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Weak operator topology convergence of sequence of bounded operators to a bounded operator
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I need to prove the following theorem.
Let $X, Y$ be Banach spaces, with $Y$ weakly sequentially complete. Let $T_n subset mathscrL(X,Y)$ with $Lambda(T_nx)$ converges for all $x in X$, $Lambda in Y^*$. Prove there is a $T in mathscrL(X,Y)$ such that $T_n to T$ in the weak operator topology.
Proof so far: $Lambda(T_nx)$ converges for all $x in X$, $Lambda in Y^*$. Since $Y^*$ is weakly sequentially complete, there is a $y in Y$ such that $Lambda(T_nx) to Lambda y_x$ for all $Lambda in Y^*$. Define $Tx :=y_x$. We need to show $T$ is bounded. Note that because $Lambda(T_nx)$ converges, it is most certainly true that $Lambda(T_nx) : T_n in T_n < infty$. Hence, by a Corollary of the PUB, $|T_n|$ is uniformly bounded by say $M > 0$.
So here now I want to prove that $|T|$ is bounded. Using reverse triangle inequality, we can get that $|Lambda(Tx)| leq M |Lambda| |x|$, but the fact that we have $Lambda$ on the left hand side does not seem to allow me to prove that $T$ is bounded. Reed and Simon do a similar proof with $X = Y$ Hilbert spaces, and they refrain from defining $T$ until later, but they do it using a very clever result of the Riesz Representation Theorem. I'm not sure maybe how to construct an analogous $T$ here instead of choosing the seemingly obvious choice from the fact that $Y$ is weakly sequentially complete. Any hints/help would be much appreciated.
functional-analysis banach-spaces weak-convergence
asked Jul 24 at 22:51
mathishard.butweloveit
1069
add a comment |Â
up vote
1
down vote
favorite
I need to prove the following theorem.
Let $X, Y$ be Banach spaces, with $Y$ weakly sequentially complete. Let $T_n subset mathscrL(X,Y)$ with $Lambda(T_nx)$ converges for all $x in X$, $Lambda in Y^*$. Prove there is a $T in mathscrL(X,Y)$ such that $T_n to T$ in the weak operator topology.
Proof so far: $Lambda(T_nx)$ converges for all $x in X$, $Lambda in Y^*$. Since $Y^*$ is weakly sequentially complete, there is a $y in Y$ such that $Lambda(T_nx) to Lambda y_x$ for all $Lambda in Y^*$. Define $Tx :=y_x$. We need to show $T$ is bounded. Note that because $Lambda(T_nx)$ converges, it is most certainly true that $Lambda(T_nx) : T_n in T_n < infty$. Hence, by a Corollary of the PUB, $|T_n|$ is uniformly bounded by say $M > 0$.
So here now I want to prove that $|T|$ is bounded. Using reverse triangle inequality, we can get that $|Lambda(Tx)| leq M |Lambda| |x|$, but the fact that we have $Lambda$ on the left hand side does not seem to allow me to prove that $T$ is bounded. Reed and Simon do a similar proof with $X = Y$ Hilbert spaces, and they refrain from defining $T$ until later, but they do it using a very clever result of the Riesz Representation Theorem. I'm not sure maybe how to construct an analogous $T$ here instead of choosing the seemingly obvious choice from the fact that $Y$ is weakly sequentially complete. Any hints/help would be much appreciated.
functional-analysis banach-spaces weak-convergence
asked Jul 24 at 22:51
mathishard.butweloveit
1069
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
I need to prove the following theorem.
Let $X, Y$ be Banach spaces, with $Y$ weakly sequentially complete. Let $T_n subset mathscrL(X,Y)$ with $Lambda(T_nx)$ converges for all $x in X$, $Lambda in Y^*$. Prove there is a $T in mathscrL(X,Y)$ such that $T_n to T$ in the weak operator topology.
Proof so far: $Lambda(T_nx)$ converges for all $x in X$, $Lambda in Y^*$. Since $Y^*$ is weakly sequentially complete, there is a $y in Y$ such that $Lambda(T_nx) to Lambda y_x$ for all $Lambda in Y^*$. Define $Tx :=y_x$. We need to show $T$ is bounded. Note that because $Lambda(T_nx)$ converges, it is most certainly true that $Lambda(T_nx) : T_n in T_n < infty$. Hence, by a Corollary of the PUB, $|T_n|$ is uniformly bounded by say $M > 0$.
