$operatorname{Ker}(B) subset operatorname{Ker(A)}$ if and only if ${left| {Ax} right|_X} leqslant alpha...
Is this statement is correct:
Let $X$ be a Banach space, and let $A$ and $B$ two continuous operators on $X$ , do we have for some constant $alpha$ the following $${left| {Ax} right|_X} leqslant alpha {left| {Bx} right|_X} Leftrightarrow operatorname{Ker}(B) subset operatorname{Ker}(A)$$.
Thank you.
real-analysis functional-analysis operator-theory banach-spaces normed-spaces
edited Nov 25 at 20:55
Bernard
118k638111
asked Nov 25 at 20:42
Gustave
702211
|
show 1 more comment
Is this statement is correct:
Let $X$ be a Banach space, and let $A$ and $B$ two continuous operators on $X$ , do we have for some constant $alpha$ the following $${left| {Ax} right|_X} leqslant alpha {left| {Bx} right|_X} Leftrightarrow operatorname{Ker}(B) subset operatorname{Ker}(A)$$.
Thank you.
real-analysis functional-analysis operator-theory banach-spaces normed-spaces
edited Nov 25 at 20:55
Bernard
118k638111
asked Nov 25 at 20:42
Gustave
702211
1
I believe that the inclusion should be reversed: If $xin ker B$, then $Vert Ax Vert leq alpha Vert Bx Vert = 0,$ so $Ax = 0$ and $x in ker A$, i.e. $ker B subset ker A$.
– MisterRiemann
Nov 25 at 20:47
Thanks. I will edit the statement. is it a sufficient and necessary condition?
– Gustave
Nov 25 at 20:50
Consider two injective operators for the reverse statement.
– Bartosz Malman
Nov 25 at 21:01
@BartoszMalman so it is correct. What about that, is there any difference? thank you sir.math.stackexchange.com/questions/3013126/…
– Gustave
Nov 25 at 21:03
@Gustave I don't think the implication holds. If it did, that would mean for any two operators with the same kernel the quantities $|Ax|$ and $|Bx|$ are comparable uniformly for $x in X$. Such a thing certainly does not hold.
– Bartosz Malman
Nov 25 at 21:06
|
show 1 more comment
Is this statement is correct:
Let $X$ be a Banach space, and let $A$ and $B$ two continuous operators on $X$ , do we have for some constant $alpha$ the following $${left| {Ax} right|_X} leqslant alpha {left| {Bx} right|_X} Leftrightarrow operatorname{Ker}(B) subset operatorname{Ker}(A)$$.
Thank you.
real-analysis functional-analysis operator-theory banach-spaces normed-spaces
edited Nov 25 at 20:55
Bernard
118k638111
asked Nov 25 at 20:42
Gustave
702211
Is this statement is correct:
Let $X$ be a Banach space, and let $A$ and $B$ two continuous operators on $X$ , do we have for some constant $alpha$ the following $${left| {Ax} right|_X} leqslant alpha {left| {Bx} right|_X} Leftrightarrow operatorname{Ker}(B) subset operatorname{Ker}(A)$$.
Thank you.
real-analysis functional-analysis operator-theory banach-spaces normed-spaces
real-analysis functional-analysis operator-theory banach-spaces normed-spaces
edited Nov 25 at 20:55
Bernard
118k638111
asked Nov 25 at 20:42
Gustave
702211
edited Nov 25 at 20:55
Bernard
118k638111
asked Nov 25 at 20:42
Gustave
702211
edited Nov 25 at 20:55
Bernard
118k638111
edited Nov 25 at 20:55
Bernard
118k638111
edited Nov 25 at 20:55
Bernard
118k638111
118k638111
asked Nov 25 at 20:42
Gustave
702211
asked Nov 25 at 20:42
Gustave
702211
asked Nov 25 at 20:42
Gustave
702211
702211
1
I believe that the inclusion should be reversed: If $xin ker B$, then $Vert Ax Vert leq alpha Vert Bx Vert = 0,$ so $Ax = 0$ and $x in ker A$, i.e. $ker B subset ker A$.
