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Prove that there exists sequence $(u_n)$ in $E$ such that $| u_n |_E = 1$ and $| Tu_n|_F rightarrow 0.$
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Let $E$ and $F$ be two Banach spaces, and let $T in K(E,F)$. Assume $dim E = infty$. Prove that there exists sequence $(u_n)$ in $E$ such that $| u_n |_E = 1$ and $| Tu_n|_F rightarrow 0.$
Comments: I'm trying to do it by contradiction.
I'm trying to show that there exists $C> 0$ such that $| Tu|_F geq C | u |_E $, for all $u in E$. With this I can show that $R(T)$ is closed and I can use the Open Application Theorem.
functional-analysis compact-operators
asked Jul 21 at 0:52
Croos
789415
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up vote
1
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Let $E$ and $F$ be two Banach spaces, and let $T in K(E,F)$. Assume $dim E = infty$. Prove that there exists sequence $(u_n)$ in $E$ such that $| u_n |_E = 1$ and $| Tu_n|_F rightarrow 0.$
Comments: I'm trying to do it by contradiction.
I'm trying to show that there exists $C> 0$ such that $| Tu|_F geq C | u |_E $, for all $u in E$. With this I can show that $R(T)$ is closed and I can use the Open Application Theorem.
functional-analysis compact-operators
asked Jul 21 at 0:52
Croos
789415
Hint; Use the Spectral Theorem for compact maps
– TheOscillator
Jul 21 at 1:43
add a comment |Â
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Let $E$ and $F$ be two Banach spaces, and let $T in K(E,F)$. Assume $dim E = infty$. Prove that there exists sequence $(u_n)$ in $E$ such that $| u_n |_E = 1$ and $| Tu_n|_F rightarrow 0.$
Comments: I'm trying to do it by contradiction.
I'm trying to show that there exists $C> 0$ such that $| Tu|_F geq C | u |_E $, for all $u in E$. With this I can show that $R(T)$ is closed and I can use the Open Application Theorem.
functional-analysis compact-operators
asked Jul 21 at 0:52
Croos
789415
Let $E$ and $F$ be two Banach spaces, and let $T in K(E,F)$. Assume $dim E = infty$. Prove that there exists sequence $(u_n)$ in $E$ such that $| u_n |_E = 1$ and $| Tu_n|_F rightarrow 0.$
Comments: I'm trying to do it by contradiction.
I'm trying to show that there exists $C> 0$ such that $| Tu|_F geq C | u |_E $, for all $u in E$. With this I can show that $R(T)$ is closed and I can use the Open Application Theorem.
functional-analysis compact-operators
asked Jul 21 at 0:52
Croos
789415
asked Jul 21 at 0:52
Croos
789415
asked Jul 21 at 0:52
Croos
789415
asked Jul 21 at 0:52
Croos
789415
789415
Hint; Use the Spectral Theorem for compact maps
– TheOscillator
Jul 21 at 1:43
add a comment |Â
Hint; Use the Spectral Theorem for compact maps
– TheOscillator
Jul 21 at 1:43
Hint; Use the Spectral Theorem for compact maps
– TheOscillator
Jul 21 at 1:43
Hint; Use the Spectral Theorem for compact maps
– TheOscillator
Jul 21 at 1:43
add a comment |Â
2 Answers
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Consider $inf :$. If this is $0$ we are done.If it is a positive number $a$ then $||Tu||geq a$ whenever $||u||=1$. By simple scaling this gives $||Tu||geq a||u||$ for all $u$. If $x_n$ is any sequence in the unit ball of $E$ than $Tx_n$ has a convergent subsequence and the inequality $||Tx_n-Tx_m||geq a||x_n-x_m||$ (applied to the subsequence) shows that $x_n$ has a convegent subsequence. Thus the unit ball of $E$ is compact which implies $E$ is finite dimensional.
answered Jul 21 at 12:34
Kavi Rama Murthy
20.6k2830
I have realized that DanielWainfleet had the same basic idea but his presentation looks a bit more complicated than what is necessary.
– Kavi Rama Murthy
Jul 21 at 12:37
Agreed. I often give a complicated A............+1
– DanielWainfleet
Jul 22 at 18:31
add a comment |Â
up vote
1
down vote
Let $G=Tx:xin E.$ Suppose $inf _F: 1==1/k>0$. Then T is injective, and $T^-1:Gto E$ is a bounded (Lipschitz-continuous) linear function (with norm $ k $) from the normed vector-space $G$, into $E.$
Let $A=(g_n)_n$ be any Cauchy sequence in $G$. Since $T^-1$ is Lipschitz-continuous on $G,$ the sequence $B=(T^-1g_n)_n$ is Cauchy in $E$. So $B$ has a limit $xin E.$ Now $T$ is continuous, so the sequence $(T(T^-1g_n))_n= A$ converges to $Txin G.$ So $G$ is closed in $F.$
Let $C= x.$ Then $cl_F (C)subset G.$ But $cl_F(C)$ is compact and $T^-1$ is continuous on $G,$ so $T^-1(cl_F(C))$ is also compact.
