Is a linear operator $M$ bounded given that $M^2$ is bounded?
$begingroup$
Proposition (?) Let $M$ be a linear map of a Banach space $X$ into itself. Then $M$ is bounded if and only if $M^2$ is bounded.
If $M$ is bounded, it is not difficult to show that $M^2$ is bounded.
What about the converse?
Aiming to apply Closed graph, I tried assuming that $x_n to x$ and $Mx_n to y$, and showing that $Mx = y$, but the best I was able to get from that is that, by the boundedness of $M^2$,
$$ M^2 x_n to M^2x, quad M^3x_n to M^2y, $$
which does not seem very fruitful.
I've also tried assuming that $M$ is unbounded in hope of reaching a contradiction, but to no avail.
I'm even not sure if the statement holds, but I can't even imagine what a counterexample might look like.
Could someone give a hint on how to proceed with the proof, or present a counterexample otherwise?
functional-analysis operator-theory
asked Dec 6 '18 at 20:03
MisterRiemannMisterRiemann
5,8451624
$endgroup$
add a comment |
$begingroup$
Proposition (?) Let $M$ be a linear map of a Banach space $X$ into itself. Then $M$ is bounded if and only if $M^2$ is bounded.
If $M$ is bounded, it is not difficult to show that $M^2$ is bounded.
What about the converse?
Aiming to apply Closed graph, I tried assuming that $x_n to x$ and $Mx_n to y$, and showing that $Mx = y$, but the best I was able to get from that is that, by the boundedness of $M^2$,
$$ M^2 x_n to M^2x, quad M^3x_n to M^2y, $$
which does not seem very fruitful.
I've also tried assuming that $M$ is unbounded in hope of reaching a contradiction, but to no avail.
I'm even not sure if the statement holds, but I can't even imagine what a counterexample might look like.
Could someone give a hint on how to proceed with the proof, or present a counterexample otherwise?
functional-analysis operator-theory
asked Dec 6 '18 at 20:03
MisterRiemannMisterRiemann
5,8451624
$endgroup$
add a comment |
$begingroup$
Proposition (?) Let $M$ be a linear map of a Banach space $X$ into itself. Then $M$ is bounded if and only if $M^2$ is bounded.
If $M$ is bounded, it is not difficult to show that $M^2$ is bounded.
What about the converse?
Aiming to apply Closed graph, I tried assuming that $x_n to x$ and $Mx_n to y$, and showing that $Mx = y$, but the best I was able to get from that is that, by the boundedness of $M^2$,
$$ M^2 x_n to M^2x, quad M^3x_n to M^2y, $$
which does not seem very fruitful.
I've also tried assuming that $M$ is unbounded in hope of reaching a contradiction, but to no avail.
I'm even not sure if the statement holds, but I can't even imagine what a counterexample might look like.
Could someone give a hint on how to proceed with the proof, or present a counterexample otherwise?
functional-analysis operator-theory
asked Dec 6 '18 at 20:03
MisterRiemannMisterRiemann
5,8451624
$endgroup$
Proposition (?) Let $M$ be a linear map of a Banach space $X$ into itself. Then $M$ is bounded if and only if $M^2$ is bounded.
If $M$ is bounded, it is not difficult to show that $M^2$ is bounded.
What about the converse?
Aiming to apply Closed graph, I tried assuming that $x_n to x$ and $Mx_n to y$, and showing that $Mx = y$, but the best I was able to get from that is that, by the boundedness of $M^2$,
$$ M^2 x_n to M^2x, quad M^3x_n to M^2y, $$
which does not seem very fruitful.
I've also tried assuming that $M$ is unbounded in hope of reaching a contradiction, but to no avail.
I'm even not sure if the statement holds, but I can't even imagine what a counterexample might look like.
