How can prove this inequality $|tau|^{2gamma} leq c_5(gamma) frac{1+|tau|}{1+|tau|^{1-2gamma}}$?
I'm reading a demonstration that uses this following inequality. For a fixed $gamma<1/4$, exists a $c_5(gamma)$, such that
$|tau|^{2gamma} leq c_5(gamma) frac{1+|tau|}{1+|tau|^{1-2gamma}}, forall tau in mathbb{R}$.
I tried to deduce that using the fact that $0<frac{1}{1+|tau|}leq1$, but i couldn't get in anywhere.
real-analysis abstract-algebra
asked Nov 23 at 19:53
João Paulo Andrade
315
add a comment |
I'm reading a demonstration that uses this following inequality. For a fixed $gamma<1/4$, exists a $c_5(gamma)$, such that
$|tau|^{2gamma} leq c_5(gamma) frac{1+|tau|}{1+|tau|^{1-2gamma}}, forall tau in mathbb{R}$.
I tried to deduce that using the fact that $0<frac{1}{1+|tau|}leq1$, but i couldn't get in anywhere.
real-analysis abstract-algebra
asked Nov 23 at 19:53
João Paulo Andrade
315
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Clement C.
Nov 24 at 22:02
add a comment |
I'm reading a demonstration that uses this following inequality. For a fixed $gamma<1/4$, exists a $c_5(gamma)$, such that
$|tau|^{2gamma} leq c_5(gamma) frac{1+|tau|}{1+|tau|^{1-2gamma}}, forall tau in mathbb{R}$.
I tried to deduce that using the fact that $0<frac{1}{1+|tau|}leq1$, but i couldn't get in anywhere.
real-analysis abstract-algebra
asked Nov 23 at 19:53
João Paulo Andrade
315
I'm reading a demonstration that uses this following inequality. For a fixed $gamma<1/4$, exists a $c_5(gamma)$, such that
$|tau|^{2gamma} leq c_5(gamma) frac{1+|tau|}{1+|tau|^{1-2gamma}}, forall tau in mathbb{R}$.
I tried to deduce that using the fact that $0<frac{1}{1+|tau|}leq1$, but i couldn't get in anywhere.
real-analysis abstract-algebra
real-analysis abstract-algebra
asked Nov 23 at 19:53
João Paulo Andrade
315
asked Nov 23 at 19:53
João Paulo Andrade
315
asked Nov 23 at 19:53
João Paulo Andrade
315
asked Nov 23 at 19:53
João Paulo Andrade
315
asked Nov 23 at 19:53
João Paulo Andrade
315
315
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Clement C.
Nov 24 at 22:02
add a comment |
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Clement C.
Nov 24 at 22:02
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Clement C.
Nov 24 at 22:02
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Clement C.
Nov 24 at 22:02
add a comment |
2 Answers
2
active
oldest
votes
Reorganizing, this is equivalent to proving that the function $fcolonmathbb{R}tomathbb{R}$ given by
$$
f(tau) = frac{|tau|+|tau|^{2gamma}} {|tau|+1}
$$
is bounded. Note that it is continuous, and $lim_{+infty} f = lim_{-infty} f = 1$ (using the fact that $2gamma <1$); therefore, it is bounded.
answered Nov 23 at 19:57
Clement C.
49.6k33886
(also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
– Clement C.
Nov 23 at 19:59
Actually it will not hold for $gamma < 1/2$, the reason why it holds is because $2gamma<1-2gamma$ when $gamma < 1/4$. I figured this, but it wasn't suffice to prove that. But thank you so much for help me.
– João Paulo Andrade
Nov 23 at 22:38
@JoãoPauloAndrade Glad it helped, but -- it does hold for all $gamma leq 1/2$ (the case $1/2$ is proven similarly, but the limit of $f$ at $infty$ is $2$, not 1).
– Clement C.
Nov 23 at 22:40
For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
– Clement C.