– MisterRiemann
Nov 25 at 20:47
Thanks. I will edit the statement. is it a sufficient and necessary condition?
– Gustave
Nov 25 at 20:50
Consider two injective operators for the reverse statement.
– Bartosz Malman
Nov 25 at 21:01
@BartoszMalman so it is correct. What about that, is there any difference? thank you sir.math.stackexchange.com/questions/3013126/…
– Gustave
Nov 25 at 21:03
@Gustave I don't think the implication holds. If it did, that would mean for any two operators with the same kernel the quantities $|Ax|$ and $|Bx|$ are comparable uniformly for $x in X$. Such a thing certainly does not hold.
– Bartosz Malman
Nov 25 at 21:06
|
show 1 more comment
1
I believe that the inclusion should be reversed: If $xin ker B$, then $Vert Ax Vert leq alpha Vert Bx Vert = 0,$ so $Ax = 0$ and $x in ker A$, i.e. $ker B subset ker A$.
– MisterRiemann
Nov 25 at 20:47
Thanks. I will edit the statement. is it a sufficient and necessary condition?
– Gustave
Nov 25 at 20:50
Consider two injective operators for the reverse statement.
– Bartosz Malman
Nov 25 at 21:01
@BartoszMalman so it is correct. What about that, is there any difference? thank you sir.math.stackexchange.com/questions/3013126/…
– Gustave
Nov 25 at 21:03
@Gustave I don't think the implication holds. If it did, that would mean for any two operators with the same kernel the quantities $|Ax|$ and $|Bx|$ are comparable uniformly for $x in X$. Such a thing certainly does not hold.
– Bartosz Malman
Nov 25 at 21:06
1
1
I believe that the inclusion should be reversed: If $xin ker B$, then $Vert Ax Vert leq alpha Vert Bx Vert = 0,$ so $Ax = 0$ and $x in ker A$, i.e. $ker B subset ker A$.
– MisterRiemann
Nov 25 at 20:47
I believe that the inclusion should be reversed: If $xin ker B$, then $Vert Ax Vert leq alpha Vert Bx Vert = 0,$ so $Ax = 0$ and $x in ker A$, i.e. $ker B subset ker A$.
– MisterRiemann
Nov 25 at 20:47
Thanks. I will edit the statement. is it a sufficient and necessary condition?
– Gustave
Nov 25 at 20:50
Thanks. I will edit the statement. is it a sufficient and necessary condition?
– Gustave
Nov 25 at 20:50
Consider two injective operators for the reverse statement.
– Bartosz Malman
Nov 25 at 21:01
Consider two injective operators for the reverse statement.
– Bartosz Malman
Nov 25 at 21:01
@BartoszMalman so it is correct. What about that, is there any difference? thank you sir.math.stackexchange.com/questions/3013126/…
– Gustave
Nov 25 at 21:03
@BartoszMalman so it is correct. What about that, is there any difference? thank you sir.math.stackexchange.com/questions/3013126/…
– Gustave
Nov 25 at 21:03
@Gustave I don't think the implication holds. If it did, that would mean for any two operators with the same kernel the quantities $|Ax|$ and $|Bx|$ are comparable uniformly for $x in X$. Such a thing certainly does not hold.
– Bartosz Malman
Nov 25 at 21:06
@Gustave I don't think the implication holds. If it did, that would mean for any two operators with the same kernel the quantities $|Ax|$ and $|Bx|$ are comparable uniformly for $x in X$. Such a thing certainly does not hold.
– Bartosz Malman
Nov 25 at 21:06
|
show 1 more comment
2 Answers
2
active
oldest
votes
An example in the infinite-dimensional case can be constructed as follows: Let $A=Id$ and $B:Xto X$ compact and injective. Thus $ker A=ker B$. If there would be
$alpha>0$ such that $|Ax|=|x|le alpha |Bx|$ then the image of $B$ would be closed, thus $B$ would be bijective and continuously invertible, which is a contradiction.