So here now I want to prove that $|T|$ is bounded. Using reverse triangle inequality, we can get that $|Lambda(Tx)| leq M |Lambda| |x|$, but the fact that we have $Lambda$ on the left hand side does not seem to allow me to prove that $T$ is bounded. Reed and Simon do a similar proof with $X = Y$ Hilbert spaces, and they refrain from defining $T$ until later, but they do it using a very clever result of the Riesz Representation Theorem. I'm not sure maybe how to construct an analogous $T$ here instead of choosing the seemingly obvious choice from the fact that $Y$ is weakly sequentially complete. Any hints/help would be much appreciated.
functional-analysis banach-spaces weak-convergence
asked Jul 24 at 22:51
mathishard.butweloveit
1069
I need to prove the following theorem.
Let $X, Y$ be Banach spaces, with $Y$ weakly sequentially complete. Let $T_n subset mathscrL(X,Y)$ with $Lambda(T_nx)$ converges for all $x in X$, $Lambda in Y^*$. Prove there is a $T in mathscrL(X,Y)$ such that $T_n to T$ in the weak operator topology.
Proof so far: $Lambda(T_nx)$ converges for all $x in X$, $Lambda in Y^*$. Since $Y^*$ is weakly sequentially complete, there is a $y in Y$ such that $Lambda(T_nx) to Lambda y_x$ for all $Lambda in Y^*$. Define $Tx :=y_x$. We need to show $T$ is bounded. Note that because $Lambda(T_nx)$ converges, it is most certainly true that $Lambda(T_nx) : T_n in T_n < infty$. Hence, by a Corollary of the PUB, $|T_n|$ is uniformly bounded by say $M > 0$.
So here now I want to prove that $|T|$ is bounded. Using reverse triangle inequality, we can get that $|Lambda(Tx)| leq M |Lambda| |x|$, but the fact that we have $Lambda$ on the left hand side does not seem to allow me to prove that $T$ is bounded. Reed and Simon do a similar proof with $X = Y$ Hilbert spaces, and they refrain from defining $T$ until later, but they do it using a very clever result of the Riesz Representation Theorem. I'm not sure maybe how to construct an analogous $T$ here instead of choosing the seemingly obvious choice from the fact that $Y$ is weakly sequentially complete. Any hints/help would be much appreciated.
functional-analysis banach-spaces weak-convergence
asked Jul 24 at 22:51
mathishard.butweloveit
1069
asked Jul 24 at 22:51
mathishard.butweloveit
1069
asked Jul 24 at 22:51
mathishard.butweloveit
1069
asked Jul 24 at 22:51
mathishard.butweloveit
1069
1069
add a comment |Â
add a comment |Â
1 Answer
1
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1
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As you said, we have
$$|Lambda(Tx)| = left|lim_ntoinfty Lambda(T_n x)right|= lim_ntoinfty|Lambda(T_n x)| le limsup_ntoinfty |Lambda| |T_n||x| le M |Lambda| |x|$$
so for $widehatTx in Y^**$, the element of $Y^**$ represented by $Tx$ we have
$$left|widehatTx(Lambda)right| = |Lambda(Tx)| le M |Lambda| |x|$$
so $|Tx| = left|widehatTxright| le M|x|$.
Therefore $T$ is bounded and $|T| le M$.
Finally, for any $x in X$ we have $Lambda(T_n x) to Lambda(Tx), forall Lambdain Y^*$ so $T_nx xrightarroww Tx$. We conclude $T_n to T$ in the weak operator topology.
A counterexample when $Y$ is only a Banach space:
Consider $T_n : c_0 to c_0$ given by $T_n(x_k)_k = (overbracex_1, x_1, ldots, x_1^n, 0, 0ldots)$.
