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?










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












$begingroup$



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?










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?










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






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









$endgroup$













    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
    });


    }
    });














    draft saved

    draft discarded


















    StackExchange.ready(
    function () {
    StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020493%2fsuppose-f-0-1-rightarrow-0-1-and-fx-leq-int-0x-sqrtftdt-sh%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









    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, $f(x)le a_nx^{b_n}to frac 14x^2$.






      share|cite|improve this answer









      $endgroup$
















        1












        1








        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









        $endgroup$



        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






























            draft saved

            draft discarded




















































            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.




            draft saved


            draft discarded














            StackExchange.ready(
            function () {
            StackExchange.openid.initPostLogin('.new-post-login', 'https%3a%2f%2fmath.stackexchange.com%2fquestions%2f3020493%2fsuppose-f-0-1-rightarrow-0-1-and-fx-leq-int-0x-sqrtftdt-sh%23new-answer', 'question_page');
            }
            );

            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







            Popular posts from this blog

            Ellipse (mathématiques)

            Quarter-circle Tiles

            Mont Emei