Important applications of the Uniform Boundedness Principle
$begingroup$
There's like three applications of the uniform boundedness principle in wikipedia:
1) If a sequence of bounded operators converges pointwise to an operator, then the limit operator is also bounded, and the convergence is uniform on compact sets.
2) Any weakly bounded subset of a normed space is bounded.
3) A result in pointwise convergence of Fourier series.
I am just asking if there's more interesting applications of the uniform boundedness principle.
general-topology functional-analysis banach-spaces big-list
asked Jul 4 '14 at 23:24
user50618user50618
1,8901518
$endgroup$
add a comment |
$begingroup$
There's like three applications of the uniform boundedness principle in wikipedia:
1) If a sequence of bounded operators converges pointwise to an operator, then the limit operator is also bounded, and the convergence is uniform on compact sets.
2) Any weakly bounded subset of a normed space is bounded.
3) A result in pointwise convergence of Fourier series.
I am just asking if there's more interesting applications of the uniform boundedness principle.
general-topology functional-analysis banach-spaces big-list
asked Jul 4 '14 at 23:24
user50618user50618
1,8901518
$endgroup$
1
$begingroup$
It can be used to show that at most finitely many of coefficient functionals corresponding to a Hamel basis can be continuous; see this post.
$endgroup$
– Martin Sleziak
Jul 6 '14 at 8:04
add a comment |
$begingroup$
There's like three applications of the uniform boundedness principle in wikipedia:
1) If a sequence of bounded operators converges pointwise to an operator, then the limit operator is also bounded, and the convergence is uniform on compact sets.
2) Any weakly bounded subset of a normed space is bounded.
3) A result in pointwise convergence of Fourier series.
I am just asking if there's more interesting applications of the uniform boundedness principle.
general-topology functional-analysis banach-spaces big-list
asked Jul 4 '14 at 23:24
user50618user50618
1,8901518
$endgroup$
There's like three applications of the uniform boundedness principle in wikipedia:
1) If a sequence of bounded operators converges pointwise to an operator, then the limit operator is also bounded, and the convergence is uniform on compact sets.
2) Any weakly bounded subset of a normed space is bounded.
3) A result in pointwise convergence of Fourier series.
I am just asking if there's more interesting applications of the uniform boundedness principle.
general-topology functional-analysis banach-spaces big-list
general-topology functional-analysis banach-spaces big-list
asked Jul 4 '14 at 23:24
user50618user50618
1,8901518
asked Jul 4 '14 at 23:24
user50618user50618
1,8901518
edited Jul 4 '14 at 23:32
user50618
asked Jul 4 '14 at 23:24
user50618user50618
1,8901518
asked Jul 4 '14 at 23:24
user50618user50618
1,8901518
asked Jul 4 '14 at 23:24
user50618user50618
1,8901518
1,8901518
1
$begingroup$
It can be used to show that at most finitely many of coefficient functionals corresponding to a Hamel basis can be continuous; see this post.
$endgroup$
– Martin Sleziak
Jul 6 '14 at 8:04
add a comment |
1
$begingroup$
It can be used to show that at most finitely many of coefficient functionals corresponding to a Hamel basis can be continuous; see this post.
$endgroup$
– Martin Sleziak
Jul 6 '14 at 8:04
1
1
$begingroup$
It can be used to show that at most finitely many of coefficient functionals corresponding to a Hamel basis can be continuous; see this post.
$endgroup$
– Martin Sleziak
Jul 6 '14 at 8:04
$begingroup$
It can be used to show that at most finitely many of coefficient functionals corresponding to a Hamel basis can be continuous; see this post.
$endgroup$
– Martin Sleziak
Jul 6 '14 at 8:04
add a comment |
3 Answers
3
active
oldest
votes
$begingroup$
Let $fin L^{p}(mathbb{T})$, for some $1leq p<infty$, where $mathbb{T}$ denotes the one-dimensional torus. Let $(a_{m})_{minmathbb{Z}}$ be a bounded complex sequence. For $Rgeq 0$, let $(a_{m}(R))_{m=1}^{infty}$ be a compactly supported sequence such that $a_{m}(R)=a_{m}$ for all $left|mright|leq R$. Define
$$S_{R}(f)(x):=sum_{minmathbb{Z}}a_{m}(R)widehat{f}(m)e^{2pi imcdot x}, forall xinmathbb{T}$$
For each $R$, the above expressions defines an operator $L^{p}rightarrow L^{p}$ that maps a function $f$ to the $lfloor{R}rfloor^{th}$ symmetric partial sum of its Fourier series.
