If $F_Lambda(x)-F_Lambda(mu)$ is regular in $x$ and $mu$ for $Lambdato+infty$, then the divergent part of...
$begingroup$
While reading a physics paper I came across this claim:
If $F_Lambda(x)-F_Lambda(mu)$ is regular in $x$ and $mu$ for $Lambdato+infty$, then the divergent part of $F_Lambda(x)$ must be a constant, that is $x$-independent.
[I removed some unnecessary details.]
The author quickly dismisses this, saying that if it were not so, then $F_Lambda(x)-F_Lambda(mu)$ would still be divergent $forall xnemu$.
Now, I don't get how this could be proved in a more precise manner: this is my attempt.
Consider a function $F$ of two variables, $(x,Lambda)inmathbf{R}^2$.
Split the function $F$ in a regular part and a singular part (with respect to $Lambda$), say $F^{(r)}$ and $F^{(s)}$ respectively, and let $c(x)$ be $lim_{Lambdato+infty}F^{(r)}_Lambda(x)$.
Fix any $muinmathbf{R}$.
Then
begin{equation*}
begin{split}
lim_{Lambdato+infty}[F_Lambda(x)-F_Lambda(mu)]&=
lim_{Lambdato+infty}[F^{(r)}_Lambda(x)-F^{(r)}_Lambda(mu)+F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)]=\ &=
c(x)-c(mu)+lim_{Lambdato+infty}[F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)]
end{split}
end{equation*}
which means that $lim_{Lambdato+infty}[F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)]$ must exist and be finite.
Now, since in this limit $F^{(s)}_Lambda(x)$ and $F^{(s)}_Lambda(mu)$ must coincide, I guessed that
begin{equation*}
F^{(s)}_Lambda(x)=
F^{(s)}_Lambda(mu)+text{some other terms that are negligible in the limit},
end{equation*}
but then I tried taking $F^{(s)}_Lambda(x)=Lambda^2+log(Lambda+x)$ (restricting the domains of the variables as needed; this shouldn't cause loss of generality).
Surely it is divergent when $Lambdato+infty$, and
begin{equation*}
F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)=
logfrac{Lambda+x}{Lambda+mu}
end{equation*}
whose limit is zero, regardless of $x$ or $mu$... this seems to contradict the claim.
Have I done something wrong? Is the claim really true?
limits
asked Jan 16 at 14:08
yellowquarkyellowquark
661411
$endgroup$
add a comment |
$begingroup$
While reading a physics paper I came across this claim:
If $F_Lambda(x)-F_Lambda(mu)$ is regular in $x$ and $mu$ for $Lambdato+infty$, then the divergent part of $F_Lambda(x)$ must be a constant, that is $x$-independent.
[I removed some unnecessary details.]
The author quickly dismisses this, saying that if it were not so, then $F_Lambda(x)-F_Lambda(mu)$ would still be divergent $forall xnemu$.
Now, I don't get how this could be proved in a more precise manner: this is my attempt.
Consider a function $F$ of two variables, $(x,Lambda)inmathbf{R}^2$.
Split the function $F$ in a regular part and a singular part (with respect to $Lambda$), say $F^{(r)}$ and $F^{(s)}$ respectively, and let $c(x)$ be $lim_{Lambdato+infty}F^{(r)}_Lambda(x)$.
Fix any $muinmathbf{R}$.
Then
begin{equation*}
begin{split}
lim_{Lambdato+infty}[F_Lambda(x)-F_Lambda(mu)]&=
lim_{Lambdato+infty}[F^{(r)}_Lambda(x)-F^{(r)}_Lambda(mu)+F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)]=\ &=
c(x)-c(mu)+lim_{Lambdato+infty}[F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)]
end{split}
end{equation*}
which means that $lim_{Lambdato+infty}[F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)]$ must exist and be finite.
Now, since in this limit $F^{(s)}_Lambda(x)$ and $F^{(s)}_Lambda(mu)$ must coincide, I guessed that
begin{equation*}
F^{(s)}_Lambda(x)=
F^{(s)}_Lambda(mu)+text{some other terms that are negligible in the limit},
end{equation*}
but then I tried taking $F^{(s)}_Lambda(x)=Lambda^2+log(Lambda+x)$ (restricting the domains of the variables as needed; this shouldn't cause loss of generality).
Surely it is divergent when $Lambdato+infty$, and
begin{equation*}
F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)=
logfrac{Lambda+x}{Lambda+mu}
end{equation*}
whose limit is zero, regardless of $x$ or $mu$... this seems to contradict the claim.
Have I done something wrong? Is the claim really true?
limits
asked Jan 16 at 14:08
yellowquarkyellowquark
661411
$endgroup$
add a comment |
$begingroup$
While reading a physics paper I came across this claim:
If $F_Lambda(x)-F_Lambda(mu)$ is regular in $x$ and $mu$ for $Lambdato+infty$, then the divergent part of $F_Lambda(x)$ must be a constant, that is $x$-independent.
[I removed some unnecessary details.]
