If $f:[0,1]to[0,1]$ is such that $f(x) leq 2 int_0^x f(t) dt$, show that $f(x)=0$ on its domain












1














From an SNS exam:




Let $f: [0,1] rightarrow [0,1]$ such that $f(x) leq 2 int_0^x f(t) dt$. How can I show that $f(x)$ is identically $0$ in its domain?




My attempt: for every $t$ in $[0,1]$ consider $x_{max,t}$ such that the maximum value of $f$ in $[0,t]$ if $f(x_{max,t})$. Now, for the geometric meaning of the integral, $int_0^t f(x)dx leq x_{max}$ so $f(t) leq 2x_{max,t}$. How can I conclude?






real-analysis integration inequality







share|cite|improve this question




















  • 1




    Is $f$ continuous.
    – hamam_Abdallah
    Nov 29 '18 at 20:06










  • Do you need to show that $f$ has a zero (i.e., there exists $x in [0,1]$ such that $f(x)=0$, or that $f$ is identically zero (i.e., for all $x in [0,1]$, $f(x)=0$? Without quantifiers, it's hard for me to interpret your question.
    – MMASRP63
    Nov 29 '18 at 20:07








  • 2




    The keyword here is Grönwall's inequality. It has probably been discussed on this site before.
    – MaoWao
    Nov 29 '18 at 20:12










  • identically zero, sorry
    – Lance
    Nov 29 '18 at 20:14










  • Here is an idea. Define the set $mathscr{I}$ of the points $x in [0. 1]$ such that the result is true on the interval $[0, x].$ Show $mathscr{I} neq varnothing$ and that $1 = sup mathscr{I} in mathscr{I}.$
    – Will M.
    Nov 29 '18 at 20:15
















1














From an SNS exam:




Let $f: [0,1] rightarrow [0,1]$ such that $f(x) leq 2 int_0^x f(t) dt$. How can I show that $f(x)$ is identically $0$ in its domain?




My attempt: for every $t$ in $[0,1]$ consider $x_{max,t}$ such that the maximum value of $f$ in $[0,t]$ if $f(x_{max,t})$. Now, for the geometric meaning of the integral, $int_0^t f(x)dx leq x_{max}$ so $f(t) leq 2x_{max,t}$. How can I conclude?






real-analysis integration inequality







share|cite|improve this question




















  • 1




    Is $f$ continuous.
    – hamam_Abdallah
    Nov 29 '18 at 20:06










  • Do you need to show that $f$ has a zero (i.e., there exists $x in [0,1]$ such that $f(x)=0$, or that $f$ is identically zero (i.e., for all $x in [0,1]$, $f(x)=0$? Without quantifiers, it's hard for me to interpret your question.
    – MMASRP63
    Nov 29 '18 at 20:07








  • 2




    The keyword here is Grönwall's inequality. It has probably been discussed on this site before.
    – MaoWao
    Nov 29 '18 at 20:12










  • identically zero, sorry
    – Lance
    Nov 29 '18 at 20:14










  • Here is an idea. Define the set $mathscr{I}$ of the points $x in [0. 1]$ such that the result is true on the interval $[0, x].$ Show $mathscr{I} neq varnothing$ and that $1 = sup mathscr{I} in mathscr{I}.$
    – Will M.
    Nov 29 '18 at 20:15














1












1








1







From an SNS exam:




Let $f: [0,1] rightarrow [0,1]$ such that $f(x) leq 2 int_0^x f(t) dt$. How can I show that $f(x)$ is identically $0$ in its domain?




My attempt: for every $t$ in $[0,1]$ consider $x_{max,t}$ such that the maximum value of $f$ in $[0,t]$ if $f(x_{max,t})$. Now, for the geometric meaning of the integral, $int_0^t f(x)dx leq x_{max}$ so $f(t) leq 2x_{max,t}$. How can I conclude?






real-analysis integration inequality







share|cite|improve this question















From an SNS exam:




Let $f: [0,1] rightarrow [0,1]$ such that $f(x) leq 2 int_0^x f(t) dt$. How can I show that $f(x)$ is identically $0$ in its domain?




