Suppose $f: [0,1] rightarrow [0,1] $ and $f(x) leq int_0^x sqrt{f(t)}dt$. Show that $f(x) leq x^2$ for all $x...












6












$begingroup$



Let $f(x): [0,1] rightarrow [0,1] $ such that $f(x) leq int_0^x sqrt{f(t)},dt$. Show that $f(x) leq x^2$ for all $x in [0,1]$.




I tried reiterating the inequality, obtaining $f(x) leq int_0^x1dt = x; f(x) leq int_0^x sqrt{t}dt = frac{2}{3}x^{3/2}$ etc... While it's easy to see that the exponent of $x$ tends to $2$, it's more difficult to show that the coefficient is $1$. Can somebody help me?






calculus real-analysis integration inequality







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$endgroup$












  • $begingroup$
    This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $alpha = 1/2$. It follows that $f(x) le x^2/4$.
    $endgroup$
    – Martin R
    Nov 30 '18 at 20:31












  • $begingroup$
    @MartinR $f$ is not necessarily conitnuous here
    $endgroup$
    – Thinking
    Nov 30 '18 at 20:38










  • $begingroup$
    @Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that.
    $endgroup$
    – Martin R
    Nov 30 '18 at 20:43










  • $begingroup$
    @MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix.
    $endgroup$
    – Thinking
    Nov 30 '18 at 20:46
















6












$begingroup$



Let $f(x): [0,1] rightarrow [0,1] $ such that $f(x) leq int_0^x sqrt{f(t)},dt$. Show that $f(x) leq x^2$ for all $x in [0,1]$.




I tried reiterating the inequality, obtaining $f(x) leq int_0^x1dt = x; f(x) leq int_0^x sqrt{t}dt = frac{2}{3}x^{3/2}$ etc... While it's easy to see that the exponent of $x$ tends to $2$, it's more difficult to show that the coefficient is $1$. Can somebody help me?






calculus real-analysis integration inequality







share|cite|improve this question











$endgroup$












  • $begingroup$
    This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $alpha = 1/2$. It follows that $f(x) le x^2/4$.
    $endgroup$
    – Martin R
    Nov 30 '18 at 20:31












  • $begingroup$
    @MartinR $f$ is not necessarily conitnuous here
    $endgroup$
    – Thinking
    Nov 30 '18 at 20:38










  • $begingroup$
    @Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that.
    $endgroup$
    – Martin R
    Nov 30 '18 at 20:43










  • $begingroup$
    @MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix.
    $endgroup$
    – Thinking
    Nov 30 '18 at 20:46














6












6








6


4



$begingroup$



Let $f(x): [0,1] rightarrow [0,1] $ such that $f(x) leq int_0^x sqrt{f(t)},dt$. Show that $f(x) leq x^2$ for all $x in [0,1]$.




I tried reiterating the inequality, obtaining $f(x) leq int_0^x1dt = x; f(x) leq int_0^x sqrt{t}dt = frac{2}{3}x^{3/2}$ etc... While it's easy to see that the exponent of $x$ tends to $2$, it's more difficult to show that the coefficient is $1$. Can somebody help me?






calculus real-analysis integration inequality







share|cite|improve this question











$endgroup$





Let $f(x): [0,1] rightarrow [0,1] $ such that $f(x) leq int_0^x sqrt{f(t)},dt$. Show that $f(x) leq x^2$ for all $x in [0,1]$.




I tried reiterating the inequality, obtaining $f(x) leq int_0^x1dt = x; f(x) leq int_0^x sqrt{t}dt = frac{2}{3}x^{3/2}$ etc... While it's easy to see that the exponent of $x$ tends to $2$, it's more difficult to show that the coefficient is $1$. Can somebody help me?






calculus real-analysis integration inequality




calculus real-analysis integration inequality






share|cite|improve this question















share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 30 '18 at 21:44









zhw.

