A counterexample of Banach Steinhaus Theorem
$begingroup$
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
- $displaystyle sup_{nin mathbb{N}} || T_n || <infty$
$T$ is a bounded linear operator
- $displaystyle || T || leq liminf_{nto infty} ||T_n ||$
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
asked Dec 9 '18 at 3:04
JeffJeff
185
$endgroup$
add a comment |
$begingroup$
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
- $displaystyle sup_{nin mathbb{N}} || T_n || <infty$
$T$ is a bounded linear operator
- $displaystyle || T || leq liminf_{nto infty} ||T_n ||$
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
asked Dec 9 '18 at 3:04
JeffJeff
185
$endgroup$
$begingroup$
Please edit the question to include what you were reading this from.
$endgroup$
– Shaun
Dec 9 '18 at 3:11
$begingroup$
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?
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:44
$begingroup$
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.
$endgroup$
– Jeff
Dec 9 '18 at 3:59
1
$begingroup$
Isn't $k leq C$ for all $k$ a contradiction?
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 4:54
$begingroup$
it is, but I already solve this problem with your help, thank you for helping me.
$endgroup$
– Jeff
Dec 16 '18 at 1:36
add a comment |
$begingroup$
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
- $displaystyle sup_{nin mathbb{N}} || T_n || <infty$
$T$ is a bounded linear operator
- $displaystyle || T || leq liminf_{nto infty} ||T_n ||$
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
asked Dec 9 '18 at 3:04
JeffJeff
185
$endgroup$
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
- $displaystyle sup_{nin mathbb{N}} || T_n || <infty$
$T$ is a bounded linear operator
- $displaystyle || T || leq liminf_{nto infty} ||T_n ||$
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
asked Dec 9 '18 at 3:04
JeffJeff
185
asked Dec 9 '18 at 3:04
JeffJeff
185
edited Dec 9 '18 at 3:51
Jeff
asked Dec 9 '18 at 3:04
JeffJeff
185
asked Dec 9 '18 at 3:04
JeffJeff
185
asked Dec 9 '18 at 3:04
JeffJeff
185
185
$begingroup$
Please edit the question to include what you were reading this from.
$endgroup$
– Shaun
Dec 9 '18 at 3:11
$begingroup$
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?
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:44
$begingroup$
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.
$endgroup$
– Jeff
Dec 9 '18 at 3:59
1
$begingroup$
Isn't $k leq C$ for all $k$ a contradiction?
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 4:54
$begingroup$
it is, but I already solve this problem with your help, thank you for helping me.
$endgroup$
– Jeff
Dec 16 '18 at 1:36
add a comment |
$begingroup$
Please edit the question to include what you were reading this from.
$endgroup$
– Shaun
Dec 9 '18 at 3:11
$begingroup$
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?
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:44
$begingroup$
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.
$endgroup$
– Jeff
Dec 9 '18 at 3:59
1
$begingroup$
Isn't $k leq C$ for all $k$ a contradiction?
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 4:54
$begingroup$
it is, but I already solve this problem with your help, thank you for helping me.
$endgroup$
– Jeff
Dec 16 '18 at 1:36
$begingroup$
Please edit the question to include what you were reading this from.
$endgroup$
– Shaun
Dec 9 '18 at 3:11
$begingroup$
Please edit the question to include what you were reading this from.
$endgroup$
– Shaun
Dec 9 '18 at 3:11
$begingroup$
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?
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:44
$begingroup$
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?
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:44
$begingroup$
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.
$endgroup$
– Jeff
Dec 9 '18 at 3:59
$begingroup$
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.
$endgroup$
– Jeff
Dec 9 '18 at 3:59
1
1
$begingroup$
Isn't $k leq C$ for all $k$ a contradiction?
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 4:54
$begingroup$
Isn't $k leq C$ for all $k$ a contradiction?
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 4:54
$begingroup$
it is, but I already solve this problem with your help, thank you for helping me.
$endgroup$
– Jeff
Dec 16 '18 at 1:36
$begingroup$
it is, but I already solve this problem with your help, thank you for helping me.
$endgroup$
– Jeff
Dec 16 '18 at 1:36
add a comment |
1 Answer
1
active
oldest
votes
$begingroup$
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 $(e_k)_{k=1}^infty$.
answered Dec 9 '18 at 3:38
RileyRiley
1625
$endgroup$
$begingroup$
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.
$endgroup$
– Jeff
Dec 16 '18 at 1:37
add a comment |
Your Answer
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1 Answer
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1 Answer
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$begingroup$
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 $(e_k)_{k=1}^infty$.
answered Dec 9 '18 at 3:38
RileyRiley
1625
$endgroup$
$begingroup$
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.
$endgroup$
– Jeff
Dec 16 '18 at 1:37
add a comment |
$begingroup$
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 $(e_k)_{k=1}^infty$.
answered Dec 9 '18 at 3:38
RileyRiley
1625
$endgroup$
$begingroup$
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.
$endgroup$
– Jeff
Dec 16 '18 at 1:37
add a comment |
$begingroup$
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 $(e_k)_{k=1}^infty$.
answered Dec 9 '18 at 3:38
RileyRiley
1625
$endgroup$
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 $(e_k)_{k=1}^infty$.
answered Dec 9 '18 at 3:38
RileyRiley
1625
answered Dec 9 '18 at 3:38
RileyRiley
1625
answered Dec 9 '18 at 3:38
RileyRiley
1625
answered Dec 9 '18 at 3:38
RileyRiley
1625
1625
$begingroup$
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.
$endgroup$
– Jeff
Dec 16 '18 at 1:37
add a comment |
$begingroup$
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.
$endgroup$
– Jeff
Dec 16 '18 at 1:37
$begingroup$
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.
$endgroup$
– Jeff
Dec 16 '18 at 1:37
$begingroup$
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.
$endgroup$
– Jeff
Dec 16 '18 at 1:37
add a comment |
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$begingroup$
Please edit the question to include what you were reading this from.
$endgroup$
– Shaun
Dec 9 '18 at 3:11
$begingroup$
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?
$endgroup$
– DanielWainfleet
Dec 9 '18 at 3:44
$begingroup$
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.
$endgroup$
– Jeff
Dec 9 '18 at 3:59
1
$begingroup$
Isn't $k leq C$ for all $k$ a contradiction?
$endgroup$
– Kavi Rama Murthy
Dec 9 '18 at 4:54
$begingroup$
it is, but I already solve this problem with your help, thank you for helping me.
$endgroup$
– Jeff
Dec 16 '18 at 1:36