Could someone give a hint on how to proceed with the proof, or present a counterexample otherwise?
functional-analysis operator-theory
functional-analysis operator-theory
asked Dec 6 '18 at 20:03
MisterRiemannMisterRiemann
5,8451624
asked Dec 6 '18 at 20:03
MisterRiemannMisterRiemann
5,8451624
asked Dec 6 '18 at 20:03
MisterRiemannMisterRiemann
5,8451624
asked Dec 6 '18 at 20:03
MisterRiemannMisterRiemann
5,8451624
asked Dec 6 '18 at 20:03
MisterRiemannMisterRiemann
5,8451624
5,8451624
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
Could one argue as follows:
Let ${x_i}_{iin I}$ be a normalised Hamel basis for $X$. Let ${y_n}_{n=1}^{infty}$ and ${z_n}_{n=1}^{infty}$ be two countable subsets of the basis which are disjoint and such that $y_nneq y_m$ unless $n=m$, and similarly for $z_n$. Define $M$ as follows:
$$My_n = nz_n,quad Mz_n = 0$$
and $Mx_i = 0$ if $x_ineq z_n$ or $x_ineq y_n$. Then $M$ is linear and unbounded however $M^2$ is $0$.
answered Dec 6 '18 at 20:33
Olof RubinOlof Rubin
1,131316
$endgroup$
$begingroup$
${z_n}$ is not the complete Hamel basis
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:44
$begingroup$
Yup. You are right.
$endgroup$
– Maksim
Dec 6 '18 at 21:46
$begingroup$
..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:47
add a comment |
$begingroup$
Hint: There are unbounded operators $M$ such that $M^2$ is the null operator.
answered Dec 6 '18 at 20:05
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
Thanks for a quick response. Could you provide an example of one such?
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:06
1
$begingroup$
@MisterRiemann You may look up here, Example 3.3.
$endgroup$
– A.Γ.
Dec 6 '18 at 20:42
$begingroup$
@A.Γ. Thank you.
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:44
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
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$begingroup$
Could one argue as follows:
Let ${x_i}_{iin I}$ be a normalised Hamel basis for $X$. Let ${y_n}_{n=1}^{infty}$ and ${z_n}_{n=1}^{infty}$ be two countable subsets of the basis which are disjoint and such that $y_nneq y_m$ unless $n=m$, and similarly for $z_n$. Define $M$ as follows:
$$My_n = nz_n,quad Mz_n = 0$$
and $Mx_i = 0$ if $x_ineq z_n$ or $x_ineq y_n$. Then $M$ is linear and unbounded however $M^2$ is $0$.
answered Dec 6 '18 at 20:33
Olof RubinOlof Rubin
1,131316
$endgroup$
$begingroup$
${z_n}$ is not the complete Hamel basis
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:44
$begingroup$
Yup. You are right.
$endgroup$
– Maksim
Dec 6 '18 at 21:46
$begingroup$
..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:47
add a comment |
$begingroup$
Could one argue as follows:
Let ${x_i}_{iin I}$ be a normalised Hamel basis for $X$. Let ${y_n}_{n=1}^{infty}$ and ${z_n}_{n=1}^{infty}$ be two countable subsets of the basis which are disjoint and such that $y_nneq y_m$ unless $n=m$, and similarly for $z_n$. Define $M$ as follows:
$$My_n = nz_n,quad Mz_n = 0$$
and $Mx_i = 0$ if $x_ineq z_n$ or $x_ineq y_n$. Then $M$ is linear and unbounded however $M^2$ is $0$.
answered Dec 6 '18 at 20:33
Olof RubinOlof Rubin
1,131316
$endgroup$
$begingroup$
${z_n}$ is not the complete Hamel basis
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:44
$begingroup$
Yup. You are right.
$endgroup$
– Maksim
Dec 6 '18 at 21:46
$begingroup$
..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:47
add a comment |
$begingroup$
Could one argue as follows:
Let ${x_i}_{iin I}$ be a normalised Hamel basis for $X$. Let ${y_n}_{n=1}^{infty}$ and ${z_n}_{n=1}^{infty}$ be two countable subsets of the basis which are disjoint and such that $y_nneq y_m$ unless $n=m$, and similarly for $z_n$. Define $M$ as follows:
$$My_n = nz_n,quad Mz_n = 0$$
and $Mx_i = 0$ if $x_ineq z_n$ or $x_ineq y_n$. Then $M$ is linear and unbounded however $M^2$ is $0$.
answered Dec 6 '18 at 20:33
Olof RubinOlof Rubin
1,131316
$endgroup$
Could one argue as follows:
Let ${x_i}_{iin I}$ be a normalised Hamel basis for $X$. Let ${y_n}_{n=1}^{infty}$ and ${z_n}_{n=1}^{infty}$ be two countable subsets of the basis which are disjoint and such that $y_nneq y_m$ unless $n=m$, and similarly for $z_n$. Define $M$ as follows:
$$My_n = nz_n,quad Mz_n = 0$$
and $Mx_i = 0$ if $x_ineq z_n$ or $x_ineq y_n$. Then $M$ is linear and unbounded however $M^2$ is $0$.
