A function with a bounded derivative
up vote
1
down vote
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Suppose there exist $delta$, $M$ two positive real numbers.
We have a function $f$ that satisfies the following :
$f(0)=0$ , $M<frac{f(x)}{x}, (xne0)$. and $|f'(x))|<delta$.
Can we have a function $f$ such that : $M>delta$?
real-analysis
asked Nov 14 at 16:04
halo anes
62
add a comment |
up vote
1
down vote
favorite
Suppose there exist $delta$, $M$ two positive real numbers.
We have a function $f$ that satisfies the following :
$f(0)=0$ , $M<frac{f(x)}{x}, (xne0)$. and $|f'(x))|<delta$.
Can we have a function $f$ such that : $M>delta$?
real-analysis
asked Nov 14 at 16:04
halo anes
62
add a comment |
up vote
1
down vote
favorite
up vote
1
down vote
favorite
Suppose there exist $delta$, $M$ two positive real numbers.
We have a function $f$ that satisfies the following :
$f(0)=0$ , $M<frac{f(x)}{x}, (xne0)$. and $|f'(x))|<delta$.
Can we have a function $f$ such that : $M>delta$?
real-analysis
asked Nov 14 at 16:04
halo anes
62
Suppose there exist $delta$, $M$ two positive real numbers.
We have a function $f$ that satisfies the following :
$f(0)=0$ , $M<frac{f(x)}{x}, (xne0)$. and $|f'(x))|<delta$.
Can we have a function $f$ such that : $M>delta$?
real-analysis
real-analysis
asked Nov 14 at 16:04
halo anes
62
asked Nov 14 at 16:04
halo anes
62
asked Nov 14 at 16:04
halo anes
62
asked Nov 14 at 16:04
halo anes
62
asked Nov 14 at 16:04
halo anes
62
62
add a comment |
add a comment |
1 Answer
1
active
oldest
votes
up vote
1
down vote
accepted
Suppose $M > delta$, and take $x > 0$ such that $M < f(x)/x$, or equivalently, $f(x) > Mx$. By the mean value theorem, there is a point $c in (0, x)$ such that $f'(c) > (Mx - 0)/(x - 0) = M$.
Thus $M < delta$.
answered Nov 14 at 16:13
davidlowryduda♦
73.7k7116248
Thanks for the answer.
– halo anes
Nov 14 at 16:32
add a comment |
1 Answer
1
active
oldest
votes
1 Answer
1
active
oldest
votes
active
oldest
votes
active
oldest
votes
up vote
1
down vote
accepted
Suppose $M > delta$, and take $x > 0$ such that $M < f(x)/x$, or equivalently, $f(x) > Mx$. By the mean value theorem, there is a point $c in (0, x)$ such that $f'(c) > (Mx - 0)/(x - 0) = M$.
Thus $M < delta$.
answered Nov 14 at 16:13
davidlowryduda♦
73.7k7116248
Thanks for the answer.
– halo anes
Nov 14 at 16:32
add a comment |
up vote
1
down vote
accepted
Suppose $M > delta$, and take $x > 0$ such that $M < f(x)/x$, or equivalently, $f(x) > Mx$. By the mean value theorem, there is a point $c in (0, x)$ such that $f'(c) > (Mx - 0)/(x - 0) = M$.
Thus $M < delta$.
answered Nov 14 at 16:13
davidlowryduda♦
73.7k7116248
Thanks for the answer.
– halo anes
Nov 14 at 16:32
add a comment |
up vote
1
down vote
accepted
up vote
1
down vote
accepted
Suppose $M > delta$, and take $x > 0$ such that $M < f(x)/x$, or equivalently, $f(x) > Mx$. By the mean value theorem, there is a point $c in (0, x)$ such that $f'(c) > (Mx - 0)/(x - 0) = M$.
Thus $M < delta$.
answered Nov 14 at 16:13
davidlowryduda♦
73.7k7116248
Suppose $M > delta$, and take $x > 0$ such that $M < f(x)/x$, or equivalently, $f(x) > Mx$. By the mean value theorem, there is a point $c in (0, x)$ such that $f'(c) > (Mx - 0)/(x - 0) = M$.
Thus $M < delta$.
answered Nov 14 at 16:13
davidlowryduda♦
73.7k7116248
answered Nov 14 at 16:13
davidlowryduda♦
73.7k7116248
answered Nov 14 at 16:13
davidlowryduda♦
73.7k7116248
answered Nov 14 at 16:13
davidlowryduda♦
73.7k7116248
73.7k7116248
Thanks for the answer.
– halo anes
Nov 14 at 16:32
add a comment |
Thanks for the answer.
– halo anes
Nov 14 at 16:32
Thanks for the answer.
– halo anes
Nov 14 at 16:32
Thanks for the answer.
– halo anes
Nov 14 at 16:32
add a comment |
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