Mixing
Isn't $Econg E^**iff dim(E)<infty $ wrong?
Clash Royale CLAN TAG#URR8PPP
up vote
3
down vote
favorite
Let $E$ a $K$-vector space and denote $E^**$ its bidual. Let $$Phi: Elongrightarrow E^**$$
defined by $$Phi(x)=left<f,xright>,quad fin E^*.$$ A theorem in my course says
$$Phitext is an isomorphismiff Etext has finite dimension.$$
The implication (i.e. $Rightarrow$) is not correct, no ? For example, if $pin (1,infty )$ we have that $L^p(mathbb R)$ is reflexive, which mean exactly that $Phi$ is bijective, right ? So I have a truble with this theorem. Moreover, the proof looks correct :
The fact that $Phi$ is injective is fine. We show the contrapositive, i.e. we suppose $E$ has infinite dimension and we prove that $Phi$ is not surjective. Fo the surjectivity, let $F=textSpan(e_1^*,...,e_n^*)neq E$. I know that there is a non zero linear form $f:E^*to K$ s.t. $Fsubset ker(f)$. For all $x=(x_i)_iin Iin E$, $$Phi(x)(e_i^*)=x_i.$$
Therefore, if $Phi (x)(F)=0$, then $x=0$ and thus $Phi (x)=0$. Therefore, $f$ is not in the range of $Phi$ and thus its not surjective.
What do you think ? By the way, I don't understand why $Phi(x)(F)=0$ implies that $Phi(x)=0$... Indeed, we can have $gnotin F$ s.t. $Phi(x)(g)neq 0$ no ?
functional-analysis
asked Jul 25 at 12:16
user386627
714214
add a comment |Â
up vote
3
down vote
favorite
Let $E$ a $K$-vector space and denote $E^**$ its bidual. Let $$Phi: Elongrightarrow E^**$$
defined by $$Phi(x)=left<f,xright>,quad fin E^*.$$ A theorem in my course says
$$Phitext is an isomorphismiff Etext has finite dimension.$$
The implication (i.e. $Rightarrow$) is not correct, no ? For example, if $pin (1,infty )$ we have that $L^p(mathbb R)$ is reflexive, which mean exactly that $Phi$ is bijective, right ? So I have a truble with this theorem. Moreover, the proof looks correct :
The fact that $Phi$ is injective is fine. We show the contrapositive, i.e. we suppose $E$ has infinite dimension and we prove that $Phi$ is not surjective. Fo the surjectivity, let $F=textSpan(e_1^*,...,e_n^*)neq E$. I know that there is a non zero linear form $f:E^*to K$ s.t. $Fsubset ker(f)$. For all $x=(x_i)_iin Iin E$, $$Phi(x)(e_i^*)=x_i.$$
Therefore, if $Phi (x)(F)=0$, then $x=0$ and thus $Phi (x)=0$. Therefore, $f$ is not in the range of $Phi$ and thus its not surjective.
What do you think ? By the way, I don't understand why $Phi(x)(F)=0$ implies that $Phi(x)=0$... Indeed, we can have $gnotin F$ s.t. $Phi(x)(g)neq 0$ no ?
functional-analysis
asked Jul 25 at 12:16
user386627
714214
9
Does $E^*$ denote the algebraic dual or the continuous dual?
– Omnomnomnom
Jul 25 at 12:21
3
Reflexivity is about the dual space of a Banach or Hilbert space comprising just the continuous linear functionals not the vector space comprising all linear functions.
– Rob Arthan
Jul 25 at 12:21
3
The problem here is that you are making confusion with the continuou dual and the alebraic dual. The algebraic dual of $L^p$ contains also unbounded linear functionals.
– Crostul
Jul 25 at 12:22
@Omnomnomnom: Algebraic I guess.