If dim($E)=infty$ this is absurd because it means the compact Hausdorff space $T^-1(cl_f(C))$ has a closed non-compact subset $x$.
answered Jul 21 at 4:00
DanielWainfleet
31.6k31643
add a comment |Â
2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
5
down vote
accepted
Consider $inf :$. If this is $0$ we are done.If it is a positive number $a$ then $||Tu||geq a$ whenever $||u||=1$. By simple scaling this gives $||Tu||geq a||u||$ for all $u$. If $x_n$ is any sequence in the unit ball of $E$ than $Tx_n$ has a convergent subsequence and the inequality $||Tx_n-Tx_m||geq a||x_n-x_m||$ (applied to the subsequence) shows that $x_n$ has a convegent subsequence. Thus the unit ball of $E$ is compact which implies $E$ is finite dimensional.
answered Jul 21 at 12:34
Kavi Rama Murthy
20.6k2830
I have realized that DanielWainfleet had the same basic idea but his presentation looks a bit more complicated than what is necessary.
– Kavi Rama Murthy
Jul 21 at 12:37
Agreed. I often give a complicated A............+1
– DanielWainfleet
Jul 22 at 18:31
add a comment |Â
up vote
5
down vote
accepted
Consider $inf :$. If this is $0$ we are done.If it is a positive number $a$ then $||Tu||geq a$ whenever $||u||=1$. By simple scaling this gives $||Tu||geq a||u||$ for all $u$. If $x_n$ is any sequence in the unit ball of $E$ than $Tx_n$ has a convergent subsequence and the inequality $||Tx_n-Tx_m||geq a||x_n-x_m||$ (applied to the subsequence) shows that $x_n$ has a convegent subsequence. Thus the unit ball of $E$ is compact which implies $E$ is finite dimensional.
answered Jul 21 at 12:34
Kavi Rama Murthy
20.6k2830
I have realized that DanielWainfleet had the same basic idea but his presentation looks a bit more complicated than what is necessary.
– Kavi Rama Murthy
Jul 21 at 12:37
Agreed. I often give a complicated A............+1
– DanielWainfleet
Jul 22 at 18:31
add a comment |Â
up vote
5
down vote
accepted
up vote
5
down vote
accepted
Consider $inf :$. If this is $0$ we are done.If it is a positive number $a$ then $||Tu||geq a$ whenever $||u||=1$. By simple scaling this gives $||Tu||geq a||u||$ for all $u$. If $x_n$ is any sequence in the unit ball of $E$ than $Tx_n$ has a convergent subsequence and the inequality $||Tx_n-Tx_m||geq a||x_n-x_m||$ (applied to the subsequence) shows that $x_n$ has a convegent subsequence. Thus the unit ball of $E$ is compact which implies $E$ is finite dimensional.
answered Jul 21 at 12:34
Kavi Rama Murthy
20.6k2830
Consider $inf :$. If this is $0$ we are done.If it is a positive number $a$ then $||Tu||geq a$ whenever $||u||=1$. By simple scaling this gives $||Tu||geq a||u||$ for all $u$. If $x_n$ is any sequence in the unit ball of $E$ than $Tx_n$ has a convergent subsequence and the inequality $||Tx_n-Tx_m||geq a||x_n-x_m||$ (applied to the subsequence) shows that $x_n$ has a convegent subsequence. Thus the unit ball of $E$ is compact which implies $E$ is finite dimensional.
answered Jul 21 at 12:34
Kavi Rama Murthy
20.6k2830
answered Jul 21 at 12:34
Kavi Rama Murthy
20.6k2830
answered Jul 21 at 12:34
Kavi Rama Murthy
20.6k2830
answered Jul 21 at 12:34
Kavi Rama Murthy
20.6k2830
20.6k2830
I have realized that DanielWainfleet had the same basic idea but his presentation looks a bit more complicated than what is necessary.
– Kavi Rama Murthy
Jul 21 at 12:37
Agreed. I often give a complicated A............+1
– DanielWainfleet
Jul 22 at 18:31
add a comment |Â
I have realized that DanielWainfleet had the same basic idea but his presentation looks a bit more complicated than what is necessary.
– Kavi Rama Murthy
Jul 21 at 12:37
Agreed. I often give a complicated A............+1
– DanielWainfleet
Jul 22 at 18:31
I have realized that DanielWainfleet had the same basic idea but his presentation looks a bit more complicated than what is necessary.
– Kavi Rama Murthy
Jul 21 at 12:37
I have realized that DanielWainfleet had the same basic idea but his presentation looks a bit more complicated than what is necessary.