The author quickly dismisses this, saying that if it were not so, then $F_Lambda(x)-F_Lambda(mu)$ would still be divergent $forall xnemu$.
Now, I don't get how this could be proved in a more precise manner: this is my attempt.
Consider a function $F$ of two variables, $(x,Lambda)inmathbf{R}^2$.
Split the function $F$ in a regular part and a singular part (with respect to $Lambda$), say $F^{(r)}$ and $F^{(s)}$ respectively, and let $c(x)$ be $lim_{Lambdato+infty}F^{(r)}_Lambda(x)$.
Fix any $muinmathbf{R}$.
Then
begin{equation*}
begin{split}
lim_{Lambdato+infty}[F_Lambda(x)-F_Lambda(mu)]&=
lim_{Lambdato+infty}[F^{(r)}_Lambda(x)-F^{(r)}_Lambda(mu)+F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)]=\ &=
c(x)-c(mu)+lim_{Lambdato+infty}[F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)]
end{split}
end{equation*}
which means that $lim_{Lambdato+infty}[F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)]$ must exist and be finite.
Now, since in this limit $F^{(s)}_Lambda(x)$ and $F^{(s)}_Lambda(mu)$ must coincide, I guessed that
begin{equation*}
F^{(s)}_Lambda(x)=
F^{(s)}_Lambda(mu)+text{some other terms that are negligible in the limit},
end{equation*}
but then I tried taking $F^{(s)}_Lambda(x)=Lambda^2+log(Lambda+x)$ (restricting the domains of the variables as needed; this shouldn't cause loss of generality).
Surely it is divergent when $Lambdato+infty$, and
begin{equation*}
F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)=
logfrac{Lambda+x}{Lambda+mu}
end{equation*}
whose limit is zero, regardless of $x$ or $mu$... this seems to contradict the claim.
Have I done something wrong? Is the claim really true?
limits
asked Jan 16 at 14:08
yellowquarkyellowquark
661411
$endgroup$
While reading a physics paper I came across this claim:
If $F_Lambda(x)-F_Lambda(mu)$ is regular in $x$ and $mu$ for $Lambdato+infty$, then the divergent part of $F_Lambda(x)$ must be a constant, that is $x$-independent.
[I removed some unnecessary details.]
The author quickly dismisses this, saying that if it were not so, then $F_Lambda(x)-F_Lambda(mu)$ would still be divergent $forall xnemu$.
Now, I don't get how this could be proved in a more precise manner: this is my attempt.
Consider a function $F$ of two variables, $(x,Lambda)inmathbf{R}^2$.
Split the function $F$ in a regular part and a singular part (with respect to $Lambda$), say $F^{(r)}$ and $F^{(s)}$ respectively, and let $c(x)$ be $lim_{Lambdato+infty}F^{(r)}_Lambda(x)$.
Fix any $muinmathbf{R}$.
Then
begin{equation*}
begin{split}
lim_{Lambdato+infty}[F_Lambda(x)-F_Lambda(mu)]&=
lim_{Lambdato+infty}[F^{(r)}_Lambda(x)-F^{(r)}_Lambda(mu)+F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)]=\ &=
c(x)-c(mu)+lim_{Lambdato+infty}[F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)]
end{split}
end{equation*}
which means that $lim_{Lambdato+infty}[F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)]$ must exist and be finite.
Now, since in this limit $F^{(s)}_Lambda(x)$ and $F^{(s)}_Lambda(mu)$ must coincide, I guessed that
begin{equation*}
F^{(s)}_Lambda(x)=
F^{(s)}_Lambda(mu)+text{some other terms that are negligible in the limit},
end{equation*}
but then I tried taking $F^{(s)}_Lambda(x)=Lambda^2+log(Lambda+x)$ (restricting the domains of the variables as needed; this shouldn't cause loss of generality).
Surely it is divergent when $Lambdato+infty$, and
begin{equation*}
F^{(s)}_Lambda(x)-F^{(s)}_Lambda(mu)=
logfrac{Lambda+x}{Lambda+mu}
end{equation*}
whose limit is zero, regardless of $x$ or $mu$... this seems to contradict the claim.
Have I done something wrong? Is the claim really true?
limits
limits
asked Jan 16 at 14:08
yellowquarkyellowquark
661411
asked Jan 16 at 14:08
yellowquarkyellowquark
661411
asked Jan 16 at 14:08
yellowquarkyellowquark
661411
asked Jan 16 at 14:08
yellowquarkyellowquark
661411
asked Jan 16 at 14:08
yellowquarkyellowquark
661411
661411
add a comment |
add a comment |
0
active
oldest
votes
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
});
}
});
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%2f3075771%2fif-f-lambdax-f-lambda-mu-is-regular-in-x-and-mu-for-lambda-to-in%23new-answer', 'question_page');
}
);
Post as a guest
Required, but never shown
0
active
oldest
votes
0
active
oldest
votes
active
oldest
votes
active
oldest
votes
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%2f3075771%2fif-f-lambdax-f-lambda-mu-is-regular-in-x-and-mu-for-lambda-to-in%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
