Does the linear bounded operator A = √B exist? [closed] Announcing the arrival of Valued Associate #679: Cesar Manara Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Extending two bounded linear functionals whose sum is dominated by the vector normIf $D$ is a dense linear subspace of $X$ then $Dto Y$ extends to $Xto Y$ uniquelyHow to apply Theorem 4.3-3 in the proof of Theorem 4.5-2 in Kreyszig's functional analysis book?Prob. 8, Sec. 4.5 in Kreyszig's functional analysis book: The inverse of the adjoint operator is the adjoint of the inverse operatorProving a linear operator is always boundedAny examples of unbounded linear operators between $ell^infty$ and $ell^infty$? $ell^p$ and $ell^p$? $ell^p$ and $ell^infty$?Any Examples Of Unbounded Linear Maps Between Normed Spaces Apart From The Differentiation Operator?There exists a unique extension $hatT$ of a bounded linear operator $T$.Closed operator, closed graphShowing an linear operator is bounded
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Does the linear bounded operator A = √B exist? [closed]
Announcing the arrival of Valued Associate #679: Cesar Manara
Planned maintenance scheduled April 23, 2019 at 00:00UTC (8:00pm US/Eastern)Extending two bounded linear functionals whose sum is dominated by the vector normIf $D$ is a dense linear subspace of $X$ then $Dto Y$ extends to $Xto Y$ uniquelyHow to apply Theorem 4.3-3 in the proof of Theorem 4.5-2 in Kreyszig's functional analysis book?Prob. 8, Sec. 4.5 in Kreyszig's functional analysis book: The inverse of the adjoint operator is the adjoint of the inverse operatorProving a linear operator is always boundedAny examples of unbounded linear operators between $ell^infty$ and $ell^infty$? $ell^p$ and $ell^p$? $ell^p$ and $ell^infty$?Any Examples Of Unbounded Linear Maps Between Normed Spaces Apart From The Differentiation Operator?There exists a unique extension $hatT$ of a bounded linear operator $T$.Closed operator, closed graphShowing an linear operator is bounded
$begingroup$
I came across this problem in my real analysis course, but I am not sure how to solve it.
Let Y be a normed linear space, and let M denote the space of
bounded linear operators from Y to itself. Let L: M → M be the map defined
by:
L(A) := A^2
Show that there exists an Ɛ > 0 such that for B ∈ M with |B − I|<Ɛ,
there exists a bounded linear operator A = √B i.e., a bounded linear
operator A satisfying A^2 = B. Is such an A unique?
I suppose that if B goes towards I then the condition |B − I|<Ɛ, which means that A^2 = I, now I have to prove that this is linear and bounded, but how do I start with the problem and when is A unique?
real-analysis
asked Apr 1 at 19:10
Apple PieApple Pie
63
$endgroup$
closed as off-topic by Connor Harris, Umberto P., Xander Henderson, Shailesh, YiFan Apr 2 at 0:44
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Connor Harris, Umberto P., Xander Henderson, Shailesh, YiFan
add a comment |
$begingroup$
I came across this problem in my real analysis course, but I am not sure how to solve it.
Let Y be a normed linear space, and let M denote the space of
bounded linear operators from Y to itself. Let L: M → M be the map defined
by:
L(A) := A^2
Show that there exists an Ɛ > 0 such that for B ∈ M with |B − I|<Ɛ,
there exists a bounded linear operator A = √B i.e., a bounded linear
operator A satisfying A^2 = B. Is such an A unique?
I suppose that if B goes towards I then the condition |B − I|<Ɛ, which means that A^2 = I, now I have to prove that this is linear and bounded, but how do I start with the problem and when is A unique?
real-analysis
asked Apr 1 at 19:10
Apple PieApple Pie
63
$endgroup$
closed as off-topic by Connor Harris, Umberto P., Xander Henderson, Shailesh, YiFan Apr 2 at 0:44
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Connor Harris, Umberto P., Xander Henderson, Shailesh, YiFan
add a comment |
$begingroup$
I came across this problem in my real analysis course, but I am not sure how to solve it.
Let Y be a normed linear space, and let M denote the space of
bounded linear operators from Y to itself. Let L: M → M be the map defined
by:
L(A) := A^2
Show that there exists an Ɛ > 0 such that for B ∈ M with |B − I|<Ɛ,
there exists a bounded linear operator A = √B i.e., a bounded linear
operator A satisfying A^2 = B. Is such an A unique?
I suppose that if B goes towards I then the condition |B − I|<Ɛ, which means that A^2 = I, now I have to prove that this is linear and bounded, but how do I start with the problem and when is A unique?
real-analysis
asked Apr 1 at 19:10
Apple PieApple Pie
63
$endgroup$
I came across this problem in my real analysis course, but I am not sure how to solve it.
Let Y be a normed linear space, and let M denote the space of
bounded linear operators from Y to itself. Let L: M → M be the map defined
by:
L(A) := A^2
Show that there exists an Ɛ > 0 such that for B ∈ M with |B − I|<Ɛ,
there exists a bounded linear operator A = √B i.e., a bounded linear
operator A satisfying A^2 = B. Is such an A unique?
I suppose that if B goes towards I then the condition |B − I|<Ɛ, which means that A^2 = I, now I have to prove that this is linear and bounded, but how do I start with the problem and when is A unique?
real-analysis
real-analysis
asked Apr 1 at 19:10
Apple PieApple Pie
63
asked Apr 1 at 19:10
Apple PieApple Pie
63
edited Apr 2 at 15:26
Apple Pie
asked Apr 1 at 19:10
Apple PieApple Pie
63
asked Apr 1 at 19:10
Apple PieApple Pie
63
asked Apr 1 at 19:10
Apple PieApple Pie
63
63
closed as off-topic by Connor Harris, Umberto P., Xander Henderson, Shailesh, YiFan Apr 2 at 0:44
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Connor Harris, Umberto P., Xander Henderson, Shailesh, YiFan
closed as off-topic by Connor Harris, Umberto P., Xander Henderson, Shailesh, YiFan Apr 2 at 0:44
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 provide additional context, which ideally explains why the question is relevant to you and our community. Some forms of context include: background and motivation, relevant definitions, source, possible strategies, your current progress, why the question is interesting or important, etc." – Connor Harris, Umberto P., Xander Henderson, Shailesh, YiFan
add a comment |
add a comment |
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