Using, in part, a simple application of the Uniform Boundedness Principle, we can turn the question of $L^{p}$ convergence of $S_{R}(f)$ to $f$ as $Rrightarrowinfty$, for arbitrary $fin L^{p}$, into a question of the uniform boundedness of the operators $S_{R}$, $Rgeq 0$.
answered Jul 5 '14 at 0:29
Matt RosenzweigMatt Rosenzweig
2,48911029
$endgroup$
add a comment |
$begingroup$
Here's a couple more examples.
If $f : Omegasubseteqmathbb{C} rightarrow mathcal{L}(X)$ is a function from an open subset $Omega$ of the complex plane into the bounded linear operators $mathcal{L}(X)$ on a complex Banach space $X$, then $f$ is holomorphic iff $lambdamapsto x^{star}(f(lambda)x)$ is holomorphic on $Omega$ for all $x in X$, $x^{star}in X^{star}$.
If $f$ is a holomorphic vector function on the punctured disk $0 < |lambda| < delta$ with values in a Banach space $X$, then $f$ has an essential singularity at $0$ iff there exists $x^{star}in X^{star}$ such that $x^{star}circ f$ has an essential singularity at $0$.
answered Jul 6 '14 at 3:58
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
$begingroup$
Thanks, can I have a reference where I can find the proofs ?
$endgroup$
– user50618
Jul 6 '14 at 22:58
1
$begingroup$
Many texts on Functional Analysis deal with the first issue, including Rudin. You can start by lifting from $lambda mapsto x^{star}(f(lambda)x)$ being holomorphic to $lambda mapsto f(lambda)x$ being holomorphic. Then you can lift again to $lambdamapsto f(lambda)$ being holomorphic. As for the second part, I think that's actually an application of the Baire category theorem, which is used to establish uniform boundedness.
$endgroup$
– DisintegratingByParts
Jul 8 '14 at 0:03
add a comment |
$begingroup$
One other application I know is in the proof of the spectral radius formula: given a Banach space $X$ and $Tin B(X)$ a bounded linear opeartor,
$$
operatorname{spr}(T)=lim_{ntoinfty}|T^n|^{1/n},
$$
where $operatorname{spr}(T)=max{|lambda|: lambdainsigma(T)}$.
answered Jan 11 at 0:02
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
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3 Answers
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active
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3 Answers
3
active
oldest
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active
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oldest
votes
$begingroup$
Let $fin L^{p}(mathbb{T})$, for some $1leq p<infty$, where $mathbb{T}$ denotes the one-dimensional torus. Let $(a_{m})_{minmathbb{Z}}$ be a bounded complex sequence. For $Rgeq 0$, let $(a_{m}(R))_{m=1}^{infty}$ be a compactly supported sequence such that $a_{m}(R)=a_{m}$ for all $left|mright|leq R$. Define
$$S_{R}(f)(x):=sum_{minmathbb{Z}}a_{m}(R)widehat{f}(m)e^{2pi imcdot x}, forall xinmathbb{T}$$
For each $R$, the above expressions defines an operator $L^{p}rightarrow L^{p}$ that maps a function $f$ to the $lfloor{R}rfloor^{th}$ symmetric partial sum of its Fourier series.
Using, in part, a simple application of the Uniform Boundedness Principle, we can turn the question of $L^{p}$ convergence of $S_{R}(f)$ to $f$ as $Rrightarrowinfty$, for arbitrary $fin L^{p}$, into a question of the uniform boundedness of the operators $S_{R}$, $Rgeq 0$.
answered Jul 5 '14 at 0:29
Matt RosenzweigMatt Rosenzweig
2,48911029
$endgroup$
add a comment |
$begingroup$
Let $fin L^{p}(mathbb{T})$, for some $1leq p<infty$, where $mathbb{T}$ denotes the one-dimensional torus. Let $(a_{m})_{minmathbb{Z}}$ be a bounded complex sequence. For $Rgeq 0$, let $(a_{m}(R))_{m=1}^{infty}$ be a compactly supported sequence such that $a_{m}(R)=a_{m}$ for all $left|mright|leq R$. Define
$$S_{R}(f)(x):=sum_{minmathbb{Z}}a_{m}(R)widehat{f}(m)e^{2pi imcdot x}, forall xinmathbb{T}$$
For each $R$, the above expressions defines an operator $L^{p}rightarrow L^{p}$ that maps a function $f$ to the $lfloor{R}rfloor^{th}$ symmetric partial sum of its Fourier series.
