Why does the second derivative of convex function exist almost everywhere?
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Let $f: Ito mathbb{R}$ be a convex function.
Why does the second derivative of $f$ exist almost everywhere?
By searching I knew that is Alexandrov theorem, but I didn't find the proof...
My try
For $x_1<x<x_2$ where function is defined , we have
$$frac{f(x)-f(x_1)}{x-x_1}lefrac{f(x_2)-f(x_1)}{x_2-x_1}lefrac{f(x_2)-f(x)}{x_2-x}$$
Considering $x_1to x^-$ and $x_2to x^+$,
we can conclude that $f_-'(x)$ and $f_+'(x)$ exist and $f_-'(x)le f_+'(x)$. (in this process we only need $frac{f(x)-f(x_1)}{x-x_1}lefrac{f(x_2)-f(x)}{x_2-x}$)
Considering $xto x_1$ and $xto x_2$,
we can conclude that $$f_+'(x_1)le frac{f(x_2)-f(x_1)}{x_2-x_1} le f_-'(x_2)$$
So we get that $f_-(x)$ and $f_+(x)$ exist and both are monotonically increasing.
In addition, $f_-'(x)le f_+'(x)$.
Because monotone functions only have jump discontinuities and are continuous everywhere except countably many points, we can conclude that $f'$ exists everywhere except countably many points.
How to move on? Any hints? Thank you in advance!
real-analysis derivatives convex-analysis
asked Dec 24 '18 at 15:01
ZeroZero
592111
$endgroup$
|
$begingroup$
Let $f: Ito mathbb{R}$ be a convex function.
Why does the second derivative of $f$ exist almost everywhere?
By searching I knew that is Alexandrov theorem, but I didn't find the proof...
My try
For $x_1<x<x_2$ where function is defined , we have
$$frac{f(x)-f(x_1)}{x-x_1}lefrac{f(x_2)-f(x_1)}{x_2-x_1}lefrac{f(x_2)-f(x)}{x_2-x}$$
Considering $x_1to x^-$ and $x_2to x^+$,
we can conclude that $f_-'(x)$ and $f_+'(x)$ exist and $f_-'(x)le f_+'(x)$. (in this process we only need $frac{f(x)-f(x_1)}{x-x_1}lefrac{f(x_2)-f(x)}{x_2-x}$)
Considering $xto x_1$ and $xto x_2$,
we can conclude that $$f_+'(x_1)le frac{f(x_2)-f(x_1)}{x_2-x_1} le f_-'(x_2)$$
So we get that $f_-(x)$ and $f_+(x)$ exist and both are monotonically increasing.
In addition, $f_-'(x)le f_+'(x)$.
Because monotone functions only have jump discontinuities and are continuous everywhere except countably many points, we can conclude that $f'$ exists everywhere except countably many points.
How to move on? Any hints? Thank you in advance!
real-analysis derivatives convex-analysis
asked Dec 24 '18 at 15:01
ZeroZero
592111
$endgroup$
$begingroup$
I believe you want to show that the closure of the set of points where $f$ is not differentiable is still countable. Then using the monotonicity of $f'$ you can probably show that $f''$ exists almost everywhere.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:29
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@SmileyCraft Could you give me some hints about how to show $f''$ exists almost everywhere through using the monotonicity of $f'$? I got stuck on that...
$endgroup$
– Zero
Dec 24 '18 at 15:50
2
$begingroup$
Have you tried to show already that the closure of the set of points where $f$ is not differentiable is countable? Because then I believe you can find a countable set of open intervals where $f$ is differentiable, such that the union has a countable complement. Using the monotonicity of $f'$ you can show that on every interval the set of points where $f'$ is not differentiable is countable.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:55
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@SmileyCraft Thank you very much! I just realized that if $f$ is a monotonic function defined on an interval $I$ then $f$ is differentiable almost everywhere on $I$ , from which I can arrive at the conclusion
$endgroup$
– Zero
Dec 24 '18 at 16:04
$begingroup$
This is not correct , convex functions can be nondifferentiable on a dense subset of $R$.
$endgroup$
– Red shoes
Dec 25 '18 at 2:01
|
$begingroup$
Let $f: Ito mathbb{R}$ be a convex function.
Why does the second derivative of $f$ exist almost everywhere?
By searching I knew that is Alexandrov theorem, but I didn't find the proof...
