Mixing
Bound for Double Sum.
$begingroup$
For $i = 1,cdots,n$ and $k=1,cdots,m$, $a_{ij} > 0$ and $b_{ij} geq 0$. Do we have the following bound,
$$ sum_{i=1}^n sum_{j=1}^m a_{ij} b_{ij} leq Big( max_i sum_j a_{ij} Big) Big( max_j sum_i b_{ij} Big) ?$$
inequality summation
asked Feb 2 at 18:42
jwyaojwyao
16311
$endgroup$
add a comment |
$begingroup$
For $i = 1,cdots,n$ and $k=1,cdots,m$, $a_{ij} > 0$ and $b_{ij} geq 0$. Do we have the following bound,
$$ sum_{i=1}^n sum_{j=1}^m a_{ij} b_{ij} leq Big( max_i sum_j a_{ij} Big) Big( max_j sum_i b_{ij} Big) ?$$
inequality summation
asked Feb 2 at 18:42
jwyaojwyao
16311
$endgroup$
add a comment |
$begingroup$
For $i = 1,cdots,n$ and $k=1,cdots,m$, $a_{ij} > 0$ and $b_{ij} geq 0$. Do we have the following bound,
$$ sum_{i=1}^n sum_{j=1}^m a_{ij} b_{ij} leq Big( max_i sum_j a_{ij} Big) Big( max_j sum_i b_{ij} Big) ?$$
inequality summation
asked Feb 2 at 18:42
jwyaojwyao
16311
$endgroup$
For $i = 1,cdots,n$ and $k=1,cdots,m$, $a_{ij} > 0$ and $b_{ij} geq 0$. Do we have the following bound,
$$ sum_{i=1}^n sum_{j=1}^m a_{ij} b_{ij} leq Big( max_i sum_j a_{ij} Big) Big( max_j sum_i b_{ij} Big) ?$$
inequality summation
inequality summation
asked Feb 2 at 18:42
jwyaojwyao
16311
asked Feb 2 at 18:42
jwyaojwyao
16311
edited Feb 2 at 18:52
jwyao
asked Feb 2 at 18:42
jwyaojwyao
16311
asked Feb 2 at 18:42
jwyaojwyao
16311
asked Feb 2 at 18:42
jwyaojwyao
16311
16311
add a comment |
add a comment |
2 Answers
2
active
oldest
votes
$begingroup$
No. For $n=m=2$ and
$$
A = B =
begin{pmatrix}
1 & x \
x & 1
end{pmatrix}
quad
$$
the left-hand side is equal to $2 + 2x^2$, and the right-hand side is $(1+x)^2$, so that the estimate does not hold for small positive $x$.
answered Feb 2 at 18:50
Martin RMartin R
31k33561
$endgroup$
$begingroup$
Thank you! What if $a_{ij}$ is strictly positive.
$endgroup$
– jwyao
Feb 2 at 18:53
$begingroup$
You changed the question after I posted the answer. But you can simply replace $0$ by a small positive number.
$endgroup$
– Martin R
Feb 2 at 18:55
$begingroup$
Makes no difference, @jwyao. Obviously if $a_{12}$ and $a_{21}$ are small enough, like $0.001$, then the left-hand side will be close enough to $2$, and the right-hand side will be close enough to $1$, so that your inequality is still violated.
$endgroup$
– TonyK
Feb 2 at 18:55
$begingroup$
@MartinR I made up the question from a problem I am studying. I tried to make it general but obviously I made a wrong generalization.
$endgroup$
– jwyao
Feb 2 at 18:56
add a comment |
$begingroup$
No, since $a_{ij}=b_{ij}=delta_{ij}+epsilon$ for small $epsilon>0$ is an obvious counterexample when $n=m$. (The LHS tends to $n$ while the RHS tends to $1$ as $epsilon to 0$.) To make your statement true, it should be
$$
sum_{i,j} a_{ij}b_{ij} le color{red}{sqrt{nm}} cdotmax_isum_j a_{ij} max_j sum_i b_{ij}.
$$
answered Feb 2 at 18:53
SongSong
18.6k21651
$endgroup$
$begingroup$
What is $n$ here? Is it the dimension of matrix?