Using, in part, a simple application of the Uniform Boundedness Principle, we can turn the question of $L^{p}$ convergence of $S_{R}(f)$ to $f$ as $Rrightarrowinfty$, for arbitrary $fin L^{p}$, into a question of the uniform boundedness of the operators $S_{R}$, $Rgeq 0$.
answered Jul 5 '14 at 0:29
Matt RosenzweigMatt Rosenzweig
2,48911029
$endgroup$
add a comment |
$begingroup$
Let $fin L^{p}(mathbb{T})$, for some $1leq p<infty$, where $mathbb{T}$ denotes the one-dimensional torus. Let $(a_{m})_{minmathbb{Z}}$ be a bounded complex sequence. For $Rgeq 0$, let $(a_{m}(R))_{m=1}^{infty}$ be a compactly supported sequence such that $a_{m}(R)=a_{m}$ for all $left|mright|leq R$. Define
$$S_{R}(f)(x):=sum_{minmathbb{Z}}a_{m}(R)widehat{f}(m)e^{2pi imcdot x}, forall xinmathbb{T}$$
For each $R$, the above expressions defines an operator $L^{p}rightarrow L^{p}$ that maps a function $f$ to the $lfloor{R}rfloor^{th}$ symmetric partial sum of its Fourier series.
Using, in part, a simple application of the Uniform Boundedness Principle, we can turn the question of $L^{p}$ convergence of $S_{R}(f)$ to $f$ as $Rrightarrowinfty$, for arbitrary $fin L^{p}$, into a question of the uniform boundedness of the operators $S_{R}$, $Rgeq 0$.
answered Jul 5 '14 at 0:29
Matt RosenzweigMatt Rosenzweig
2,48911029
$endgroup$
Let $fin L^{p}(mathbb{T})$, for some $1leq p<infty$, where $mathbb{T}$ denotes the one-dimensional torus. Let $(a_{m})_{minmathbb{Z}}$ be a bounded complex sequence. For $Rgeq 0$, let $(a_{m}(R))_{m=1}^{infty}$ be a compactly supported sequence such that $a_{m}(R)=a_{m}$ for all $left|mright|leq R$. Define
$$S_{R}(f)(x):=sum_{minmathbb{Z}}a_{m}(R)widehat{f}(m)e^{2pi imcdot x}, forall xinmathbb{T}$$
For each $R$, the above expressions defines an operator $L^{p}rightarrow L^{p}$ that maps a function $f$ to the $lfloor{R}rfloor^{th}$ symmetric partial sum of its Fourier series.
Using, in part, a simple application of the Uniform Boundedness Principle, we can turn the question of $L^{p}$ convergence of $S_{R}(f)$ to $f$ as $Rrightarrowinfty$, for arbitrary $fin L^{p}$, into a question of the uniform boundedness of the operators $S_{R}$, $Rgeq 0$.
answered Jul 5 '14 at 0:29
Matt RosenzweigMatt Rosenzweig
2,48911029
answered Jul 5 '14 at 0:29
Matt RosenzweigMatt Rosenzweig
2,48911029
answered Jul 5 '14 at 0:29
Matt RosenzweigMatt Rosenzweig
2,48911029
answered Jul 5 '14 at 0:29
Matt RosenzweigMatt Rosenzweig
2,48911029
2,48911029
add a comment |
add a comment |
$begingroup$
Here's a couple more examples.
If $f : Omegasubseteqmathbb{C} rightarrow mathcal{L}(X)$ is a function from an open subset $Omega$ of the complex plane into the bounded linear operators $mathcal{L}(X)$ on a complex Banach space $X$, then $f$ is holomorphic iff $lambdamapsto x^{star}(f(lambda)x)$ is holomorphic on $Omega$ for all $x in X$, $x^{star}in X^{star}$.
If $f$ is a holomorphic vector function on the punctured disk $0 < |lambda| < delta$ with values in a Banach space $X$, then $f$ has an essential singularity at $0$ iff there exists $x^{star}in X^{star}$ such that $x^{star}circ f$ has an essential singularity at $0$.
answered Jul 6 '14 at 3:58
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
$begingroup$
Thanks, can I have a reference where I can find the proofs ?
