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Given $X$ Hilbert space, $Tin X^*$, $y$ projection of $x_0$ on $Y=text{Ker}T$, why does $x-frac{T(x)}{T(x_0-y)}(x_0-y)in text{Ker}T$?

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0

$begingroup$

Let

• 
  $X$ Hilbert space


• 
  $Tin X^*$, ie $T:Xtomathbb R$ linear


• 
  $Y=text{Ker}T={xin X:T(x)=0}$ closed subspace of $X$
  


• $x_0in Xsetminus Y$


• 
  $yin Y$ orthogonal projection of $x_0$ on $Y$, ie $langle x_0-y,z rangle=0$ for every $zin Y$
  


• 
  $w=x-frac{T(x)}{T(x_0-y)}(x_0-y)$, with $xin X$
  

Show that

1. 
   $Y$ is closed


2. 
   $win Y$ for every $xin X$
   

(1) Hints?

(2) In the case $xin Y$, then $T(x)=0$ and so $w=x-0=xin Y$.

In the case $xin Xsetminus Y$, by linearity of $T$, it is $T(x_0-y)=T(x_0)-(y)=T(x_0)$, since $yin Y$.

so $w=x-frac{T(x)}{T(x_0)}(x_0-y)$.

We can write $X=Yoplus Y^perp$, so we can see $xin X$ as sum of an element of $Y$ with an element of $Y^perp$. Since $(x_0-y)in Y^perp$ and $frac{T(x)}{T(x_0)}$ is just a scalar, I suppose that the product of them is still an element of $Y^perp$.

So we can see $w$ as "an element of $Y$" $+$ "an element of $Y^perp$" $-$ "an element of $Y^perp$". I suppose that whatever $x$ will be, we can choose the second term in such a way that it will be equal to the third one, so that they will cancel out.

How to prove this in a rigorous way?

functional-analysis hilbert-spaces

share|cite|improve this question

asked 15 hours ago

sound wavesound wave

28819

$endgroup$

add a comment | 

0

$begingroup$

Let

• 
  $X$ Hilbert space


• 
  $Tin X^*$, ie $T:Xtomathbb R$ linear


• 
  $Y=text{Ker}T={xin X:T(x)=0}$ closed subspace of $X$
  


• $x_0in Xsetminus Y$


• 
  $yin Y$ orthogonal projection of $x_0$ on $Y$, ie $langle x_0-y,z rangle=0$ for every $zin Y$
  


• 
  $w=x-frac{T(x)}{T(x_0-y)}(x_0-y)$, with $xin X$
  

Show that

1. 
   $Y$ is closed


2. 
   $win Y$ for every $xin X$
   

(1) Hints?

(2) In the case $xin Y$, then $T(x)=0$ and so $w=x-0=xin Y$.

In the case $xin Xsetminus Y$, by linearity of $T$, it is $T(x_0-y)=T(x_0)-(y)=T(x_0)$, since $yin Y$.

so $w=x-frac{T(x)}{T(x_0)}(x_0-y)$.

We can write $X=Yoplus Y^perp$, so we can see $xin X$ as sum of an element of $Y$ with an element of $Y^perp$. Since $(x_0-y)in Y^perp$ and $frac{T(x)}{T(x_0)}$ is just a scalar, I suppose that the product of them is still an element of $Y^perp$.

So we can see $w$ as "an element of $Y$" $+$ "an element of $Y^perp$" $-$ "an element of $Y^perp$". I suppose that whatever $x$ will be, we can choose the second term in such a way that it will be equal to the third one, so that they will cancel out.

How to prove this in a rigorous way?

functional-analysis hilbert-spaces

share|cite|improve this question

asked 15 hours ago

sound wavesound wave

28819

$endgroup$

add a comment | 

$begingroup$

Let

• 
  $X$ Hilbert space


• 
  $Tin X^*$, ie $T:Xtomathbb R$ linear


• 
  $Y=text{Ker}T={xin X:T(x)=0}$ closed subspace of $X$
  


• $x_0in Xsetminus Y$


• 
  $yin Y$ orthogonal projection of $x_0$ on $Y$, ie $langle x_0-y,z rangle=0$ for every $zin Y$
  


• 
  $w=x-frac{T(x)}{T(x_0-y)}(x_0-y)$, with $xin X$
  

Show that

1. 
   $Y$ is closed


2. 
   $win Y$ for every $xin X$
   

(1) Hints?

(2) In the case $xin Y$, then $T(x)=0$ and so $w=x-0=xin Y$.

In the case $xin Xsetminus Y$, by linearity of $T$, it is $T(x_0-y)=T(x_0)-(y)=T(x_0)$, since $yin Y$.

so $w=x-frac{T(x)}{T(x_0)}(x_0-y)$.

We can write $X=Yoplus Y^perp$, so we can see $xin X$ as sum of an element of $Y$ with an element of $Y^perp$. Since $(x_0-y)in Y^perp$ and $frac{T(x)}{T(x_0)}$ is just a scalar, I suppose that the product of them is still an element of $Y^perp$.

