$f : mathbb{R}^+ → mathbb{R}$ with $f(0) = f'(0) = 0$ and $f(x) 0$?












18












$begingroup$


I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.



Here is the statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.




I appreciate any help!










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$endgroup$








  • 1




    $begingroup$
    $f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
    $endgroup$
    – Eelvex
    Jan 17 at 18:14






  • 16




    $begingroup$
    @Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
    $endgroup$
    – SvanN
    Jan 17 at 18:36
















18












$begingroup$


I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.



Here is the statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.




I appreciate any help!










share|cite|improve this question











$endgroup$








  • 1




    $begingroup$
    $f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
    $endgroup$
    – Eelvex
    Jan 17 at 18:14






  • 16




    $begingroup$
    @Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
    $endgroup$
    – SvanN
    Jan 17 at 18:36














18












18








18


2



$begingroup$


I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.



Here is the statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.




I appreciate any help!










share|cite|improve this question











$endgroup$




I want to intuitively argue that there is no function with some properties, and find it tricky to explain it to someone who just understands that derivatives are representative of increase rates of a function.



Here is the statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the first, second and third derivative of $f(x)$ are strictly positive on $x>0$.




I appreciate any help!







real-analysis functions derivatives






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share|cite|improve this question













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share|cite|improve this question








edited 2 days ago









user21820

38.8k543153




38.8k543153










asked Jan 17 at 18:04









Math-funMath-fun

7,1081527




7,1081527








  • 1




    $begingroup$
    $f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
    $endgroup$
    – Eelvex
    Jan 17 at 18:14






  • 16




    $begingroup$
    @Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
    $endgroup$
    – SvanN
    Jan 17 at 18:36














  • 1




    $begingroup$
    $f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
    $endgroup$
    – Eelvex
    Jan 17 at 18:14






  • 16




    $begingroup$
    @Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
    $endgroup$
    – SvanN
    Jan 17 at 18:36








1




1




$begingroup$
$f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
$endgroup$
– Eelvex
Jan 17 at 18:14




$begingroup$
$f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
$endgroup$
– Eelvex
Jan 17 at 18:14




16




16




$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
Jan 17 at 18:36




$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
Jan 17 at 18:36










3 Answers
3






active

oldest

votes


















41












$begingroup$

I do not think it is true. Take for example




  • $f(x) = 1 - x + frac12x^2 - e^{-x}$

  • $f'(x) = - 1 + x + e^{-x}$

  • $f''(x) = 1 - e^{-x}$

  • $f'''(x) = e^{-x}$


Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases



enter image description here






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
    $endgroup$
    – Daniel Schepler
    Jan 17 at 19:06






  • 1




    $begingroup$
    @DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
    $endgroup$
    – Henry
    Jan 17 at 19:15






  • 2




    $begingroup$
    Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
    $endgroup$
    – Math-fun
    Jan 17 at 20:44










  • $begingroup$
    How did you conclude that the second derivative of $f$ at $0$ is $0$, and at $x>0$ is $1$?
    $endgroup$
    – D777
    2 days ago








  • 1




    $begingroup$
    @D777 - You do not need the second derivative to be $0$ at $0$, but you do need it to be less than $2$ everywhere and to be positive and increasing for $x gt 0$ so at least $0$ at $0$, and an easy counterexample comes from having it go from $0$ towards $1$.
    $endgroup$
    – Henry
    2 days ago





















14












$begingroup$

Your intuition is correct for a slightly different statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the third derivative of $f(x)$ is bounded below by some $epsilon > 0$ on $x>0$.




The argument is as follows: Suppose that such a function exists. Then we have $f'''(x) geq epsilon$, which means that (integrating both sides)
$$
f''(x) - f''(0) geq epsilon x.
$$

Denote $f''(0) = a$, so that $f''(x) geq epsilon x + a$. Integrating two more times, we have
$$
f'(x) geq frac{1}{2} epsilon x^2 + a x + f'(0) = frac{1}{2} epsilon x^2 + a x
$$
$$
f(x) geq frac{1}{6} epsilon x^3 + a x^2 + f(0) = frac{1}{6} epsilon x^3 + frac{1}{2} a x^2
$$

But if $f(x) < x^2$, we have
$$
x^2 > f(x) geq frac{1}{6} epsilon x^3 + frac{1}{2} a x^2
$$

for all $x > 0$, which reduces to
$$
frac{6(1 - a/2)}{epsilon} > x.
$$

for all $x > 0$. For any value of $epsilon > 0$ and $a in mathbb{R}$, this will be violated by some $x > 0$, and we have a contradiction.