71.8k43075




71.8k43075










asked Nov 30 '18 at 18:55









LanceLance

8912




8912












  • $begingroup$
    This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $alpha = 1/2$. It follows that $f(x) le x^2/4$.
    $endgroup$
    – Martin R
    Nov 30 '18 at 20:31












  • $begingroup$
    @MartinR $f$ is not necessarily conitnuous here
    $endgroup$
    – Thinking
    Nov 30 '18 at 20:38










  • $begingroup$
    @Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that.
    $endgroup$
    – Martin R
    Nov 30 '18 at 20:43










  • $begingroup$
    @MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix.
    $endgroup$
    – Thinking
    Nov 30 '18 at 20:46


















  • $begingroup$
    This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $alpha = 1/2$. It follows that $f(x) le x^2/4$.
    $endgroup$
    – Martin R
    Nov 30 '18 at 20:31












  • $begingroup$
    @MartinR $f$ is not necessarily conitnuous here
    $endgroup$
    – Thinking
    Nov 30 '18 at 20:38










  • $begingroup$
    @Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that.
    $endgroup$
    – Martin R
    Nov 30 '18 at 20:43










  • $begingroup$
    @MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix.
    $endgroup$
    – Thinking
    Nov 30 '18 at 20:46
















$begingroup$
This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $alpha = 1/2$. It follows that $f(x) le x^2/4$.
$endgroup$
– Martin R
Nov 30 '18 at 20:31






$begingroup$
This is a special case of math.stackexchange.com/q/2050893/42969 – choose $C=1$ and $alpha = 1/2$. It follows that $f(x) le x^2/4$.
$endgroup$
– Martin R
Nov 30 '18 at 20:31














$begingroup$
@MartinR $f$ is not necessarily conitnuous here
$endgroup$
– Thinking
Nov 30 '18 at 20:38




$begingroup$
@MartinR $f$ is not necessarily conitnuous here
$endgroup$
– Thinking
Nov 30 '18 at 20:38












$begingroup$
@Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that.
$endgroup$
– Martin R
Nov 30 '18 at 20:43




$begingroup$
@Thinking: You are right. – I assume that a similar argument works if $f$ is only integrable, but I am not 100% sure about that.
$endgroup$
– Martin R
Nov 30 '18 at 20:43












$begingroup$
@MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix.
$endgroup$
– Thinking
Nov 30 '18 at 20:46




$begingroup$
@MartinR maybe... but in this case $v$ is not necessarily differentiable so there might be several problems to fix.
$endgroup$
– Thinking
Nov 30 '18 at 20:46










2 Answers
2






active

oldest

votes


















4












$begingroup$

I assume that $f$ is Lebesgue integrable along with the other hypotheses. Let $xin [0,1].$ Let $M_x$ be the supremum of $f$ over $[0,x].$ If $M_x=0,$ there is nothing to prove. So assume $M_x>0.$ Let $0<epsilon< M_x.$ Choose $x_epsilonin [0,x]$ such that $f(x_epsilon) > M_x-epsilon.$ Then



$$M_x-epsilon < f(x_epsilon) le int_0^{x_epsilon}sqrt {f(t)},dt le xsqrt{M_x}.$$



Now let $epsilonto 0^+$ to see $M_xle xsqrt{M_x},$ which implies $sqrt{M_x}le x.$ Squaring, we see $f(x)le M_xle x^2,$ giving the result.






share|cite|improve this answer











$endgroup$













  • $begingroup$
    There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
    $endgroup$
    – Thinking
    Nov 30 '18 at 20:54








  • 1




    $begingroup$
    @Thinking I don't see the problem. Where exactly do you think my proof breaks down?
    $endgroup$
    – zhw.
    Nov 30 '18 at 21:04










  • $begingroup$
    You are right, I completely missounderstood the proof. It's actually clever. (+1)
    $endgroup$
    – Thinking
    Nov 30 '18 at 21:08










  • $begingroup$
    @Thinking I've edited to remove the continuity hypothesis.
    $endgroup$
    – zhw.
    Nov 30 '18 at 21:24










  • $begingroup$
    @Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
    $endgroup$
    – zhw.
    Nov 30 '18 at 21:52



