My attempt: for every $t$ in $[0,1]$ consider $x_{max,t}$ such that the maximum value of $f$ in $[0,t]$ if $f(x_{max,t})$. Now, for the geometric meaning of the integral, $int_0^t f(x)dx leq x_{max}$ so $f(t) leq 2x_{max,t}$. How can I conclude?






real-analysis integration inequality




real-analysis integration inequality






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share|cite|improve this question













share|cite|improve this question




share|cite|improve this question








edited Nov 29 '18 at 20:23









davidlowryduda♦

74.4k7117251




74.4k7117251










asked Nov 29 '18 at 19:55









LanceLance

8912




8912








  • 1




    Is $f$ continuous.
    – hamam_Abdallah
    Nov 29 '18 at 20:06










  • Do you need to show that $f$ has a zero (i.e., there exists $x in [0,1]$ such that $f(x)=0$, or that $f$ is identically zero (i.e., for all $x in [0,1]$, $f(x)=0$? Without quantifiers, it's hard for me to interpret your question.
    – MMASRP63
    Nov 29 '18 at 20:07








  • 2




    The keyword here is Grönwall's inequality. It has probably been discussed on this site before.
    – MaoWao
    Nov 29 '18 at 20:12










  • identically zero, sorry
    – Lance
    Nov 29 '18 at 20:14










  • Here is an idea. Define the set $mathscr{I}$ of the points $x in [0. 1]$ such that the result is true on the interval $[0, x].$ Show $mathscr{I} neq varnothing$ and that $1 = sup mathscr{I} in mathscr{I}.$
    – Will M.
    Nov 29 '18 at 20:15














  • 1




    Is $f$ continuous.
    – hamam_Abdallah
    Nov 29 '18 at 20:06










  • Do you need to show that $f$ has a zero (i.e., there exists $x in [0,1]$ such that $f(x)=0$, or that $f$ is identically zero (i.e., for all $x in [0,1]$, $f(x)=0$? Without quantifiers, it's hard for me to interpret your question.
    – MMASRP63
    Nov 29 '18 at 20:07








  • 2




    The keyword here is Grönwall's inequality. It has probably been discussed on this site before.
    – MaoWao
    Nov 29 '18 at 20:12










  • identically zero, sorry
    – Lance
    Nov 29 '18 at 20:14










  • Here is an idea. Define the set $mathscr{I}$ of the points $x in [0. 1]$ such that the result is true on the interval $[0, x].$ Show $mathscr{I} neq varnothing$ and that $1 = sup mathscr{I} in mathscr{I}.$
    – Will M.
    Nov 29 '18 at 20:15








1




1




Is $f$ continuous.
– hamam_Abdallah
Nov 29 '18 at 20:06




Is $f$ continuous.
– hamam_Abdallah
Nov 29 '18 at 20:06












Do you need to show that $f$ has a zero (i.e., there exists $x in [0,1]$ such that $f(x)=0$, or that $f$ is identically zero (i.e., for all $x in [0,1]$, $f(x)=0$? Without quantifiers, it's hard for me to interpret your question.
– MMASRP63
Nov 29 '18 at 20:07






Do you need to show that $f$ has a zero (i.e., there exists $x in [0,1]$ such that $f(x)=0$, or that $f$ is identically zero (i.e., for all $x in [0,1]$, $f(x)=0$? Without quantifiers, it's hard for me to interpret your question.
– MMASRP63
Nov 29 '18 at 20:07






2




2




The keyword here is Grönwall's inequality. It has probably been discussed on this site before.
– MaoWao
Nov 29 '18 at 20:12




The keyword here is Grönwall's inequality. It has probably been discussed on this site before.
– MaoWao
Nov 29 '18 at 20:12












identically zero, sorry
– Lance
Nov 29 '18 at 20:14




identically zero, sorry
– Lance
Nov 29 '18 at 20:14












Here is an idea. Define the set $mathscr{I}$ of the points $x in [0. 1]$ such that the result is true on the interval $[0, x].$ Show $mathscr{I} neq varnothing$ and that $1 = sup mathscr{I} in mathscr{I}.$
– Will M.
Nov 29 '18 at 20:15




Here is an idea. Define the set $mathscr{I}$ of the points $x in [0. 1]$ such that the result is true on the interval $[0, x].$ Show $mathscr{I} neq varnothing$ and that $1 = sup mathscr{I} in mathscr{I}.$
– Will M.
Nov 29 '18 at 20:15










3 Answers
3






active

oldest

votes


















2














As mentioned in the comments, this is a special case of Grönwall's inequality. One can proceed directly as follows:



Define $g:[0, 1] to Bbb R$ as
$$
g(x) = e^{-2x} int_0^x f(t) , dt , .
$$
Then $g$ is non-negative with $g(0) = 0$ and decreasing:
$$
g'(x) = e^{-2x} left( f(x) - 2 int_0^x f(t) , dt right) le 0 , .
$$
It follows that $g$ (and consequently $f$) is identically zero
on the interval.