$endgroup$
– jwyao
Feb 2 at 18:54
$begingroup$
@jwyao I thought $A,B$ were square matrices...
$endgroup$
– Song
Feb 2 at 18:55
add a comment |
Your Answer
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
});
}
});
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097628%2fbound-for-double-sum%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
$begingroup$
No. For $n=m=2$ and
$$
A = B =
begin{pmatrix}
1 & x \
x & 1
end{pmatrix}
quad
$$
the left-hand side is equal to $2 + 2x^2$, and the right-hand side is $(1+x)^2$, so that the estimate does not hold for small positive $x$.
answered Feb 2 at 18:50
Martin RMartin R
31k33561
$endgroup$
$begingroup$
Thank you! What if $a_{ij}$ is strictly positive.
$endgroup$
– jwyao
Feb 2 at 18:53
$begingroup$
You changed the question after I posted the answer. But you can simply replace $0$ by a small positive number.
$endgroup$
– Martin R
Feb 2 at 18:55
$begingroup$
Makes no difference, @jwyao. Obviously if $a_{12}$ and $a_{21}$ are small enough, like $0.001$, then the left-hand side will be close enough to $2$, and the right-hand side will be close enough to $1$, so that your inequality is still violated.
$endgroup$
– TonyK
Feb 2 at 18:55
$begingroup$
@MartinR I made up the question from a problem I am studying. I tried to make it general but obviously I made a wrong generalization.
$endgroup$
– jwyao
Feb 2 at 18:56
add a comment |
$begingroup$
No. For $n=m=2$ and
$$
A = B =
begin{pmatrix}
1 & x \
x & 1
end{pmatrix}
quad
$$
the left-hand side is equal to $2 + 2x^2$, and the right-hand side is $(1+x)^2$, so that the estimate does not hold for small positive $x$.
answered Feb 2 at 18:50
Martin RMartin R
31k33561
$endgroup$
$begingroup$
Thank you! What if $a_{ij}$ is strictly positive.
$endgroup$
– jwyao
Feb 2 at 18:53
$begingroup$
You changed the question after I posted the answer. But you can simply replace $0$ by a small positive number.
$endgroup$
– Martin R
Feb 2 at 18:55
$begingroup$
Makes no difference, @jwyao. Obviously if $a_{12}$ and $a_{21}$ are small enough, like $0.001$, then the left-hand side will be close enough to $2$, and the right-hand side will be close enough to $1$, so that your inequality is still violated.
$endgroup$
– TonyK
Feb 2 at 18:55
$begingroup$
@MartinR I made up the question from a problem I am studying. I tried to make it general but obviously I made a wrong generalization.
$endgroup$
– jwyao
Feb 2 at 18:56
add a comment |
$begingroup$
No. For $n=m=2$ and
$$
A = B =
begin{pmatrix}
1 & x \
x & 1
end{pmatrix}
quad
$$
the left-hand side is equal to $2 + 2x^2$, and the right-hand side is $(1+x)^2$, so that the estimate does not hold for small positive $x$.
answered Feb 2 at 18:50
Martin RMartin R
31k33561
$endgroup$
No. For $n=m=2$ and
$$
A = B =
begin{pmatrix}
1 & x \
x & 1
end{pmatrix}
quad
$$
the left-hand side is equal to $2 + 2x^2$, and the right-hand side is $(1+x)^2$, so that the estimate does not hold for small positive $x$.
answered Feb 2 at 18:50
Martin RMartin R
31k33561
edited Feb 2 at 18:56
answered Feb 2 at 18:50
Martin RMartin R
31k33561
answered Feb 2 at 18:50
Martin RMartin R
31k33561
answered Feb 2 at 18:50
Martin RMartin R
31k33561
31k33561
$begingroup$
Thank you! What if $a_{ij}$ is strictly positive.
