$lim_{h to 0^+} frac{mathcal L (f^{-1}([x,x+h)))}{h}$ where $f^{-1}$ is the preimage of a Lipschitz function
$begingroup$
Motivated by the question Compute derivative of $g(x)=mathcal L({y in [a,b]: x >f(y)})$, I'd like to ask:
Let $f:mathbb{R} to mathbb{R}$. How can one compute the limit $$lim_{h to 0^+} frac{mathcal L(f^{-1}([x,x+h))}{h},$$
where $f^{-1}$ denotes the pre-image of $f$; $mathcal L$ the Lebesgue measure; and
assuming only that $f$ is Lipschitz continuous?
In the special case $f$ monotone, the question is much easier, since
$$f^{-1}([x,x+h)) = [f^{-1}(x), f^{-1}(x+h)).$$
real-analysis calculus functional-analysis measure-theory
asked Dec 22 '18 at 14:32
HiroHiro
193
$endgroup$
add a comment |
$begingroup$
Motivated by the question Compute derivative of $g(x)=mathcal L({y in [a,b]: x >f(y)})$, I'd like to ask:
Let $f:mathbb{R} to mathbb{R}$. How can one compute the limit $$lim_{h to 0^+} frac{mathcal L(f^{-1}([x,x+h))}{h},$$
where $f^{-1}$ denotes the pre-image of $f$; $mathcal L$ the Lebesgue measure; and
assuming only that $f$ is Lipschitz continuous?
In the special case $f$ monotone, the question is much easier, since
$$f^{-1}([x,x+h)) = [f^{-1}(x), f^{-1}(x+h)).$$
real-analysis calculus functional-analysis measure-theory
asked Dec 22 '18 at 14:32
HiroHiro
193
$endgroup$
$begingroup$
Why is that not nonsense? Given a set A of real numbers and a nonzero h, what is A/h? If for all nonzero h, g(h) is a set of real numbers, what is the limit of g(h) as h approaches zero? What happens when h is negative?
$endgroup$
– William Elliot
Dec 23 '18 at 3:31
1
$begingroup$
@WilliamElliot I ment the measure of the set, of course. There was a typo.
$endgroup$
– Hiro
Dec 23 '18 at 9:05
$begingroup$
Limit as h -> 0 indicates h can be positive or negative. For h < 0, [x,x+h) is empty, giving a liminf of 0. Thus, if there is a limit, it has to be zero.
$endgroup$
– William Elliot
Dec 23 '18 at 20:41
$begingroup$
@WilliamElliot I've edited the question to address this ambiguity: we consider only $h to 0^+$.
$endgroup$
– Hiro
Dec 24 '18 at 0:07
$begingroup$
@Hiro you are probably after Lebesgue differentiation theorem: en.wikipedia.org/wiki/Lebesgue_differentiation_theorem
$endgroup$
– Will M.
Dec 24 '18 at 0:42
add a comment |
$begingroup$
Motivated by the question Compute derivative of $g(x)=mathcal L({y in [a,b]: x >f(y)})$, I'd like to ask:
Let $f:mathbb{R} to mathbb{R}$. How can one compute the limit $$lim_{h to 0^+} frac{mathcal L(f^{-1}([x,x+h))}{h},$$
where $f^{-1}$ denotes the pre-image of $f$; $mathcal L$ the Lebesgue measure; and
assuming only that $f$ is Lipschitz continuous?
In the special case $f$ monotone, the question is much easier, since
$$f^{-1}([x,x+h)) = [f^{-1}(x), f^{-1}(x+h)).$$
real-analysis calculus functional-analysis measure-theory
asked Dec 22 '18 at 14:32
HiroHiro
193
$endgroup$
Motivated by the question Compute derivative of $g(x)=mathcal L({y in [a,b]: x >f(y)})$, I'd like to ask:
Let $f:mathbb{R} to mathbb{R}$. How can one compute the limit $$lim_{h to 0^+} frac{mathcal L(f^{-1}([x,x+h))}{h},$$
where $f^{-1}$ denotes the pre-image of $f$; $mathcal L$ the Lebesgue measure; and
assuming only that $f$ is Lipschitz continuous?
In the special case $f$ monotone, the question is much easier, since
$$f^{-1}([x,x+h)) = [f^{-1}(x), f^{-1}(x+h)).$$
real-analysis calculus functional-analysis measure-theory
real-analysis calculus functional-analysis measure-theory
asked Dec 22 '18 at 14:32
HiroHiro
193
asked Dec 22 '18 at 14:32
HiroHiro
193
edited Dec 24 '18 at 0:06
Hiro
asked Dec 22 '18 at 14:32
HiroHiro
193
asked Dec 22 '18 at 14:32
HiroHiro
193
asked Dec 22 '18 at 14:32
HiroHiro
193
193
$begingroup$
Why is that not nonsense? Given a set A of real numbers and a nonzero h, what is A/h? If for all nonzero h, g(h) is a set of real numbers, what is the limit of g(h) as h approaches zero? What happens when h is negative?
$endgroup$
– William Elliot
Dec 23 '18 at 3:31
1
$begingroup$
@WilliamElliot I ment the measure of the set, of course. There was a typo.