Note that the upper bound for $f$ is not needed for this conclusion, only the
non-negativity.






share|cite|improve this answer





















  • To evaluate $g'$ you use FTC and thus continuity of $f$ is used implicitly. But +1 for the trick of multiplication by the exponential factor.
    – Paramanand Singh
    Nov 30 '18 at 17:27










  • @ParamanandSingh: $f$ needs to be integrable in some sense (Riemann or Lebesgue), otherwise the given inequality makes no sense. Now my knowledge of Lebesgue integration theory is a bit rusty. Isn't $int_0^x f(t) dt$ (and thus $g$) differentiable almost everywhere, and equal to the integral of its derivative? Would that be sufficient to draw the same conclusion?
    – Martin R
    Nov 30 '18 at 19:55



















6














One way to proceed is to iterate the inequality.



Note that since $f(t) leq 1$, we have that
$$ f(x) leq 2 int_0^x 1 dt = frac{2^1}{1!}x = 2x.$$
Thus
$$ f(x) leq 2 int_0^x 2t dt = frac{2^2}{2!}x^2 = 2x^2.$$
Iterating again shows that
$$ f(x) leq frac{2^3}{3!} x^3 = frac{4}{3} x^3,$$
and more generally that
$$ f(x) leq frac{2^n}{n!} x^n$$
for any $n geq 1$.



But for any $x in [0,1]$, this tends to $0$.






share|cite|improve this answer

















  • 1




    This does not use any assumption of continuity of $f$ like the other answer implicitly does. +1
    – Paramanand Singh
    Nov 30 '18 at 17:24



















0














hint



We have that $$forall x,Xin[0,1]$$



$$0le f(x)le 1$$



and



$$f(X)le 2int_0^Xf(t)dt$$



if $X=x=0$ we find that



$$0le f(0)le 2times 0$$






share|cite|improve this answer





















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    3 Answers
    3






    active

    oldest

    votes








    3 Answers
    3






    active

    oldest

    votes









    active

    oldest

    votes






    active

    oldest

    votes









    2














    As mentioned in the comments, this is a special case of Grönwall's inequality. One can proceed directly as follows:



    Define $g:[0, 1] to Bbb R$ as
    $$
    g(x) = e^{-2x} int_0^x f(t) , dt , .
    $$
    Then $g$ is non-negative with $g(0) = 0$ and decreasing:
    $$
    g'(x) = e^{-2x} left( f(x) - 2 int_0^x f(t) , dt right) le 0 , .
    $$
    It follows that $g$ (and consequently $f$) is identically zero
    on the interval.



    Note that the upper bound for $f$ is not needed for this conclusion, only the
    non-negativity.






    share|cite|improve this answer





















    • To evaluate $g'$ you use FTC and thus continuity of $f$ is used implicitly. But +1 for the trick of multiplication by the exponential factor.
      – Paramanand Singh
      Nov 30 '18 at 17:27










    • @ParamanandSingh: $f$ needs to be integrable in some sense (Riemann or Lebesgue), otherwise the given inequality makes no sense. Now my knowledge of Lebesgue integration theory is a bit rusty. Isn't $int_0^x f(t) dt$ (and thus $g$) differentiable almost everywhere, and equal to the integral of its derivative? Would that be sufficient to draw the same conclusion?
      – Martin R
      Nov 30 '18 at 19:55
















    2














    As mentioned in the comments, this is a special case of Grönwall's inequality. One can proceed directly as follows:



    Define $g:[0, 1] to Bbb R$ as
    $$
    g(x) = e^{-2x} int_0^x f(t) , dt , .
    $$
    Then $g$ is non-negative with $g(0) = 0$ and decreasing:
    $$
    g'(x) = e^{-2x} left( f(x) - 2 int_0^x f(t) , dt right) le 0 , .
    $$
    It follows that $g$ (and consequently $f$) is identically zero
    on the interval.



    Note that the upper bound for $f$ is not needed for this conclusion, only the
    non-negativity.






    share|cite|improve this answer





















    • To evaluate $g'$ you use FTC and thus continuity of $f$ is used implicitly. But +1 for the trick of multiplication by the exponential factor.
      – Paramanand Singh
      Nov 30 '18 at 17:27










    • @ParamanandSingh: $f$ needs to be integrable in some sense (Riemann or Lebesgue), otherwise the given inequality makes no sense. Now my knowledge of Lebesgue integration theory is a bit rusty. Isn't $int_0^x f(t) dt$ (and thus $g$) differentiable almost everywhere, and equal to the integral of its derivative? Would that be sufficient to draw the same conclusion?
      – Martin R
      Nov 30 '18 at 19:55














    2












    2








    2






    As mentioned in the comments, this is a special case of Grönwall's inequality. One can proceed directly as follows:



    Define $g:[0, 1] to Bbb R$ as
    $$
    g(x) = e^{-2x} int_0^x f(t) , dt , .
    $$
    Then $g$ is non-negative with $g(0) = 0$ and decreasing:
    $$
    g'(x) = e^{-2x} left( f(x) - 2 int_0^x f(t) , dt right) le 0 , .
    $$
    It follows that $g$ (and consequently $f$) is identically zero
    on the interval.