Intuitionally: if the third derivative is always at least some value, then we know that the function has to grow at least as fast as $frac{1}{6} x^3$ times that value. But any function that grows as fast as $x^3$ (times any positive number) must have a greater value than $x^2$ at some point.






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$endgroup$













  • $begingroup$
    You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
    $endgroup$
    – Math-fun
    Jan 17 at 20:46



















9












$begingroup$

As in the first answer and comment by Daniel, we can choose any $f'''$ that is positive as long as its definite integral out to infinity is finite and smaller than 2. then we back up, each function is the definite integral from $0.$



$$ f'''(x) = frac{1}{1+x^2} $$
$$ f''(x) = arctan x $$
$$ f'(x) = x arctan x - frac{1}{2} log left(1+x^2 right) $$
$$ f(x) = left( frac{x^2 -1}{2} right) arctan x - frac{x}{2} log left(1+x^2 right) + frac{x}{2}$$



As $f'' < frac{pi}{2},$ we get $$f'(x) = int_0^x f''(t) dt < int_0^x frac{pi }{2} dt = frac{pi x}{2}$$
$$f(x) = int_0^x f'(t) dt < int_0^x frac{pi t}{2} dt = frac{pi x^2}{4}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I will check the functional forms soon, this looks quite interesting! +1
    $endgroup$
    – Math-fun
    Jan 17 at 20:45











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3 Answers
3






active

oldest

votes








3 Answers
3






active

oldest

votes









active

oldest

votes






active

oldest

votes









41












$begingroup$

I do not think it is true. Take for example




  • $f(x) = 1 - x + frac12x^2 - e^{-x}$

  • $f'(x) = - 1 + x + e^{-x}$

  • $f''(x) = 1 - e^{-x}$

  • $f'''(x) = e^{-x}$


Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases



enter image description here






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
    $endgroup$
    – Daniel Schepler
    Jan 17 at 19:06






  • 1




    $begingroup$
    @DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
    $endgroup$
    – Henry
    Jan 17 at 19:15






  • 2




    $begingroup$
    Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
    $endgroup$
    – Math-fun
    Jan 17 at 20:44










  • $begingroup$
    How did you conclude that the second derivative of $f$ at $0$ is $0$, and at $x>0$ is $1$?
    $endgroup$
    – D777
    2 days ago








  • 1




    $begingroup$
    @D777 - You do not need the second derivative to be $0$ at $0$, but you do need it to be less than $2$ everywhere and to be positive and increasing for $x gt 0$ so at least $0$ at $0$, and an easy counterexample comes from having it go from $0$ towards $1$.
    $endgroup$
    – Henry
    2 days ago


















41












$begingroup$

I do not think it is true. Take for example




  • $f(x) = 1 - x + frac12x^2 - e^{-x}$

  • $f'(x) = - 1 + x + e^{-x}$

  • $f''(x) = 1 - e^{-x}$

  • $f'''(x) = e^{-x}$


Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases



enter image description here






share|cite|improve this answer











$endgroup$









  • 3




    $begingroup$
    Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
    $endgroup$
    – Daniel Schepler
    Jan 17 at 19:06






  • 1




    $begingroup$
    @DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
    $endgroup$
    – Henry
    Jan 17 at 19:15






  • 2




    $begingroup$
    Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
    $endgroup$
    – Math-fun
    Jan 17 at 20:44










  • $begingroup$
    How did you conclude that the second derivative of $f$ at $0$ is $0$, and at $x>0$ is $1$?
    $endgroup$
    – D777
    2 days ago