– Kavi Rama Murthy
Jul 21 at 12:37
Agreed. I often give a complicated A............+1
– DanielWainfleet
Jul 22 at 18:31
Agreed. I often give a complicated A............+1
– DanielWainfleet
Jul 22 at 18:31
add a comment |Â
up vote
1
down vote
Let $G=Tx:xin E.$ Suppose $inf _F: 1==1/k>0$. Then T is injective, and $T^-1:Gto E$ is a bounded (Lipschitz-continuous) linear function (with norm $ k $) from the normed vector-space $G$, into $E.$
Let $A=(g_n)_n$ be any Cauchy sequence in $G$. Since $T^-1$ is Lipschitz-continuous on $G,$ the sequence $B=(T^-1g_n)_n$ is Cauchy in $E$. So $B$ has a limit $xin E.$ Now $T$ is continuous, so the sequence $(T(T^-1g_n))_n= A$ converges to $Txin G.$ So $G$ is closed in $F.$
Let $C= x.$ Then $cl_F (C)subset G.$ But $cl_F(C)$ is compact and $T^-1$ is continuous on $G,$ so $T^-1(cl_F(C))$ is also compact.
If dim($E)=infty$ this is absurd because it means the compact Hausdorff space $T^-1(cl_f(C))$ has a closed non-compact subset $x$.
answered Jul 21 at 4:00
DanielWainfleet
31.6k31643
add a comment |Â
up vote
1
down vote
Let $G=Tx:xin E.$ Suppose $inf _F: 1==1/k>0$. Then T is injective, and $T^-1:Gto E$ is a bounded (Lipschitz-continuous) linear function (with norm $ k $) from the normed vector-space $G$, into $E.$
Let $A=(g_n)_n$ be any Cauchy sequence in $G$. Since $T^-1$ is Lipschitz-continuous on $G,$ the sequence $B=(T^-1g_n)_n$ is Cauchy in $E$. So $B$ has a limit $xin E.$ Now $T$ is continuous, so the sequence $(T(T^-1g_n))_n= A$ converges to $Txin G.$ So $G$ is closed in $F.$
Let $C= x.$ Then $cl_F (C)subset G.$ But $cl_F(C)$ is compact and $T^-1$ is continuous on $G,$ so $T^-1(cl_F(C))$ is also compact.
If dim($E)=infty$ this is absurd because it means the compact Hausdorff space $T^-1(cl_f(C))$ has a closed non-compact subset $x$.
answered Jul 21 at 4:00
DanielWainfleet
31.6k31643
add a comment |Â
up vote
1
down vote
up vote
1
down vote
Let $G=Tx:xin E.$ Suppose $inf _F: 1==1/k>0$. Then T is injective, and $T^-1:Gto E$ is a bounded (Lipschitz-continuous) linear function (with norm $ k $) from the normed vector-space $G$, into $E.$
Let $A=(g_n)_n$ be any Cauchy sequence in $G$. Since $T^-1$ is Lipschitz-continuous on $G,$ the sequence $B=(T^-1g_n)_n$ is Cauchy in $E$. So $B$ has a limit $xin E.$ Now $T$ is continuous, so the sequence $(T(T^-1g_n))_n= A$ converges to $Txin G.$ So $G$ is closed in $F.$
Let $C= x.$ Then $cl_F (C)subset G.$ But $cl_F(C)$ is compact and $T^-1$ is continuous on $G,$ so $T^-1(cl_F(C))$ is also compact.
If dim($E)=infty$ this is absurd because it means the compact Hausdorff space $T^-1(cl_f(C))$ has a closed non-compact subset $x$.
answered Jul 21 at 4:00
DanielWainfleet
31.6k31643
Let $G=Tx:xin E.$ Suppose $inf _F: 1==1/k>0$. Then T is injective, and $T^-1:Gto E$ is a bounded (Lipschitz-continuous) linear function (with norm $ k $) from the normed vector-space $G$, into $E.$
Let $A=(g_n)_n$ be any Cauchy sequence in $G$. Since $T^-1$ is Lipschitz-continuous on $G,$ the sequence $B=(T^-1g_n)_n$ is Cauchy in $E$. So $B$ has a limit $xin E.$ Now $T$ is continuous, so the sequence $(T(T^-1g_n))_n= A$ converges to $Txin G.$ So $G$ is closed in $F.$
Let $C= x.$ Then $cl_F (C)subset G.$ But $cl_F(C)$ is compact and $T^-1$ is continuous on $G,$ so $T^-1(cl_F(C))$ is also compact.
If dim($E)=infty$ this is absurd because it means the compact Hausdorff space $T^-1(cl_f(C))$ has a closed non-compact subset $x$.
answered Jul 21 at 4:00
DanielWainfleet
31.6k31643
edited Jul 21 at 4:05
answered Jul 21 at 4:00
DanielWainfleet
31.6k31643
answered Jul 21 at 4:00
DanielWainfleet
31.6k31643
answered Jul 21 at 4:00
DanielWainfleet
31.6k31643
31.6k31643
add a comment |Â
add a comment |Â
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Hint; Use the Spectral Theorem for compact maps
– TheOscillator
Jul 21 at 1:43