Proof of Lemma preceding Principle of Condensation of Singularities The Next CEO of Stack OverflowThe Principle of Condensation of SingularitiesUniform boundedness principle statementThe Principle of Condensation of SingularitiesRequirements for the principle of uniform boundednessImportance of the uniform boundedness principleWhy is this “proof” of the Uniform Boundedness Principle via the contrapositive erroneous?Is the uniform boundedness principle not trivially obvious?Alternative proof of Uniform Boundedness PrincipleDirect proof of Closed Graph Theorem (or Bounded Inverse Theorem) from Uniform Boundedness PrincipleUniform boundedness principle and closed graph TheoremRephrasing the “if” part of the statement of the Principle of Uniform Boundedness
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Proof of Lemma preceding Principle of Condensation of Singularities
The Next CEO of Stack OverflowThe Principle of Condensation of SingularitiesUniform boundedness principle statementThe Principle of Condensation of SingularitiesRequirements for the principle of uniform boundednessImportance of the uniform boundedness principleWhy is this “proof” of the Uniform Boundedness Principle via the contrapositive erroneous?Is the uniform boundedness principle not trivially obvious?Alternative proof of Uniform Boundedness PrincipleDirect proof of Closed Graph Theorem (or Bounded Inverse Theorem) from Uniform Boundedness PrincipleUniform boundedness principle and closed graph TheoremRephrasing the “if” part of the statement of the Principle of Uniform Boundedness
$begingroup$
Under the Wikipedia page for the Principle of Uniform Boundedness, we have the Corollaries of the Uniform Boundedness Principle. The third of these relates to the Principle of Condensation of Singularities (a question on which was asked here), however, the following Lemma is given as part of a motivation to achieving that result:
[Let $X,Y$ be Banach Spaces and] Let $L(X, Y)$ denote the continuous operators from $X$ to $Y$, with the operator norm. If the collection $F$ is unbounded in $L(X, Y)$, then by the uniform boundedness principle we have:
$$R = x ∈ X : sup_ T ∈ F ‖ T x ‖_Y = ∞ ≠ ∅$$
I'm just wondering how exactly the Principle of Uniform Boundedness is being applied here; I was thinking to try and prove the result by contradiction and invoking PUB, but I don't see a way to do this without needing to make two deliberate incorrect hypotheses - in particular to assume that what we want to show holds for all $xin X$ and to assume that for these $x$ we have that $sup_ T ∈ F ‖ T x ‖_Y < ∞$. However, that would only tell us that either one of the hypotheses were wrong.
I wonder if instead one should work directly from Baire Category theory to prove the result - and indeed, perhaps, if that is what is meant by saying that UBP ensures this.
Could somebody give me a hint to set me on the right path to proving this result - as to whether I should use PUB directly, or work from Baire?
functional-analysis operator-theory banach-spaces baire-category
asked Mar 18 at 12:20
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
$endgroup$
add a comment |
$begingroup$
Under the Wikipedia page for the Principle of Uniform Boundedness, we have the Corollaries of the Uniform Boundedness Principle. The third of these relates to the Principle of Condensation of Singularities (a question on which was asked here), however, the following Lemma is given as part of a motivation to achieving that result:
[Let $X,Y$ be Banach Spaces and] Let $L(X, Y)$ denote the continuous operators from $X$ to $Y$, with the operator norm. If the collection $F$ is unbounded in $L(X, Y)$, then by the uniform boundedness principle we have:
$$R = x ∈ X : sup_ T ∈ F ‖ T x ‖_Y = ∞ ≠ ∅$$
I'm just wondering how exactly the Principle of Uniform Boundedness is being applied here; I was thinking to try and prove the result by contradiction and invoking PUB, but I don't see a way to do this without needing to make two deliberate incorrect hypotheses - in particular to assume that what we want to show holds for all $xin X$ and to assume that for these $x$ we have that $sup_ T ∈ F ‖ T x ‖_Y < ∞$. However, that would only tell us that either one of the hypotheses were wrong.
I wonder if instead one should work directly from Baire Category theory to prove the result - and indeed, perhaps, if that is what is meant by saying that UBP ensures this.
Could somebody give me a hint to set me on the right path to proving this result - as to whether I should use PUB directly, or work from Baire?
functional-analysis operator-theory banach-spaces baire-category
asked Mar 18 at 12:20
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
$endgroup$
add a comment |
$begingroup$
Under the Wikipedia page for the Principle of Uniform Boundedness, we have the Corollaries of the Uniform Boundedness Principle. The third of these relates to the Principle of Condensation of Singularities (a question on which was asked here), however, the following Lemma is given as part of a motivation to achieving that result:
[Let $X,Y$ be Banach Spaces and] Let $L(X, Y)$ denote the continuous operators from $X$ to $Y$, with the operator norm. If the collection $F$ is unbounded in $L(X, Y)$, then by the uniform boundedness principle we have:
$$R = x ∈ X : sup_ T ∈ F ‖ T x ‖_Y = ∞ ≠ ∅$$
I'm just wondering how exactly the Principle of Uniform Boundedness is being applied here; I was thinking to try and prove the result by contradiction and invoking PUB, but I don't see a way to do this without needing to make two deliberate incorrect hypotheses - in particular to assume that what we want to show holds for all $xin X$ and to assume that for these $x$ we have that $sup_ T ∈ F ‖ T x ‖_Y < ∞$. However, that would only tell us that either one of the hypotheses were wrong.
