Lipschitzness of derivatives
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Let $f in C^infty_b(mathbb{R}^d; mathbb{R}^d)$, so bounded, infinitely differentiable with bounded derivatives mapping $mathbb{R}^d$ to $mathbb{R}^d$. I'll write $|cdot|$ for the norm on $mathbb{R}^d$ as well as the induced operator norm. By boundedness, we have a $B_1 > 0$ such that $forall x in mathbb{R}^d$
$$|Df(x)|leq B_1$$
this immediately gives that $forall x,y_1,y_2 in mathbb{R}^d$
$$|Df(x)y_1 - Df(x)y_2| leq B_1 |y_1 - y_2|.$$
Is it possible (maybe using boundedness of $D^2f$ + MVE) to obtain that there exists $L_1 > 0$ such that $forall x_1,x_2,y_1,y_2 in mathbb{R}^d$
$$|Df(x_1)y_1 - Df(x_2)y_2| leq L_1 |y_1 - y_2|.$$
Similarly can this be done for higher order derivatives say
$$|Df^2(x_1)[y_1,y_1] - Df^2(x_2)[y_2,y_2]| leq L_2 |y_1 - y_2|^2.$$
Any help is appreciated, thanks!
calculus real-analysis functional-analysis derivatives lipschitz-functions
asked Nov 14 at 20:36
John Doe
375
add a comment |
up vote
0
down vote
favorite
Let $f in C^infty_b(mathbb{R}^d; mathbb{R}^d)$, so bounded, infinitely differentiable with bounded derivatives mapping $mathbb{R}^d$ to $mathbb{R}^d$. I'll write $|cdot|$ for the norm on $mathbb{R}^d$ as well as the induced operator norm. By boundedness, we have a $B_1 > 0$ such that $forall x in mathbb{R}^d$
$$|Df(x)|leq B_1$$
this immediately gives that $forall x,y_1,y_2 in mathbb{R}^d$
$$|Df(x)y_1 - Df(x)y_2| leq B_1 |y_1 - y_2|.$$
Is it possible (maybe using boundedness of $D^2f$ + MVE) to obtain that there exists $L_1 > 0$ such that $forall x_1,x_2,y_1,y_2 in mathbb{R}^d$
$$|Df(x_1)y_1 - Df(x_2)y_2| leq L_1 |y_1 - y_2|.$$
Similarly can this be done for higher order derivatives say
$$|Df^2(x_1)[y_1,y_1] - Df^2(x_2)[y_2,y_2]| leq L_2 |y_1 - y_2|^2.$$
Any help is appreciated, thanks!
calculus real-analysis functional-analysis derivatives lipschitz-functions
asked Nov 14 at 20:36
John Doe
375
1
That would be quite peculiar: then $Df(x_1)=Df(x_2)$ for all $x_1,x_2$... :)
– John B
Nov 14 at 20:44
Yeah... hummm. Maybe something less like $|Df(x_1)y_1 - Df(x_2)y_2| leq L_1 |x_1-x_2||y_1-y_2|$?
– John Doe
Nov 14 at 21:05
First: you should tell us what you have in mind to ask this. Second: you should make some effort, really, does it look like a norm what you just wrote? You see, we know well what should be expected, you need to think about it instead of guessing.
– John B
Nov 14 at 21:22
What do you mean? I'm guessing because I don't know the answer; this is not some homework problem. What I want is a bound on $|Df(x_1)y_1 - Df(x_2)y_2|$ in terms of $x_1,x_2,y_1,y_2$ and constants with w/e necessary assumptions on $f$. What I can get, for example, is by boundedness of $D^2f$, $Df$ is Lipschitz so there is an $L_1$ such that $|Df(x_1)-Df(x_2)| leq L_1 |x_1-x_2|$ then $|(Df(x_1)-Df(x_2))(y_1-y_2)|leq L_1 |x_1-x_2||y_1-y_2|$, but this not quite what I want. I was thinking of adding and subtracting something to the original to get this, but I can't quite see it.
