Mixing
Show that $varphi in E'$ and if $E$ is a Banach space then $varphi in E$
$begingroup$
Problem: Let $E$ be a normed space over field $mathbb{C}$. Fix a continuous function $f: left[ a,b right] rightarrow E$ with $left[ a,b right] subset mathbb{R}$. Consider $varphi: E' rightarrow mathbb{C}$ given by $varphi(y) := displaystyleint_a^b (y circ f)(t)dt, forall y in E'$. Show that $varphi in E''$ and if $E$ is a Banach space then $varphi in E$, in which $E'$ is the dual space of $E$.
Could you give me some hint to solve the problem.
functional-analysis
asked Jan 16 at 14:39
MinhMinh
1959
$endgroup$
|
show 2 more comments
$begingroup$
Problem: Let $E$ be a normed space over field $mathbb{C}$. Fix a continuous function $f: left[ a,b right] rightarrow E$ with $left[ a,b right] subset mathbb{R}$. Consider $varphi: E' rightarrow mathbb{C}$ given by $varphi(y) := displaystyleint_a^b (y circ f)(t)dt, forall y in E'$. Show that $varphi in E''$ and if $E$ is a Banach space then $varphi in E$, in which $E'$ is the dual space of $E$.
Could you give me some hint to solve the problem.
functional-analysis
asked Jan 16 at 14:39
MinhMinh
1959
$endgroup$
1
$begingroup$
Likely the first part is $E’’$ instead of $E’$. And I suggest reproducing the construction of Riemann integral in $E$.
$endgroup$
– Mindlack
Jan 16 at 14:45
$begingroup$
I agree.$varphiin E''$ instead of $varphiin E'$
$endgroup$
– Peter Melech
Jan 16 at 14:47
$begingroup$
I have edited!!!
$endgroup$
– Minh
Jan 16 at 14:49
1
$begingroup$
1) is just definition manipulation, and 2) is about proving exactly as in $mathbb{R}$ that a continuous function can be integrated.
$endgroup$
– Mindlack
Jan 16 at 14:52
$begingroup$
@Mindlack The field we are considering is $mathbb{C}$, it must be different from $mathbb{R}$.
$endgroup$
– Minh
Jan 16 at 15:10
|
show 2 more comments
$begingroup$
Problem: Let $E$ be a normed space over field $mathbb{C}$. Fix a continuous function $f: left[ a,b right] rightarrow E$ with $left[ a,b right] subset mathbb{R}$. Consider $varphi: E' rightarrow mathbb{C}$ given by $varphi(y) := displaystyleint_a^b (y circ f)(t)dt, forall y in E'$. Show that $varphi in E''$ and if $E$ is a Banach space then $varphi in E$, in which $E'$ is the dual space of $E$.
Could you give me some hint to solve the problem.
functional-analysis
asked Jan 16 at 14:39
MinhMinh
1959
$endgroup$
Problem: Let $E$ be a normed space over field $mathbb{C}$. Fix a continuous function $f: left[ a,b right] rightarrow E$ with $left[ a,b right] subset mathbb{R}$. Consider $varphi: E' rightarrow mathbb{C}$ given by $varphi(y) := displaystyleint_a^b (y circ f)(t)dt, forall y in E'$. Show that $varphi in E''$ and if $E$ is a Banach space then $varphi in E$, in which $E'$ is the dual space of $E$.
Could you give me some hint to solve the problem.
functional-analysis
functional-analysis
asked Jan 16 at 14:39
MinhMinh
1959
asked Jan 16 at 14:39
MinhMinh
1959
edited Jan 16 at 14:48
Minh
asked Jan 16 at 14:39
MinhMinh
1959
asked Jan 16 at 14:39
MinhMinh
1959
asked Jan 16 at 14:39
MinhMinh
1959
1959
1
$begingroup$
Likely the first part is $E’’$ instead of $E’$. And I suggest reproducing the construction of Riemann integral in $E$.
$endgroup$
– Mindlack
Jan 16 at 14:45
$begingroup$
I agree.$varphiin E''$ instead of $varphiin E'$
$endgroup$
– Peter Melech
Jan 16 at 14:47
$begingroup$
I have edited!!!
$endgroup$
– Minh
Jan 16 at 14:49
1
$begingroup$
1) is just definition manipulation, and 2) is about proving exactly as in $mathbb{R}$ that a continuous function can be integrated.
$endgroup$
– Mindlack
Jan 16 at 14:52
$begingroup$
@Mindlack The field we are considering is $mathbb{C}$, it must be different from $mathbb{R}$.
$endgroup$
– Minh
Jan 16 at 15:10
|
show 2 more comments
1
$begingroup$
Likely the first part is $E’’$ instead of $E’$. And I suggest reproducing the construction of Riemann integral in $E$.
