Mixing
Existence of a linear map using Hahn-Banach theorem [closed]
Clash Royale CLAN TAG#URR8PPP
up vote
-2
down vote
favorite
Let V and W be Banach spaces and let $T: V rightarrow W$ and $S: W' rightarrow V'$ be maps with $f(T(v)) = S(f)(v)$ for all $f in W', v in V$. ($V',W'$ are the dual spaces of V and W)
Prove that T is linear and bounded, using the Hahn-Banach theorem.
This exercise appeared in my exam and I had no idea how to prove it. Somehow I find this exercise a little bit strange, but I don't know why.
functional-analysis
asked Aug 6 at 6:19
AM88
96
closed as off-topic by Morgan Rodgers, uniquesolution, amWhy, Jendrik Stelzner, Adrian Keister Aug 6 at 13:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, uniquesolution, amWhy, Jendrik Stelzner, Adrian Keister
add a comment |Â
up vote
-2
down vote
favorite
Let V and W be Banach spaces and let $T: V rightarrow W$ and $S: W' rightarrow V'$ be maps with $f(T(v)) = S(f)(v)$ for all $f in W', v in V$. ($V',W'$ are the dual spaces of V and W)
Prove that T is linear and bounded, using the Hahn-Banach theorem.
This exercise appeared in my exam and I had no idea how to prove it. Somehow I find this exercise a little bit strange, but I don't know why.
functional-analysis
asked Aug 6 at 6:19
AM88
96
closed as off-topic by Morgan Rodgers, uniquesolution, amWhy, Jendrik Stelzner, Adrian Keister Aug 6 at 13:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, uniquesolution, amWhy, Jendrik Stelzner, Adrian Keister
It seems that you also need Uniform Boundedness Principle.
– Danny Pak-Keung Chan
Aug 6 at 7:37
Good question! Who (and why) deducted score?
– Danny Pak-Keung Chan
Aug 6 at 7:40
add a comment |Â
up vote
-2
down vote
favorite
up vote
-2
down vote
favorite
Let V and W be Banach spaces and let $T: V rightarrow W$ and $S: W' rightarrow V'$ be maps with $f(T(v)) = S(f)(v)$ for all $f in W', v in V$. ($V',W'$ are the dual spaces of V and W)
Prove that T is linear and bounded, using the Hahn-Banach theorem.
This exercise appeared in my exam and I had no idea how to prove it. Somehow I find this exercise a little bit strange, but I don't know why.
functional-analysis
asked Aug 6 at 6:19
AM88
96
Let V and W be Banach spaces and let $T: V rightarrow W$ and $S: W' rightarrow V'$ be maps with $f(T(v)) = S(f)(v)$ for all $f in W', v in V$. ($V',W'$ are the dual spaces of V and W)
Prove that T is linear and bounded, using the Hahn-Banach theorem.
This exercise appeared in my exam and I had no idea how to prove it. Somehow I find this exercise a little bit strange, but I don't know why.
functional-analysis
asked Aug 6 at 6:19
AM88
96
edited Aug 8 at 11:22
asked Aug 6 at 6:19
AM88
96
asked Aug 6 at 6:19
AM88
96
asked Aug 6 at 6:19
AM88
96
96
closed as off-topic by Morgan Rodgers, uniquesolution, amWhy, Jendrik Stelzner, Adrian Keister Aug 6 at 13:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, uniquesolution, amWhy, Jendrik Stelzner, Adrian Keister
closed as off-topic by Morgan Rodgers, uniquesolution, amWhy, Jendrik Stelzner, Adrian Keister Aug 6 at 13:11
This question appears to be off-topic. The users who voted to close gave this specific reason:
- "This question is missing context or other details: Please improve the question by providing additional context, which ideally includes your thoughts on the problem and any attempts you have made to solve it. This information helps others identify where you have difficulties and helps them write answers appropriate to your experience level." – Morgan Rodgers, uniquesolution, amWhy, Jendrik Stelzner, Adrian Keister
It seems that you also need Uniform Boundedness Principle.
– Danny Pak-Keung Chan
Aug 6 at 7:37
Good question! Who (and why) deducted score?
– Danny Pak-Keung Chan
Aug 6 at 7:40
add a comment |Â
It seems that you also need Uniform Boundedness Principle.