– John Doe
Nov 14 at 21:38
add a comment |
up vote
0
down vote
favorite
up vote
0
down vote
favorite
Let $f in C^infty_b(mathbb{R}^d; mathbb{R}^d)$, so bounded, infinitely differentiable with bounded derivatives mapping $mathbb{R}^d$ to $mathbb{R}^d$. I'll write $|cdot|$ for the norm on $mathbb{R}^d$ as well as the induced operator norm. By boundedness, we have a $B_1 > 0$ such that $forall x in mathbb{R}^d$
$$|Df(x)|leq B_1$$
this immediately gives that $forall x,y_1,y_2 in mathbb{R}^d$
$$|Df(x)y_1 - Df(x)y_2| leq B_1 |y_1 - y_2|.$$
Is it possible (maybe using boundedness of $D^2f$ + MVE) to obtain that there exists $L_1 > 0$ such that $forall x_1,x_2,y_1,y_2 in mathbb{R}^d$
$$|Df(x_1)y_1 - Df(x_2)y_2| leq L_1 |y_1 - y_2|.$$
Similarly can this be done for higher order derivatives say
$$|Df^2(x_1)[y_1,y_1] - Df^2(x_2)[y_2,y_2]| leq L_2 |y_1 - y_2|^2.$$
Any help is appreciated, thanks!
calculus real-analysis functional-analysis derivatives lipschitz-functions
asked Nov 14 at 20:36
John Doe
375
Let $f in C^infty_b(mathbb{R}^d; mathbb{R}^d)$, so bounded, infinitely differentiable with bounded derivatives mapping $mathbb{R}^d$ to $mathbb{R}^d$. I'll write $|cdot|$ for the norm on $mathbb{R}^d$ as well as the induced operator norm. By boundedness, we have a $B_1 > 0$ such that $forall x in mathbb{R}^d$
$$|Df(x)|leq B_1$$
this immediately gives that $forall x,y_1,y_2 in mathbb{R}^d$
$$|Df(x)y_1 - Df(x)y_2| leq B_1 |y_1 - y_2|.$$
Is it possible (maybe using boundedness of $D^2f$ + MVE) to obtain that there exists $L_1 > 0$ such that $forall x_1,x_2,y_1,y_2 in mathbb{R}^d$
$$|Df(x_1)y_1 - Df(x_2)y_2| leq L_1 |y_1 - y_2|.$$
Similarly can this be done for higher order derivatives say
$$|Df^2(x_1)[y_1,y_1] - Df^2(x_2)[y_2,y_2]| leq L_2 |y_1 - y_2|^2.$$
Any help is appreciated, thanks!
calculus real-analysis functional-analysis derivatives lipschitz-functions
calculus real-analysis functional-analysis derivatives lipschitz-functions
asked Nov 14 at 20:36
John Doe
375
asked Nov 14 at 20:36
John Doe
375
asked Nov 14 at 20:36
John Doe
375
asked Nov 14 at 20:36
John Doe
375
asked Nov 14 at 20:36
John Doe
375
375
1
That would be quite peculiar: then $Df(x_1)=Df(x_2)$ for all $x_1,x_2$... :)
– John B
Nov 14 at 20:44
Yeah... hummm. Maybe something less like $|Df(x_1)y_1 - Df(x_2)y_2| leq L_1 |x_1-x_2||y_1-y_2|$?
– John Doe
Nov 14 at 21:05
First: you should tell us what you have in mind to ask this. Second: you should make some effort, really, does it look like a norm what you just wrote? You see, we know well what should be expected, you need to think about it instead of guessing.
– John B
Nov 14 at 21:22
What do you mean? I'm guessing because I don't know the answer; this is not some homework problem. What I want is a bound on $|Df(x_1)y_1 - Df(x_2)y_2|$ in terms of $x_1,x_2,y_1,y_2$ and constants with w/e necessary assumptions on $f$. What I can get, for example, is by boundedness of $D^2f$, $Df$ is Lipschitz so there is an $L_1$ such that $|Df(x_1)-Df(x_2)| leq L_1 |x_1-x_2|$ then $|(Df(x_1)-Df(x_2))(y_1-y_2)|leq L_1 |x_1-x_2||y_1-y_2|$, but this not quite what I want. I was thinking of adding and subtracting something to the original to get this, but I can't quite see it.
– John Doe
Nov 14 at 21:38
add a comment |
1
That would be quite peculiar: then $Df(x_1)=Df(x_2)$ for all $x_1,x_2$... :)
– John B
Nov 14 at 20:44
Yeah... hummm. Maybe something less like $|Df(x_1)y_1 - Df(x_2)y_2| leq L_1 |x_1-x_2||y_1-y_2|$?
– John Doe
Nov 14 at 21:05
First: you should tell us what you have in mind to ask this. Second: you should make some effort, really, does it look like a norm what you just wrote? You see, we know well what should be expected, you need to think about it instead of guessing.