answered Dec 6 '18 at 20:33
Olof RubinOlof Rubin
1,131316
edited Dec 6 '18 at 20:44
answered Dec 6 '18 at 20:33
Olof RubinOlof Rubin
1,131316
answered Dec 6 '18 at 20:33
Olof RubinOlof Rubin
1,131316
answered Dec 6 '18 at 20:33
Olof RubinOlof Rubin
1,131316
1,131316
$begingroup$
${z_n}$ is not the complete Hamel basis
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:44
$begingroup$
Yup. You are right.
$endgroup$
– Maksim
Dec 6 '18 at 21:46
$begingroup$
..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:47
add a comment |
$begingroup$
${z_n}$ is not the complete Hamel basis
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:44
$begingroup$
Yup. You are right.
$endgroup$
– Maksim
Dec 6 '18 at 21:46
$begingroup$
..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:47
$begingroup$
${z_n}$ is not the complete Hamel basis
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:44
$begingroup$
${z_n}$ is not the complete Hamel basis
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:44
$begingroup$
Yup. You are right.
$endgroup$
– Maksim
Dec 6 '18 at 21:46
$begingroup$
Yup. You are right.
$endgroup$
– Maksim
Dec 6 '18 at 21:46
$begingroup$
..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:47
$begingroup$
..unless $X$ is finite dimensional but then I guess $M$ being linear is continuous by default :)
$endgroup$
– Olof Rubin
Dec 6 '18 at 21:47
add a comment |
$begingroup$
Hint: There are unbounded operators $M$ such that $M^2$ is the null operator.
answered Dec 6 '18 at 20:05
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
Thanks for a quick response. Could you provide an example of one such?
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:06
1
$begingroup$
@MisterRiemann You may look up here, Example 3.3.
$endgroup$
– A.Γ.
Dec 6 '18 at 20:42
$begingroup$
@A.Γ. Thank you.
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:44
add a comment |
$begingroup$
Hint: There are unbounded operators $M$ such that $M^2$ is the null operator.
answered Dec 6 '18 at 20:05
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
$begingroup$
Thanks for a quick response. Could you provide an example of one such?
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:06
1
$begingroup$
@MisterRiemann You may look up here, Example 3.3.
$endgroup$
– A.Γ.
Dec 6 '18 at 20:42
$begingroup$
@A.Γ. Thank you.
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:44
add a comment |
$begingroup$
Hint: There are unbounded operators $M$ such that $M^2$ is the null operator.
answered Dec 6 '18 at 20:05
José Carlos SantosJosé Carlos Santos
155k22124227
$endgroup$
Hint: There are unbounded operators $M$ such that $M^2$ is the null operator.
answered Dec 6 '18 at 20:05
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 6 '18 at 20:05
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 6 '18 at 20:05
José Carlos SantosJosé Carlos Santos
155k22124227
answered Dec 6 '18 at 20:05
José Carlos SantosJosé Carlos Santos
155k22124227
155k22124227
$begingroup$
Thanks for a quick response. Could you provide an example of one such?
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:06
1
$begingroup$
@MisterRiemann You may look up here, Example 3.3.
$endgroup$
– A.Γ.
Dec 6 '18 at 20:42
$begingroup$
@A.Γ. Thank you.
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:44
add a comment |
$begingroup$
Thanks for a quick response. Could you provide an example of one such?
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:06
1
$begingroup$
@MisterRiemann You may look up here, Example 3.3.
$endgroup$
– A.Γ.
Dec 6 '18 at 20:42
$begingroup$
@A.Γ. Thank you.
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:44
$begingroup$
Thanks for a quick response. Could you provide an example of one such?
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:06
$begingroup$
Thanks for a quick response. Could you provide an example of one such?
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:06
1
1
$begingroup$
@MisterRiemann You may look up here, Example 3.3.
$endgroup$
– A.Γ.
Dec 6 '18 at 20:42
$begingroup$
@MisterRiemann You may look up here, Example 3.3.
$endgroup$
– A.Γ.
Dec 6 '18 at 20:42
$begingroup$
@A.Γ. Thank you.
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:44
$begingroup$
@A.Γ. Thank you.
$endgroup$
– MisterRiemann
Dec 6 '18 at 20:44
add a comment |
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