– Danny Pak-Keung Chan
Aug 6 at 7:37
Good question! Who (and why) deducted score?
– Danny Pak-Keung Chan
Aug 6 at 7:40
It seems that you also need Uniform Boundedness Principle.
– Danny Pak-Keung Chan
Aug 6 at 7:37
It seems that you also need Uniform Boundedness Principle.
– Danny Pak-Keung Chan
Aug 6 at 7:37
Good question! Who (and why) deducted score?
– Danny Pak-Keung Chan
Aug 6 at 7:40
Good question! Who (and why) deducted score?
– Danny Pak-Keung Chan
Aug 6 at 7:40
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
- To show that $T(v_1+v_2)=Tv_1+Tv_2$ for any $v_1,v_2in V$:
Prove by contradiction. Suppose the contrary that there exist $v_1,v_2in V$
such that $T(v_1+v_2)neq Tv_1+Tv_2$. By Hahn-Banach Theorem,
there exists $fin W'$ such that $f(T(v_1+v_2))neq f(Tv_1+Tv_2)$.
On the other hand, $LHS=(Sf)(v_1+v_2)=(Sf)(v_1)+(Sf)(v_2)=f(Tv_1)+f(Tv_2)=f(Tv_1+Tv_2)=RHS$,
which is a contradiction.
- To show that $T(alpha v)=alpha T(v)$ for any scalar $alpha$
(i.e., $alphainmathbbR$ or $alphainmathbbC$, depending
on the scalar field of $V$ and $W$) and $vin V$:
Prove by contradiction. Suppose the contrary that there exist scalar
$alpha$ and $vin V$ such that $T(alpha v)neqalpha T(v)$. By
Hahn-Banach Theorem, there exists $fin W'$ such that $f(T(alpha v))neq f(alpha T(v))$.
On the other hand, $f(T(alpha v))=(Sf)(alpha v)=alphacdot(Sf)(v)=alphacdot f(Tv)=f(alphacdot Tv)$,
which is a contradiction.
By 1 and 2, $T$ is a linear map. Lastly, we show that $T$ is bounded.
Firstly, we show that $supleq1<infty$.
Let $vin V$ be arbitrary, then
begineqnarray*
sup_fin W',|Sf(v)| & = & sup_fin W',|f(Tv)|\
& leq & sup_fin W',|f|cdot|Tv|\
& = & |Tv|\
& < & infty
endeqnarray*
By Uniform Boundedness Principle, we have that $M:=supleq1<infty$.
Now, we are ready to show that $T$ is bounded and $|T|leq M$.
Let $vin V$ with $|v|leq1$ be arbitrary. If $Tv=0$, we have
$|Tv|leq M$. Suppose that $Tvneq0$. Consider the one dimensional
space spanned by $Tv$: $V_0=alphacdotwidehatTvmidalphainmathbbC$
(or $alphainmathbbR$), where $widehatTv=Tv/|Tv|$. Define
$f_0:V_0rightarrowmathbbC$ by $f(alphacdotwidehatTv)=alpha$.
Clearly $f_0$ is linear with $|f_0|=1$. By Hahn-Banach Theorem,
there exists $fin V'$ such that $|f|=|f_0|=1$ and $fvert_V_0=f_0$.
We have that
begineqnarray*
|Tv| & = & f(|Tv|cdotwidehatTv)\
& = & |f(|Tv|cdotwidehatTv)|\
& = & |f(Tv)|\
& = & |(Sf)(v)|\
& leq & |Sf|cdot|v|\
& leq & M.
endeqnarray*
It follows that $|T|leq M$.
edited Aug 6 at 12:31
Delta-u
4,742518
answered Aug 6 at 7:36
Danny Pak-Keung Chan
1,87628
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
- To show that $T(v_1+v_2)=Tv_1+Tv_2$ for any $v_1,v_2in V$:
Prove by contradiction. Suppose the contrary that there exist $v_1,v_2in V$
such that $T(v_1+v_2)neq Tv_1+Tv_2$. By Hahn-Banach Theorem,
there exists $fin W'$ such that $f(T(v_1+v_2))neq f(Tv_1+Tv_2)$.