We have $(c_0)^* = ell^1$ so for any $x in c_0, Lambda = (lambda_n)_n in ell^1$ we have
$$Lambda(T_n x) = Lambda(overbracex_1, x_1, ldots, x_1^n, 0, 0ldots) = x_1cdot sum_k=1^n lambda_n xrightarrowntoinfty x_1cdot sum_k=1^infty lambda_n$$
so $(Lambda(T_n x))_n$ certainly converges. But $(T_n)_n$ doesn't converge in the weak operator topology. Indeed, for $x = (1, 0, 0, ldots) in c_0$ we have $T_nx = (overbrace1, ldots, 1^n, 0, 0ldots)$. The coordinate-wise limit of this sequence is $(1, 1, ldots) notin c_0$ so $(T_nx)_n$ cannot converge weakly in $c_0$.
answered Jul 25 at 0:27
mechanodroid
22.2k52041
Thanks for your reply, I am in the middle of reading the solution, but I have a question about the first statement. I also struggled as to whether or not $y_x$ would be dependent upon $Lambda$, but I didn't think so because it said every weakly Cauchy sequence had a weak limit. In other words, weakly Cauchy means weakly convergent, so $y_x$ should be independent of $Lambda$. Is my reasoning incorrect?
– mathishard.butweloveit
Jul 25 at 0:49
@user707959 Yes, $(Lambda (T_n x))_n$ is weakly convergent but I'm not sure its clear that the limit does not depentdon $Lambda$. Anyway, I added a possible counterexample, have a look.
– mechanodroid
Jul 25 at 0:55
@user707959 Wait, actually I agree with you, it should be independent of $Lambda$.
– mechanodroid
Jul 25 at 0:57
1
@user707959 It turns out the proof is exactly the same when $Y$ is weakly sequentially complete, because the functional $L_x$ indeed was in $Y$, it was represented precisely by your vector $y_x$. Have a look now.
– mechanodroid
Jul 25 at 10:37
1
Technicality for future readers: note that $widehatTx$ is not yet in $Y^**$ but it is most certainly a linear operator, and we prove it is bounded hence in $Y^**$.
– mathishard.butweloveit
Jul 25 at 23:00
 |Â
show 3 more comments
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
As you said, we have
$$|Lambda(Tx)| = left|lim_ntoinfty Lambda(T_n x)right|= lim_ntoinfty|Lambda(T_n x)| le limsup_ntoinfty |Lambda| |T_n||x| le M |Lambda| |x|$$
so for $widehatTx in Y^**$, the element of $Y^**$ represented by $Tx$ we have
$$left|widehatTx(Lambda)right| = |Lambda(Tx)| le M |Lambda| |x|$$
so $|Tx| = left|widehatTxright| le M|x|$.
Therefore $T$ is bounded and $|T| le M$.
Finally, for any $x in X$ we have $Lambda(T_n x) to Lambda(Tx), forall Lambdain Y^*$ so $T_nx xrightarroww Tx$. We conclude $T_n to T$ in the weak operator topology.
A counterexample when $Y$ is only a Banach space:
Consider $T_n : c_0 to c_0$ given by $T_n(x_k)_k = (overbracex_1, x_1, ldots, x_1^n, 0, 0ldots)$.
We have $(c_0)^* = ell^1$ so for any $x in c_0, Lambda = (lambda_n)_n in ell^1$ we have
$$Lambda(T_n x) = Lambda(overbracex_1, x_1, ldots, x_1^n, 0, 0ldots) = x_1cdot sum_k=1^n lambda_n xrightarrowntoinfty x_1cdot sum_k=1^infty lambda_n$$
so $(Lambda(T_n x))_n$ certainly converges. But $(T_n)_n$ doesn't converge in the weak operator topology. Indeed, for $x = (1, 0, 0, ldots) in c_0$ we have $T_nx = (overbrace1, ldots, 1^n, 0, 0ldots)$. The coordinate-wise limit of this sequence is $(1, 1, ldots) notin c_0$ so $(T_nx)_n$ cannot converge weakly in $c_0$.
answered Jul 25 at 0:27
mechanodroid
22.2k52041
Thanks for your reply, I am in the middle of reading the solution, but I have a question about the first statement. I also struggled as to whether or not $y_x$ would be dependent upon $Lambda$, but I didn't think so because it said every weakly Cauchy sequence had a weak limit. In other words, weakly Cauchy means weakly convergent, so $y_x$ should be independent of $Lambda$. Is my reasoning incorrect?
– mathishard.butweloveit
Jul 25 at 0:49
@user707959 Yes, $(Lambda (T_n x))_n$ is weakly convergent but I'm not sure its clear that the limit does not depentdon $Lambda$. Anyway, I added a possible counterexample, have a look.
– mechanodroid
Jul 25 at 0:55
@user707959 Wait, actually I agree with you, it should be independent of $Lambda$.