An easy example is given by $X=l^2$ and
$$
Bx = (x_1,x_2/2,x_3/3,dots,x_n/n,dots).
$$
Here you could prove by hand that no such $alpha$ exists (without recurring to compactness).
answered Nov 26 at 9:56
daw
24k1544
Thank you sir. For the prove of sir gerw it is not detailes, can you explain me why does he take the $B^{-1}$, and how the inclusion of kernels works because I have some confusion here. Thanks
– Gustave
Nov 26 at 11:10
add a comment |
In finite dimensions, it should be true. Let us denote by $R$ the range of $B$ and by $S$ the range of $A$. Then, you can define the linear operator
$$A , B^{-1} : R to S.$$
Due to the assumptions on the kernels, this operator is well defined and linear. Thus, it is continuous, i.e., there is $alpha ge 0$ with
$$ |A , B^{-1} y | le alpha , |y| qquadforall y in R.$$
Now, setting $y = B , x$ yields the claim.
answered Nov 26 at 7:20
gerw
19k11133
Thanks. But How did you know that $B$ is invertible?
– Gustave
Nov 26 at 7:32
1
I do not use that $B$ is invertible. But for every $y in R$, there exist at least one $x$ with $B , x = y$ and the value of $A x$ is independent for all such $x$. Hence, you can define $(A B^{-1})(y) := A x$.
– gerw
Nov 26 at 7:40
Thank you sir..
– Gustave
Nov 26 at 7:41
@daw: Ups. I removed this part of my answer.
– gerw
Nov 26 at 11:40
add a comment |
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2 Answers
2
active
oldest
votes
2 Answers
2
active
oldest
votes
active
oldest
votes
active
oldest
votes
An example in the infinite-dimensional case can be constructed as follows: Let $A=Id$ and $B:Xto X$ compact and injective. Thus $ker A=ker B$. If there would be
$alpha>0$ such that $|Ax|=|x|le alpha |Bx|$ then the image of $B$ would be closed, thus $B$ would be bijective and continuously invertible, which is a contradiction.
An easy example is given by $X=l^2$ and
$$
Bx = (x_1,x_2/2,x_3/3,dots,x_n/n,dots).
$$
Here you could prove by hand that no such $alpha$ exists (without recurring to compactness).
answered Nov 26 at 9:56
daw
24k1544
Thank you sir. For the prove of sir gerw it is not detailes, can you explain me why does he take the $B^{-1}$, and how the inclusion of kernels works because I have some confusion here. Thanks
– Gustave
Nov 26 at 11:10
add a comment |
An example in the infinite-dimensional case can be constructed as follows: Let $A=Id$ and $B:Xto X$ compact and injective. Thus $ker A=ker B$. If there would be
$alpha>0$ such that $|Ax|=|x|le alpha |Bx|$ then the image of $B$ would be closed, thus $B$ would be bijective and continuously invertible, which is a contradiction.
An easy example is given by $X=l^2$ and
$$
Bx = (x_1,x_2/2,x_3/3,dots,x_n/n,dots).
$$
Here you could prove by hand that no such $alpha$ exists (without recurring to compactness).
answered Nov 26 at 9:56
daw
24k1544
Thank you sir. For the prove of sir gerw it is not detailes, can you explain me why does he take the $B^{-1}$, and how the inclusion of kernels works because I have some confusion here. Thanks
– Gustave
Nov 26 at 11:10
add a comment |
An example in the infinite-dimensional case can be constructed as follows: Let $A=Id$ and $B:Xto X$ compact and injective. Thus $ker A=ker B$. If there would be
$alpha>0$ such that $|Ax|=|x|le alpha |Bx|$ then the image of $B$ would be closed, thus $B$ would be bijective and continuously invertible, which is a contradiction.