  • 1




    $begingroup$
    @D777 - You do not need the second derivative to be $0$ at $0$, but you do need it to be less than $2$ everywhere and to be positive and increasing for $x gt 0$ so at least $0$ at $0$, and an easy counterexample comes from having it go from $0$ towards $1$.
    $endgroup$
    – Henry
    2 days ago
















41












41








41





$begingroup$

I do not think it is true. Take for example




  • $f(x) = 1 - x + frac12x^2 - e^{-x}$

  • $f'(x) = - 1 + x + e^{-x}$

  • $f''(x) = 1 - e^{-x}$

  • $f'''(x) = e^{-x}$


Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases



enter image description here






share|cite|improve this answer











$endgroup$



I do not think it is true. Take for example




  • $f(x) = 1 - x + frac12x^2 - e^{-x}$

  • $f'(x) = - 1 + x + e^{-x}$

  • $f''(x) = 1 - e^{-x}$

  • $f'''(x) = e^{-x}$


Here is a graph: note that the black line for $x^2$ is above the dark blue line for $f(x)$, while the green line for $f''(x)$ stays below $1$ while the second derivative of $x^2$ is $2$, and the red line for $f'''(x)$ tends towards $0$ from above as $x$ increases



enter image description here







share|cite|improve this answer














share|cite|improve this answer



share|cite|improve this answer








edited Jan 17 at 18:56

























answered Jan 17 at 18:32









HenryHenry

99.3k479165




99.3k479165








  • 3




    $begingroup$
    Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
    $endgroup$
    – Daniel Schepler
    Jan 17 at 19:06






  • 1




    $begingroup$
    @DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
    $endgroup$
    – Henry
    Jan 17 at 19:15






  • 2




    $begingroup$
    Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
    $endgroup$
    – Math-fun
    Jan 17 at 20:44










  • $begingroup$
    How did you conclude that the second derivative of $f$ at $0$ is $0$, and at $x>0$ is $1$?
    $endgroup$
    – D777
    2 days ago








  • 1




    $begingroup$
    @D777 - You do not need the second derivative to be $0$ at $0$, but you do need it to be less than $2$ everywhere and to be positive and increasing for $x gt 0$ so at least $0$ at $0$, and an easy counterexample comes from having it go from $0$ towards $1$.
    $endgroup$
    – Henry
    2 days ago
















  • 3




    $begingroup$
    Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
    $endgroup$
    – Daniel Schepler
    Jan 17 at 19:06






  • 1




    $begingroup$
    @DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
    $endgroup$
    – Henry
    Jan 17 at 19:15






  • 2




    $begingroup$
    Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
    $endgroup$
    – Math-fun
    Jan 17 at 20:44










  • $begingroup$
    How did you conclude that the second derivative of $f$ at $0$ is $0$, and at $x>0$ is $1$?
    $endgroup$
    – D777
    2 days ago








  • 1




    $begingroup$
    @D777 - You do not need the second derivative to be $0$ at $0$, but you do need it to be less than $2$ everywhere and to be positive and increasing for $x gt 0$ so at least $0$ at $0$, and an easy counterexample comes from having it go from $0$ towards $1$.
    $endgroup$
    – Henry
    2 days ago










3




3




$begingroup$
Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
$endgroup$
– Daniel Schepler
Jan 17 at 19:06




$begingroup$
Analytically, once you verify $f(0) = f'(0) = f''(0) = 0$ then $f'''(x) > 0$ for $x>0$ implies $f''(x) > 0$ for $x > 0$, which implies $f'(x) > 0$ for $x > 0$, which implies $f(x) > 0$ for $x > 0$. Whereas $f''(x) < 1$ for $x > 0$ along with the initial conditions implies $f'(x) < x$ for $x > 0$, implying $f(x) < frac{1}{2} x^2$ for $x > 0$.
$endgroup$
– Daniel Schepler
Jan 17 at 19:06




1




1




$begingroup$
@DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
$endgroup$
– Henry
Jan 17 at 19:15




$begingroup$
@DanielSchepler - yes indeed. Choosing $f''(x)=1-e^{-x}$ was the key to the counterexample, though there will be others with $f''(0)=0$ then increasing to less than $2$
$endgroup$
– Henry
Jan 17 at 19:15