On the other hand, $LHS=(Sf)(v_1+v_2)=(Sf)(v_1)+(Sf)(v_2)=f(Tv_1)+f(Tv_2)=f(Tv_1+Tv_2)=RHS$,
which is a contradiction.
- To show that $T(alpha v)=alpha T(v)$ for any scalar $alpha$
(i.e., $alphainmathbbR$ or $alphainmathbbC$, depending
on the scalar field of $V$ and $W$) and $vin V$:
Prove by contradiction. Suppose the contrary that there exist scalar
$alpha$ and $vin V$ such that $T(alpha v)neqalpha T(v)$. By
Hahn-Banach Theorem, there exists $fin W'$ such that $f(T(alpha v))neq f(alpha T(v))$.
On the other hand, $f(T(alpha v))=(Sf)(alpha v)=alphacdot(Sf)(v)=alphacdot f(Tv)=f(alphacdot Tv)$,
which is a contradiction.
By 1 and 2, $T$ is a linear map. Lastly, we show that $T$ is bounded.
Firstly, we show that $supleq1<infty$.
Let $vin V$ be arbitrary, then
begineqnarray*
sup_fin W',|Sf(v)| & = & sup_fin W',|f(Tv)|\
& leq & sup_fin W',|f|cdot|Tv|\
& = & |Tv|\
& < & infty
endeqnarray*
By Uniform Boundedness Principle, we have that $M:=supleq1<infty$.
Now, we are ready to show that $T$ is bounded and $|T|leq M$.
Let $vin V$ with $|v|leq1$ be arbitrary. If $Tv=0$, we have
$|Tv|leq M$. Suppose that $Tvneq0$. Consider the one dimensional
space spanned by $Tv$: $V_0=alphacdotwidehatTvmidalphainmathbbC$
(or $alphainmathbbR$), where $widehatTv=Tv/|Tv|$. Define
$f_0:V_0rightarrowmathbbC$ by $f(alphacdotwidehatTv)=alpha$.
Clearly $f_0$ is linear with $|f_0|=1$. By Hahn-Banach Theorem,
there exists $fin V'$ such that $|f|=|f_0|=1$ and $fvert_V_0=f_0$.
We have that
begineqnarray*
|Tv| & = & f(|Tv|cdotwidehatTv)\
& = & |f(|Tv|cdotwidehatTv)|\
& = & |f(Tv)|\
& = & |(Sf)(v)|\
& leq & |Sf|cdot|v|\
& leq & M.
endeqnarray*
It follows that $|T|leq M$.
edited Aug 6 at 12:31
Delta-u
4,742518
answered Aug 6 at 7:36
Danny Pak-Keung Chan
1,87628
add a comment |Â
up vote
1
down vote
accepted
- To show that $T(v_1+v_2)=Tv_1+Tv_2$ for any $v_1,v_2in V$:
Prove by contradiction. Suppose the contrary that there exist $v_1,v_2in V$
such that $T(v_1+v_2)neq Tv_1+Tv_2$. By Hahn-Banach Theorem,
there exists $fin W'$ such that $f(T(v_1+v_2))neq f(Tv_1+Tv_2)$.
On the other hand, $LHS=(Sf)(v_1+v_2)=(Sf)(v_1)+(Sf)(v_2)=f(Tv_1)+f(Tv_2)=f(Tv_1+Tv_2)=RHS$,
which is a contradiction.
- To show that $T(alpha v)=alpha T(v)$ for any scalar $alpha$
(i.e., $alphainmathbbR$ or $alphainmathbbC$, depending
on the scalar field of $V$ and $W$) and $vin V$:
Prove by contradiction. Suppose the contrary that there exist scalar
$alpha$ and $vin V$ such that $T(alpha v)neqalpha T(v)$. By
Hahn-Banach Theorem, there exists $fin W'$ such that $f(T(alpha v))neq f(alpha T(v))$.
On the other hand, $f(T(alpha v))=(Sf)(alpha v)=alphacdot(Sf)(v)=alphacdot f(Tv)=f(alphacdot Tv)$,
which is a contradiction.
By 1 and 2, $T$ is a linear map. Lastly, we show that $T$ is bounded.
Firstly, we show that $supleq1<infty$.