An easy example is given by $X=l^2$ and
$$
Bx = (x_1,x_2/2,x_3/3,dots,x_n/n,dots).
$$
Here you could prove by hand that no such $alpha$ exists (without recurring to compactness).
answered Nov 26 at 9:56
daw
24k1544
An example in the infinite-dimensional case can be constructed as follows: Let $A=Id$ and $B:Xto X$ compact and injective. Thus $ker A=ker B$. If there would be
$alpha>0$ such that $|Ax|=|x|le alpha |Bx|$ then the image of $B$ would be closed, thus $B$ would be bijective and continuously invertible, which is a contradiction.
An easy example is given by $X=l^2$ and
$$
Bx = (x_1,x_2/2,x_3/3,dots,x_n/n,dots).
$$
Here you could prove by hand that no such $alpha$ exists (without recurring to compactness).
answered Nov 26 at 9:56
daw
24k1544
answered Nov 26 at 9:56
daw
24k1544
answered Nov 26 at 9:56
daw
24k1544
answered Nov 26 at 9:56
daw
24k1544
24k1544
Thank you sir. For the prove of sir gerw it is not detailes, can you explain me why does he take the $B^{-1}$, and how the inclusion of kernels works because I have some confusion here. Thanks
– Gustave
Nov 26 at 11:10
add a comment |
Thank you sir. For the prove of sir gerw it is not detailes, can you explain me why does he take the $B^{-1}$, and how the inclusion of kernels works because I have some confusion here. Thanks
– Gustave
Nov 26 at 11:10
Thank you sir. For the prove of sir gerw it is not detailes, can you explain me why does he take the $B^{-1}$, and how the inclusion of kernels works because I have some confusion here. Thanks
– Gustave
Nov 26 at 11:10
Thank you sir. For the prove of sir gerw it is not detailes, can you explain me why does he take the $B^{-1}$, and how the inclusion of kernels works because I have some confusion here. Thanks
– Gustave
Nov 26 at 11:10
add a comment |
In finite dimensions, it should be true. Let us denote by $R$ the range of $B$ and by $S$ the range of $A$. Then, you can define the linear operator
$$A , B^{-1} : R to S.$$
Due to the assumptions on the kernels, this operator is well defined and linear. Thus, it is continuous, i.e., there is $alpha ge 0$ with
$$ |A , B^{-1} y | le alpha , |y| qquadforall y in R.$$
Now, setting $y = B , x$ yields the claim.
answered Nov 26 at 7:20
gerw
19k11133
Thanks. But How did you know that $B$ is invertible?
– Gustave
Nov 26 at 7:32
1
I do not use that $B$ is invertible. But for every $y in R$, there exist at least one $x$ with $B , x = y$ and the value of $A x$ is independent for all such $x$. Hence, you can define $(A B^{-1})(y) := A x$.
– gerw
Nov 26 at 7:40
Thank you sir..
– Gustave
Nov 26 at 7:41
@daw: Ups. I removed this part of my answer.
– gerw
Nov 26 at 11:40
add a comment |
In finite dimensions, it should be true. Let us denote by $R$ the range of $B$ and by $S$ the range of $A$. Then, you can define the linear operator
$$A , B^{-1} : R to S.$$
Due to the assumptions on the kernels, this operator is well defined and linear. Thus, it is continuous, i.e., there is $alpha ge 0$ with
$$ |A , B^{-1} y | le alpha , |y| qquadforall y in R.$$
Now, setting $y = B , x$ yields the claim.
answered Nov 26 at 7:20
gerw
19k11133
Thanks. But How did you know that $B$ is invertible?
– Gustave
Nov 26 at 7:32
1
I do not use that $B$ is invertible. But for every $y in R$, there exist at least one $x$ with $B , x = y$ and the value of $A x$ is independent for all such $x$. Hence, you can define $(A B^{-1})(y) := A x$.
– gerw
Nov 26 at 7:40
Thank you sir..
– Gustave
Nov 26 at 7:41
@daw: Ups. I removed this part of my answer.