– user386627
Jul 25 at 12:23
add a comment |Â
up vote
3
down vote
favorite
up vote
3
down vote
favorite
Let $E$ a $K$-vector space and denote $E^**$ its bidual. Let $$Phi: Elongrightarrow E^**$$
defined by $$Phi(x)=left<f,xright>,quad fin E^*.$$ A theorem in my course says
$$Phitext is an isomorphismiff Etext has finite dimension.$$
The implication (i.e. $Rightarrow$) is not correct, no ? For example, if $pin (1,infty )$ we have that $L^p(mathbb R)$ is reflexive, which mean exactly that $Phi$ is bijective, right ? So I have a truble with this theorem. Moreover, the proof looks correct :
The fact that $Phi$ is injective is fine. We show the contrapositive, i.e. we suppose $E$ has infinite dimension and we prove that $Phi$ is not surjective. Fo the surjectivity, let $F=textSpan(e_1^*,...,e_n^*)neq E$. I know that there is a non zero linear form $f:E^*to K$ s.t. $Fsubset ker(f)$. For all $x=(x_i)_iin Iin E$, $$Phi(x)(e_i^*)=x_i.$$
Therefore, if $Phi (x)(F)=0$, then $x=0$ and thus $Phi (x)=0$. Therefore, $f$ is not in the range of $Phi$ and thus its not surjective.
What do you think ? By the way, I don't understand why $Phi(x)(F)=0$ implies that $Phi(x)=0$... Indeed, we can have $gnotin F$ s.t. $Phi(x)(g)neq 0$ no ?
functional-analysis
asked Jul 25 at 12:16
user386627
714214
Let $E$ a $K$-vector space and denote $E^**$ its bidual. Let $$Phi: Elongrightarrow E^**$$
defined by $$Phi(x)=left<f,xright>,quad fin E^*.$$ A theorem in my course says
$$Phitext is an isomorphismiff Etext has finite dimension.$$
The implication (i.e. $Rightarrow$) is not correct, no ? For example, if $pin (1,infty )$ we have that $L^p(mathbb R)$ is reflexive, which mean exactly that $Phi$ is bijective, right ? So I have a truble with this theorem. Moreover, the proof looks correct :
The fact that $Phi$ is injective is fine. We show the contrapositive, i.e. we suppose $E$ has infinite dimension and we prove that $Phi$ is not surjective. Fo the surjectivity, let $F=textSpan(e_1^*,...,e_n^*)neq E$. I know that there is a non zero linear form $f:E^*to K$ s.t. $Fsubset ker(f)$. For all $x=(x_i)_iin Iin E$, $$Phi(x)(e_i^*)=x_i.$$
Therefore, if $Phi (x)(F)=0$, then $x=0$ and thus $Phi (x)=0$. Therefore, $f$ is not in the range of $Phi$ and thus its not surjective.
What do you think ? By the way, I don't understand why $Phi(x)(F)=0$ implies that $Phi(x)=0$... Indeed, we can have $gnotin F$ s.t. $Phi(x)(g)neq 0$ no ?
functional-analysis
asked Jul 25 at 12:16
user386627
714214
edited Jul 25 at 12:25
asked Jul 25 at 12:16
user386627
714214
asked Jul 25 at 12:16
user386627
714214
asked Jul 25 at 12:16
user386627
714214
714214
9
Does $E^*$ denote the algebraic dual or the continuous dual?
– Omnomnomnom
Jul 25 at 12:21
3
Reflexivity is about the dual space of a Banach or Hilbert space comprising just the continuous linear functionals not the vector space comprising all linear functions.
– Rob Arthan
Jul 25 at 12:21
3
The problem here is that you are making confusion with the continuou dual and the alebraic dual. The algebraic dual of $L^p$ contains also unbounded linear functionals.
– Crostul
Jul 25 at 12:22
@Omnomnomnom: Algebraic I guess.
– user386627
Jul 25 at 12:23
add a comment |Â
9
Does $E^*$ denote the algebraic dual or the continuous dual?
– Omnomnomnom
Jul 25 at 12:21
3
Reflexivity is about the dual space of a Banach or Hilbert space comprising just the continuous linear functionals not the vector space comprising all linear functions.
– Rob Arthan
Jul 25 at 12:21
3
The problem here is that you are making confusion with the continuou dual and the alebraic dual. The algebraic dual of $L^p$ contains also unbounded linear functionals.
– Crostul
Jul 25 at 12:22
@Omnomnomnom: Algebraic I guess.
– user386627
Jul 25 at 12:23
9
9
Does $E^*$ denote the algebraic dual or the continuous dual?