2




2




$begingroup$
Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
$endgroup$
– Math-fun
Jan 17 at 20:44




$begingroup$
Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
$endgroup$
– Math-fun
Jan 17 at 20:44












$begingroup$
How did you conclude that the second derivative of $f$ at $0$ is $0$, and at $x>0$ is $1$?
$endgroup$
– D777
2 days ago






$begingroup$
How did you conclude that the second derivative of $f$ at $0$ is $0$, and at $x>0$ is $1$?
$endgroup$
– D777
2 days ago






1




1




$begingroup$
@D777 - You do not need the second derivative to be $0$ at $0$, but you do need it to be less than $2$ everywhere and to be positive and increasing for $x gt 0$ so at least $0$ at $0$, and an easy counterexample comes from having it go from $0$ towards $1$.
$endgroup$
– Henry
2 days ago






$begingroup$
@D777 - You do not need the second derivative to be $0$ at $0$, but you do need it to be less than $2$ everywhere and to be positive and increasing for $x gt 0$ so at least $0$ at $0$, and an easy counterexample comes from having it go from $0$ towards $1$.
$endgroup$
– Henry
2 days ago













14












$begingroup$

Your intuition is correct for a slightly different statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the third derivative of $f(x)$ is bounded below by some $epsilon > 0$ on $x>0$.




The argument is as follows: Suppose that such a function exists. Then we have $f'''(x) geq epsilon$, which means that (integrating both sides)
$$
f''(x) - f''(0) geq epsilon x.
$$

Denote $f''(0) = a$, so that $f''(x) geq epsilon x + a$. Integrating two more times, we have
$$
f'(x) geq frac{1}{2} epsilon x^2 + a x + f'(0) = frac{1}{2} epsilon x^2 + a x
$$
$$
f(x) geq frac{1}{6} epsilon x^3 + a x^2 + f(0) = frac{1}{6} epsilon x^3 + frac{1}{2} a x^2
$$

But if $f(x) < x^2$, we have
$$
x^2 > f(x) geq frac{1}{6} epsilon x^3 + frac{1}{2} a x^2
$$

for all $x > 0$, which reduces to
$$
frac{6(1 - a/2)}{epsilon} > x.
$$

for all $x > 0$. For any value of $epsilon > 0$ and $a in mathbb{R}$, this will be violated by some $x > 0$, and we have a contradiction.



Intuitionally: if the third derivative is always at least some value, then we know that the function has to grow at least as fast as $frac{1}{6} x^3$ times that value. But any function that grows as fast as $x^3$ (times any positive number) must have a greater value than $x^2$ at some point.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
    $endgroup$
    – Math-fun
    Jan 17 at 20:46
















14












$begingroup$

Your intuition is correct for a slightly different statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the third derivative of $f(x)$ is bounded below by some $epsilon > 0$ on $x>0$.




The argument is as follows: Suppose that such a function exists. Then we have $f'''(x) geq epsilon$, which means that (integrating both sides)
$$
f''(x) - f''(0) geq epsilon x.
$$

Denote $f''(0) = a$, so that $f''(x) geq epsilon x + a$. Integrating two more times, we have
$$
f'(x) geq frac{1}{2} epsilon x^2 + a x + f'(0) = frac{1}{2} epsilon x^2 + a x
$$
$$
f(x) geq frac{1}{6} epsilon x^3 + a x^2 + f(0) = frac{1}{6} epsilon x^3 + frac{1}{2} a x^2
$$

But if $f(x) < x^2$, we have
$$
x^2 > f(x) geq frac{1}{6} epsilon x^3 + frac{1}{2} a x^2
$$

for all $x > 0$, which reduces to
$$
frac{6(1 - a/2)}{epsilon} > x.
$$

for all $x > 0$. For any value of $epsilon > 0$ and $a in mathbb{R}$, this will be violated by some $x > 0$, and we have a contradiction.



Intuitionally: if the third derivative is always at least some value, then we know that the function has to grow at least as fast as $frac{1}{6} x^3$ times that value. But any function that grows as fast as $x^3$ (times any positive number) must have a greater value than $x^2$ at some point.






share|cite|improve this answer









$endgroup$













  • $begingroup$
    You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
    $endgroup$
    – Math-fun
    Jan 17 at 20:46














14












14








14





$begingroup$

Your intuition is correct for a slightly different statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the third derivative of $f(x)$ is bounded below by some $epsilon > 0$ on $x>0$.




