A counterexample of Banach Steinhaus Theorem












1














I was reading about a consequence of Banach-Steinhaus theorem which states that:




Let $E$ be a Banach space and $F$ be a normed space, and let ${T_n}_{nin mathbb{N}}$ be a sequence of bounded linear operators from $E$ to $F$, if the sequence ${T_n x}_{nin mathbb{N}}$ converges for each $xin E$, then if we define:
$$
T: Elongrightarrow F
$$
$$
x mapsto Tx = lim_{nto infty} T_n x
$$
then




  1. $displaystyle sup_{nin mathbb{N}} || T_n || <infty$


  2. $T$ is a bounded linear operator

  3. $displaystyle || T || leq liminf_{nto infty} ||T_n ||$






So, I was wondering when this doesn't hold.





I tried the following example: Let $E=F=c_{00}$ the space of bounded sequences with a finite number of non-zero terms. Obviously $c_{00}$ is not a Banach space, so there is the reason the statement above is not verified, but in order to see that, I defined a sequence of bounded linear operators as follows:



For each $nin mathbb{N}$, let $ T_n: Elongrightarrow F
$ such that
$$x=(x_1, x_2, ..., x_n, 0,0,...) mapsto T_n x = (x_1,2 x_2,..., n x_n, 0, 0,..)
$$
then $T_n$ is a bounded linear operator for every $nin mathbb{N}$, but if we define $T$ as above, $T$ is a linear unbounded operator.




I tried to see why is an unbounded operator, this was my attempt:



Suppose by contradiction that $T$ is a bounded operator, then exist $C>0$ such that



$$
||Tx ||leq C ||x ||
$$
for every $xin E$



if we consider $x=e_k=(0,0,...,0,1,0,0,..)$, $1$ on the $k$-th position. We have that



$$
T e_k = lim_{nto infty} T_n e_k = k e_k
$$
then



$$
|| T e_k || =k leq C
$$
but this says that $T$ is bounded.




Did I miss something in this proof?.







functional-analysis proof-verification banach-spaces examples-counterexamples normed-spaces







share|cite|improve this question
























  • Please edit the question to include what you were reading this from.
    – Shaun
    Dec 9 at 3:11










  • It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction?
    – DanielWainfleet
    Dec 9 at 3:44










  • You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly.
    – Jeff
    Dec 9 at 3:59








  • 1




    Isn't $k leq C$ for all $k$ a contradiction?
    – Kavi Rama Murthy
    Dec 9 at 4:54












  • it is, but I already solve this problem with your help, thank you for helping me.
    – Jeff
    Dec 16 at 1:36
















1














I was reading about a consequence of Banach-Steinhaus theorem which states that:




Let $E$ be a Banach space and $F$ be a normed space, and let ${T_n}_{nin mathbb{N}}$ be a sequence of bounded linear operators from $E$ to $F$, if the sequence ${T_n x}_{nin mathbb{N}}$ converges for each $xin E$, then if we define:
$$
T: Elongrightarrow F
$$
$$
x mapsto Tx = lim_{nto infty} T_n x
$$
then




  1. $displaystyle sup_{nin mathbb{N}} || T_n || <infty$


  2. $T$ is a bounded linear operator

  3. $displaystyle || T || leq liminf_{nto infty} ||T_n ||$






So, I was wondering when this doesn't hold.





I tried the following example: Let $E=F=c_{00}$ the space of bounded sequences with a finite number of non-zero terms. Obviously $c_{00}$ is not a Banach space, so there is the reason the statement above is not verified, but in order to see that, I defined a sequence of bounded linear operators as follows:



For each $nin mathbb{N}$, let $ T_n: Elongrightarrow F
$ such that
$$x=(x_1, x_2, ..., x_n, 0,0,...) mapsto T_n x = (x_1,2 x_2,..., n x_n, 0, 0,..)
$$
then $T_n$ is a bounded linear operator for every $nin mathbb{N}$, but if we define $T$ as above, $T$ is a linear unbounded operator.




I tried to see why is an unbounded operator, this was my attempt:



Suppose by contradiction that $T$ is a bounded operator, then exist $C>0$ such that



$$
||Tx ||leq C ||x ||
$$
for every $xin E$



if we consider $x=e_k=(0,0,...,0,1,0,0,..)$, $1$ on the $k$-th position. We have that



$$
T e_k = lim_{nto infty} T_n e_k = k e_k
$$
then



$$
|| T e_k || =k leq C
$$
but this says that $T$ is bounded.