– Omnomnomnom
Jul 25 at 12:21
Does $E^*$ denote the algebraic dual or the continuous dual?
– Omnomnomnom
Jul 25 at 12:21
3
3
Reflexivity is about the dual space of a Banach or Hilbert space comprising just the continuous linear functionals not the vector space comprising all linear functions.
– Rob Arthan
Jul 25 at 12:21
Reflexivity is about the dual space of a Banach or Hilbert space comprising just the continuous linear functionals not the vector space comprising all linear functions.
– Rob Arthan
Jul 25 at 12:21
3
3
The problem here is that you are making confusion with the continuou dual and the alebraic dual. The algebraic dual of $L^p$ contains also unbounded linear functionals.
– Crostul
Jul 25 at 12:22
The problem here is that you are making confusion with the continuou dual and the alebraic dual. The algebraic dual of $L^p$ contains also unbounded linear functionals.
– Crostul
Jul 25 at 12:22
@Omnomnomnom: Algebraic I guess.
– user386627
Jul 25 at 12:23
@Omnomnomnom: Algebraic I guess.
– user386627
Jul 25 at 12:23
add a comment |Â
1 Answer
1
active
oldest
votes
up vote
3
down vote
I also don't understand why $Phi(x)(F)=0$ implies $Phi(x)=0$.
Consider this argument instead: let $(e_i)_iin I$ be a basis for $E$ and let $(e_i^*)_iin I$ be the dual functionals in $E^*$ (defined by $e_i^*(e_j) = delta_ij$).
Let $f$ be a linear functional in $E^**$ such that $f(e_i^*) = 1, forall i in I$. To convince yourself that such $f$ exists, note that $(e_i^*)_iin I$ is linearly independent in $E^*$ so it can be extended to a basis $B$ for $E^*$. Now define $f$ on $B$ as $f(e_i^*) = 1, forall i in I$ and something arbitrary on the rest of $B$.
Assume that there exists $x = sum_iin Ix_ie_i in E$ with only finitely many $x_i ne 0$ such that $Phi(x) = f$.
Then in particular $$1 = f(e_i^*) = Phi(x)(e_i^*) = e_i^*(x) = x_i$$
which is a contradiction because now $x_i ne 0, forall iin I$, and $I$ is infinite.
Hence $f$ is not in the range of $Phi$.
answered Jul 25 at 14:09
mechanodroid
22.2k52041
add a comment |Â
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
3
down vote
I also don't understand why $Phi(x)(F)=0$ implies $Phi(x)=0$.
Consider this argument instead: let $(e_i)_iin I$ be a basis for $E$ and let $(e_i^*)_iin I$ be the dual functionals in $E^*$ (defined by $e_i^*(e_j) = delta_ij$).
Let $f$ be a linear functional in $E^**$ such that $f(e_i^*) = 1, forall i in I$. To convince yourself that such $f$ exists, note that $(e_i^*)_iin I$ is linearly independent in $E^*$ so it can be extended to a basis $B$ for $E^*$. Now define $f$ on $B$ as $f(e_i^*) = 1, forall i in I$ and something arbitrary on the rest of $B$.
Assume that there exists $x = sum_iin Ix_ie_i in E$ with only finitely many $x_i ne 0$ such that $Phi(x) = f$.
Then in particular $$1 = f(e_i^*) = Phi(x)(e_i^*) = e_i^*(x) = x_i$$
which is a contradiction because now $x_i ne 0, forall iin I$, and $I$ is infinite.
Hence $f$ is not in the range of $Phi$.
answered Jul 25 at 14:09
mechanodroid
22.2k52041
add a comment |Â
up vote
3
down vote
I also don't understand why $Phi(x)(F)=0$ implies $Phi(x)=0$.
Consider this argument instead: let $(e_i)_iin I$ be a basis for $E$ and let $(e_i^*)_iin I$ be the dual functionals in $E^*$ (defined by $e_i^*(e_j) = delta_ij$).
Let $f$ be a linear functional in $E^**$ such that $f(e_i^*) = 1, forall i in I$. To convince yourself that such $f$ exists, note that $(e_i^*)_iin I$ is linearly independent in $E^*$ so it can be extended to a basis $B$ for $E^*$. Now define $f$ on $B$ as $f(e_i^*) = 1, forall i in I$ and something arbitrary on the rest of $B$.