Nov 23 at 22:46
add a comment |
Let $|tau| = x geq 0$ and Consider:
$$f(x) = dfrac{x^{2gamma}+x}{1+x}$$
and prove this has a maximum by taking derivative.
answered Nov 23 at 19:58
dezdichado
6,1781929
add a comment |
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%2f3010766%2fhow-can-prove-this-inequality-tau2-gamma-leq-c-5-gamma-frac1-tau%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
Reorganizing, this is equivalent to proving that the function $fcolonmathbb{R}tomathbb{R}$ given by
$$
f(tau) = frac{|tau|+|tau|^{2gamma}} {|tau|+1}
$$
is bounded. Note that it is continuous, and $lim_{+infty} f = lim_{-infty} f = 1$ (using the fact that $2gamma <1$); therefore, it is bounded.
answered Nov 23 at 19:57
Clement C.
49.6k33886
(also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
– Clement C.
Nov 23 at 19:59
Actually it will not hold for $gamma < 1/2$, the reason why it holds is because $2gamma<1-2gamma$ when $gamma < 1/4$. I figured this, but it wasn't suffice to prove that. But thank you so much for help me.
– João Paulo Andrade
Nov 23 at 22:38
@JoãoPauloAndrade Glad it helped, but -- it does hold for all $gamma leq 1/2$ (the case $1/2$ is proven similarly, but the limit of $f$ at $infty$ is $2$, not 1).
– Clement C.
Nov 23 at 22:40
For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
– Clement C.
Nov 23 at 22:46
add a comment |
Reorganizing, this is equivalent to proving that the function $fcolonmathbb{R}tomathbb{R}$ given by
$$
f(tau) = frac{|tau|+|tau|^{2gamma}} {|tau|+1}
$$
is bounded. Note that it is continuous, and $lim_{+infty} f = lim_{-infty} f = 1$ (using the fact that $2gamma <1$); therefore, it is bounded.
answered Nov 23 at 19:57
Clement C.
49.6k33886
(also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
– Clement C.
Nov 23 at 19:59
Actually it will not hold for $gamma < 1/2$, the reason why it holds is because $2gamma<1-2gamma$ when $gamma < 1/4$. I figured this, but it wasn't suffice to prove that. But thank you so much for help me.
– João Paulo Andrade
Nov 23 at 22:38
@JoãoPauloAndrade Glad it helped, but -- it does hold for all $gamma leq 1/2$ (the case $1/2$ is proven similarly, but the limit of $f$ at $infty$ is $2$, not 1).
– Clement C.
Nov 23 at 22:40
For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
– Clement C.
Nov 23 at 22:46
add a comment |
Reorganizing, this is equivalent to proving that the function $fcolonmathbb{R}tomathbb{R}$ given by
$$
f(tau) = frac{|tau|+|tau|^{2gamma}} {|tau|+1}
$$
is bounded. Note that it is continuous, and $lim_{+infty} f = lim_{-infty} f = 1$ (using the fact that $2gamma <1$); therefore, it is bounded.
answered Nov 23 at 19:57
Clement C.
49.6k33886
Reorganizing, this is equivalent to proving that the function $fcolonmathbb{R}tomathbb{R}$ given by
$$
f(tau) = frac{|tau|+|tau|^{2gamma}} {|tau|+1}
$$
is bounded. Note that it is continuous, and $lim_{+infty} f = lim_{-infty} f = 1$ (using the fact that $2gamma <1$); therefore, it is bounded.
answered Nov 23 at 19:57
Clement C.
49.6k33886
answered Nov 23 at 19:57
Clement C.
49.6k33886
answered Nov 23 at 19:57
Clement C.
49.6k33886
answered Nov 23 at 19:57
Clement C.
49.6k33886
49.6k33886
(also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
– Clement C.
Nov 23 at 19:59
Actually it will not hold for $gamma < 1/2$, the reason why it holds is because $2gamma<1-2gamma$ when $gamma < 1/4$. I figured this, but it wasn't suffice to prove that. But thank you so much for help me.