My try
For $x_1<x<x_2$ where function is defined , we have
$$frac{f(x)-f(x_1)}{x-x_1}lefrac{f(x_2)-f(x_1)}{x_2-x_1}lefrac{f(x_2)-f(x)}{x_2-x}$$
Considering $x_1to x^-$ and $x_2to x^+$,
we can conclude that $f_-'(x)$ and $f_+'(x)$ exist and $f_-'(x)le f_+'(x)$. (in this process we only need $frac{f(x)-f(x_1)}{x-x_1}lefrac{f(x_2)-f(x)}{x_2-x}$)
Considering $xto x_1$ and $xto x_2$,
we can conclude that $$f_+'(x_1)le frac{f(x_2)-f(x_1)}{x_2-x_1} le f_-'(x_2)$$
So we get that $f_-(x)$ and $f_+(x)$ exist and both are monotonically increasing.
In addition, $f_-'(x)le f_+'(x)$.
Because monotone functions only have jump discontinuities and are continuous everywhere except countably many points, we can conclude that $f'$ exists everywhere except countably many points.
How to move on? Any hints? Thank you in advance!
real-analysis derivatives convex-analysis
asked Dec 24 '18 at 15:01
ZeroZero
592111
$endgroup$
Let $f: Ito mathbb{R}$ be a convex function.
Why does the second derivative of $f$ exist almost everywhere?
By searching I knew that is Alexandrov theorem, but I didn't find the proof...
My try
For $x_1<x<x_2$ where function is defined , we have
$$frac{f(x)-f(x_1)}{x-x_1}lefrac{f(x_2)-f(x_1)}{x_2-x_1}lefrac{f(x_2)-f(x)}{x_2-x}$$
Considering $x_1to x^-$ and $x_2to x^+$,
we can conclude that $f_-'(x)$ and $f_+'(x)$ exist and $f_-'(x)le f_+'(x)$. (in this process we only need $frac{f(x)-f(x_1)}{x-x_1}lefrac{f(x_2)-f(x)}{x_2-x}$)
Considering $xto x_1$ and $xto x_2$,
we can conclude that $$f_+'(x_1)le frac{f(x_2)-f(x_1)}{x_2-x_1} le f_-'(x_2)$$
So we get that $f_-(x)$ and $f_+(x)$ exist and both are monotonically increasing.
In addition, $f_-'(x)le f_+'(x)$.
Because monotone functions only have jump discontinuities and are continuous everywhere except countably many points, we can conclude that $f'$ exists everywhere except countably many points.
How to move on? Any hints? Thank you in advance!
real-analysis derivatives convex-analysis
real-analysis derivatives convex-analysis
asked Dec 24 '18 at 15:01
ZeroZero
592111
asked Dec 24 '18 at 15:01
ZeroZero
592111
asked Dec 24 '18 at 15:01
ZeroZero
592111
asked Dec 24 '18 at 15:01
ZeroZero
592111
asked Dec 24 '18 at 15:01
ZeroZero
592111
592111
$begingroup$
I believe you want to show that the closure of the set of points where $f$ is not differentiable is still countable. Then using the monotonicity of $f'$ you can probably show that $f''$ exists almost everywhere.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:29
$begingroup$
@SmileyCraft Could you give me some hints about how to show $f''$ exists almost everywhere through using the monotonicity of $f'$? I got stuck on that...
$endgroup$
– Zero
Dec 24 '18 at 15:50
2
$begingroup$
Have you tried to show already that the closure of the set of points where $f$ is not differentiable is countable? Because then I believe you can find a countable set of open intervals where $f$ is differentiable, such that the union has a countable complement. Using the monotonicity of $f'$ you can show that on every interval the set of points where $f'$ is not differentiable is countable.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:55
$begingroup$
@SmileyCraft Thank you very much! I just realized that if $f$ is a monotonic function defined on an interval $I$ then $f$ is differentiable almost everywhere on $I$ , from which I can arrive at the conclusion
$endgroup$
– Zero
Dec 24 '18 at 16:04
$begingroup$
This is not correct , convex functions can be nondifferentiable on a dense subset of $R$.
$endgroup$
– Red shoes
Dec 25 '18 at 2:01
|
$begingroup$
I believe you want to show that the closure of the set of points where $f$ is not differentiable is still countable. Then using the monotonicity of $f'$ you can probably show that $f''$ exists almost everywhere.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:29
$begingroup$
@SmileyCraft Could you give me some hints about how to show $f''$ exists almost everywhere through using the monotonicity of $f'$? I got stuck on that...