Assume that there exists $x = sum_iin Ix_ie_i in E$ with only finitely many $x_i ne 0$ such that $Phi(x) = f$.
Then in particular $$1 = f(e_i^*) = Phi(x)(e_i^*) = e_i^*(x) = x_i$$
which is a contradiction because now $x_i ne 0, forall iin I$, and $I$ is infinite.
Hence $f$ is not in the range of $Phi$.
answered Jul 25 at 14:09
mechanodroid
22.2k52041
add a comment |Â
up vote
3
down vote
up vote
3
down vote
I also don't understand why $Phi(x)(F)=0$ implies $Phi(x)=0$.
Consider this argument instead: let $(e_i)_iin I$ be a basis for $E$ and let $(e_i^*)_iin I$ be the dual functionals in $E^*$ (defined by $e_i^*(e_j) = delta_ij$).
Let $f$ be a linear functional in $E^**$ such that $f(e_i^*) = 1, forall i in I$. To convince yourself that such $f$ exists, note that $(e_i^*)_iin I$ is linearly independent in $E^*$ so it can be extended to a basis $B$ for $E^*$. Now define $f$ on $B$ as $f(e_i^*) = 1, forall i in I$ and something arbitrary on the rest of $B$.
Assume that there exists $x = sum_iin Ix_ie_i in E$ with only finitely many $x_i ne 0$ such that $Phi(x) = f$.
Then in particular $$1 = f(e_i^*) = Phi(x)(e_i^*) = e_i^*(x) = x_i$$
which is a contradiction because now $x_i ne 0, forall iin I$, and $I$ is infinite.
Hence $f$ is not in the range of $Phi$.
answered Jul 25 at 14:09
mechanodroid
22.2k52041
I also don't understand why $Phi(x)(F)=0$ implies $Phi(x)=0$.
Consider this argument instead: let $(e_i)_iin I$ be a basis for $E$ and let $(e_i^*)_iin I$ be the dual functionals in $E^*$ (defined by $e_i^*(e_j) = delta_ij$).
Let $f$ be a linear functional in $E^**$ such that $f(e_i^*) = 1, forall i in I$. To convince yourself that such $f$ exists, note that $(e_i^*)_iin I$ is linearly independent in $E^*$ so it can be extended to a basis $B$ for $E^*$. Now define $f$ on $B$ as $f(e_i^*) = 1, forall i in I$ and something arbitrary on the rest of $B$.
Assume that there exists $x = sum_iin Ix_ie_i in E$ with only finitely many $x_i ne 0$ such that $Phi(x) = f$.
Then in particular $$1 = f(e_i^*) = Phi(x)(e_i^*) = e_i^*(x) = x_i$$
which is a contradiction because now $x_i ne 0, forall iin I$, and $I$ is infinite.
Hence $f$ is not in the range of $Phi$.
answered Jul 25 at 14:09
mechanodroid
22.2k52041
answered Jul 25 at 14:09
mechanodroid
22.2k52041
answered Jul 25 at 14:09
mechanodroid
22.2k52041
answered Jul 25 at 14:09
mechanodroid
22.2k52041
22.2k52041
add a comment |Â
add a comment |Â
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
StackExchange.ready(
function ()
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f2862368%2fisnt-e-cong-e-iff-dime-infty-wrong%23new-answer', 'question_page');
);
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Sign up or log in
StackExchange.ready(function ()
StackExchange.helpers.onClickDraftSave('#login-link');
);
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
9
Does $E^*$ denote the algebraic dual or the continuous dual?
– Omnomnomnom
Jul 25 at 12:21
3
Reflexivity is about the dual space of a Banach or Hilbert space comprising just the continuous linear functionals not the vector space comprising all linear functions.
– Rob Arthan
Jul 25 at 12:21
3
The problem here is that you are making confusion with the continuou dual and the alebraic dual. The algebraic dual of $L^p$ contains also unbounded linear functionals.
– Crostul
Jul 25 at 12:22
@Omnomnomnom: Algebraic I guess.
– user386627
Jul 25 at 12:23