$endgroup$
– jwyao
Feb 2 at 18:53
$begingroup$
You changed the question after I posted the answer. But you can simply replace $0$ by a small positive number.
$endgroup$
– Martin R
Feb 2 at 18:55
$begingroup$
Makes no difference, @jwyao. Obviously if $a_{12}$ and $a_{21}$ are small enough, like $0.001$, then the left-hand side will be close enough to $2$, and the right-hand side will be close enough to $1$, so that your inequality is still violated.
$endgroup$
– TonyK
Feb 2 at 18:55
$begingroup$
@MartinR I made up the question from a problem I am studying. I tried to make it general but obviously I made a wrong generalization.
$endgroup$
– jwyao
Feb 2 at 18:56
add a comment |
$begingroup$
Thank you! What if $a_{ij}$ is strictly positive.
$endgroup$
– jwyao
Feb 2 at 18:53
$begingroup$
You changed the question after I posted the answer. But you can simply replace $0$ by a small positive number.
$endgroup$
– Martin R
Feb 2 at 18:55
$begingroup$
Makes no difference, @jwyao. Obviously if $a_{12}$ and $a_{21}$ are small enough, like $0.001$, then the left-hand side will be close enough to $2$, and the right-hand side will be close enough to $1$, so that your inequality is still violated.
$endgroup$
– TonyK
Feb 2 at 18:55
$begingroup$
@MartinR I made up the question from a problem I am studying. I tried to make it general but obviously I made a wrong generalization.
$endgroup$
– jwyao
Feb 2 at 18:56
$begingroup$
Thank you! What if $a_{ij}$ is strictly positive.
$endgroup$
– jwyao
Feb 2 at 18:53
$begingroup$
Thank you! What if $a_{ij}$ is strictly positive.
$endgroup$
– jwyao
Feb 2 at 18:53
$begingroup$
You changed the question after I posted the answer. But you can simply replace $0$ by a small positive number.
$endgroup$
– Martin R
Feb 2 at 18:55
$begingroup$
You changed the question after I posted the answer. But you can simply replace $0$ by a small positive number.
$endgroup$
– Martin R
Feb 2 at 18:55
$begingroup$
Makes no difference, @jwyao. Obviously if $a_{12}$ and $a_{21}$ are small enough, like $0.001$, then the left-hand side will be close enough to $2$, and the right-hand side will be close enough to $1$, so that your inequality is still violated.
$endgroup$
– TonyK
Feb 2 at 18:55
$begingroup$
Makes no difference, @jwyao. Obviously if $a_{12}$ and $a_{21}$ are small enough, like $0.001$, then the left-hand side will be close enough to $2$, and the right-hand side will be close enough to $1$, so that your inequality is still violated.
$endgroup$
– TonyK
Feb 2 at 18:55
$begingroup$
@MartinR I made up the question from a problem I am studying. I tried to make it general but obviously I made a wrong generalization.
$endgroup$
– jwyao
Feb 2 at 18:56
$begingroup$
@MartinR I made up the question from a problem I am studying. I tried to make it general but obviously I made a wrong generalization.
$endgroup$
– jwyao
Feb 2 at 18:56
add a comment |
$begingroup$
No, since $a_{ij}=b_{ij}=delta_{ij}+epsilon$ for small $epsilon>0$ is an obvious counterexample when $n=m$. (The LHS tends to $n$ while the RHS tends to $1$ as $epsilon to 0$.) To make your statement true, it should be
$$
sum_{i,j} a_{ij}b_{ij} le color{red}{sqrt{nm}} cdotmax_isum_j a_{ij} max_j sum_i b_{ij}.
$$
answered Feb 2 at 18:53
SongSong
18.6k21651
$endgroup$
$begingroup$
What is $n$ here? Is it the dimension of matrix?