– gerw
Nov 26 at 11:40
add a comment |
In finite dimensions, it should be true. Let us denote by $R$ the range of $B$ and by $S$ the range of $A$. Then, you can define the linear operator
$$A , B^{-1} : R to S.$$
Due to the assumptions on the kernels, this operator is well defined and linear. Thus, it is continuous, i.e., there is $alpha ge 0$ with
$$ |A , B^{-1} y | le alpha , |y| qquadforall y in R.$$
Now, setting $y = B , x$ yields the claim.
answered Nov 26 at 7:20
gerw
19k11133
In finite dimensions, it should be true. Let us denote by $R$ the range of $B$ and by $S$ the range of $A$. Then, you can define the linear operator
$$A , B^{-1} : R to S.$$
Due to the assumptions on the kernels, this operator is well defined and linear. Thus, it is continuous, i.e., there is $alpha ge 0$ with
$$ |A , B^{-1} y | le alpha , |y| qquadforall y in R.$$
Now, setting $y = B , x$ yields the claim.
answered Nov 26 at 7:20
gerw
19k11133
edited Nov 26 at 11:40
answered Nov 26 at 7:20
gerw
19k11133
answered Nov 26 at 7:20
gerw
19k11133
answered Nov 26 at 7:20
gerw
19k11133
19k11133
Thanks. But How did you know that $B$ is invertible?
– Gustave
Nov 26 at 7:32
1
I do not use that $B$ is invertible. But for every $y in R$, there exist at least one $x$ with $B , x = y$ and the value of $A x$ is independent for all such $x$. Hence, you can define $(A B^{-1})(y) := A x$.
– gerw
Nov 26 at 7:40
Thank you sir..
– Gustave
Nov 26 at 7:41
@daw: Ups. I removed this part of my answer.
– gerw
Nov 26 at 11:40
add a comment |
Thanks. But How did you know that $B$ is invertible?
– Gustave
Nov 26 at 7:32
1
I do not use that $B$ is invertible. But for every $y in R$, there exist at least one $x$ with $B , x = y$ and the value of $A x$ is independent for all such $x$. Hence, you can define $(A B^{-1})(y) := A x$.
– gerw
Nov 26 at 7:40
Thank you sir..
– Gustave
Nov 26 at 7:41
@daw: Ups. I removed this part of my answer.
– gerw
Nov 26 at 11:40
Thanks. But How did you know that $B$ is invertible?
– Gustave
Nov 26 at 7:32
Thanks. But How did you know that $B$ is invertible?
– Gustave
Nov 26 at 7:32
1
1
I do not use that $B$ is invertible. But for every $y in R$, there exist at least one $x$ with $B , x = y$ and the value of $A x$ is independent for all such $x$. Hence, you can define $(A B^{-1})(y) := A x$.
– gerw
Nov 26 at 7:40
I do not use that $B$ is invertible. But for every $y in R$, there exist at least one $x$ with $B , x = y$ and the value of $A x$ is independent for all such $x$. Hence, you can define $(A B^{-1})(y) := A x$.
– gerw
Nov 26 at 7:40
Thank you sir..
– Gustave
Nov 26 at 7:41
Thank you sir..
– Gustave
Nov 26 at 7:41
@daw: Ups. I removed this part of my answer.
– gerw
Nov 26 at 11:40
@daw: Ups. I removed this part of my answer.
– gerw
Nov 26 at 11:40
add a comment |
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1
I believe that the inclusion should be reversed: If $xin ker B$, then $Vert Ax Vert leq alpha Vert Bx Vert = 0,$ so $Ax = 0$ and $x in ker A$, i.e. $ker B subset ker A$.
– MisterRiemann
Nov 25 at 20:47
Thanks. I will edit the statement. is it a sufficient and necessary condition?
– Gustave
Nov 25 at 20:50
Consider two injective operators for the reverse statement.
– Bartosz Malman
Nov 25 at 21:01
@BartoszMalman so it is correct. What about that, is there any difference? thank you sir.math.stackexchange.com/questions/3013126/…
– Gustave
Nov 25 at 21:03
@Gustave I don't think the implication holds. If it did, that would mean for any two operators with the same kernel the quantities $|Ax|$ and $|Bx|$ are comparable uniformly for $x in X$. Such a thing certainly does not hold.
– Bartosz Malman
Nov 25 at 21:06