$endgroup$
– Hiro
Dec 23 '18 at 9:05
$begingroup$
Limit as h -> 0 indicates h can be positive or negative. For h < 0, [x,x+h) is empty, giving a liminf of 0. Thus, if there is a limit, it has to be zero.
$endgroup$
– William Elliot
Dec 23 '18 at 20:41
$begingroup$
@WilliamElliot I've edited the question to address this ambiguity: we consider only $h to 0^+$.
$endgroup$
– Hiro
Dec 24 '18 at 0:07
$begingroup$
@Hiro you are probably after Lebesgue differentiation theorem: en.wikipedia.org/wiki/Lebesgue_differentiation_theorem
$endgroup$
– Will M.
Dec 24 '18 at 0:42
add a comment |
$begingroup$
Why is that not nonsense? Given a set A of real numbers and a nonzero h, what is A/h? If for all nonzero h, g(h) is a set of real numbers, what is the limit of g(h) as h approaches zero? What happens when h is negative?
$endgroup$
– William Elliot
Dec 23 '18 at 3:31
1
$begingroup$
@WilliamElliot I ment the measure of the set, of course. There was a typo.
$endgroup$
– Hiro
Dec 23 '18 at 9:05
$begingroup$
Limit as h -> 0 indicates h can be positive or negative. For h < 0, [x,x+h) is empty, giving a liminf of 0. Thus, if there is a limit, it has to be zero.
$endgroup$
– William Elliot
Dec 23 '18 at 20:41
$begingroup$
@WilliamElliot I've edited the question to address this ambiguity: we consider only $h to 0^+$.
$endgroup$
– Hiro
Dec 24 '18 at 0:07
$begingroup$
@Hiro you are probably after Lebesgue differentiation theorem: en.wikipedia.org/wiki/Lebesgue_differentiation_theorem
$endgroup$
– Will M.
Dec 24 '18 at 0:42
$begingroup$
Why is that not nonsense? Given a set A of real numbers and a nonzero h, what is A/h? If for all nonzero h, g(h) is a set of real numbers, what is the limit of g(h) as h approaches zero? What happens when h is negative?
$endgroup$
– William Elliot
Dec 23 '18 at 3:31
$begingroup$
Why is that not nonsense? Given a set A of real numbers and a nonzero h, what is A/h? If for all nonzero h, g(h) is a set of real numbers, what is the limit of g(h) as h approaches zero? What happens when h is negative?
$endgroup$
– William Elliot
Dec 23 '18 at 3:31
1
1
$begingroup$
@WilliamElliot I ment the measure of the set, of course. There was a typo.
$endgroup$
– Hiro
Dec 23 '18 at 9:05
$begingroup$
@WilliamElliot I ment the measure of the set, of course. There was a typo.
$endgroup$
– Hiro
Dec 23 '18 at 9:05
$begingroup$
Limit as h -> 0 indicates h can be positive or negative. For h < 0, [x,x+h) is empty, giving a liminf of 0. Thus, if there is a limit, it has to be zero.
$endgroup$
– William Elliot
Dec 23 '18 at 20:41
$begingroup$
Limit as h -> 0 indicates h can be positive or negative. For h < 0, [x,x+h) is empty, giving a liminf of 0. Thus, if there is a limit, it has to be zero.
$endgroup$
– William Elliot
Dec 23 '18 at 20:41
$begingroup$
@WilliamElliot I've edited the question to address this ambiguity: we consider only $h to 0^+$.
$endgroup$
– Hiro
Dec 24 '18 at 0:07
$begingroup$
@WilliamElliot I've edited the question to address this ambiguity: we consider only $h to 0^+$.
$endgroup$
– Hiro
Dec 24 '18 at 0:07
$begingroup$
@Hiro you are probably after Lebesgue differentiation theorem: en.wikipedia.org/wiki/Lebesgue_differentiation_theorem
$endgroup$
– Will M.
Dec 24 '18 at 0:42
$begingroup$
@Hiro you are probably after Lebesgue differentiation theorem: en.wikipedia.org/wiki/Lebesgue_differentiation_theorem
$endgroup$
– Will M.
Dec 24 '18 at 0:42
add a comment |
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$begingroup$
Why is that not nonsense? Given a set A of real numbers and a nonzero h, what is A/h? If for all nonzero h, g(h) is a set of real numbers, what is the limit of g(h) as h approaches zero? What happens when h is negative?
$endgroup$
– William Elliot
Dec 23 '18 at 3:31
1
$begingroup$
@WilliamElliot I ment the measure of the set, of course. There was a typo.
$endgroup$
– Hiro
Dec 23 '18 at 9:05
$begingroup$
Limit as h -> 0 indicates h can be positive or negative. For h < 0, [x,x+h) is empty, giving a liminf of 0. Thus, if there is a limit, it has to be zero.
$endgroup$
– William Elliot
Dec 23 '18 at 20:41
$begingroup$
@WilliamElliot I've edited the question to address this ambiguity: we consider only $h to 0^+$.
$endgroup$
– Hiro
Dec 24 '18 at 0:07
$begingroup$
@Hiro you are probably after Lebesgue differentiation theorem: en.wikipedia.org/wiki/Lebesgue_differentiation_theorem
$endgroup$
– Will M.
Dec 24 '18 at 0:42