    Note that the upper bound for $f$ is not needed for this conclusion, only the
    non-negativity.






    share|cite|improve this answer












    As mentioned in the comments, this is a special case of Grönwall's inequality. One can proceed directly as follows:



    Define $g:[0, 1] to Bbb R$ as
    $$
    g(x) = e^{-2x} int_0^x f(t) , dt , .
    $$
    Then $g$ is non-negative with $g(0) = 0$ and decreasing:
    $$
    g'(x) = e^{-2x} left( f(x) - 2 int_0^x f(t) , dt right) le 0 , .
    $$
    It follows that $g$ (and consequently $f$) is identically zero
    on the interval.



    Note that the upper bound for $f$ is not needed for this conclusion, only the
    non-negativity.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 30 '18 at 6:10









    Martin RMartin R

    27.3k33254




    27.3k33254












    • To evaluate $g'$ you use FTC and thus continuity of $f$ is used implicitly. But +1 for the trick of multiplication by the exponential factor.
      – Paramanand Singh
      Nov 30 '18 at 17:27










    • @ParamanandSingh: $f$ needs to be integrable in some sense (Riemann or Lebesgue), otherwise the given inequality makes no sense. Now my knowledge of Lebesgue integration theory is a bit rusty. Isn't $int_0^x f(t) dt$ (and thus $g$) differentiable almost everywhere, and equal to the integral of its derivative? Would that be sufficient to draw the same conclusion?
      – Martin R
      Nov 30 '18 at 19:55


















    • To evaluate $g'$ you use FTC and thus continuity of $f$ is used implicitly. But +1 for the trick of multiplication by the exponential factor.
      – Paramanand Singh
      Nov 30 '18 at 17:27










    • @ParamanandSingh: $f$ needs to be integrable in some sense (Riemann or Lebesgue), otherwise the given inequality makes no sense. Now my knowledge of Lebesgue integration theory is a bit rusty. Isn't $int_0^x f(t) dt$ (and thus $g$) differentiable almost everywhere, and equal to the integral of its derivative? Would that be sufficient to draw the same conclusion?
      – Martin R
      Nov 30 '18 at 19:55
















    To evaluate $g'$ you use FTC and thus continuity of $f$ is used implicitly. But +1 for the trick of multiplication by the exponential factor.
    – Paramanand Singh
    Nov 30 '18 at 17:27




    To evaluate $g'$ you use FTC and thus continuity of $f$ is used implicitly. But +1 for the trick of multiplication by the exponential factor.
    – Paramanand Singh
    Nov 30 '18 at 17:27












    @ParamanandSingh: $f$ needs to be integrable in some sense (Riemann or Lebesgue), otherwise the given inequality makes no sense. Now my knowledge of Lebesgue integration theory is a bit rusty. Isn't $int_0^x f(t) dt$ (and thus $g$) differentiable almost everywhere, and equal to the integral of its derivative? Would that be sufficient to draw the same conclusion?
    – Martin R
    Nov 30 '18 at 19:55




    @ParamanandSingh: $f$ needs to be integrable in some sense (Riemann or Lebesgue), otherwise the given inequality makes no sense. Now my knowledge of Lebesgue integration theory is a bit rusty. Isn't $int_0^x f(t) dt$ (and thus $g$) differentiable almost everywhere, and equal to the integral of its derivative? Would that be sufficient to draw the same conclusion?
    – Martin R
    Nov 30 '18 at 19:55











    6














    One way to proceed is to iterate the inequality.



    Note that since $f(t) leq 1$, we have that
    $$ f(x) leq 2 int_0^x 1 dt = frac{2^1}{1!}x = 2x.$$
    Thus
    $$ f(x) leq 2 int_0^x 2t dt = frac{2^2}{2!}x^2 = 2x^2.$$
    Iterating again shows that
    $$ f(x) leq frac{2^3}{3!} x^3 = frac{4}{3} x^3,$$
    and more generally that
    $$ f(x) leq frac{2^n}{n!} x^n$$
    for any $n geq 1$.