1












$begingroup$

Suppose $b_{n+1}=frac 12(b_n+2)$ and $a_{n+1}=frac{a_n^{1/2}}{b_{n+1}}$, then $f(x) le a_nx^{b_n}Rightarrow f(x)le a_{n+1}x^{b_{n+1}}$. As you showed, $f(x)le a_0x^{b_0}$ where $a_0=b_0=1$. To solve the recurrence for $b_n$ put $B_n=2^{n}b_n$ and notice $b_{n+1}-frac 12b_n=1Rightarrow 2^{n+1}b_{n+1}-2^{n}b_n=B_{n+1}-B_n=2^{n+1}$, summing over $n$ gives $B_{N}-B_0=sum_{n=0}^{N-1} 2^{n+1}=2^{N+1}-2$ and therefore $b_{N}=2^{-N}(2^{N+1}-2+b_0)=2-2^{-N}$. The $b_n$'s are increasing, so $frac 1{b_n}$ is a decreasing sequence. Assume $a_{n+1}le a_n$ so $a_{n+1}^{1/2}le a_n^{1/2}$ and hence $a_{n+2}=frac{a_{n+1}^{1/2}}{b_{n+2}} le frac{a_{n}^{1/2}}{b_{n+1}}=a_{n+1}$. Because $a_1=frac 23le 1 =a_0$, this holds for $n=0$ and induction shows $a_n$ is a decreasing sequence bounded below by $0$, thus has a limit $a$. Taking this limit shows $a=a^{1/2}/2$ because $b_nto 2$, and therefore $a=1/4$. Finally, $f(x)le a_nx^{b_n}to frac 14x^2$.






share|cite|improve this answer









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    2 Answers
    2






    active

    oldest

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    2 Answers
    2






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    4












    $begingroup$

    I assume that $f$ is Lebesgue integrable along with the other hypotheses. Let $xin [0,1].$ Let $M_x$ be the supremum of $f$ over $[0,x].$ If $M_x=0,$ there is nothing to prove. So assume $M_x>0.$ Let $0<epsilon< M_x.$ Choose $x_epsilonin [0,x]$ such that $f(x_epsilon) > M_x-epsilon.$ Then



    $$M_x-epsilon < f(x_epsilon) le int_0^{x_epsilon}sqrt {f(t)},dt le xsqrt{M_x}.$$



    Now let $epsilonto 0^+$ to see $M_xle xsqrt{M_x},$ which implies $sqrt{M_x}le x.$ Squaring, we see $f(x)le M_xle x^2,$ giving the result.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
      $endgroup$
      – Thinking
      Nov 30 '18 at 20:54








    • 1




      $begingroup$
      @Thinking I don't see the problem. Where exactly do you think my proof breaks down?
      $endgroup$
      – zhw.
      Nov 30 '18 at 21:04










    • $begingroup$
      You are right, I completely missounderstood the proof. It's actually clever. (+1)
      $endgroup$
      – Thinking
      Nov 30 '18 at 21:08










    • $begingroup$
      @Thinking I've edited to remove the continuity hypothesis.
      $endgroup$
      – zhw.
      Nov 30 '18 at 21:24










    • $begingroup$
      @Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
      $endgroup$
      – zhw.
      Nov 30 '18 at 21:52
















    4












    $begingroup$

    I assume that $f$ is Lebesgue integrable along with the other hypotheses. Let $xin [0,1].$ Let $M_x$ be the supremum of $f$ over $[0,x].$ If $M_x=0,$ there is nothing to prove. So assume $M_x>0.$ Let $0<epsilon< M_x.$ Choose $x_epsilonin [0,x]$ such that $f(x_epsilon) > M_x-epsilon.$ Then



    $$M_x-epsilon < f(x_epsilon) le int_0^{x_epsilon}sqrt {f(t)},dt le xsqrt{M_x}.$$



    Now let $epsilonto 0^+$ to see $M_xle xsqrt{M_x},$ which implies $sqrt{M_x}le x.$ Squaring, we see $f(x)le M_xle x^2,$ giving the result.






    share|cite|improve this answer











    $endgroup$













    • $begingroup$
      There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
      $endgroup$
      – Thinking
      Nov 30 '18 at 20:54








    • 1




      $begingroup$
      @Thinking I don't see the problem. Where exactly do you think my proof breaks down?
      $endgroup$
      – zhw.
      Nov 30 '18 at 21:04