So we can see $w$ as "an element of $Y$" $+$ "an element of $Y^perp$" $-$ "an element of $Y^perp$". I suppose that whatever $x$ will be, we can choose the second term in such a way that it will be equal to the third one, so that they will cancel out.

How to prove this in a rigorous way?

functional-analysis hilbert-spaces

share|cite|improve this question

asked 15 hours ago

sound wavesound wave

28819

$endgroup$

share|cite|improve this question

asked 15 hours ago

sound wavesound wave

28819

share|cite|improve this question

asked 15 hours ago

sound wavesound wave

28819

asked 15 hours ago

sound wavesound wave

28819

asked 15 hours ago

sound wavesound wave

28819

1 Answer
1

active

oldest

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1

$begingroup$

$T$ is continous and $Y$ is the inverse image of ${0}$ under $T$ so it is closed.

Second part follows from linearity of $T$: $Tw=Tx-frac {Tx} {T(x_0-y)} {T(x_0-y)}=Tx-Tx=0$.

share|cite|improve this answer

answered 15 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. I understand the second part. About the first one, are you saying that since ${0}$ is closed, also $Y=T^{-1}({0})$ is closed ?
  $endgroup$
  – sound wave
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that is what I meant
  $endgroup$
  – Kavi Rama Murthy
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, but I don't understand why from continuity of $T$ and the fact that ${0}$ is closed, it follows that $T^{-1}({0})$ is closed.
  $endgroup$
  – sound wave
  13 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $tiny{ciao!}$ @soundwave if $T: X rightarrow mathbb{R}$ is continuous, and ${ 0 }$ is closed, then $X setminus T^{-1}({ 0 })=T^{-1}(mathbb{R} setminus {0 })$ is an open set. So $T^{-1}({0})$ is closed (since its complement is open)
  $endgroup$
  – VoB
  12 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @VoB grazie! ;)
  $endgroup$
  – sound wave
  12 hours ago
  

  
  
  
  

add a comment | 

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1

$begingroup$

$T$ is continous and $Y$ is the inverse image of ${0}$ under $T$ so it is closed.

Second part follows from linearity of $T$: $Tw=Tx-frac {Tx} {T(x_0-y)} {T(x_0-y)}=Tx-Tx=0$.

share|cite|improve this answer

answered 15 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. I understand the second part. About the first one, are you saying that since ${0}$ is closed, also $Y=T^{-1}({0})$ is closed ?
  $endgroup$
  – sound wave
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that is what I meant
  $endgroup$
  – Kavi Rama Murthy
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, but I don't understand why from continuity of $T$ and the fact that ${0}$ is closed, it follows that $T^{-1}({0})$ is closed.
  $endgroup$
  – sound wave
  13 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $tiny{ciao!}$ @soundwave if $T: X rightarrow mathbb{R}$ is continuous, and ${ 0 }$ is closed, then $X setminus T^{-1}({ 0 })=T^{-1}(mathbb{R} setminus {0 })$ is an open set. So $T^{-1}({0})$ is closed (since its complement is open)
  $endgroup$
  – VoB
  12 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @VoB grazie! ;)
  $endgroup$
  – sound wave
  12 hours ago
  

  
  
  
  

add a comment | 

1

$begingroup$

$T$ is continous and $Y$ is the inverse image of ${0}$ under $T$ so it is closed.

Second part follows from linearity of $T$: $Tw=Tx-frac {Tx} {T(x_0-y)} {T(x_0-y)}=Tx-Tx=0$.

share|cite|improve this answer

answered 15 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Thank you very much. I understand the second part. About the first one, are you saying that since ${0}$ is closed, also $Y=T^{-1}({0})$ is closed ?
  $endgroup$
  – sound wave
  14 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, that is what I meant
  $endgroup$
  – Kavi Rama Murthy
  13 hours ago
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Ok, but I don't understand why from continuity of $T$ and the fact that ${0}$ is closed, it follows that $T^{-1}({0})$ is closed.
  $endgroup$
  – sound wave
  13 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  $tiny{ciao!}$ @soundwave if $T: X rightarrow mathbb{R}$ is continuous, and ${ 0 }$ is closed, then $X setminus T^{-1}({ 0 })=T^{-1}(mathbb{R} setminus {0 })$ is an open set. So $T^{-1}({0})$ is closed (since its complement is open)
  $endgroup$
  – VoB
  12 hours ago
  
  
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  @VoB grazie! ;)
  $endgroup$
  – sound wave
  12 hours ago
  

  
  
  
  

add a comment | 

$begingroup$

$T$ is continous and $Y$ is the inverse image of ${0}$ under $T$ so it is closed.

Second part follows from linearity of $T$: $Tw=Tx-frac {Tx} {T(x_0-y)} {T(x_0-y)}=Tx-Tx=0$.

share|cite|improve this answer

answered 15 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

$endgroup$

share|cite|improve this answer

answered 15 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

answered 15 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766

answered 15 hours ago

Kavi Rama MurthyKavi Rama Murthy

65.4k42766