Let $vin V$ be arbitrary, then
begineqnarray*
sup_fin W',|Sf(v)| & = & sup_fin W',|f(Tv)|\
& leq & sup_fin W',|f|cdot|Tv|\
& = & |Tv|\
& < & infty
endeqnarray*
By Uniform Boundedness Principle, we have that $M:=supleq1<infty$.
Now, we are ready to show that $T$ is bounded and $|T|leq M$.
Let $vin V$ with $|v|leq1$ be arbitrary. If $Tv=0$, we have
$|Tv|leq M$. Suppose that $Tvneq0$. Consider the one dimensional
space spanned by $Tv$: $V_0=alphacdotwidehatTvmidalphainmathbbC$
(or $alphainmathbbR$), where $widehatTv=Tv/|Tv|$. Define
$f_0:V_0rightarrowmathbbC$ by $f(alphacdotwidehatTv)=alpha$.
Clearly $f_0$ is linear with $|f_0|=1$. By Hahn-Banach Theorem,
there exists $fin V'$ such that $|f|=|f_0|=1$ and $fvert_V_0=f_0$.
We have that
begineqnarray*
|Tv| & = & f(|Tv|cdotwidehatTv)\
& = & |f(|Tv|cdotwidehatTv)|\
& = & |f(Tv)|\
& = & |(Sf)(v)|\
& leq & |Sf|cdot|v|\
& leq & M.
endeqnarray*
It follows that $|T|leq M$.
edited Aug 6 at 12:31
Delta-u
4,742518
answered Aug 6 at 7:36
Danny Pak-Keung Chan
1,87628
add a comment |Â
up vote
1
down vote
accepted
up vote
1
down vote
accepted
- To show that $T(v_1+v_2)=Tv_1+Tv_2$ for any $v_1,v_2in V$:
Prove by contradiction. Suppose the contrary that there exist $v_1,v_2in V$
such that $T(v_1+v_2)neq Tv_1+Tv_2$. By Hahn-Banach Theorem,
there exists $fin W'$ such that $f(T(v_1+v_2))neq f(Tv_1+Tv_2)$.
On the other hand, $LHS=(Sf)(v_1+v_2)=(Sf)(v_1)+(Sf)(v_2)=f(Tv_1)+f(Tv_2)=f(Tv_1+Tv_2)=RHS$,
which is a contradiction.
- To show that $T(alpha v)=alpha T(v)$ for any scalar $alpha$
(i.e., $alphainmathbbR$ or $alphainmathbbC$, depending
on the scalar field of $V$ and $W$) and $vin V$:
Prove by contradiction. Suppose the contrary that there exist scalar
$alpha$ and $vin V$ such that $T(alpha v)neqalpha T(v)$. By
Hahn-Banach Theorem, there exists $fin W'$ such that $f(T(alpha v))neq f(alpha T(v))$.
On the other hand, $f(T(alpha v))=(Sf)(alpha v)=alphacdot(Sf)(v)=alphacdot f(Tv)=f(alphacdot Tv)$,
which is a contradiction.
By 1 and 2, $T$ is a linear map. Lastly, we show that $T$ is bounded.
Firstly, we show that $supleq1<infty$.
Let $vin V$ be arbitrary, then
begineqnarray*
sup_fin W',|Sf(v)| & = & sup_fin W',|f(Tv)|\
& leq & sup_fin W',|f|cdot|Tv|\
& = & |Tv|\
& < & infty
endeqnarray*
By Uniform Boundedness Principle, we have that $M:=supleq1<infty$.
Now, we are ready to show that $T$ is bounded and $|T|leq M$.
Let $vin V$ with $|v|leq1$ be arbitrary. If $Tv=0$, we have
$|Tv|leq M$. Suppose that $Tvneq0$. Consider the one dimensional
space spanned by $Tv$: $V_0=alphacdotwidehatTvmidalphainmathbbC$
(or $alphainmathbbR$), where $widehatTv=Tv/|Tv|$. Define
$f_0:V_0rightarrowmathbbC$ by $f(alphacdotwidehatTv)=alpha$.
Clearly $f_0$ is linear with $|f_0|=1$. By Hahn-Banach Theorem,
there exists $fin V'$ such that $|f|=|f_0|=1$ and $fvert_V_0=f_0$.