$endgroup$
– Mindlack
Jan 16 at 14:45
$begingroup$
I agree.$varphiin E''$ instead of $varphiin E'$
$endgroup$
– Peter Melech
Jan 16 at 14:47
$begingroup$
I have edited!!!
$endgroup$
– Minh
Jan 16 at 14:49
1
$begingroup$
1) is just definition manipulation, and 2) is about proving exactly as in $mathbb{R}$ that a continuous function can be integrated.
$endgroup$
– Mindlack
Jan 16 at 14:52
$begingroup$
@Mindlack The field we are considering is $mathbb{C}$, it must be different from $mathbb{R}$.
$endgroup$
– Minh
Jan 16 at 15:10
1
1
$begingroup$
Likely the first part is $E’’$ instead of $E’$. And I suggest reproducing the construction of Riemann integral in $E$.
$endgroup$
– Mindlack
Jan 16 at 14:45
$begingroup$
Likely the first part is $E’’$ instead of $E’$. And I suggest reproducing the construction of Riemann integral in $E$.
$endgroup$
– Mindlack
Jan 16 at 14:45
$begingroup$
I agree.$varphiin E''$ instead of $varphiin E'$
$endgroup$
– Peter Melech
Jan 16 at 14:47
$begingroup$
I agree.$varphiin E''$ instead of $varphiin E'$
$endgroup$
– Peter Melech
Jan 16 at 14:47
$begingroup$
I have edited!!!
$endgroup$
– Minh
Jan 16 at 14:49
$begingroup$
I have edited!!!
$endgroup$
– Minh
Jan 16 at 14:49
1
1
$begingroup$
1) is just definition manipulation, and 2) is about proving exactly as in $mathbb{R}$ that a continuous function can be integrated.
$endgroup$
– Mindlack
Jan 16 at 14:52
$begingroup$
1) is just definition manipulation, and 2) is about proving exactly as in $mathbb{R}$ that a continuous function can be integrated.
$endgroup$
– Mindlack
Jan 16 at 14:52
$begingroup$
@Mindlack The field we are considering is $mathbb{C}$, it must be different from $mathbb{R}$.
$endgroup$
– Minh
Jan 16 at 15:10
$begingroup$
@Mindlack The field we are considering is $mathbb{C}$, it must be different from $mathbb{R}$.
$endgroup$
– Minh
Jan 16 at 15:10
|
show 2 more comments
2 Answers
2
active
oldest
votes
$begingroup$
If the integral is a Lebesgue integral then
$$
phi(y) = int_a^b (ycirc f)(t) dt = yleft( int_a^b f(t)dt right) ,
$$
where the integral on the right is a Bochner integral. Then $int_a^b f(t)dtin E$ is the element you are looking for.
answered Jan 16 at 17:27
dawdaw
24.5k1645
$endgroup$
$begingroup$
Could you give me a solve using Riemann or Lesbegue integral because I don't study Bochner integral.
$endgroup$
– Minh
Jan 17 at 3:43
add a comment |
$begingroup$
$ymapsto varphi(y)=int_a^b(ycirc f)(t)dt$ is linear by the linearity of the integral. It is continuous because $y_nto y$ in $E'$ means uniform convergence on the unit ball of $E$ and hence on every multiple on the unit ball, and continuity of $f:[a,b]to E$ implies that the range is compact and hence bounded in $E$. This shows that $varphiin E''$.
To show that $varphi in E$, if $E$ is complete, you can either use that the dual of $(E',tau)$ is $E$ (where $tau$ is the topology of uniform convergence on absolutely convex compact sets -- note that the argumet for the continuity of $varphi$ actually shows that it is $tau$-continuous), or you can show more directly that $varphi$ is in the closure of $E$ in $E''$. This could be done by approximating $f$ uniformly on $[a,b]$ by linear combinations of functions of the form $tmapsto I_{[alpha,beta]}(t) e$ for $ein E$ and $ale alphale betale b$ (these are the $E$-valued step functions for which you can calculate the corresponding $varphi$ explicitely).
answered Jan 17 at 13:59
JochenJochen
6,8131023
$endgroup$
$begingroup$
I don't understand what are $E-$valued step functions.
$endgroup$
– Minh
Jan 17 at 15:18
$begingroup$
As I wrote: linear combinations of indicator functions times vectors in $E $.
$endgroup$
– Jochen
Jan 17 at 17:06
add a comment |
Your Answer
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
$begingroup$
If the integral is a Lebesgue integral then
$$
phi(y) = int_a^b (ycirc f)(t) dt = yleft( int_a^b f(t)dt right) ,
$$
where the integral on the right is a Bochner integral. Then $int_a^b f(t)dtin E$ is the element you are looking for.
answered Jan 16 at 17:27
dawdaw
24.5k1645
$endgroup$
$begingroup$
Could you give me a solve using Riemann or Lesbegue integral because I don't study Bochner integral.