$endgroup$
– user50618
Jul 6 '14 at 22:58
1
$begingroup$
Many texts on Functional Analysis deal with the first issue, including Rudin. You can start by lifting from $lambda mapsto x^{star}(f(lambda)x)$ being holomorphic to $lambda mapsto f(lambda)x$ being holomorphic. Then you can lift again to $lambdamapsto f(lambda)$ being holomorphic. As for the second part, I think that's actually an application of the Baire category theorem, which is used to establish uniform boundedness.
$endgroup$
– DisintegratingByParts
Jul 8 '14 at 0:03
add a comment |
$begingroup$
Here's a couple more examples.
If $f : Omegasubseteqmathbb{C} rightarrow mathcal{L}(X)$ is a function from an open subset $Omega$ of the complex plane into the bounded linear operators $mathcal{L}(X)$ on a complex Banach space $X$, then $f$ is holomorphic iff $lambdamapsto x^{star}(f(lambda)x)$ is holomorphic on $Omega$ for all $x in X$, $x^{star}in X^{star}$.
If $f$ is a holomorphic vector function on the punctured disk $0 < |lambda| < delta$ with values in a Banach space $X$, then $f$ has an essential singularity at $0$ iff there exists $x^{star}in X^{star}$ such that $x^{star}circ f$ has an essential singularity at $0$.
answered Jul 6 '14 at 3:58
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
$begingroup$
Thanks, can I have a reference where I can find the proofs ?
$endgroup$
– user50618
Jul 6 '14 at 22:58
1
$begingroup$
Many texts on Functional Analysis deal with the first issue, including Rudin. You can start by lifting from $lambda mapsto x^{star}(f(lambda)x)$ being holomorphic to $lambda mapsto f(lambda)x$ being holomorphic. Then you can lift again to $lambdamapsto f(lambda)$ being holomorphic. As for the second part, I think that's actually an application of the Baire category theorem, which is used to establish uniform boundedness.
$endgroup$
– DisintegratingByParts
Jul 8 '14 at 0:03
add a comment |
$begingroup$
Here's a couple more examples.
If $f : Omegasubseteqmathbb{C} rightarrow mathcal{L}(X)$ is a function from an open subset $Omega$ of the complex plane into the bounded linear operators $mathcal{L}(X)$ on a complex Banach space $X$, then $f$ is holomorphic iff $lambdamapsto x^{star}(f(lambda)x)$ is holomorphic on $Omega$ for all $x in X$, $x^{star}in X^{star}$.
If $f$ is a holomorphic vector function on the punctured disk $0 < |lambda| < delta$ with values in a Banach space $X$, then $f$ has an essential singularity at $0$ iff there exists $x^{star}in X^{star}$ such that $x^{star}circ f$ has an essential singularity at $0$.
answered Jul 6 '14 at 3:58
DisintegratingByPartsDisintegratingByParts
60.4k42681
$endgroup$
Here's a couple more examples.
If $f : Omegasubseteqmathbb{C} rightarrow mathcal{L}(X)$ is a function from an open subset $Omega$ of the complex plane into the bounded linear operators $mathcal{L}(X)$ on a complex Banach space $X$, then $f$ is holomorphic iff $lambdamapsto x^{star}(f(lambda)x)$ is holomorphic on $Omega$ for all $x in X$, $x^{star}in X^{star}$.
If $f$ is a holomorphic vector function on the punctured disk $0 < |lambda| < delta$ with values in a Banach space $X$, then $f$ has an essential singularity at $0$ iff there exists $x^{star}in X^{star}$ such that $x^{star}circ f$ has an essential singularity at $0$.
answered Jul 6 '14 at 3:58
DisintegratingByPartsDisintegratingByParts
60.4k42681
answered Jul 6 '14 at 3:58
DisintegratingByPartsDisintegratingByParts
60.4k42681
answered Jul 6 '14 at 3:58
DisintegratingByPartsDisintegratingByParts
60.4k42681
answered Jul 6 '14 at 3:58
DisintegratingByPartsDisintegratingByParts
60.4k42681
60.4k42681
$begingroup$
Thanks, can I have a reference where I can find the proofs ?
