How can prove this inequality $|tau|^{2gamma} leq c_5(gamma) frac{1+|tau|}{1+|tau|^{1-2gamma}}$?












1












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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







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    – Clement C.
    Nov 24 '18 at 22:02
















1












$begingroup$


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







share|cite|improve this question









$endgroup$












  • $begingroup$
    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?.
    $endgroup$
    – Clement C.
    Nov 24 '18 at 22:02














1












1








1





$begingroup$


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







share|cite|improve this question









$endgroup$




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






share|cite|improve this question













share|cite|improve this question











share|cite|improve this question




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asked Nov 23 '18 at 19:53









João Paulo AndradeJoão Paulo Andrade

315




315












  • $begingroup$
    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?.
    $endgroup$
    – Clement C.
    Nov 24 '18 at 22:02


















  • $begingroup$
    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?.
    $endgroup$
    – Clement C.
    Nov 24 '18 at 22:02
















$begingroup$
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?.
$endgroup$
– Clement C.
Nov 24 '18 at 22:02




$begingroup$
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?.
$endgroup$
– Clement C.
Nov 24 '18 at 22:02










2 Answers
2






active

oldest

votes


















1












$begingroup$

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.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    (also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
    $endgroup$
    – Clement C.
    Nov 23 '18 at 19:59










  • $begingroup$
    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.
    $endgroup$
    – João Paulo Andrade
    Nov 23 '18 at 22:38










  • $begingroup$
    @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).
    $endgroup$
    – Clement C.
    Nov 23 '18 at 22:40










  • $begingroup$
    For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
    $endgroup$
    – Clement C.
    Nov 23 '18 at 22:46



















1












$begingroup$

Let $|tau| = x geq 0$ and Consider:
$$f(x) = dfrac{x^{2gamma}+x}{1+x}$$
and prove this has a maximum by taking derivative.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

    votes








    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      (also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
      $endgroup$
      – Clement C.
      Nov 23 '18 at 19:59










    • $begingroup$
      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.
      $endgroup$
      – João Paulo Andrade
      Nov 23 '18 at 22:38










    • $begingroup$
      @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).
      $endgroup$
      – Clement C.
      Nov 23 '18 at 22:40










    • $begingroup$
      For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
      $endgroup$
      – Clement C.
      Nov 23 '18 at 22:46
















    1












    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$













    • $begingroup$
      (also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
      $endgroup$
      – Clement C.
      Nov 23 '18 at 19:59










    • $begingroup$
      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.
      $endgroup$
      – João Paulo Andrade
      Nov 23 '18 at 22:38










    • $begingroup$
      @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).
      $endgroup$
      – Clement C.
      Nov 23 '18 at 22:40










    • $begingroup$
      For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
      $endgroup$
      – Clement C.
      Nov 23 '18 at 22:46














    1












    1








    1





    $begingroup$

    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.






    share|cite|improve this answer









    $endgroup$



    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.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 23 '18 at 19:57









    Clement C.Clement C.

    49.8k33886




    49.8k33886












    • $begingroup$
      (also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
      $endgroup$
      – Clement C.
      Nov 23 '18 at 19:59










    • $begingroup$
      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.
      $endgroup$
      – João Paulo Andrade
      Nov 23 '18 at 22:38










    • $begingroup$
      @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).
      $endgroup$
      – Clement C.
      Nov 23 '18 at 22:40










    • $begingroup$
      For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
      $endgroup$
      – Clement C.
      Nov 23 '18 at 22:46


















    • $begingroup$
      (also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
      $endgroup$
      – Clement C.
      Nov 23 '18 at 19:59










    • $begingroup$
      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.
      $endgroup$
      – João Paulo Andrade
      Nov 23 '18 at 22:38










    • $begingroup$
      @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).
      $endgroup$
      – Clement C.
      Nov 23 '18 at 22:40










    • $begingroup$
      For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
      $endgroup$
      – Clement C.
      Nov 23 '18 at 22:46
















    $begingroup$
    (also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
    $endgroup$
    – Clement C.
    Nov 23 '18 at 19:59




    $begingroup$
    (also, as shown in this proof, the result holds for $gamma < 1/2$, not only $gamma<1/4$.)
    $endgroup$
    – Clement C.
    Nov 23 '18 at 19:59












    $begingroup$
    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.
    $endgroup$
    – João Paulo Andrade
    Nov 23 '18 at 22:38




    $begingroup$
    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.
    $endgroup$
    – João Paulo Andrade
    Nov 23 '18 at 22:38












    $begingroup$
    @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).
    $endgroup$
    – Clement C.
    Nov 23 '18 at 22:40




    $begingroup$
    @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).
    $endgroup$
    – Clement C.
    Nov 23 '18 at 22:40












    $begingroup$
    For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
    $endgroup$
    – Clement C.
    Nov 23 '18 at 22:46




    $begingroup$
    For instance, for $gamma=1/2$, take $c_5(1/2)=2$. You have $$|tau| leq 2cdot frac{1+|tau|}{2}$$ for all $tau$.
    $endgroup$
    – Clement C.
    Nov 23 '18 at 22:46











    1












    $begingroup$

    Let $|tau| = x geq 0$ and Consider:
    $$f(x) = dfrac{x^{2gamma}+x}{1+x}$$
    and prove this has a maximum by taking derivative.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Let $|tau| = x geq 0$ and Consider:
      $$f(x) = dfrac{x^{2gamma}+x}{1+x}$$
      and prove this has a maximum by taking derivative.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        1





        $begingroup$

        Let $|tau| = x geq 0$ and Consider:
        $$f(x) = dfrac{x^{2gamma}+x}{1+x}$$
        and prove this has a maximum by taking derivative.






        share|cite|improve this answer









        $endgroup$



        Let $|tau| = x geq 0$ and Consider:
        $$f(x) = dfrac{x^{2gamma}+x}{1+x}$$
        and prove this has a maximum by taking derivative.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 23 '18 at 19:58









        dezdichadodezdichado

        6,3001929




        6,3001929






























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