– mechanodroid
Jul 25 at 0:57
1
@user707959 It turns out the proof is exactly the same when $Y$ is weakly sequentially complete, because the functional $L_x$ indeed was in $Y$, it was represented precisely by your vector $y_x$. Have a look now.
– mechanodroid
Jul 25 at 10:37
1
Technicality for future readers: note that $widehatTx$ is not yet in $Y^**$ but it is most certainly a linear operator, and we prove it is bounded hence in $Y^**$.
– mathishard.butweloveit
Jul 25 at 23:00
 |Â
show 3 more comments
up vote
1
down vote
accepted
As you said, we have
$$|Lambda(Tx)| = left|lim_ntoinfty Lambda(T_n x)right|= lim_ntoinfty|Lambda(T_n x)| le limsup_ntoinfty |Lambda| |T_n||x| le M |Lambda| |x|$$
so for $widehatTx in Y^**$, the element of $Y^**$ represented by $Tx$ we have
$$left|widehatTx(Lambda)right| = |Lambda(Tx)| le M |Lambda| |x|$$
so $|Tx| = left|widehatTxright| le M|x|$.
Therefore $T$ is bounded and $|T| le M$.
Finally, for any $x in X$ we have $Lambda(T_n x) to Lambda(Tx), forall Lambdain Y^*$ so $T_nx xrightarroww Tx$. We conclude $T_n to T$ in the weak operator topology.
A counterexample when $Y$ is only a Banach space:
Consider $T_n : c_0 to c_0$ given by $T_n(x_k)_k = (overbracex_1, x_1, ldots, x_1^n, 0, 0ldots)$.
We have $(c_0)^* = ell^1$ so for any $x in c_0, Lambda = (lambda_n)_n in ell^1$ we have
$$Lambda(T_n x) = Lambda(overbracex_1, x_1, ldots, x_1^n, 0, 0ldots) = x_1cdot sum_k=1^n lambda_n xrightarrowntoinfty x_1cdot sum_k=1^infty lambda_n$$
so $(Lambda(T_n x))_n$ certainly converges. But $(T_n)_n$ doesn't converge in the weak operator topology. Indeed, for $x = (1, 0, 0, ldots) in c_0$ we have $T_nx = (overbrace1, ldots, 1^n, 0, 0ldots)$. The coordinate-wise limit of this sequence is $(1, 1, ldots) notin c_0$ so $(T_nx)_n$ cannot converge weakly in $c_0$.
answered Jul 25 at 0:27
mechanodroid
22.2k52041
Thanks for your reply, I am in the middle of reading the solution, but I have a question about the first statement. I also struggled as to whether or not $y_x$ would be dependent upon $Lambda$, but I didn't think so because it said every weakly Cauchy sequence had a weak limit. In other words, weakly Cauchy means weakly convergent, so $y_x$ should be independent of $Lambda$. Is my reasoning incorrect?
– mathishard.butweloveit
Jul 25 at 0:49
@user707959 Yes, $(Lambda (T_n x))_n$ is weakly convergent but I'm not sure its clear that the limit does not depentdon $Lambda$. Anyway, I added a possible counterexample, have a look.
– mechanodroid
Jul 25 at 0:55
@user707959 Wait, actually I agree with you, it should be independent of $Lambda$.
– mechanodroid
Jul 25 at 0:57
1
@user707959 It turns out the proof is exactly the same when $Y$ is weakly sequentially complete, because the functional $L_x$ indeed was in $Y$, it was represented precisely by your vector $y_x$. Have a look now.
– mechanodroid
Jul 25 at 10:37
1
Technicality for future readers: note that $widehatTx$ is not yet in $Y^**$ but it is most certainly a linear operator, and we prove it is bounded hence in $Y^**$.
– mathishard.butweloveit
Jul 25 at 23:00
 |Â
show 3 more comments
up vote
1
down vote
accepted
up vote
1
down vote
accepted
As you said, we have
$$|Lambda(Tx)| = left|lim_ntoinfty Lambda(T_n x)right|= lim_ntoinfty|Lambda(T_n x)| le limsup_ntoinfty |Lambda| |T_n||x| le M |Lambda| |x|$$
so for $widehatTx in Y^**$, the element of $Y^**$ represented by $Tx$ we have
$$left|widehatTx(Lambda)right| = |Lambda(Tx)| le M |Lambda| |x|$$
so $|Tx| = left|widehatTxright| le M|x|$.