    • $begingroup$
      You are right, I completely missounderstood the proof. It's actually clever. (+1)
      $endgroup$
      – Thinking
      Nov 30 '18 at 21:08










    • $begingroup$
      @Thinking I've edited to remove the continuity hypothesis.
      $endgroup$
      – zhw.
      Nov 30 '18 at 21:24










    • $begingroup$
      @Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
      $endgroup$
      – zhw.
      Nov 30 '18 at 21:52














    4












    4








    4





    $begingroup$

    I assume that $f$ is Lebesgue integrable along with the other hypotheses. Let $xin [0,1].$ Let $M_x$ be the supremum of $f$ over $[0,x].$ If $M_x=0,$ there is nothing to prove. So assume $M_x>0.$ Let $0<epsilon< M_x.$ Choose $x_epsilonin [0,x]$ such that $f(x_epsilon) > M_x-epsilon.$ Then



    $$M_x-epsilon < f(x_epsilon) le int_0^{x_epsilon}sqrt {f(t)},dt le xsqrt{M_x}.$$



    Now let $epsilonto 0^+$ to see $M_xle xsqrt{M_x},$ which implies $sqrt{M_x}le x.$ Squaring, we see $f(x)le M_xle x^2,$ giving the result.






    share|cite|improve this answer











    $endgroup$



    I assume that $f$ is Lebesgue integrable along with the other hypotheses. Let $xin [0,1].$ Let $M_x$ be the supremum of $f$ over $[0,x].$ If $M_x=0,$ there is nothing to prove. So assume $M_x>0.$ Let $0<epsilon< M_x.$ Choose $x_epsilonin [0,x]$ such that $f(x_epsilon) > M_x-epsilon.$ Then



    $$M_x-epsilon < f(x_epsilon) le int_0^{x_epsilon}sqrt {f(t)},dt le xsqrt{M_x}.$$



    Now let $epsilonto 0^+$ to see $M_xle xsqrt{M_x},$ which implies $sqrt{M_x}le x.$ Squaring, we see $f(x)le M_xle x^2,$ giving the result.







    share|cite|improve this answer














    share|cite|improve this answer



    share|cite|improve this answer








    edited Dec 1 '18 at 16:42

























    answered Nov 30 '18 at 20:50









    zhw.zhw.

    71.8k43075




    71.8k43075












    • $begingroup$
      There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
      $endgroup$
      – Thinking
      Nov 30 '18 at 20:54








    • 1




      $begingroup$
      @Thinking I don't see the problem. Where exactly do you think my proof breaks down?
      $endgroup$
      – zhw.
      Nov 30 '18 at 21:04










    • $begingroup$
      You are right, I completely missounderstood the proof. It's actually clever. (+1)
      $endgroup$
      – Thinking
      Nov 30 '18 at 21:08










    • $begingroup$
      @Thinking I've edited to remove the continuity hypothesis.
      $endgroup$
      – zhw.
      Nov 30 '18 at 21:24










    • $begingroup$
      @Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
      $endgroup$
      – zhw.
      Nov 30 '18 at 21:52


















    • $begingroup$
      There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
      $endgroup$
      – Thinking
      Nov 30 '18 at 20:54








    • 1




      $begingroup$
      @Thinking I don't see the problem. Where exactly do you think my proof breaks down?
      $endgroup$
      – zhw.
      Nov 30 '18 at 21:04










    • $begingroup$
      You are right, I completely missounderstood the proof. It's actually clever. (+1)
      $endgroup$
      – Thinking
      Nov 30 '18 at 21:08










    • $begingroup$
      @Thinking I've edited to remove the continuity hypothesis.
      $endgroup$
      – zhw.
      Nov 30 '18 at 21:24










    • $begingroup$
      @Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
      $endgroup$
      – zhw.
      Nov 30 '18 at 21:52
















    $begingroup$
    There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
    $endgroup$
    – Thinking
    Nov 30 '18 at 20:54






    $begingroup$
    There's a problem, if the number of local maxima is dense in $[0,1]$ no ? $x_0$ can be really close to $0$, then the next one $x_1$ can be really close to $x_0$... and we never reach the whole interval $[0,1]$
    $endgroup$
    – Thinking
    Nov 30 '18 at 20:54