Did I miss something in this proof?.







functional-analysis proof-verification banach-spaces examples-counterexamples normed-spaces







share|cite|improve this question
























  • Please edit the question to include what you were reading this from.
    – Shaun
    Dec 9 at 3:11










  • It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction?
    – DanielWainfleet
    Dec 9 at 3:44










  • You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly.
    – Jeff
    Dec 9 at 3:59








  • 1




    Isn't $k leq C$ for all $k$ a contradiction?
    – Kavi Rama Murthy
    Dec 9 at 4:54












  • it is, but I already solve this problem with your help, thank you for helping me.
    – Jeff
    Dec 16 at 1:36














1












1








1


2





I was reading about a consequence of Banach-Steinhaus theorem which states that:




Let $E$ be a Banach space and $F$ be a normed space, and let ${T_n}_{nin mathbb{N}}$ be a sequence of bounded linear operators from $E$ to $F$, if the sequence ${T_n x}_{nin mathbb{N}}$ converges for each $xin E$, then if we define:
$$
T: Elongrightarrow F
$$
$$
x mapsto Tx = lim_{nto infty} T_n x
$$
then




  1. $displaystyle sup_{nin mathbb{N}} || T_n || <infty$


  2. $T$ is a bounded linear operator

  3. $displaystyle || T || leq liminf_{nto infty} ||T_n ||$






So, I was wondering when this doesn't hold.





I tried the following example: Let $E=F=c_{00}$ the space of bounded sequences with a finite number of non-zero terms. Obviously $c_{00}$ is not a Banach space, so there is the reason the statement above is not verified, but in order to see that, I defined a sequence of bounded linear operators as follows:



For each $nin mathbb{N}$, let $ T_n: Elongrightarrow F
$ such that
$$x=(x_1, x_2, ..., x_n, 0,0,...) mapsto T_n x = (x_1,2 x_2,..., n x_n, 0, 0,..)
$$
then $T_n$ is a bounded linear operator for every $nin mathbb{N}$, but if we define $T$ as above, $T$ is a linear unbounded operator.




I tried to see why is an unbounded operator, this was my attempt:



Suppose by contradiction that $T$ is a bounded operator, then exist $C>0$ such that



$$
||Tx ||leq C ||x ||
$$
for every $xin E$



if we consider $x=e_k=(0,0,...,0,1,0,0,..)$, $1$ on the $k$-th position. We have that



$$
T e_k = lim_{nto infty} T_n e_k = k e_k
$$
then



$$
|| T e_k || =k leq C
$$
but this says that $T$ is bounded.




Did I miss something in this proof?.







functional-analysis proof-verification banach-spaces examples-counterexamples normed-spaces







share|cite|improve this question















I was reading about a consequence of Banach-Steinhaus theorem which states that:




Let $E$ be a Banach space and $F$ be a normed space, and let ${T_n}_{nin mathbb{N}}$ be a sequence of bounded linear operators from $E$ to $F$, if the sequence ${T_n x}_{nin mathbb{N}}$ converges for each $xin E$, then if we define:
$$
T: Elongrightarrow F
$$
$$
x mapsto Tx = lim_{nto infty} T_n x
$$
then




  1. $displaystyle sup_{nin mathbb{N}} || T_n || <infty$


  2. $T$ is a bounded linear operator

  3. $displaystyle || T || leq liminf_{nto infty} ||T_n ||$






So, I was wondering when this doesn't hold.





I tried the following example: Let $E=F=c_{00}$ the space of bounded sequences with a finite number of non-zero terms. Obviously $c_{00}$ is not a Banach space, so there is the reason the statement above is not verified, but in order to see that, I defined a sequence of bounded linear operators as follows:



For each $nin mathbb{N}$, let $ T_n: Elongrightarrow F
$ such that
$$x=(x_1, x_2, ..., x_n, 0,0,...) mapsto T_n x = (x_1,2 x_2,..., n x_n, 0, 0,..)
$$
then $T_n$ is a bounded linear operator for every $nin mathbb{N}$, but if we define $T$ as above, $T$ is a linear unbounded operator.




I tried to see why is an unbounded operator, this was my attempt:



Suppose by contradiction that $T$ is a bounded operator, then exist $C>0$ such that



$$
||Tx ||leq C ||x ||
$$
for every $xin E$



if we consider $x=e_k=(0,0,...,0,1,0,0,..)$, $1$ on the $k$-th position. We have that



$$
T e_k = lim_{nto infty} T_n e_k = k e_k
$$
then



$$
|| T e_k || =k leq C
$$
but this says that $T$ is bounded.