$endgroup$
– Zero
Dec 24 '18 at 15:50
2
$begingroup$
Have you tried to show already that the closure of the set of points where $f$ is not differentiable is countable? Because then I believe you can find a countable set of open intervals where $f$ is differentiable, such that the union has a countable complement. Using the monotonicity of $f'$ you can show that on every interval the set of points where $f'$ is not differentiable is countable.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:55
$begingroup$
@SmileyCraft Thank you very much! I just realized that if $f$ is a monotonic function defined on an interval $I$ then $f$ is differentiable almost everywhere on $I$ , from which I can arrive at the conclusion
$endgroup$
– Zero
Dec 24 '18 at 16:04
$begingroup$
This is not correct , convex functions can be nondifferentiable on a dense subset of $R$.
$endgroup$
– Red shoes
Dec 25 '18 at 2:01
$begingroup$
I believe you want to show that the closure of the set of points where $f$ is not differentiable is still countable. Then using the monotonicity of $f'$ you can probably show that $f''$ exists almost everywhere.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:29
$begingroup$
I believe you want to show that the closure of the set of points where $f$ is not differentiable is still countable. Then using the monotonicity of $f'$ you can probably show that $f''$ exists almost everywhere.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:29
$begingroup$
@SmileyCraft Could you give me some hints about how to show $f''$ exists almost everywhere through using the monotonicity of $f'$? I got stuck on that...
$endgroup$
– Zero
Dec 24 '18 at 15:50
$begingroup$
@SmileyCraft Could you give me some hints about how to show $f''$ exists almost everywhere through using the monotonicity of $f'$? I got stuck on that...
$endgroup$
– Zero
Dec 24 '18 at 15:50
2
2
$begingroup$
Have you tried to show already that the closure of the set of points where $f$ is not differentiable is countable? Because then I believe you can find a countable set of open intervals where $f$ is differentiable, such that the union has a countable complement. Using the monotonicity of $f'$ you can show that on every interval the set of points where $f'$ is not differentiable is countable.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:55
$begingroup$
Have you tried to show already that the closure of the set of points where $f$ is not differentiable is countable? Because then I believe you can find a countable set of open intervals where $f$ is differentiable, such that the union has a countable complement. Using the monotonicity of $f'$ you can show that on every interval the set of points where $f'$ is not differentiable is countable.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:55
$begingroup$
@SmileyCraft Thank you very much! I just realized that if $f$ is a monotonic function defined on an interval $I$ then $f$ is differentiable almost everywhere on $I$ , from which I can arrive at the conclusion
$endgroup$
– Zero
Dec 24 '18 at 16:04
$begingroup$
@SmileyCraft Thank you very much! I just realized that if $f$ is a monotonic function defined on an interval $I$ then $f$ is differentiable almost everywhere on $I$ , from which I can arrive at the conclusion
$endgroup$
– Zero
Dec 24 '18 at 16:04
$begingroup$
This is not correct , convex functions can be nondifferentiable on a dense subset of $R$.
$endgroup$
– Red shoes
Dec 25 '18 at 2:01
$begingroup$
This is not correct , convex functions can be nondifferentiable on a dense subset of $R$.
$endgroup$
– Red shoes
Dec 25 '18 at 2:01
|
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$begingroup$
I believe you want to show that the closure of the set of points where $f$ is not differentiable is still countable. Then using the monotonicity of $f'$ you can probably show that $f''$ exists almost everywhere.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:29
$begingroup$
@SmileyCraft Could you give me some hints about how to show $f''$ exists almost everywhere through using the monotonicity of $f'$? I got stuck on that...
$endgroup$
– Zero
Dec 24 '18 at 15:50
2
$begingroup$
Have you tried to show already that the closure of the set of points where $f$ is not differentiable is countable? Because then I believe you can find a countable set of open intervals where $f$ is differentiable, such that the union has a countable complement. Using the monotonicity of $f'$ you can show that on every interval the set of points where $f'$ is not differentiable is countable.
$endgroup$
– SmileyCraft
Dec 24 '18 at 15:55
$begingroup$
@SmileyCraft Thank you very much! I just realized that if $f$ is a monotonic function defined on an interval $I$ then $f$ is differentiable almost everywhere on $I$ , from which I can arrive at the conclusion
$endgroup$
– Zero
Dec 24 '18 at 16:04
$begingroup$
This is not correct , convex functions can be nondifferentiable on a dense subset of $R$.
$endgroup$
– Red shoes
Dec 25 '18 at 2:01