How can prove this inequality $|tau|^{2gamma} leq c_5(gamma) frac{1+|tau|}{1+|tau|^{1-2gamma}}$?
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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
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up vote
1
down vote
favorite
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
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up vote
1
down vote
favorite
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
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2 Answers
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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.
49.2k33785
(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 |
up vote
1
down vote
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
6,0841929
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
up vote
1
down vote
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.
49.2k33785
(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 |
up vote
1
down vote
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.
49.2k33785
(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 |
up vote
1
down vote
up vote
1
down vote
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.
49.2k33785
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.
49.2k33785
answered Nov 23 at 19:57
Clement C.
49.2k33785
answered Nov 23 at 19:57
Clement C.
49.2k33785
answered Nov 23 at 19:57
Clement C.
49.2k33785
49.2k33785
(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 |
up vote
1
down vote
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
6,0841929
add a comment |
up vote
1
down vote
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
6,0841929
add a comment |
up vote
1
down vote
up vote
1
down vote
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
6,0841929
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
6,0841929
answered Nov 23 at 19:58
dezdichado
6,0841929
answered Nov 23 at 19:58
dezdichado
6,0841929
answered Nov 23 at 19:58
dezdichado
6,0841929
6,0841929
add a comment |
add a comment |
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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