– John B
Nov 14 at 21:22
What do you mean? I'm guessing because I don't know the answer; this is not some homework problem. What I want is a bound on $|Df(x_1)y_1 - Df(x_2)y_2|$ in terms of $x_1,x_2,y_1,y_2$ and constants with w/e necessary assumptions on $f$. What I can get, for example, is by boundedness of $D^2f$, $Df$ is Lipschitz so there is an $L_1$ such that $|Df(x_1)-Df(x_2)| leq L_1 |x_1-x_2|$ then $|(Df(x_1)-Df(x_2))(y_1-y_2)|leq L_1 |x_1-x_2||y_1-y_2|$, but this not quite what I want. I was thinking of adding and subtracting something to the original to get this, but I can't quite see it.
– John Doe
Nov 14 at 21:38
1
1
That would be quite peculiar: then $Df(x_1)=Df(x_2)$ for all $x_1,x_2$... :)
– John B
Nov 14 at 20:44
That would be quite peculiar: then $Df(x_1)=Df(x_2)$ for all $x_1,x_2$... :)
– John B
Nov 14 at 20:44
Yeah... hummm. Maybe something less like $|Df(x_1)y_1 - Df(x_2)y_2| leq L_1 |x_1-x_2||y_1-y_2|$?
– John Doe
Nov 14 at 21:05
Yeah... hummm. Maybe something less like $|Df(x_1)y_1 - Df(x_2)y_2| leq L_1 |x_1-x_2||y_1-y_2|$?
– John Doe
Nov 14 at 21:05
First: you should tell us what you have in mind to ask this. Second: you should make some effort, really, does it look like a norm what you just wrote? You see, we know well what should be expected, you need to think about it instead of guessing.
– John B
Nov 14 at 21:22
First: you should tell us what you have in mind to ask this. Second: you should make some effort, really, does it look like a norm what you just wrote? You see, we know well what should be expected, you need to think about it instead of guessing.
– John B
Nov 14 at 21:22
What do you mean? I'm guessing because I don't know the answer; this is not some homework problem. What I want is a bound on $|Df(x_1)y_1 - Df(x_2)y_2|$ in terms of $x_1,x_2,y_1,y_2$ and constants with w/e necessary assumptions on $f$. What I can get, for example, is by boundedness of $D^2f$, $Df$ is Lipschitz so there is an $L_1$ such that $|Df(x_1)-Df(x_2)| leq L_1 |x_1-x_2|$ then $|(Df(x_1)-Df(x_2))(y_1-y_2)|leq L_1 |x_1-x_2||y_1-y_2|$, but this not quite what I want. I was thinking of adding and subtracting something to the original to get this, but I can't quite see it.
– John Doe
Nov 14 at 21:38
What do you mean? I'm guessing because I don't know the answer; this is not some homework problem. What I want is a bound on $|Df(x_1)y_1 - Df(x_2)y_2|$ in terms of $x_1,x_2,y_1,y_2$ and constants with w/e necessary assumptions on $f$. What I can get, for example, is by boundedness of $D^2f$, $Df$ is Lipschitz so there is an $L_1$ such that $|Df(x_1)-Df(x_2)| leq L_1 |x_1-x_2|$ then $|(Df(x_1)-Df(x_2))(y_1-y_2)|leq L_1 |x_1-x_2||y_1-y_2|$, but this not quite what I want. I was thinking of adding and subtracting something to the original to get this, but I can't quite see it.
– John Doe
Nov 14 at 21:38
add a comment |
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That would be quite peculiar: then $Df(x_1)=Df(x_2)$ for all $x_1,x_2$... :)
– John B
Nov 14 at 20:44
Yeah... hummm. Maybe something less like $|Df(x_1)y_1 - Df(x_2)y_2| leq L_1 |x_1-x_2||y_1-y_2|$?
– John Doe
Nov 14 at 21:05
First: you should tell us what you have in mind to ask this. Second: you should make some effort, really, does it look like a norm what you just wrote? You see, we know well what should be expected, you need to think about it instead of guessing.
– John B
Nov 14 at 21:22
What do you mean? I'm guessing because I don't know the answer; this is not some homework problem. What I want is a bound on $|Df(x_1)y_1 - Df(x_2)y_2|$ in terms of $x_1,x_2,y_1,y_2$ and constants with w/e necessary assumptions on $f$. What I can get, for example, is by boundedness of $D^2f$, $Df$ is Lipschitz so there is an $L_1$ such that $|Df(x_1)-Df(x_2)| leq L_1 |x_1-x_2|$ then $|(Df(x_1)-Df(x_2))(y_1-y_2)|leq L_1 |x_1-x_2||y_1-y_2|$, but this not quite what I want. I was thinking of adding and subtracting something to the original to get this, but I can't quite see it.
– John Doe
Nov 14 at 21:38