    1




    1




    $begingroup$
    @Thinking I don't see the problem. Where exactly do you think my proof breaks down?
    $endgroup$
    – zhw.
    Nov 30 '18 at 21:04




    $begingroup$
    @Thinking I don't see the problem. Where exactly do you think my proof breaks down?
    $endgroup$
    – zhw.
    Nov 30 '18 at 21:04












    $begingroup$
    You are right, I completely missounderstood the proof. It's actually clever. (+1)
    $endgroup$
    – Thinking
    Nov 30 '18 at 21:08




    $begingroup$
    You are right, I completely missounderstood the proof. It's actually clever. (+1)
    $endgroup$
    – Thinking
    Nov 30 '18 at 21:08












    $begingroup$
    @Thinking I've edited to remove the continuity hypothesis.
    $endgroup$
    – zhw.
    Nov 30 '18 at 21:24




    $begingroup$
    @Thinking I've edited to remove the continuity hypothesis.
    $endgroup$
    – zhw.
    Nov 30 '18 at 21:24












    $begingroup$
    @Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
    $endgroup$
    – zhw.
    Nov 30 '18 at 21:52




    $begingroup$
    @Thinking Yes of course. Otherwise $M_x-epsilon$ is an upper bound.
    $endgroup$
    – zhw.
    Nov 30 '18 at 21:52











    1












    $begingroup$

    Suppose $b_{n+1}=frac 12(b_n+2)$ and $a_{n+1}=frac{a_n^{1/2}}{b_{n+1}}$, then $f(x) le a_nx^{b_n}Rightarrow f(x)le a_{n+1}x^{b_{n+1}}$. As you showed, $f(x)le a_0x^{b_0}$ where $a_0=b_0=1$. To solve the recurrence for $b_n$ put $B_n=2^{n}b_n$ and notice $b_{n+1}-frac 12b_n=1Rightarrow 2^{n+1}b_{n+1}-2^{n}b_n=B_{n+1}-B_n=2^{n+1}$, summing over $n$ gives $B_{N}-B_0=sum_{n=0}^{N-1} 2^{n+1}=2^{N+1}-2$ and therefore $b_{N}=2^{-N}(2^{N+1}-2+b_0)=2-2^{-N}$. The $b_n$'s are increasing, so $frac 1{b_n}$ is a decreasing sequence. Assume $a_{n+1}le a_n$ so $a_{n+1}^{1/2}le a_n^{1/2}$ and hence $a_{n+2}=frac{a_{n+1}^{1/2}}{b_{n+2}} le frac{a_{n}^{1/2}}{b_{n+1}}=a_{n+1}$. Because $a_1=frac 23le 1 =a_0$, this holds for $n=0$ and induction shows $a_n$ is a decreasing sequence bounded below by $0$, thus has a limit $a$. Taking this limit shows $a=a^{1/2}/2$ because $b_nto 2$, and therefore $a=1/4$. Finally, $f(x)le a_nx^{b_n}to frac 14x^2$.






    share|cite|improve this answer









    $endgroup$


















      1












      $begingroup$

      Suppose $b_{n+1}=frac 12(b_n+2)$ and $a_{n+1}=frac{a_n^{1/2}}{b_{n+1}}$, then $f(x) le a_nx^{b_n}Rightarrow f(x)le a_{n+1}x^{b_{n+1}}$. As you showed, $f(x)le a_0x^{b_0}$ where $a_0=b_0=1$. To solve the recurrence for $b_n$ put $B_n=2^{n}b_n$ and notice $b_{n+1}-frac 12b_n=1Rightarrow 2^{n+1}b_{n+1}-2^{n}b_n=B_{n+1}-B_n=2^{n+1}$, summing over $n$ gives $B_{N}-B_0=sum_{n=0}^{N-1} 2^{n+1}=2^{N+1}-2$ and therefore $b_{N}=2^{-N}(2^{N+1}-2+b_0)=2-2^{-N}$. The $b_n$'s are increasing, so $frac 1{b_n}$ is a decreasing sequence. Assume $a_{n+1}le a_n$ so $a_{n+1}^{1/2}le a_n^{1/2}$ and hence $a_{n+2}=frac{a_{n+1}^{1/2}}{b_{n+2}} le frac{a_{n}^{1/2}}{b_{n+1}}=a_{n+1}$. Because $a_1=frac 23le 1 =a_0$, this holds for $n=0$ and induction shows $a_n$ is a decreasing sequence bounded below by $0$, thus has a limit $a$. Taking this limit shows $a=a^{1/2}/2$ because $b_nto 2$, and therefore $a=1/4$. Finally, $f(x)le a_nx^{b_n}to frac 14x^2$.