The argument is as follows: Suppose that such a function exists. Then we have $f'''(x) geq epsilon$, which means that (integrating both sides)
$$
f''(x) - f''(0) geq epsilon x.
$$

Denote $f''(0) = a$, so that $f''(x) geq epsilon x + a$. Integrating two more times, we have
$$
f'(x) geq frac{1}{2} epsilon x^2 + a x + f'(0) = frac{1}{2} epsilon x^2 + a x
$$
$$
f(x) geq frac{1}{6} epsilon x^3 + a x^2 + f(0) = frac{1}{6} epsilon x^3 + frac{1}{2} a x^2
$$

But if $f(x) < x^2$, we have
$$
x^2 > f(x) geq frac{1}{6} epsilon x^3 + frac{1}{2} a x^2
$$

for all $x > 0$, which reduces to
$$
frac{6(1 - a/2)}{epsilon} > x.
$$

for all $x > 0$. For any value of $epsilon > 0$ and $a in mathbb{R}$, this will be violated by some $x > 0$, and we have a contradiction.



Intuitionally: if the third derivative is always at least some value, then we know that the function has to grow at least as fast as $frac{1}{6} x^3$ times that value. But any function that grows as fast as $x^3$ (times any positive number) must have a greater value than $x^2$ at some point.






share|cite|improve this answer









$endgroup$



Your intuition is correct for a slightly different statement:




There is no function $f(x)$ on $x ge 0$ such that $f(0)=0$, $f'(0)=0$, $f(x)<x^2$ for $x>0$ and that the third derivative of $f(x)$ is bounded below by some $epsilon > 0$ on $x>0$.




The argument is as follows: Suppose that such a function exists. Then we have $f'''(x) geq epsilon$, which means that (integrating both sides)
$$
f''(x) - f''(0) geq epsilon x.
$$

Denote $f''(0) = a$, so that $f''(x) geq epsilon x + a$. Integrating two more times, we have
$$
f'(x) geq frac{1}{2} epsilon x^2 + a x + f'(0) = frac{1}{2} epsilon x^2 + a x
$$
$$
f(x) geq frac{1}{6} epsilon x^3 + a x^2 + f(0) = frac{1}{6} epsilon x^3 + frac{1}{2} a x^2
$$

But if $f(x) < x^2$, we have
$$
x^2 > f(x) geq frac{1}{6} epsilon x^3 + frac{1}{2} a x^2
$$

for all $x > 0$, which reduces to
$$
frac{6(1 - a/2)}{epsilon} > x.
$$

for all $x > 0$. For any value of $epsilon > 0$ and $a in mathbb{R}$, this will be violated by some $x > 0$, and we have a contradiction.



Intuitionally: if the third derivative is always at least some value, then we know that the function has to grow at least as fast as $frac{1}{6} x^3$ times that value. But any function that grows as fast as $x^3$ (times any positive number) must have a greater value than $x^2$ at some point.







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 20:38









Michael SeifertMichael Seifert

4,897625




4,897625












  • $begingroup$
    You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
    $endgroup$
    – Math-fun
    Jan 17 at 20:46


















  • $begingroup$
    You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
    $endgroup$
    – Math-fun
    Jan 17 at 20:46
















$begingroup$
You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
$endgroup$
– Math-fun
Jan 17 at 20:46




$begingroup$
You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
$endgroup$
– Math-fun
Jan 17 at 20:46











9












$begingroup$

As in the first answer and comment by Daniel, we can choose any $f'''$ that is positive as long as its definite integral out to infinity is finite and smaller than 2. then we back up, each function is the definite integral from $0.$



$$ f'''(x) = frac{1}{1+x^2} $$
$$ f''(x) = arctan x $$
$$ f'(x) = x arctan x - frac{1}{2} log left(1+x^2 right) $$
$$ f(x) = left( frac{x^2 -1}{2} right) arctan x - frac{x}{2} log left(1+x^2 right) + frac{x}{2}$$