We have that
begineqnarray*
|Tv| & = & f(|Tv|cdotwidehatTv)\
& = & |f(|Tv|cdotwidehatTv)|\
& = & |f(Tv)|\
& = & |(Sf)(v)|\
& leq & |Sf|cdot|v|\
& leq & M.
endeqnarray*
It follows that $|T|leq M$.
edited Aug 6 at 12:31
Delta-u
4,742518
answered Aug 6 at 7:36
Danny Pak-Keung Chan
1,87628
- To show that $T(v_1+v_2)=Tv_1+Tv_2$ for any $v_1,v_2in V$:
Prove by contradiction. Suppose the contrary that there exist $v_1,v_2in V$
such that $T(v_1+v_2)neq Tv_1+Tv_2$. By Hahn-Banach Theorem,
there exists $fin W'$ such that $f(T(v_1+v_2))neq f(Tv_1+Tv_2)$.
On the other hand, $LHS=(Sf)(v_1+v_2)=(Sf)(v_1)+(Sf)(v_2)=f(Tv_1)+f(Tv_2)=f(Tv_1+Tv_2)=RHS$,
which is a contradiction.
- To show that $T(alpha v)=alpha T(v)$ for any scalar $alpha$
(i.e., $alphainmathbbR$ or $alphainmathbbC$, depending
on the scalar field of $V$ and $W$) and $vin V$:
Prove by contradiction. Suppose the contrary that there exist scalar
$alpha$ and $vin V$ such that $T(alpha v)neqalpha T(v)$. By
Hahn-Banach Theorem, there exists $fin W'$ such that $f(T(alpha v))neq f(alpha T(v))$.
On the other hand, $f(T(alpha v))=(Sf)(alpha v)=alphacdot(Sf)(v)=alphacdot f(Tv)=f(alphacdot Tv)$,
which is a contradiction.
By 1 and 2, $T$ is a linear map. Lastly, we show that $T$ is bounded.
Firstly, we show that $supleq1<infty$.
Let $vin V$ be arbitrary, then
begineqnarray*
sup_fin W',|Sf(v)| & = & sup_fin W',|f(Tv)|\
& leq & sup_fin W',|f|cdot|Tv|\
& = & |Tv|\
& < & infty
endeqnarray*
By Uniform Boundedness Principle, we have that $M:=supleq1<infty$.
Now, we are ready to show that $T$ is bounded and $|T|leq M$.
Let $vin V$ with $|v|leq1$ be arbitrary. If $Tv=0$, we have
$|Tv|leq M$. Suppose that $Tvneq0$. Consider the one dimensional
space spanned by $Tv$: $V_0=alphacdotwidehatTvmidalphainmathbbC$
(or $alphainmathbbR$), where $widehatTv=Tv/|Tv|$. Define
$f_0:V_0rightarrowmathbbC$ by $f(alphacdotwidehatTv)=alpha$.
Clearly $f_0$ is linear with $|f_0|=1$. By Hahn-Banach Theorem,
there exists $fin V'$ such that $|f|=|f_0|=1$ and $fvert_V_0=f_0$.
We have that
begineqnarray*
|Tv| & = & f(|Tv|cdotwidehatTv)\
& = & |f(|Tv|cdotwidehatTv)|\
& = & |f(Tv)|\
& = & |(Sf)(v)|\
& leq & |Sf|cdot|v|\
& leq & M.
endeqnarray*
It follows that $|T|leq M$.
edited Aug 6 at 12:31
Delta-u
4,742518
answered Aug 6 at 7:36
Danny Pak-Keung Chan
1,87628
edited Aug 6 at 12:31
Delta-u
4,742518
edited Aug 6 at 12:31
Delta-u
4,742518
edited Aug 6 at 12:31
Delta-u
4,742518
4,742518
answered Aug 6 at 7:36
Danny Pak-Keung Chan
1,87628
answered Aug 6 at 7:36
Danny Pak-Keung Chan
1,87628
answered Aug 6 at 7:36
Danny Pak-Keung Chan
1,87628
1,87628
add a comment |Â
add a comment |Â
It seems that you also need Uniform Boundedness Principle.
– Danny Pak-Keung Chan
Aug 6 at 7:37
Good question! Who (and why) deducted score?
– Danny Pak-Keung Chan
Aug 6 at 7:40