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        $begingroup$

        Suppose $b_{n+1}=frac 12(b_n+2)$ and $a_{n+1}=frac{a_n^{1/2}}{b_{n+1}}$, then $f(x) le a_nx^{b_n}Rightarrow f(x)le a_{n+1}x^{b_{n+1}}$. As you showed, $f(x)le a_0x^{b_0}$ where $a_0=b_0=1$. To solve the recurrence for $b_n$ put $B_n=2^{n}b_n$ and notice $b_{n+1}-frac 12b_n=1Rightarrow 2^{n+1}b_{n+1}-2^{n}b_n=B_{n+1}-B_n=2^{n+1}$, summing over $n$ gives $B_{N}-B_0=sum_{n=0}^{N-1} 2^{n+1}=2^{N+1}-2$ and therefore $b_{N}=2^{-N}(2^{N+1}-2+b_0)=2-2^{-N}$. The $b_n$'s are increasing, so $frac 1{b_n}$ is a decreasing sequence. Assume $a_{n+1}le a_n$ so $a_{n+1}^{1/2}le a_n^{1/2}$ and hence $a_{n+2}=frac{a_{n+1}^{1/2}}{b_{n+2}} le frac{a_{n}^{1/2}}{b_{n+1}}=a_{n+1}$. Because $a_1=frac 23le 1 =a_0$, this holds for $n=0$ and induction shows $a_n$ is a decreasing sequence bounded below by $0$, thus has a limit $a$. Taking this limit shows $a=a^{1/2}/2$ because $b_nto 2$, and therefore $a=1/4$. Finally, $f(x)le a_nx^{b_n}to frac 14x^2$.






        share|cite|improve this answer









        $endgroup$



        Suppose $b_{n+1}=frac 12(b_n+2)$ and $a_{n+1}=frac{a_n^{1/2}}{b_{n+1}}$, then $f(x) le a_nx^{b_n}Rightarrow f(x)le a_{n+1}x^{b_{n+1}}$. As you showed, $f(x)le a_0x^{b_0}$ where $a_0=b_0=1$. To solve the recurrence for $b_n$ put $B_n=2^{n}b_n$ and notice $b_{n+1}-frac 12b_n=1Rightarrow 2^{n+1}b_{n+1}-2^{n}b_n=B_{n+1}-B_n=2^{n+1}$, summing over $n$ gives $B_{N}-B_0=sum_{n=0}^{N-1} 2^{n+1}=2^{N+1}-2$ and therefore $b_{N}=2^{-N}(2^{N+1}-2+b_0)=2-2^{-N}$. The $b_n$'s are increasing, so $frac 1{b_n}$ is a decreasing sequence. Assume $a_{n+1}le a_n$ so $a_{n+1}^{1/2}le a_n^{1/2}$ and hence $a_{n+2}=frac{a_{n+1}^{1/2}}{b_{n+2}} le frac{a_{n}^{1/2}}{b_{n+1}}=a_{n+1}$. Because $a_1=frac 23le 1 =a_0$, this holds for $n=0$ and induction shows $a_n$ is a decreasing sequence bounded below by $0$, thus has a limit $a$. Taking this limit shows $a=a^{1/2}/2$ because $b_nto 2$, and therefore $a=1/4$. Finally, $f(x)le a_nx^{b_n}to frac 14x^2$.







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 30 '18 at 20:12









        Guacho PerezGuacho Perez

        3,88911132




        3,88911132






























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