As $f'' < frac{pi}{2},$ we get $$f'(x) = int_0^x f''(t) dt < int_0^x frac{pi }{2} dt = frac{pi x}{2}$$
$$f(x) = int_0^x f'(t) dt < int_0^x frac{pi t}{2} dt = frac{pi x^2}{4}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I will check the functional forms soon, this looks quite interesting! +1
    $endgroup$
    – Math-fun
    Jan 17 at 20:45
















9












$begingroup$

As in the first answer and comment by Daniel, we can choose any $f'''$ that is positive as long as its definite integral out to infinity is finite and smaller than 2. then we back up, each function is the definite integral from $0.$



$$ f'''(x) = frac{1}{1+x^2} $$
$$ f''(x) = arctan x $$
$$ f'(x) = x arctan x - frac{1}{2} log left(1+x^2 right) $$
$$ f(x) = left( frac{x^2 -1}{2} right) arctan x - frac{x}{2} log left(1+x^2 right) + frac{x}{2}$$



As $f'' < frac{pi}{2},$ we get $$f'(x) = int_0^x f''(t) dt < int_0^x frac{pi }{2} dt = frac{pi x}{2}$$
$$f(x) = int_0^x f'(t) dt < int_0^x frac{pi t}{2} dt = frac{pi x^2}{4}$$






share|cite|improve this answer









$endgroup$













  • $begingroup$
    I will check the functional forms soon, this looks quite interesting! +1
    $endgroup$
    – Math-fun
    Jan 17 at 20:45














9












9








9





$begingroup$

As in the first answer and comment by Daniel, we can choose any $f'''$ that is positive as long as its definite integral out to infinity is finite and smaller than 2. then we back up, each function is the definite integral from $0.$



$$ f'''(x) = frac{1}{1+x^2} $$
$$ f''(x) = arctan x $$
$$ f'(x) = x arctan x - frac{1}{2} log left(1+x^2 right) $$
$$ f(x) = left( frac{x^2 -1}{2} right) arctan x - frac{x}{2} log left(1+x^2 right) + frac{x}{2}$$



As $f'' < frac{pi}{2},$ we get $$f'(x) = int_0^x f''(t) dt < int_0^x frac{pi }{2} dt = frac{pi x}{2}$$
$$f(x) = int_0^x f'(t) dt < int_0^x frac{pi t}{2} dt = frac{pi x^2}{4}$$






share|cite|improve this answer









$endgroup$



As in the first answer and comment by Daniel, we can choose any $f'''$ that is positive as long as its definite integral out to infinity is finite and smaller than 2. then we back up, each function is the definite integral from $0.$



$$ f'''(x) = frac{1}{1+x^2} $$
$$ f''(x) = arctan x $$
$$ f'(x) = x arctan x - frac{1}{2} log left(1+x^2 right) $$
$$ f(x) = left( frac{x^2 -1}{2} right) arctan x - frac{x}{2} log left(1+x^2 right) + frac{x}{2}$$



As $f'' < frac{pi}{2},$ we get $$f'(x) = int_0^x f''(t) dt < int_0^x frac{pi }{2} dt = frac{pi x}{2}$$
$$f(x) = int_0^x f'(t) dt < int_0^x frac{pi t}{2} dt = frac{pi x^2}{4}$$







share|cite|improve this answer












share|cite|improve this answer



share|cite|improve this answer










answered Jan 17 at 20:37









Will JagyWill Jagy

102k5101199




102k5101199












  • $begingroup$
    I will check the functional forms soon, this looks quite interesting! +1
    $endgroup$
    – Math-fun
    Jan 17 at 20:45


















  • $begingroup$
    I will check the functional forms soon, this looks quite interesting! +1
    $endgroup$
    – Math-fun
    Jan 17 at 20:45
















$begingroup$
I will check the functional forms soon, this looks quite interesting! +1
$endgroup$
– Math-fun
Jan 17 at 20:45




$begingroup$
I will check the functional forms soon, this looks quite interesting! +1
$endgroup$
– Math-fun
Jan 17 at 20:45


















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