Mixing
Uniform Boundedness Principle for Functionals
$begingroup$
Kindly check if my proof is correct. Alternative proofs are welcome too!
Let $Delta$ be an arbitrary index set and let $E$ be a complete metric space and ${ f_alpha }_{alpha in Delta}$ be a family of real-valued continuous functionals on $E.$ Then, there exists an open set $U$ in $E$ on which ${ f_alpha }_{alpha in Delta}$ is uniformly bounded.
My Attempt
Let $ B(X,Bbb{R})$ be the space of bounded linear functionals. Then,
${ f_alpha }_{alpha in Delta}in B(X,Bbb{R})$, since $f_alpha $ is continuous for each $alpha in Delta.$
$forall;xin X$, there exists $M_xgeq 0,$ such that $;left|f_alpha(x) right|leq M_x,;forall;alpha in Delta.$
Fix $nin Bbb{N}$ and define $$gamma_{n}={xin X:; left|f_alpha(x) right| leq n,;alpha in Delta }.$$
Then, $gamma_{n}$ is closed for each $n$ and $X=bigcuplimits_{nin Bbb{N}}gamma_{n}$. Then, by Baire's Lemma, there exists $n_0$ such that $$gamma_{n_0}strut^mathrm{o}neq phi.$$
Let $x_0in gamma_{n_0}strut^mathrm{o}$, then there exists $r>0$ such that $B(x_0,r)subseteq gamma_{n_0}strut^mathrm{o}subseteq gamma_{n_0}.$ Let $zin B(0,1)$, then
$$x_0+rz in B(x_0,r).$$
This implies that $$rleft|f_alpha(z) right|-left|f_alpha(x_0) right|leq left|f_alpha(x_0)+r f_alpha(z) right| =left|f_alpha(x_0+rz) right| leq n_0$$
and so, $$left|f_alpha(z) right|leqdfrac{2n_0}{r}:=M,;forall;zin B(0,1),;;text{since};;x_0in gamma_{n_0}.$$
Take $U=B(0,1)$, then $$left|f_alpha(x) right|leq M,;;forall;zin B(0,1),;forall;alpha in Delta.$$
real-analysis general-topology functional-analysis baire-category
asked Jan 18 at 21:34
Omojola MichealOmojola Micheal
1,920324
$endgroup$
add a comment |
$begingroup$
Kindly check if my proof is correct. Alternative proofs are welcome too!
Let $Delta$ be an arbitrary index set and let $E$ be a complete metric space and ${ f_alpha }_{alpha in Delta}$ be a family of real-valued continuous functionals on $E.$ Then, there exists an open set $U$ in $E$ on which ${ f_alpha }_{alpha in Delta}$ is uniformly bounded.
My Attempt
Let $ B(X,Bbb{R})$ be the space of bounded linear functionals. Then,
${ f_alpha }_{alpha in Delta}in B(X,Bbb{R})$, since $f_alpha $ is continuous for each $alpha in Delta.$
$forall;xin X$, there exists $M_xgeq 0,$ such that $;left|f_alpha(x) right|leq M_x,;forall;alpha in Delta.$
Fix $nin Bbb{N}$ and define $$gamma_{n}={xin X:; left|f_alpha(x) right| leq n,;alpha in Delta }.$$
Then, $gamma_{n}$ is closed for each $n$ and $X=bigcuplimits_{nin Bbb{N}}gamma_{n}$. Then, by Baire's Lemma, there exists $n_0$ such that $$gamma_{n_0}strut^mathrm{o}neq phi.$$
Let $x_0in gamma_{n_0}strut^mathrm{o}$, then there exists $r>0$ such that $B(x_0,r)subseteq gamma_{n_0}strut^mathrm{o}subseteq gamma_{n_0}.$ Let $zin B(0,1)$, then
$$x_0+rz in B(x_0,r).$$
This implies that $$rleft|f_alpha(z) right|-left|f_alpha(x_0) right|leq left|f_alpha(x_0)+r f_alpha(z) right| =left|f_alpha(x_0+rz) right| leq n_0$$
and so, $$left|f_alpha(z) right|leqdfrac{2n_0}{r}:=M,;forall;zin B(0,1),;;text{since};;x_0in gamma_{n_0}.$$
Take $U=B(0,1)$, then $$left|f_alpha(x) right|leq M,;;forall;zin B(0,1),;forall;alpha in Delta.$$
real-analysis general-topology functional-analysis baire-category
asked Jan 18 at 21:34
Omojola MichealOmojola Micheal
1,920324
$endgroup$
$begingroup$
I'll try to construct a counterexample. Let $E=[0,1]$ and $f_alpha(x)=alpha$, $alphain Delta=(0,infty)$. No boundedness then.
$endgroup$
– pabodu
Feb 8 at 3:58
add a comment |
$begingroup$
Kindly check if my proof is correct. Alternative proofs are welcome too!
Let $Delta$ be an arbitrary index set and let $E$ be a complete metric space and ${ f_alpha }_{alpha in Delta}$ be a family of real-valued continuous functionals on $E.$ Then, there exists an open set $U$ in $E$ on which ${ f_alpha }_{alpha in Delta}$ is uniformly bounded.