$endgroup$
– user50618
Jul 6 '14 at 22:58
1
$begingroup$
Many texts on Functional Analysis deal with the first issue, including Rudin. You can start by lifting from $lambda mapsto x^{star}(f(lambda)x)$ being holomorphic to $lambda mapsto f(lambda)x$ being holomorphic. Then you can lift again to $lambdamapsto f(lambda)$ being holomorphic. As for the second part, I think that's actually an application of the Baire category theorem, which is used to establish uniform boundedness.
$endgroup$
– DisintegratingByParts
Jul 8 '14 at 0:03
add a comment |
$begingroup$
Thanks, can I have a reference where I can find the proofs ?
$endgroup$
– user50618
Jul 6 '14 at 22:58
1
$begingroup$
Many texts on Functional Analysis deal with the first issue, including Rudin. You can start by lifting from $lambda mapsto x^{star}(f(lambda)x)$ being holomorphic to $lambda mapsto f(lambda)x$ being holomorphic. Then you can lift again to $lambdamapsto f(lambda)$ being holomorphic. As for the second part, I think that's actually an application of the Baire category theorem, which is used to establish uniform boundedness.
$endgroup$
– DisintegratingByParts
Jul 8 '14 at 0:03
$begingroup$
Thanks, can I have a reference where I can find the proofs ?
$endgroup$
– user50618
Jul 6 '14 at 22:58
$begingroup$
Thanks, can I have a reference where I can find the proofs ?
$endgroup$
– user50618
Jul 6 '14 at 22:58
1
1
$begingroup$
Many texts on Functional Analysis deal with the first issue, including Rudin. You can start by lifting from $lambda mapsto x^{star}(f(lambda)x)$ being holomorphic to $lambda mapsto f(lambda)x$ being holomorphic. Then you can lift again to $lambdamapsto f(lambda)$ being holomorphic. As for the second part, I think that's actually an application of the Baire category theorem, which is used to establish uniform boundedness.
$endgroup$
– DisintegratingByParts
Jul 8 '14 at 0:03
$begingroup$
Many texts on Functional Analysis deal with the first issue, including Rudin. You can start by lifting from $lambda mapsto x^{star}(f(lambda)x)$ being holomorphic to $lambda mapsto f(lambda)x$ being holomorphic. Then you can lift again to $lambdamapsto f(lambda)$ being holomorphic. As for the second part, I think that's actually an application of the Baire category theorem, which is used to establish uniform boundedness.
$endgroup$
– DisintegratingByParts
Jul 8 '14 at 0:03
add a comment |
$begingroup$
One other application I know is in the proof of the spectral radius formula: given a Banach space $X$ and $Tin B(X)$ a bounded linear opeartor,
$$
operatorname{spr}(T)=lim_{ntoinfty}|T^n|^{1/n},
$$
where $operatorname{spr}(T)=max{|lambda|: lambdainsigma(T)}$.
answered Jan 11 at 0:02
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
One other application I know is in the proof of the spectral radius formula: given a Banach space $X$ and $Tin B(X)$ a bounded linear opeartor,
$$
operatorname{spr}(T)=lim_{ntoinfty}|T^n|^{1/n},
$$
where $operatorname{spr}(T)=max{|lambda|: lambdainsigma(T)}$.
answered Jan 11 at 0:02
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
add a comment |
$begingroup$
One other application I know is in the proof of the spectral radius formula: given a Banach space $X$ and $Tin B(X)$ a bounded linear opeartor,
$$
operatorname{spr}(T)=lim_{ntoinfty}|T^n|^{1/n},
$$
where $operatorname{spr}(T)=max{|lambda|: lambdainsigma(T)}$.
answered Jan 11 at 0:02
Martin ArgeramiMartin Argerami
129k1184185
$endgroup$
One other application I know is in the proof of the spectral radius formula: given a Banach space $X$ and $Tin B(X)$ a bounded linear opeartor,
$$
operatorname{spr}(T)=lim_{ntoinfty}|T^n|^{1/n},
$$
where $operatorname{spr}(T)=max{|lambda|: lambdainsigma(T)}$.
answered Jan 11 at 0:02
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 11 at 0:02
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 11 at 0:02
Martin ArgeramiMartin Argerami
129k1184185
answered Jan 11 at 0:02
Martin ArgeramiMartin Argerami
129k1184185
129k1184185
add a comment |
add a comment |
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$begingroup$
It can be used to show that at most finitely many of coefficient functionals corresponding to a Hamel basis can be continuous; see this post.
$endgroup$
– Martin Sleziak
Jul 6 '14 at 8:04