$endgroup$
– Minh
Jan 17 at 3:43
add a comment |
$begingroup$
If the integral is a Lebesgue integral then
$$
phi(y) = int_a^b (ycirc f)(t) dt = yleft( int_a^b f(t)dt right) ,
$$
where the integral on the right is a Bochner integral. Then $int_a^b f(t)dtin E$ is the element you are looking for.
answered Jan 16 at 17:27
dawdaw
24.5k1645
$endgroup$
$begingroup$
Could you give me a solve using Riemann or Lesbegue integral because I don't study Bochner integral.
$endgroup$
– Minh
Jan 17 at 3:43
add a comment |
$begingroup$
If the integral is a Lebesgue integral then
$$
phi(y) = int_a^b (ycirc f)(t) dt = yleft( int_a^b f(t)dt right) ,
$$
where the integral on the right is a Bochner integral. Then $int_a^b f(t)dtin E$ is the element you are looking for.
answered Jan 16 at 17:27
dawdaw
24.5k1645
$endgroup$
If the integral is a Lebesgue integral then
$$
phi(y) = int_a^b (ycirc f)(t) dt = yleft( int_a^b f(t)dt right) ,
$$
where the integral on the right is a Bochner integral. Then $int_a^b f(t)dtin E$ is the element you are looking for.
answered Jan 16 at 17:27
dawdaw
24.5k1645
answered Jan 16 at 17:27
dawdaw
24.5k1645
answered Jan 16 at 17:27
dawdaw
24.5k1645
answered Jan 16 at 17:27
dawdaw
24.5k1645
24.5k1645
$begingroup$
Could you give me a solve using Riemann or Lesbegue integral because I don't study Bochner integral.
$endgroup$
– Minh
Jan 17 at 3:43
add a comment |
$begingroup$
Could you give me a solve using Riemann or Lesbegue integral because I don't study Bochner integral.
$endgroup$
– Minh
Jan 17 at 3:43
$begingroup$
Could you give me a solve using Riemann or Lesbegue integral because I don't study Bochner integral.
$endgroup$
– Minh
Jan 17 at 3:43
$begingroup$
Could you give me a solve using Riemann or Lesbegue integral because I don't study Bochner integral.
$endgroup$
– Minh
Jan 17 at 3:43
add a comment |
$begingroup$
$ymapsto varphi(y)=int_a^b(ycirc f)(t)dt$ is linear by the linearity of the integral. It is continuous because $y_nto y$ in $E'$ means uniform convergence on the unit ball of $E$ and hence on every multiple on the unit ball, and continuity of $f:[a,b]to E$ implies that the range is compact and hence bounded in $E$. This shows that $varphiin E''$.
To show that $varphi in E$, if $E$ is complete, you can either use that the dual of $(E',tau)$ is $E$ (where $tau$ is the topology of uniform convergence on absolutely convex compact sets -- note that the argumet for the continuity of $varphi$ actually shows that it is $tau$-continuous), or you can show more directly that $varphi$ is in the closure of $E$ in $E''$. This could be done by approximating $f$ uniformly on $[a,b]$ by linear combinations of functions of the form $tmapsto I_{[alpha,beta]}(t) e$ for $ein E$ and $ale alphale betale b$ (these are the $E$-valued step functions for which you can calculate the corresponding $varphi$ explicitely).
answered Jan 17 at 13:59
JochenJochen
6,8131023
$endgroup$
$begingroup$
I don't understand what are $E-$valued step functions.
$endgroup$
– Minh
Jan 17 at 15:18
$begingroup$
As I wrote: linear combinations of indicator functions times vectors in $E $.
$endgroup$
– Jochen
Jan 17 at 17:06
add a comment |
$begingroup$
$ymapsto varphi(y)=int_a^b(ycirc f)(t)dt$ is linear by the linearity of the integral. It is continuous because $y_nto y$ in $E'$ means uniform convergence on the unit ball of $E$ and hence on every multiple on the unit ball, and continuity of $f:[a,b]to E$ implies that the range is compact and hence bounded in $E$. This shows that $varphiin E''$.
To show that $varphi in E$, if $E$ is complete, you can either use that the dual of $(E',tau)$ is $E$ (where $tau$ is the topology of uniform convergence on absolutely convex compact sets -- note that the argumet for the continuity of $varphi$ actually shows that it is $tau$-continuous), or you can show more directly that $varphi$ is in the closure of $E$ in $E''$. This could be done by approximating $f$ uniformly on $[a,b]$ by linear combinations of functions of the form $tmapsto I_{[alpha,beta]}(t) e$ for $ein E$ and $ale alphale betale b$ (these are the $E$-valued step functions for which you can calculate the corresponding $varphi$ explicitely).
answered Jan 17 at 13:59
JochenJochen
6,8131023
$endgroup$
$begingroup$
I don't understand what are $E-$valued step functions.