My Attempt
Let $ B(X,Bbb{R})$ be the space of bounded linear functionals. Then,
${ f_alpha }_{alpha in Delta}in B(X,Bbb{R})$, since $f_alpha $ is continuous for each $alpha in Delta.$
$forall;xin X$, there exists $M_xgeq 0,$ such that $;left|f_alpha(x) right|leq M_x,;forall;alpha in Delta.$
Fix $nin Bbb{N}$ and define $$gamma_{n}={xin X:; left|f_alpha(x) right| leq n,;alpha in Delta }.$$
Then, $gamma_{n}$ is closed for each $n$ and $X=bigcuplimits_{nin Bbb{N}}gamma_{n}$. Then, by Baire's Lemma, there exists $n_0$ such that $$gamma_{n_0}strut^mathrm{o}neq phi.$$
Let $x_0in gamma_{n_0}strut^mathrm{o}$, then there exists $r>0$ such that $B(x_0,r)subseteq gamma_{n_0}strut^mathrm{o}subseteq gamma_{n_0}.$ Let $zin B(0,1)$, then
$$x_0+rz in B(x_0,r).$$
This implies that $$rleft|f_alpha(z) right|-left|f_alpha(x_0) right|leq left|f_alpha(x_0)+r f_alpha(z) right| =left|f_alpha(x_0+rz) right| leq n_0$$
and so, $$left|f_alpha(z) right|leqdfrac{2n_0}{r}:=M,;forall;zin B(0,1),;;text{since};;x_0in gamma_{n_0}.$$
Take $U=B(0,1)$, then $$left|f_alpha(x) right|leq M,;;forall;zin B(0,1),;forall;alpha in Delta.$$
real-analysis general-topology functional-analysis baire-category
asked Jan 18 at 21:34
Omojola MichealOmojola Micheal
1,920324
$endgroup$
Kindly check if my proof is correct. Alternative proofs are welcome too!
Let $Delta$ be an arbitrary index set and let $E$ be a complete metric space and ${ f_alpha }_{alpha in Delta}$ be a family of real-valued continuous functionals on $E.$ Then, there exists an open set $U$ in $E$ on which ${ f_alpha }_{alpha in Delta}$ is uniformly bounded.
My Attempt
Let $ B(X,Bbb{R})$ be the space of bounded linear functionals. Then,
${ f_alpha }_{alpha in Delta}in B(X,Bbb{R})$, since $f_alpha $ is continuous for each $alpha in Delta.$
$forall;xin X$, there exists $M_xgeq 0,$ such that $;left|f_alpha(x) right|leq M_x,;forall;alpha in Delta.$
Fix $nin Bbb{N}$ and define $$gamma_{n}={xin X:; left|f_alpha(x) right| leq n,;alpha in Delta }.$$
Then, $gamma_{n}$ is closed for each $n$ and $X=bigcuplimits_{nin Bbb{N}}gamma_{n}$. Then, by Baire's Lemma, there exists $n_0$ such that $$gamma_{n_0}strut^mathrm{o}neq phi.$$
Let $x_0in gamma_{n_0}strut^mathrm{o}$, then there exists $r>0$ such that $B(x_0,r)subseteq gamma_{n_0}strut^mathrm{o}subseteq gamma_{n_0}.$ Let $zin B(0,1)$, then
$$x_0+rz in B(x_0,r).$$
This implies that $$rleft|f_alpha(z) right|-left|f_alpha(x_0) right|leq left|f_alpha(x_0)+r f_alpha(z) right| =left|f_alpha(x_0+rz) right| leq n_0$$
and so, $$left|f_alpha(z) right|leqdfrac{2n_0}{r}:=M,;forall;zin B(0,1),;;text{since};;x_0in gamma_{n_0}.$$
Take $U=B(0,1)$, then $$left|f_alpha(x) right|leq M,;;forall;zin B(0,1),;forall;alpha in Delta.$$
real-analysis general-topology functional-analysis baire-category
real-analysis general-topology functional-analysis baire-category
asked Jan 18 at 21:34
Omojola MichealOmojola Micheal
1,920324
asked Jan 18 at 21:34
Omojola MichealOmojola Micheal
1,920324
edited Jan 19 at 0:25
Omojola Micheal
asked Jan 18 at 21:34
Omojola MichealOmojola Micheal
1,920324
asked Jan 18 at 21:34
Omojola MichealOmojola Micheal
1,920324
asked Jan 18 at 21:34
Omojola MichealOmojola Micheal
1,920324
1,920324
$begingroup$
I'll try to construct a counterexample. Let $E=[0,1]$ and $f_alpha(x)=alpha$, $alphain Delta=(0,infty)$. No boundedness then.
$endgroup$
– pabodu
Feb 8 at 3:58
add a comment |
$begingroup$
I'll try to construct a counterexample. Let $E=[0,1]$ and $f_alpha(x)=alpha$, $alphain Delta=(0,infty)$. No boundedness then.
$endgroup$
– pabodu
Feb 8 at 3:58
$begingroup$
I'll try to construct a counterexample. Let $E=[0,1]$ and $f_alpha(x)=alpha$, $alphain Delta=(0,infty)$. No boundedness then.
$endgroup$
– pabodu
Feb 8 at 3:58
$begingroup$
I'll try to construct a counterexample. Let $E=[0,1]$ and $f_alpha(x)=alpha$, $alphain Delta=(0,infty)$. No boundedness then.
$endgroup$
– pabodu
Feb 8 at 3:58
add a comment |
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$begingroup$
I'll try to construct a counterexample. Let $E=[0,1]$ and $f_alpha(x)=alpha$, $alphain Delta=(0,infty)$. No boundedness then.
$endgroup$
– pabodu
Feb 8 at 3:58