– João Paulo Andrade
Nov 23 at 22:38
@JoãoPauloAndrade Glad it helped, but -- it does hold for all $gamma leq 1/2$ (the case $1/2$ is proven similarly, but the limit of $f$ at $infty$ is $2$, not 1).
– Clement C.
Nov 23 at 22:40
For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
– Clement C.
Nov 23 at 22:46
add a comment |
(also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
– Clement C.
Nov 23 at 19:59
Actually it will not hold for $gamma < 1/2$, the reason why it holds is because $2gamma<1-2gamma$ when $gamma < 1/4$. I figured this, but it wasn't suffice to prove that. But thank you so much for help me.
– João Paulo Andrade
Nov 23 at 22:38
@JoãoPauloAndrade Glad it helped, but -- it does hold for all $gamma leq 1/2$ (the case $1/2$ is proven similarly, but the limit of $f$ at $infty$ is $2$, not 1).
– Clement C.
Nov 23 at 22:40
For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
– Clement C.
Nov 23 at 22:46
(also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
– Clement C.
Nov 23 at 19:59
(also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
– Clement C.
Nov 23 at 19:59
Actually it will not hold for $gamma < 1/2$, the reason why it holds is because $2gamma<1-2gamma$ when $gamma < 1/4$. I figured this, but it wasn't suffice to prove that. But thank you so much for help me.
– João Paulo Andrade
Nov 23 at 22:38
Actually it will not hold for $gamma < 1/2$, the reason why it holds is because $2gamma<1-2gamma$ when $gamma < 1/4$. I figured this, but it wasn't suffice to prove that. But thank you so much for help me.
– João Paulo Andrade
Nov 23 at 22:38
@JoãoPauloAndrade Glad it helped, but -- it does hold for all $gamma leq 1/2$ (the case $1/2$ is proven similarly, but the limit of $f$ at $infty$ is $2$, not 1).
– Clement C.
Nov 23 at 22:40
@JoãoPauloAndrade Glad it helped, but -- it does hold for all $gamma leq 1/2$ (the case $1/2$ is proven similarly, but the limit of $f$ at $infty$ is $2$, not 1).
– Clement C.
Nov 23 at 22:40
For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
– Clement C.
Nov 23 at 22:46
For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
– Clement C.
Nov 23 at 22:46
add a comment |
Let $|tau| = x geq 0$ and Consider:
$$f(x) = dfrac{x^{2gamma}+x}{1+x}$$
and prove this has a maximum by taking derivative.
answered Nov 23 at 19:58
dezdichado
6,1781929
add a comment |
Let $|tau| = x geq 0$ and Consider:
$$f(x) = dfrac{x^{2gamma}+x}{1+x}$$
and prove this has a maximum by taking derivative.
answered Nov 23 at 19:58
dezdichado
6,1781929
add a comment |
Let $|tau| = x geq 0$ and Consider:
$$f(x) = dfrac{x^{2gamma}+x}{1+x}$$
and prove this has a maximum by taking derivative.
answered Nov 23 at 19:58
dezdichado
6,1781929
Let $|tau| = x geq 0$ and Consider:
$$f(x) = dfrac{x^{2gamma}+x}{1+x}$$
and prove this has a maximum by taking derivative.
answered Nov 23 at 19:58
dezdichado
6,1781929
answered Nov 23 at 19:58
dezdichado
6,1781929
answered Nov 23 at 19:58
dezdichado
6,1781929
answered Nov 23 at 19:58
dezdichado
6,1781929
6,1781929
add a comment |
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.
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.
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%2f3010766%2fhow-can-prove-this-inequality-tau2-gamma-leq-c-5-gamma-frac1-tau%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
After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark $checkmark$ next to it. This scores points for you and for the person who answered your question. You can find out more about accepting answers here: How do I accept an answer?, Why should we accept answers?, What should I do if someone answers my question?.
– Clement C.
Nov 24 at 22:02