Therefore $T$ is bounded and $|T| le M$.
Finally, for any $x in X$ we have $Lambda(T_n x) to Lambda(Tx), forall Lambdain Y^*$ so $T_nx xrightarroww Tx$. We conclude $T_n to T$ in the weak operator topology.
A counterexample when $Y$ is only a Banach space:
Consider $T_n : c_0 to c_0$ given by $T_n(x_k)_k = (overbracex_1, x_1, ldots, x_1^n, 0, 0ldots)$.
We have $(c_0)^* = ell^1$ so for any $x in c_0, Lambda = (lambda_n)_n in ell^1$ we have
$$Lambda(T_n x) = Lambda(overbracex_1, x_1, ldots, x_1^n, 0, 0ldots) = x_1cdot sum_k=1^n lambda_n xrightarrowntoinfty x_1cdot sum_k=1^infty lambda_n$$
so $(Lambda(T_n x))_n$ certainly converges. But $(T_n)_n$ doesn't converge in the weak operator topology. Indeed, for $x = (1, 0, 0, ldots) in c_0$ we have $T_nx = (overbrace1, ldots, 1^n, 0, 0ldots)$. The coordinate-wise limit of this sequence is $(1, 1, ldots) notin c_0$ so $(T_nx)_n$ cannot converge weakly in $c_0$.
answered Jul 25 at 0:27
mechanodroid
22.2k52041
As you said, we have
$$|Lambda(Tx)| = left|lim_ntoinfty Lambda(T_n x)right|= lim_ntoinfty|Lambda(T_n x)| le limsup_ntoinfty |Lambda| |T_n||x| le M |Lambda| |x|$$
so for $widehatTx in Y^**$, the element of $Y^**$ represented by $Tx$ we have
$$left|widehatTx(Lambda)right| = |Lambda(Tx)| le M |Lambda| |x|$$
so $|Tx| = left|widehatTxright| le M|x|$.
Therefore $T$ is bounded and $|T| le M$.
Finally, for any $x in X$ we have $Lambda(T_n x) to Lambda(Tx), forall Lambdain Y^*$ so $T_nx xrightarroww Tx$. We conclude $T_n to T$ in the weak operator topology.
A counterexample when $Y$ is only a Banach space:
Consider $T_n : c_0 to c_0$ given by $T_n(x_k)_k = (overbracex_1, x_1, ldots, x_1^n, 0, 0ldots)$.
We have $(c_0)^* = ell^1$ so for any $x in c_0, Lambda = (lambda_n)_n in ell^1$ we have
$$Lambda(T_n x) = Lambda(overbracex_1, x_1, ldots, x_1^n, 0, 0ldots) = x_1cdot sum_k=1^n lambda_n xrightarrowntoinfty x_1cdot sum_k=1^infty lambda_n$$
so $(Lambda(T_n x))_n$ certainly converges. But $(T_n)_n$ doesn't converge in the weak operator topology. Indeed, for $x = (1, 0, 0, ldots) in c_0$ we have $T_nx = (overbrace1, ldots, 1^n, 0, 0ldots)$. The coordinate-wise limit of this sequence is $(1, 1, ldots) notin c_0$ so $(T_nx)_n$ cannot converge weakly in $c_0$.
answered Jul 25 at 0:27
mechanodroid
22.2k52041
edited Jul 25 at 10:34
answered Jul 25 at 0:27
mechanodroid
22.2k52041
answered Jul 25 at 0:27
mechanodroid
22.2k52041
answered Jul 25 at 0:27
mechanodroid
22.2k52041
22.2k52041
Thanks for your reply, I am in the middle of reading the solution, but I have a question about the first statement. I also struggled as to whether or not $y_x$ would be dependent upon $Lambda$, but I didn't think so because it said every weakly Cauchy sequence had a weak limit. In other words, weakly Cauchy means weakly convergent, so $y_x$ should be independent of $Lambda$. Is my reasoning incorrect?
– mathishard.butweloveit
Jul 25 at 0:49
@user707959 Yes, $(Lambda (T_n x))_n$ is weakly convergent but I'm not sure its clear that the limit does not depentdon $Lambda$. Anyway, I added a possible counterexample, have a look.
– mechanodroid
Jul 25 at 0:55
@user707959 Wait, actually I agree with you, it should be independent of $Lambda$.