$endgroup$
– Minh
Jan 17 at 15:18
$begingroup$
As I wrote: linear combinations of indicator functions times vectors in $E $.
$endgroup$
– Jochen
Jan 17 at 17:06
add a comment |
$begingroup$
$ymapsto varphi(y)=int_a^b(ycirc f)(t)dt$ is linear by the linearity of the integral. It is continuous because $y_nto y$ in $E'$ means uniform convergence on the unit ball of $E$ and hence on every multiple on the unit ball, and continuity of $f:[a,b]to E$ implies that the range is compact and hence bounded in $E$. This shows that $varphiin E''$.
To show that $varphi in E$, if $E$ is complete, you can either use that the dual of $(E',tau)$ is $E$ (where $tau$ is the topology of uniform convergence on absolutely convex compact sets -- note that the argumet for the continuity of $varphi$ actually shows that it is $tau$-continuous), or you can show more directly that $varphi$ is in the closure of $E$ in $E''$. This could be done by approximating $f$ uniformly on $[a,b]$ by linear combinations of functions of the form $tmapsto I_{[alpha,beta]}(t) e$ for $ein E$ and $ale alphale betale b$ (these are the $E$-valued step functions for which you can calculate the corresponding $varphi$ explicitely).
answered Jan 17 at 13:59
JochenJochen
6,8131023
$endgroup$
$ymapsto varphi(y)=int_a^b(ycirc f)(t)dt$ is linear by the linearity of the integral. It is continuous because $y_nto y$ in $E'$ means uniform convergence on the unit ball of $E$ and hence on every multiple on the unit ball, and continuity of $f:[a,b]to E$ implies that the range is compact and hence bounded in $E$. This shows that $varphiin E''$.
To show that $varphi in E$, if $E$ is complete, you can either use that the dual of $(E',tau)$ is $E$ (where $tau$ is the topology of uniform convergence on absolutely convex compact sets -- note that the argumet for the continuity of $varphi$ actually shows that it is $tau$-continuous), or you can show more directly that $varphi$ is in the closure of $E$ in $E''$. This could be done by approximating $f$ uniformly on $[a,b]$ by linear combinations of functions of the form $tmapsto I_{[alpha,beta]}(t) e$ for $ein E$ and $ale alphale betale b$ (these are the $E$-valued step functions for which you can calculate the corresponding $varphi$ explicitely).
answered Jan 17 at 13:59
JochenJochen
6,8131023
answered Jan 17 at 13:59
JochenJochen
6,8131023
answered Jan 17 at 13:59
JochenJochen
6,8131023
answered Jan 17 at 13:59
JochenJochen
6,8131023
6,8131023
$begingroup$
I don't understand what are $E-$valued step functions.
$endgroup$
– Minh
Jan 17 at 15:18
$begingroup$
As I wrote: linear combinations of indicator functions times vectors in $E $.
$endgroup$
– Jochen
Jan 17 at 17:06
add a comment |
$begingroup$
I don't understand what are $E-$valued step functions.
$endgroup$
– Minh
Jan 17 at 15:18
$begingroup$
As I wrote: linear combinations of indicator functions times vectors in $E $.
$endgroup$
– Jochen
Jan 17 at 17:06
$begingroup$
I don't understand what are $E-$valued step functions.
$endgroup$
– Minh
Jan 17 at 15:18
$begingroup$
I don't understand what are $E-$valued step functions.
$endgroup$
– Minh
Jan 17 at 15:18
$begingroup$
As I wrote: linear combinations of indicator functions times vectors in $E $.
$endgroup$
– Jochen
Jan 17 at 17:06
$begingroup$
As I wrote: linear combinations of indicator functions times vectors in $E $.
$endgroup$
– Jochen
Jan 17 at 17:06
add a comment |
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1
$begingroup$
Likely the first part is $E’’$ instead of $E’$. And I suggest reproducing the construction of Riemann integral in $E$.
$endgroup$
– Mindlack
Jan 16 at 14:45
$begingroup$
I agree.$varphiin E''$ instead of $varphiin E'$
$endgroup$
– Peter Melech
Jan 16 at 14:47
$begingroup$
I have edited!!!
$endgroup$
– Minh
Jan 16 at 14:49
1
$begingroup$
1) is just definition manipulation, and 2) is about proving exactly as in $mathbb{R}$ that a continuous function can be integrated.
$endgroup$
– Mindlack
Jan 16 at 14:52
$begingroup$
@Mindlack The field we are considering is $mathbb{C}$, it must be different from $mathbb{R}$.
$endgroup$
– Minh
Jan 16 at 15:10