$operatorname{Ker}(B) subset operatorname{Ker(A)}$ if and only if ${left| {Ax} right|_X} leqslant alpha...












0














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.










share|cite|improve this question




















  • 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


















0














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.










share|cite|improve this question




















  • 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
















0












0








0







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.










share|cite|improve this question















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






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 25 at 20:55









Bernard

118k638111




118k638111










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
















  • 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












2 Answers
2






active

oldest

votes


















1














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).






share|cite|improve this answer





















  • 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





















1














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.






share|cite|improve this answer























  • 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











Your Answer





StackExchange.ifUsing("editor", function () {
return StackExchange.using("mathjaxEditing", function () {
StackExchange.MarkdownEditor.creationCallbacks.add(function (editor, postfix) {
StackExchange.mathjaxEditing.prepareWmdForMathJax(editor, postfix, [["$", "$"], ["\\(","\\)"]]);
});
});
}, "mathjax-editing");

StackExchange.ready(function() {
var channelOptions = {
tags: "".split(" "),
id: "69"
};
initTagRenderer("".split(" "), "".split(" "), channelOptions);

StackExchange.using("externalEditor", function() {
// Have to fire editor after snippets, if snippets enabled
if (StackExchange.settings.snippets.snippetsEnabled) {
StackExchange.using("snippets", function() {
createEditor();
});
}
else {
createEditor();
}
});

function createEditor() {
StackExchange.prepareEditor({
heartbeatType: 'answer',
autoActivateHeartbeat: false,
convertImagesToLinks: true,
noModals: true,
showLowRepImageUploadWarning: true,
reputationToPostImages: 10,
bindNavPrevention: true,
postfix: "",
imageUploader: {
brandingHtml: "Powered by u003ca class="icon-imgur-white" href="https://imgur.com/"u003eu003c/au003e",
contentPolicyHtml: "User contributions licensed under u003ca href="https://creativecommons.org/licenses/by-sa/3.0/"u003ecc by-sa 3.0 with attribution requiredu003c/au003e u003ca href="https://stackoverflow.com/legal/content-policy"u003e(content policy)u003c/au003e",
allowUrls: true
},
noCode: true, onDemand: true,
discardSelector: ".discard-answer"
,immediatelyShowMarkdownHelp:true
});


}
});














draft saved

draft discarded


















StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013363%2foperatornamekerb-subset-operatornamekera-if-and-only-if-left%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown

























2 Answers
2






active

oldest

votes








2 Answers
2






active

oldest

votes









active

oldest

votes






active

oldest

votes









1














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).






share|cite|improve this answer





















  • 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


















1














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).






share|cite|improve this answer





















  • 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
















1












1








1






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).






share|cite|improve this answer












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).







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










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




















  • 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













1














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.






share|cite|improve this answer























  • 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
















1














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.






share|cite|improve this answer























  • 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














1












1








1






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.






share|cite|improve this answer














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.







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Nov 26 at 11:40

























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


















  • 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


















draft saved

draft discarded




















































Thanks for contributing an answer to Mathematics Stack Exchange!


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


Use MathJax to format equations. MathJax reference.


To learn more, see our tips on writing great answers.





Some of your past answers have not been well-received, and you're in danger of being blocked from answering.


Please pay close attention to the following guidance:


  • Please be sure to answer the question. Provide details and share your research!

But avoid



  • Asking for help, clarification, or responding to other answers.

  • Making statements based on opinion; back them up with references or personal experience.


To learn more, see our tips on writing great answers.




draft saved


draft discarded














StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3013363%2foperatornamekerb-subset-operatornamekera-if-and-only-if-left%23new-answer', 'question_page');
}
);

Post as a guest















Required, but never shown





















































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown

































Required, but never shown














Required, but never shown












Required, but never shown







Required, but never shown







Popular posts from this blog

Probability when a professor distributes a quiz and homework assignment to a class of n students.

Aardman Animations

Are they similar matrix