Intuitive reasoning that a function can't exist
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Sorry for the title; I had no idea how to make it concise yet informative.
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
$endgroup$
add a comment |
$begingroup$
Sorry for the title; I had no idea how to make it concise yet informative.
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
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1
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$f''' > 0$ means that it grows at least as fast as $x^3$ but then it will be $f > x^2$ in either $[0,1]$ or $[1,infty]$.
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– Eelvex
8 hours ago
11
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@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
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– SvanN
8 hours ago
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@SvanN, indeed.
$endgroup$
– Eelvex
6 hours ago
add a comment |
$begingroup$
Sorry for the title; I had no idea how to make it concise yet informative.
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
$endgroup$
Sorry for the title; I had no idea how to make it concise yet informative.
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
real-analysis functions
edited 8 hours ago
SvanN
1,9921422
1,9921422
asked 8 hours ago
Math-funMath-fun
7,0581426
7,0581426
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
8 hours ago
11
$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
8 hours ago
$begingroup$
@SvanN, indeed.
$endgroup$
– Eelvex
6 hours ago
add a comment |
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
8 hours ago
11
$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
8 hours ago
$begingroup$
@SvanN, indeed.
$endgroup$
– Eelvex
6 hours ago
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
8 hours ago
$begingroup$
$f''' > 0$ 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
8 hours ago
11
11
$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
8 hours ago
$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
8 hours ago
$begingroup$
@SvanN, indeed.
$endgroup$
– Eelvex
6 hours ago
$begingroup$
@SvanN, indeed.
$endgroup$
– Eelvex
6 hours ago
add a comment |
3 Answers
3
active
oldest
votes
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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
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2
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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 $x > 0$.
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– Daniel Schepler
7 hours ago
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@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 $2$
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– Henry
7 hours ago
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Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
$endgroup$
– Math-fun
5 hours ago
add a comment |
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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 $$f'(x) = int_0^x f''(t) dt < int_0^x frac{pi }{2} dt = frac{pi x}{2}$$
$$f(x) = int_0^x f'(t) dt < int_0^x frac{pi t}{2} dt = frac{pi x^2}{4}$$
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I will check the functional forms soon, this looks quite interesting! +1
$endgroup$
– Math-fun
5 hours ago
add a comment |
$begingroup$
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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You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
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– Math-fun
5 hours ago
add a comment |
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3 Answers
3
active
oldest
votes
3 Answers
3
active
oldest
votes
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active
oldest
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$begingroup$
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
$endgroup$
2
$begingroup$
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
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
$endgroup$
– Math-fun
5 hours ago
add a comment |
$begingroup$
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
$endgroup$
2
$begingroup$
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
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
$endgroup$
– Math-fun
5 hours ago
add a comment |
$begingroup$
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
$endgroup$
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
edited 7 hours ago
answered 8 hours ago
HenryHenry
99k477164
99k477164
2
$begingroup$
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
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
$endgroup$
– Math-fun
5 hours ago
add a comment |
2
$begingroup$
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
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
$endgroup$
– Math-fun
5 hours ago
2
2
$begingroup$
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
7 hours ago
$begingroup$
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
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
@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
7 hours ago
$begingroup$
Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
$endgroup$
– Math-fun
5 hours ago
$begingroup$
Many thanks for this counter example which together with other very nice hints makes the issue clear to me!
$endgroup$
– Math-fun
5 hours ago
add a comment |
$begingroup$
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 $$f'(x) = int_0^x f''(t) dt < int_0^x frac{pi }{2} dt = frac{pi x}{2}$$
$$f(x) = int_0^x f'(t) dt < int_0^x frac{pi t}{2} dt = frac{pi x^2}{4}$$
$endgroup$
$begingroup$
I will check the functional forms soon, this looks quite interesting! +1
$endgroup$
– Math-fun
5 hours ago
add a comment |
$begingroup$
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 $$f'(x) = int_0^x f''(t) dt < int_0^x frac{pi }{2} dt = frac{pi x}{2}$$
$$f(x) = int_0^x f'(t) dt < int_0^x frac{pi t}{2} dt = frac{pi x^2}{4}$$
$endgroup$
$begingroup$
I will check the functional forms soon, this looks quite interesting! +1
$endgroup$
– Math-fun
5 hours ago
add a comment |
$begingroup$
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 $$f'(x) = int_0^x f''(t) dt < int_0^x frac{pi }{2} dt = frac{pi x}{2}$$
$$f(x) = int_0^x f'(t) dt < int_0^x frac{pi t}{2} dt = frac{pi x^2}{4}$$
$endgroup$
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 $$f'(x) = int_0^x f''(t) dt < int_0^x frac{pi }{2} dt = frac{pi x}{2}$$
$$f(x) = int_0^x f'(t) dt < int_0^x frac{pi t}{2} dt = frac{pi x^2}{4}$$
answered 6 hours ago
Will JagyWill Jagy
102k5101199
102k5101199
$begingroup$
I will check the functional forms soon, this looks quite interesting! +1
$endgroup$
– Math-fun
5 hours ago
add a comment |
$begingroup$
I will check the functional forms soon, this looks quite interesting! +1
$endgroup$
– Math-fun
5 hours ago
$begingroup$
I will check the functional forms soon, this looks quite interesting! +1
$endgroup$
– Math-fun
5 hours ago
$begingroup$
I will check the functional forms soon, this looks quite interesting! +1
$endgroup$
– Math-fun
5 hours ago
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
$endgroup$
– Math-fun
5 hours ago
add a comment |
$begingroup$
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.
$endgroup$
$begingroup$
You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
$endgroup$
– Math-fun
5 hours ago
add a comment |
$begingroup$
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.
$endgroup$
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.
answered 6 hours ago
Michael SeifertMichael Seifert
4,797624
4,797624
$begingroup$
You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
$endgroup$
– Math-fun
5 hours ago
add a comment |
$begingroup$
You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
$endgroup$
– Math-fun
5 hours ago
$begingroup$
You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
$endgroup$
– Math-fun
5 hours ago
$begingroup$
You are indeed right. Given the other counter example and the comments I guess I understood this issue :-) +1
$endgroup$
– Math-fun
5 hours ago
add a comment |
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$f''' > 0$ 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
8 hours ago
11
$begingroup$
@Eelvex I would be careful with that: $f''' > 0$ need not imply that $f''' > varepsilon$ for some constant $varepsilon > 0$.
$endgroup$
– SvanN
8 hours ago
$begingroup$
@SvanN, indeed.
$endgroup$
– Eelvex
6 hours ago