Did I miss something in this proof?.







functional-analysis proof-verification banach-spaces examples-counterexamples normed-spaces




functional-analysis proof-verification banach-spaces examples-counterexamples normed-spaces






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Dec 9 at 3:51

























asked Dec 9 at 3:04









Jeff

185




185












  • Please edit the question to include what you were reading this from.
    – Shaun
    Dec 9 at 3:11










  • It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction?
    – DanielWainfleet
    Dec 9 at 3:44










  • You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly.
    – Jeff
    Dec 9 at 3:59








  • 1




    Isn't $k leq C$ for all $k$ a contradiction?
    – Kavi Rama Murthy
    Dec 9 at 4:54












  • it is, but I already solve this problem with your help, thank you for helping me.
    – Jeff
    Dec 16 at 1:36


















  • Please edit the question to include what you were reading this from.
    – Shaun
    Dec 9 at 3:11










  • It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction?
    – DanielWainfleet
    Dec 9 at 3:44










  • You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly.
    – Jeff
    Dec 9 at 3:59








  • 1




    Isn't $k leq C$ for all $k$ a contradiction?
    – Kavi Rama Murthy
    Dec 9 at 4:54












  • it is, but I already solve this problem with your help, thank you for helping me.
    – Jeff
    Dec 16 at 1:36
















Please edit the question to include what you were reading this from.
– Shaun
Dec 9 at 3:11




Please edit the question to include what you were reading this from.
– Shaun
Dec 9 at 3:11












It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction?
– DanielWainfleet
Dec 9 at 3:44




It seems you are trying to show $T$ is unbounded by contradiction. You begin "Suppose $T$ is bounded" and you end with "but this says $T$ is bounded". Where is the contradiction?
– DanielWainfleet
Dec 9 at 3:44












You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly.
– Jeff
Dec 9 at 3:59






You right, I tried to prove that by contradiction. I forgot to include that part. Thanks. The question is that whether my attempt of proof was right or this example is not valid either. I put that T is bounded since I couldn’t het anything more from that. A partner says that contradiction occurs because T is bounded for every k and that’s all. Buy I don’t see that clearly.
– Jeff
Dec 9 at 3:59






1




1




Isn't $k leq C$ for all $k$ a contradiction?
– Kavi Rama Murthy
Dec 9 at 4:54






Isn't $k leq C$ for all $k$ a contradiction?
– Kavi Rama Murthy
Dec 9 at 4:54














it is, but I already solve this problem with your help, thank you for helping me.
– Jeff
Dec 16 at 1:36




it is, but I already solve this problem with your help, thank you for helping me.
– Jeff
Dec 16 at 1:36










1 Answer
1






active

oldest

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0














The norm of $T$ is the supremum of $lvertlvert Tx rvertrvert$ over all unit vectors $x in E$ and this is at least the supremum of $lvertlvert Te_k rvertrvert$ over all elementary basis vectors $(e_k)_{k=1}^infty$.






share|cite|improve this answer





















  • Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
    – Jeff
    Dec 16 at 1:37













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0














The norm of $T$ is the supremum of $lvertlvert Tx rvertrvert$ over all unit vectors $x in E$ and this is at least the supremum of $lvertlvert Te_k rvertrvert$ over all elementary basis vectors $(e_k)_{k=1}^infty$.






share|cite|improve this answer





















  • Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
    – Jeff
    Dec 16 at 1:37


















0














The norm of $T$ is the supremum of $lvertlvert Tx rvertrvert$ over all unit vectors $x in E$ and this is at least the supremum of $lvertlvert Te_k rvertrvert$ over all elementary basis vectors $(e_k)_{k=1}^infty$.






share|cite|improve this answer





















  • Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
    – Jeff
    Dec 16 at 1:37
















0












0








0






The norm of $T$ is the supremum of $lvertlvert Tx rvertrvert$ over all unit vectors $x in E$ and this is at least the supremum of $lvertlvert Te_k rvertrvert$ over all elementary basis vectors $(e_k)_{k=1}^infty$.






share|cite|improve this answer












The norm of $T$ is the supremum of $lvertlvert Tx rvertrvert$ over all unit vectors $x in E$ and this is at least the supremum of $lvertlvert Te_k rvertrvert$ over all elementary basis vectors $(e_k)_{k=1}^infty$.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Dec 9 at 3:38









Riley

1395




1395












  • Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
    – Jeff
    Dec 16 at 1:37




















  • Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
    – Jeff
    Dec 16 at 1:37


















Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
– Jeff
Dec 16 at 1:37






Your comment give me the answer, because if it suffices for all $xin E$ then it suffices for a basis vector $e_k$, for every $kin mathbb{N}$, then it follows the contradiction I was looking for.
– Jeff
Dec 16 at 1:37




















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