– mechanodroid
Jul 25 at 0:57
1
@user707959 It turns out the proof is exactly the same when $Y$ is weakly sequentially complete, because the functional $L_x$ indeed was in $Y$, it was represented precisely by your vector $y_x$. Have a look now.
– mechanodroid
Jul 25 at 10:37
1
Technicality for future readers: note that $widehatTx$ is not yet in $Y^**$ but it is most certainly a linear operator, and we prove it is bounded hence in $Y^**$.
– mathishard.butweloveit
Jul 25 at 23:00
 |Â
show 3 more comments
Thanks for your reply, I am in the middle of reading the solution, but I have a question about the first statement. I also struggled as to whether or not $y_x$ would be dependent upon $Lambda$, but I didn't think so because it said every weakly Cauchy sequence had a weak limit. In other words, weakly Cauchy means weakly convergent, so $y_x$ should be independent of $Lambda$. Is my reasoning incorrect?
– mathishard.butweloveit
Jul 25 at 0:49
@user707959 Yes, $(Lambda (T_n x))_n$ is weakly convergent but I'm not sure its clear that the limit does not depentdon $Lambda$. Anyway, I added a possible counterexample, have a look.
– mechanodroid
Jul 25 at 0:55
@user707959 Wait, actually I agree with you, it should be independent of $Lambda$.
– mechanodroid
Jul 25 at 0:57
1
@user707959 It turns out the proof is exactly the same when $Y$ is weakly sequentially complete, because the functional $L_x$ indeed was in $Y$, it was represented precisely by your vector $y_x$. Have a look now.
– mechanodroid
Jul 25 at 10:37
1
Technicality for future readers: note that $widehatTx$ is not yet in $Y^**$ but it is most certainly a linear operator, and we prove it is bounded hence in $Y^**$.
– mathishard.butweloveit
Jul 25 at 23:00
Thanks for your reply, I am in the middle of reading the solution, but I have a question about the first statement. I also struggled as to whether or not $y_x$ would be dependent upon $Lambda$, but I didn't think so because it said every weakly Cauchy sequence had a weak limit. In other words, weakly Cauchy means weakly convergent, so $y_x$ should be independent of $Lambda$. Is my reasoning incorrect?
– mathishard.butweloveit
Jul 25 at 0:49
Thanks for your reply, I am in the middle of reading the solution, but I have a question about the first statement. I also struggled as to whether or not $y_x$ would be dependent upon $Lambda$, but I didn't think so because it said every weakly Cauchy sequence had a weak limit. In other words, weakly Cauchy means weakly convergent, so $y_x$ should be independent of $Lambda$. Is my reasoning incorrect?
– mathishard.butweloveit
Jul 25 at 0:49
@user707959 Yes, $(Lambda (T_n x))_n$ is weakly convergent but I'm not sure its clear that the limit does not depentdon $Lambda$. Anyway, I added a possible counterexample, have a look.
– mechanodroid
Jul 25 at 0:55
@user707959 Yes, $(Lambda (T_n x))_n$ is weakly convergent but I'm not sure its clear that the limit does not depentdon $Lambda$. Anyway, I added a possible counterexample, have a look.
– mechanodroid
Jul 25 at 0:55
@user707959 Wait, actually I agree with you, it should be independent of $Lambda$.
– mechanodroid
Jul 25 at 0:57
@user707959 Wait, actually I agree with you, it should be independent of $Lambda$.
– mechanodroid
Jul 25 at 0:57
1
1
@user707959 It turns out the proof is exactly the same when $Y$ is weakly sequentially complete, because the functional $L_x$ indeed was in $Y$, it was represented precisely by your vector $y_x$. Have a look now.
– mechanodroid
Jul 25 at 10:37
@user707959 It turns out the proof is exactly the same when $Y$ is weakly sequentially complete, because the functional $L_x$ indeed was in $Y$, it was represented precisely by your vector $y_x$. Have a look now.
– mechanodroid
Jul 25 at 10:37
1
1
Technicality for future readers: note that $widehatTx$ is not yet in $Y^**$ but it is most certainly a linear operator, and we prove it is bounded hence in $Y^**$.
– mathishard.butweloveit
Jul 25 at 23:00
Technicality for future readers: note that $widehatTx$ is not yet in $Y^**$ but it is most certainly a linear operator, and we prove it is bounded hence in $Y^**$.
– mathishard.butweloveit
Jul 25 at 23:00
 |Â
show 3 more comments
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