Mixing
uniform boundness principle (Banach- Steinhaus)
$begingroup$
I am a beginner of functional analysis and I can't understand at all the Banach Steinhaus theorem:
Let $E$ and $F$ be two Banach spaces and let $(T_i)_{i in I}$ be a family (not necessarily countable) of continuous linear operators from $E$ into $F.$ Assume that
$$ sup_{i in I} lVert T_ix rVert<infty quad forall x in E. $$
then
$$ sup_{i in I} lVert T_i rVert< ∞ .$$
(from Brezis page 32)
my question is : if the family consists of a finite number of linear bdd operators the hypothesis:
$$sup_{i in I} lVert T_ix rVert < infty quad forall x in E$$
is always verified, isn't it?
and more generally, what is this theorem telling me?
I apologize for the banality of my question but I can't fully understand this theorem.
functional-analysis analysis banach-spaces
edited Jan 10 at 17:50
ktoi
2,4061617
asked Jan 10 at 17:33
whowhowhowho
152
$endgroup$
add a comment |
$begingroup$
I am a beginner of functional analysis and I can't understand at all the Banach Steinhaus theorem:
Let $E$ and $F$ be two Banach spaces and let $(T_i)_{i in I}$ be a family (not necessarily countable) of continuous linear operators from $E$ into $F.$ Assume that
$$ sup_{i in I} lVert T_ix rVert<infty quad forall x in E. $$
then
$$ sup_{i in I} lVert T_i rVert< ∞ .$$
(from Brezis page 32)
my question is : if the family consists of a finite number of linear bdd operators the hypothesis:
$$sup_{i in I} lVert T_ix rVert < infty quad forall x in E$$
is always verified, isn't it?
and more generally, what is this theorem telling me?
I apologize for the banality of my question but I can't fully understand this theorem.
functional-analysis analysis banach-spaces
edited Jan 10 at 17:50
ktoi
2,4061617
asked Jan 10 at 17:33
whowhowhowho
152
$endgroup$
$begingroup$
Welcome to MSE! I've edited you question and added mathjax to make it easier to read. For future reference, see here on how to type mathematical expressions on this site: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– ktoi
Jan 10 at 17:52
$begingroup$
There are already a lot of nice answers to this question in the related questions, but the main takeaway here is that for Banach spaces, pointwise boundedness $implies$ uniform boundedness for families of bounded linear operators; that is, for each $x in E$, there is a constant $c_x$ (dependent on $x$!) bounding all $|| T_i x ||$. Banach Steinhaus tells us that we can remove the dependence on $x$.
$endgroup$
– Rellek
Jan 10 at 17:55
add a comment |
$begingroup$
I am a beginner of functional analysis and I can't understand at all the Banach Steinhaus theorem:
Let $E$ and $F$ be two Banach spaces and let $(T_i)_{i in I}$ be a family (not necessarily countable) of continuous linear operators from $E$ into $F.$ Assume that
$$ sup_{i in I} lVert T_ix rVert<infty quad forall x in E. $$
then
$$ sup_{i in I} lVert T_i rVert< ∞ .$$
(from Brezis page 32)
my question is : if the family consists of a finite number of linear bdd operators the hypothesis:
$$sup_{i in I} lVert T_ix rVert < infty quad forall x in E$$
is always verified, isn't it?
and more generally, what is this theorem telling me?
I apologize for the banality of my question but I can't fully understand this theorem.
functional-analysis analysis banach-spaces
edited Jan 10 at 17:50
ktoi
2,4061617
asked Jan 10 at 17:33
whowhowhowho
152
$endgroup$
I am a beginner of functional analysis and I can't understand at all the Banach Steinhaus theorem:
Let $E$ and $F$ be two Banach spaces and let $(T_i)_{i in I}$ be a family (not necessarily countable) of continuous linear operators from $E$ into $F.$ Assume that
$$ sup_{i in I} lVert T_ix rVert<infty quad forall x in E. $$
then
$$ sup_{i in I} lVert T_i rVert< ∞ .$$
(from Brezis page 32)
my question is : if the family consists of a finite number of linear bdd operators the hypothesis:
$$sup_{i in I} lVert T_ix rVert < infty quad forall x in E$$
is always verified, isn't it?
and more generally, what is this theorem telling me?