$endgroup$
– jwyao
Feb 2 at 18:54
$begingroup$
@jwyao I thought $A,B$ were square matrices...
$endgroup$
– Song
Feb 2 at 18:55
add a comment |
$begingroup$
No, since $a_{ij}=b_{ij}=delta_{ij}+epsilon$ for small $epsilon>0$ is an obvious counterexample when $n=m$. (The LHS tends to $n$ while the RHS tends to $1$ as $epsilon to 0$.) To make your statement true, it should be
$$
sum_{i,j} a_{ij}b_{ij} le color{red}{sqrt{nm}} cdotmax_isum_j a_{ij} max_j sum_i b_{ij}.
$$
answered Feb 2 at 18:53
SongSong
18.6k21651
$endgroup$
$begingroup$
What is $n$ here? Is it the dimension of matrix?
$endgroup$
– jwyao
Feb 2 at 18:54
$begingroup$
@jwyao I thought $A,B$ were square matrices...
$endgroup$
– Song
Feb 2 at 18:55
add a comment |
$begingroup$
No, since $a_{ij}=b_{ij}=delta_{ij}+epsilon$ for small $epsilon>0$ is an obvious counterexample when $n=m$. (The LHS tends to $n$ while the RHS tends to $1$ as $epsilon to 0$.) To make your statement true, it should be
$$
sum_{i,j} a_{ij}b_{ij} le color{red}{sqrt{nm}} cdotmax_isum_j a_{ij} max_j sum_i b_{ij}.
$$
answered Feb 2 at 18:53
SongSong
18.6k21651
$endgroup$
No, since $a_{ij}=b_{ij}=delta_{ij}+epsilon$ for small $epsilon>0$ is an obvious counterexample when $n=m$. (The LHS tends to $n$ while the RHS tends to $1$ as $epsilon to 0$.) To make your statement true, it should be
$$
sum_{i,j} a_{ij}b_{ij} le color{red}{sqrt{nm}} cdotmax_isum_j a_{ij} max_j sum_i b_{ij}.
$$
answered Feb 2 at 18:53
SongSong
18.6k21651
edited Feb 2 at 19:00
answered Feb 2 at 18:53
SongSong
18.6k21651
answered Feb 2 at 18:53
SongSong
18.6k21651
answered Feb 2 at 18:53
SongSong
18.6k21651
18.6k21651
$begingroup$
What is $n$ here? Is it the dimension of matrix?
$endgroup$
– jwyao
Feb 2 at 18:54
$begingroup$
@jwyao I thought $A,B$ were square matrices...
$endgroup$
– Song
Feb 2 at 18:55
add a comment |
$begingroup$
What is $n$ here? Is it the dimension of matrix?
$endgroup$
– jwyao
Feb 2 at 18:54
$begingroup$
@jwyao I thought $A,B$ were square matrices...
$endgroup$
– Song
Feb 2 at 18:55
$begingroup$
What is $n$ here? Is it the dimension of matrix?
$endgroup$
– jwyao
Feb 2 at 18:54
$begingroup$
What is $n$ here? Is it the dimension of matrix?
$endgroup$
– jwyao
Feb 2 at 18:54
$begingroup$
@jwyao I thought $A,B$ were square matrices...
$endgroup$
– Song
Feb 2 at 18:55
$begingroup$
@jwyao I thought $A,B$ were square matrices...
$endgroup$
– Song
Feb 2 at 18:55
add a comment |
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.
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
StackExchange.ready(
function () {
StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3097628%2fbound-for-double-sum%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Post as a guest
Required, but never shown
Sign up or log in
StackExchange.ready(function () {
StackExchange.helpers.onClickDraftSave('#login-link');
});
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
Sign up using Google
Sign up using Facebook
Sign up using Email and Password
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