    But for any $x in [0,1]$, this tends to $0$.






    share|cite|improve this answer

















    • 1




      This does not use any assumption of continuity of $f$ like the other answer implicitly does. +1
      – Paramanand Singh
      Nov 30 '18 at 17:24
















    6














    One way to proceed is to iterate the inequality.



    Note that since $f(t) leq 1$, we have that
    $$ f(x) leq 2 int_0^x 1 dt = frac{2^1}{1!}x = 2x.$$
    Thus
    $$ f(x) leq 2 int_0^x 2t dt = frac{2^2}{2!}x^2 = 2x^2.$$
    Iterating again shows that
    $$ f(x) leq frac{2^3}{3!} x^3 = frac{4}{3} x^3,$$
    and more generally that
    $$ f(x) leq frac{2^n}{n!} x^n$$
    for any $n geq 1$.



    But for any $x in [0,1]$, this tends to $0$.






    share|cite|improve this answer

















    • 1




      This does not use any assumption of continuity of $f$ like the other answer implicitly does. +1
      – Paramanand Singh
      Nov 30 '18 at 17:24














    6












    6








    6






    One way to proceed is to iterate the inequality.



    Note that since $f(t) leq 1$, we have that
    $$ f(x) leq 2 int_0^x 1 dt = frac{2^1}{1!}x = 2x.$$
    Thus
    $$ f(x) leq 2 int_0^x 2t dt = frac{2^2}{2!}x^2 = 2x^2.$$
    Iterating again shows that
    $$ f(x) leq frac{2^3}{3!} x^3 = frac{4}{3} x^3,$$
    and more generally that
    $$ f(x) leq frac{2^n}{n!} x^n$$
    for any $n geq 1$.



    But for any $x in [0,1]$, this tends to $0$.






    share|cite|improve this answer












    One way to proceed is to iterate the inequality.



    Note that since $f(t) leq 1$, we have that
    $$ f(x) leq 2 int_0^x 1 dt = frac{2^1}{1!}x = 2x.$$
    Thus
    $$ f(x) leq 2 int_0^x 2t dt = frac{2^2}{2!}x^2 = 2x^2.$$
    Iterating again shows that
    $$ f(x) leq frac{2^3}{3!} x^3 = frac{4}{3} x^3,$$
    and more generally that
    $$ f(x) leq frac{2^n}{n!} x^n$$
    for any $n geq 1$.



    But for any $x in [0,1]$, this tends to $0$.







    share|cite|improve this answer












    share|cite|improve this answer



    share|cite|improve this answer










    answered Nov 29 '18 at 20:21









    davidlowryduda♦davidlowryduda

    74.4k7117251




    74.4k7117251








    • 1




      This does not use any assumption of continuity of $f$ like the other answer implicitly does. +1
      – Paramanand Singh
      Nov 30 '18 at 17:24














    • 1




      This does not use any assumption of continuity of $f$ like the other answer implicitly does. +1
      – Paramanand Singh
      Nov 30 '18 at 17:24








    1




    1




    This does not use any assumption of continuity of $f$ like the other answer implicitly does. +1
    – Paramanand Singh
    Nov 30 '18 at 17:24




    This does not use any assumption of continuity of $f$ like the other answer implicitly does. +1
    – Paramanand Singh
    Nov 30 '18 at 17:24











    0














    hint



    We have that $$forall x,Xin[0,1]$$



    $$0le f(x)le 1$$



    and



    $$f(X)le 2int_0^Xf(t)dt$$



    if $X=x=0$ we find that



    $$0le f(0)le 2times 0$$






    share|cite|improve this answer


























      0














      hint



      We have that $$forall x,Xin[0,1]$$



      $$0le f(x)le 1$$



      and



      $$f(X)le 2int_0^Xf(t)dt$$



      if $X=x=0$ we find that



      $$0le f(0)le 2times 0$$






      share|cite|improve this answer
























        0












        0








        0






        hint



        We have that $$forall x,Xin[0,1]$$



        $$0le f(x)le 1$$



        and



        $$f(X)le 2int_0^Xf(t)dt$$



        if $X=x=0$ we find that



        $$0le f(0)le 2times 0$$






        share|cite|improve this answer












        hint



        We have that $$forall x,Xin[0,1]$$



        $$0le f(x)le 1$$



        and



        $$f(X)le 2int_0^Xf(t)dt$$



        if $X=x=0$ we find that



        $$0le f(0)le 2times 0$$







        share|cite|improve this answer












        share|cite|improve this answer



        share|cite|improve this answer










        answered Nov 29 '18 at 20:02









        hamam_Abdallahhamam_Abdallah

        38.1k21634




        38.1k21634






























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