I apologize for the banality of my question but I can't fully understand this theorem.
functional-analysis analysis banach-spaces
functional-analysis analysis banach-spaces
edited Jan 10 at 17:50
ktoi
2,4061617
asked Jan 10 at 17:33
whowhowhowho
152
edited Jan 10 at 17:50
ktoi
2,4061617
asked Jan 10 at 17:33
whowhowhowho
152
edited Jan 10 at 17:50
ktoi
2,4061617
edited Jan 10 at 17:50
ktoi
2,4061617
edited Jan 10 at 17:50
ktoi
2,4061617
2,4061617
asked Jan 10 at 17:33
whowhowhowho
152
asked Jan 10 at 17:33
whowhowhowho
152
asked Jan 10 at 17:33
whowhowhowho
152
152
$begingroup$
Welcome to MSE! I've edited you question and added mathjax to make it easier to read. For future reference, see here on how to type mathematical expressions on this site: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– ktoi
Jan 10 at 17:52
$begingroup$
There are already a lot of nice answers to this question in the related questions, but the main takeaway here is that for Banach spaces, pointwise boundedness $implies$ uniform boundedness for families of bounded linear operators; that is, for each $x in E$, there is a constant $c_x$ (dependent on $x$!) bounding all $|| T_i x ||$. Banach Steinhaus tells us that we can remove the dependence on $x$.
$endgroup$
– Rellek
Jan 10 at 17:55
add a comment |
$begingroup$
Welcome to MSE! I've edited you question and added mathjax to make it easier to read. For future reference, see here on how to type mathematical expressions on this site: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– ktoi
Jan 10 at 17:52
$begingroup$
There are already a lot of nice answers to this question in the related questions, but the main takeaway here is that for Banach spaces, pointwise boundedness $implies$ uniform boundedness for families of bounded linear operators; that is, for each $x in E$, there is a constant $c_x$ (dependent on $x$!) bounding all $|| T_i x ||$. Banach Steinhaus tells us that we can remove the dependence on $x$.
$endgroup$
– Rellek
Jan 10 at 17:55
$begingroup$
Welcome to MSE! I've edited you question and added mathjax to make it easier to read. For future reference, see here on how to type mathematical expressions on this site: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– ktoi
Jan 10 at 17:52
$begingroup$
Welcome to MSE! I've edited you question and added mathjax to make it easier to read. For future reference, see here on how to type mathematical expressions on this site: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– ktoi
Jan 10 at 17:52
$begingroup$
There are already a lot of nice answers to this question in the related questions, but the main takeaway here is that for Banach spaces, pointwise boundedness $implies$ uniform boundedness for families of bounded linear operators; that is, for each $x in E$, there is a constant $c_x$ (dependent on $x$!) bounding all $|| T_i x ||$. Banach Steinhaus tells us that we can remove the dependence on $x$.
$endgroup$
– Rellek
Jan 10 at 17:55
$begingroup$
There are already a lot of nice answers to this question in the related questions, but the main takeaway here is that for Banach spaces, pointwise boundedness $implies$ uniform boundedness for families of bounded linear operators; that is, for each $x in E$, there is a constant $c_x$ (dependent on $x$!) bounding all $|| T_i x ||$. Banach Steinhaus tells us that we can remove the dependence on $x$.
$endgroup$
– Rellek
Jan 10 at 17:55
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
Yes, if the family is finite, then $sup_i|T_ix|=max{|T_1x|,ldots,|T_nx|}<infty$, and also $sup_i|T_i|=max{|T_1|,ldots,|T_n|}<infty$. The theorem is relevant when the family is infinite.
What they theorem says it what it says, there's no much philosophy there: if your family ${T_i}$ is bounded "pointwise" (at every $x$), then it is bounded uniformly: there exists $c>0$ such that $|T_i|<c$ for all $i$.
The theorem has many applications. Typical ones are
to prove that if ${T_nx}$ converges for all $xin E$, then $Tx=lim T_nx$ defines a bounded operator.
a weakly bounded set in a Banach space is norm-bounded; in particular, a weakly convergent sequence is bounded.
to prove the spectral radius formula: for $Tin B(E)$, $$operatorname{spr}(T)=lim_{ntoinfty}|T^n|^{1/n}.$$
answered Jan 10 at 23:43
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
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$begingroup$
Yes, if the family is finite, then $sup_i|T_ix|=max{|T_1x|,ldots,|T_nx|}<infty$, and also $sup_i|T_i|=max{|T_1|,ldots,|T_n|}<infty$. The theorem is relevant when the family is infinite.