I wonder if instead one should work directly from Baire Category theory to prove the result - and indeed, perhaps, if that is what is meant by saying that UBP ensures this.
Could somebody give me a hint to set me on the right path to proving this result - as to whether I should use PUB directly, or work from Baire?
functional-analysis operator-theory banach-spaces baire-category
asked Mar 18 at 12:20
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
$endgroup$
Under the Wikipedia page for the Principle of Uniform Boundedness, we have the Corollaries of the Uniform Boundedness Principle. The third of these relates to the Principle of Condensation of Singularities (a question on which was asked here), however, the following Lemma is given as part of a motivation to achieving that result:
[Let $X,Y$ be Banach Spaces and] Let $L(X, Y)$ denote the continuous operators from $X$ to $Y$, with the operator norm. If the collection $F$ is unbounded in $L(X, Y)$, then by the uniform boundedness principle we have:
$$R = x ∈ X : sup_ T ∈ F ‖ T x ‖_Y = ∞ ≠ ∅$$
I'm just wondering how exactly the Principle of Uniform Boundedness is being applied here; I was thinking to try and prove the result by contradiction and invoking PUB, but I don't see a way to do this without needing to make two deliberate incorrect hypotheses - in particular to assume that what we want to show holds for all $xin X$ and to assume that for these $x$ we have that $sup_ T ∈ F ‖ T x ‖_Y < ∞$. However, that would only tell us that either one of the hypotheses were wrong.
I wonder if instead one should work directly from Baire Category theory to prove the result - and indeed, perhaps, if that is what is meant by saying that UBP ensures this.
Could somebody give me a hint to set me on the right path to proving this result - as to whether I should use PUB directly, or work from Baire?
functional-analysis operator-theory banach-spaces baire-category
functional-analysis operator-theory banach-spaces baire-category
asked Mar 18 at 12:20
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
asked Mar 18 at 12:20
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
asked Mar 18 at 12:20
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
asked Mar 18 at 12:20
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
asked Mar 18 at 12:20
Jeremy Jeffrey JamesJeremy Jeffrey James
1,055717
1,055717
add a comment |
add a comment |
1 Answer
1
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$begingroup$
Suppose that $R =emptyset.$
If $x in X$, then $x notin R$, hence there is $c_x ge 0$ such that
$||Tx||_Y le c_x$ for all $T in F.$
The $PUB$ gives that $$ is bounded, a contradiction.
answered Mar 18 at 12:36
FredFred
48.8k11849
$endgroup$
$begingroup$
I see now how silly my question was ... my thanks
$endgroup$
– Jeremy Jeffrey James
Mar 18 at 12:39
add a comment |
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1 Answer
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1 Answer
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$begingroup$
Suppose that $R =emptyset.$
If $x in X$, then $x notin R$, hence there is $c_x ge 0$ such that
$||Tx||_Y le c_x$ for all $T in F.$
The $PUB$ gives that $$ is bounded, a contradiction.
answered Mar 18 at 12:36
FredFred
48.8k11849
$endgroup$
$begingroup$
I see now how silly my question was ... my thanks
$endgroup$
– Jeremy Jeffrey James
Mar 18 at 12:39
add a comment |
$begingroup$
Suppose that $R =emptyset.$
If $x in X$, then $x notin R$, hence there is $c_x ge 0$ such that
$||Tx||_Y le c_x$ for all $T in F.$
The $PUB$ gives that $$ is bounded, a contradiction.
answered Mar 18 at 12:36
FredFred
48.8k11849
$endgroup$
$begingroup$
I see now how silly my question was ... my thanks
$endgroup$
– Jeremy Jeffrey James
Mar 18 at 12:39
add a comment |
$begingroup$
Suppose that $R =emptyset.$
If $x in X$, then $x notin R$, hence there is $c_x ge 0$ such that
$||Tx||_Y le c_x$ for all $T in F.$
The $PUB$ gives that $$ is bounded, a contradiction.
answered Mar 18 at 12:36
FredFred
48.8k11849
$endgroup$
Suppose that $R =emptyset.$
If $x in X$, then $x notin R$, hence there is $c_x ge 0$ such that
$||Tx||_Y le c_x$ for all $T in F.$
The $PUB$ gives that $$ is bounded, a contradiction.
answered Mar 18 at 12:36
FredFred
48.8k11849
answered Mar 18 at 12:36
FredFred
48.8k11849
answered Mar 18 at 12:36
FredFred
48.8k11849
answered Mar 18 at 12:36
FredFred
48.8k11849
48.8k11849
$begingroup$
I see now how silly my question was ... my thanks
$endgroup$
– Jeremy Jeffrey James
Mar 18 at 12:39
add a comment |
$begingroup$
I see now how silly my question was ... my thanks
$endgroup$
– Jeremy Jeffrey James
Mar 18 at 12:39
$begingroup$
I see now how silly my question was ... my thanks
$endgroup$
– Jeremy Jeffrey James
Mar 18 at 12:39
$begingroup$
I see now how silly my question was ... my thanks
$endgroup$
– Jeremy Jeffrey James
Mar 18 at 12:39
add a comment |
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