup$
  – Severin Schraven
  Mar 31 at 15:31
  
  

  
  
  
  

add a comment | 

$begingroup$

Let $H$ be a Hilbert space and $f in H^*$. Then there is unique $y in H$ such that $$ f(x)= langle x,y rangle$$ for all $xin H. $

In the proof of this, first we use projection theorem and express $H$ as direct sum of null space of functional and it's orthogonal complement.

Then to find the value of $y$ we take non zero element say $z$ from orthogonal complement of null space of functional and we can show that the required candidate of $y$ is some scalar multiple of $z$. In particular $y=fracf(z)z^2.$

My doubt is here $y$ is dependent on choice of $z$ so if we choose another element say $z^1$ then $y$ also varies according to this, so there are more than one value for fix $f(x)$.

Can anyone explain it.

Thank you.

functional-analysis hilbert-spaces riesz-representation-theorem

share|cite|improve this question

asked Mar 31 at 14:55

user462999user462999

265

$endgroup$

share|cite|improve this question

asked Mar 31 at 14:55

user462999user462999

265

share|cite|improve this question

asked Mar 31 at 14:55

user462999user462999

265

asked Mar 31 at 14:55

user462999user462999

265

asked Mar 31 at 14:55

user462999user462999

265

1 Answer
1

active

oldest

votes

2

$begingroup$

The uniqueness statement can easily be proven.
Let's assume we have elements $y,y'$ such that $f(x)=langle x,yrangle$ and $f(x)=langle x,y'rangle$ holds true for all $x in H$.
This implies $langle x,y-y'rangle=0$ for all $x in H$.

Then, it particular it must hold for $x :=y-y'$ which implies $y=y'$.

I agree that is not clear in the first place that $y$ is independent of the choice of $z$ but we can show it this way.
I am not sure whether there is way to see this differently.

share|cite|improve this answer

answered Mar 31 at 15:29

Jonas LenzJonas Lenz

694215

$endgroup$

add a comment | 

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2

$begingroup$

The uniqueness statement can easily be proven.
Let's assume we have elements $y,y'$ such that $f(x)=langle x,yrangle$ and $f(x)=langle x,y'rangle$ holds true for all $x in H$.
This implies $langle x,y-y'rangle=0$ for all $x in H$.

Then, it particular it must hold for $x :=y-y'$ which implies $y=y'$.

I agree that is not clear in the first place that $y$ is independent of the choice of $z$ but we can show it this way.
I am not sure whether there is way to see this differently.

share|cite|improve this answer

answered Mar 31 at 15:29

Jonas LenzJonas Lenz

694215

$endgroup$

add a comment | 

2

$begingroup$

The uniqueness statement can easily be proven.
Let's assume we have elements $y,y'$ such that $f(x)=langle x,yrangle$ and $f(x)=langle x,y'rangle$ holds true for all $x in H$.
This implies $langle x,y-y'rangle=0$ for all $x in H$.

Then, it particular it must hold for $x :=y-y'$ which implies $y=y'$.

I agree that is not clear in the first place that $y$ is independent of the choice of $z$ but we can show it this way.
I am not sure whether there is way to see this differently.

share|cite|improve this answer

answered Mar 31 at 15:29

Jonas LenzJonas Lenz

694215

$endgroup$

add a comment | 

$begingroup$

The uniqueness statement can easily be proven.
Let's assume we have elements $y,y'$ such that $f(x)=langle x,yrangle$ and $f(x)=langle x,y'rangle$ holds true for all $x in H$.
This implies $langle x,y-y'rangle=0$ for all $x in H$.

Then, it particular it must hold for $x :=y-y'$ which implies $y=y'$.

I agree that is not clear in the first place that $y$ is independent of the choice of $z$ but we can show it this way.
I am not sure whether there is way to see this differently.

share|cite|improve this answer

answered Mar 31 at 15:29

Jonas LenzJonas Lenz

694215

$endgroup$

share|cite|improve this answer

answered Mar 31 at 15:29

Jonas LenzJonas Lenz

694215

answered Mar 31 at 15:29

Jonas LenzJonas Lenz

694215

answered Mar 31 at 15:29

Jonas LenzJonas Lenz

694215