What they theorem says it what it says, there's no much philosophy there: if your family ${T_i}$ is bounded "pointwise" (at every $x$), then it is bounded uniformly: there exists $c>0$ such that $|T_i|<c$ for all $i$.
The theorem has many applications. Typical ones are
to prove that if ${T_nx}$ converges for all $xin E$, then $Tx=lim T_nx$ defines a bounded operator.
a weakly bounded set in a Banach space is norm-bounded; in particular, a weakly convergent sequence is bounded.
to prove the spectral radius formula: for $Tin B(E)$, $$operatorname{spr}(T)=lim_{ntoinfty}|T^n|^{1/n}.$$
answered Jan 10 at 23:43
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
Yes, if the family is finite, then $sup_i|T_ix|=max{|T_1x|,ldots,|T_nx|}<infty$, and also $sup_i|T_i|=max{|T_1|,ldots,|T_n|}<infty$. The theorem is relevant when the family is infinite.
What they theorem says it what it says, there's no much philosophy there: if your family ${T_i}$ is bounded "pointwise" (at every $x$), then it is bounded uniformly: there exists $c>0$ such that $|T_i|<c$ for all $i$.
The theorem has many applications. Typical ones are
to prove that if ${T_nx}$ converges for all $xin E$, then $Tx=lim T_nx$ defines a bounded operator.
a weakly bounded set in a Banach space is norm-bounded; in particular, a weakly convergent sequence is bounded.
to prove the spectral radius formula: for $Tin B(E)$, $$operatorname{spr}(T)=lim_{ntoinfty}|T^n|^{1/n}.$$
answered Jan 10 at 23:43
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
add a comment |
$begingroup$
Yes, if the family is finite, then $sup_i|T_ix|=max{|T_1x|,ldots,|T_nx|}<infty$, and also $sup_i|T_i|=max{|T_1|,ldots,|T_n|}<infty$. The theorem is relevant when the family is infinite.
What they theorem says it what it says, there's no much philosophy there: if your family ${T_i}$ is bounded "pointwise" (at every $x$), then it is bounded uniformly: there exists $c>0$ such that $|T_i|<c$ for all $i$.
The theorem has many applications. Typical ones are
to prove that if ${T_nx}$ converges for all $xin E$, then $Tx=lim T_nx$ defines a bounded operator.
a weakly bounded set in a Banach space is norm-bounded; in particular, a weakly convergent sequence is bounded.
to prove the spectral radius formula: for $Tin B(E)$, $$operatorname{spr}(T)=lim_{ntoinfty}|T^n|^{1/n}.$$
answered Jan 10 at 23:43
Martin ArgeramiMartin Argerami
126k1182181
$endgroup$
Yes, if the family is finite, then $sup_i|T_ix|=max{|T_1x|,ldots,|T_nx|}<infty$, and also $sup_i|T_i|=max{|T_1|,ldots,|T_n|}<infty$. The theorem is relevant when the family is infinite.
What they theorem says it what it says, there's no much philosophy there: if your family ${T_i}$ is bounded "pointwise" (at every $x$), then it is bounded uniformly: there exists $c>0$ such that $|T_i|<c$ for all $i$.
The theorem has many applications. Typical ones are
to prove that if ${T_nx}$ converges for all $xin E$, then $Tx=lim T_nx$ defines a bounded operator.
a weakly bounded set in a Banach space is norm-bounded; in particular, a weakly convergent sequence is bounded.
to prove the spectral radius formula: for $Tin B(E)$, $$operatorname{spr}(T)=lim_{ntoinfty}|T^n|^{1/n}.$$
answered Jan 10 at 23:43
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 23:43
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 23:43
Martin ArgeramiMartin Argerami
126k1182181
answered Jan 10 at 23:43
Martin ArgeramiMartin Argerami
126k1182181
126k1182181
add a comment |
add a comment |
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$begingroup$
Welcome to MSE! I've edited you question and added mathjax to make it easier to read. For future reference, see here on how to type mathematical expressions on this site: math.meta.stackexchange.com/questions/5020/…
$endgroup$
– ktoi
Jan 10 at 17:52
$begingroup$
There are already a lot of nice answers to this question in the related questions, but the main takeaway here is that for Banach spaces, pointwise boundedness $implies$ uniform boundedness for families of bounded linear operators; that is, for each $x in E$, there is a constant $c_x$ (dependent on $x$!) bounding all $|| T_i x ||$. Banach Steinhaus tells us that we can remove the dependence on $x$.
$endgroup$
– Rellek
Jan 10 at 17:55