Extension of Uniformly Differentiable function - from open ball to $mathbbR^n$

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Consider $f: U to mathbbR^m$, where $U$ is an open ball in $mathbbR^n$.

Suppose $f$ is uniformly differentiable on $U$:

$$forall epsilon>0,,exists delta>0:|!|bf h|!|<delta,bf x,bf x+bf hin U Longrightarrow |!|f(bf x+bf h)-f(bf x)-f'(bf x)(bf h)|!|<epsilon |bf h|$$

That is, the $delta$ is valid for all $bf x$.

I wonder if the following is true:

Question: Can we always construct an extension of $barf$ of $f$, such that $barf(bf x) = f(bf x)$, for $bf x in U$, and $barf$ uniformly differentiable on $mathbbR^n$?

derivatives continuity uniform-continuity

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edited Jul 19 at 0:12

asked Jul 18 at 21:10

Guillaume F.

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Consider $f: U to mathbbR^m$, where $U$ is an open ball in $mathbbR^n$.

Suppose $f$ is uniformly differentiable on $U$:

$$forall epsilon>0,,exists delta>0:|!|bf h|!|<delta,bf x,bf x+bf hin U Longrightarrow |!|f(bf x+bf h)-f(bf x)-f'(bf x)(bf h)|!|<epsilon |bf h|$$

That is, the $delta$ is valid for all $bf x$.

I wonder if the following is true:

Question: Can we always construct an extension of $barf$ of $f$, such that $barf(bf x) = f(bf x)$, for $bf x in U$, and $barf$ uniformly differentiable on $mathbbR^n$?

derivatives continuity uniform-continuity

share|cite|improve this question

edited Jul 19 at 0:12

asked Jul 18 at 21:10

Guillaume F.

351211

add a comment | 

up vote
0
down vote

favorite

1

up vote
0
down vote

favorite

1

1

Consider $f: U to mathbbR^m$, where $U$ is an open ball in $mathbbR^n$.

Suppose $f$ is uniformly differentiable on $U$:

$$forall epsilon>0,,exists delta>0:|!|bf h|!|<delta,bf x,bf x+bf hin U Longrightarrow |!|f(bf x+bf h)-f(bf x)-f'(bf x)(bf h)|!|<epsilon |bf h|$$

That is, the $delta$ is valid for all $bf x$.

I wonder if the following is true:

Question: Can we always construct an extension of $barf$ of $f$, such that $barf(bf x) = f(bf x)$, for $bf x in U$, and $barf$ uniformly differentiable on $mathbbR^n$?

derivatives continuity uniform-continuity

share|cite|improve this question

edited Jul 19 at 0:12

asked Jul 18 at 21:10

Guillaume F.

351211

Consider $f: U to mathbbR^m$, where $U$ is an open ball in $mathbbR^n$.

Suppose $f$ is uniformly differentiable on $U$:

$$forall epsilon>0,,exists delta>0:|!|bf h|!|<delta,bf x,bf x+bf hin U Longrightarrow |!|f(bf x+bf h)-f(bf x)-f'(bf x)(bf h)|!|<epsilon |bf h|$$

That is, the $delta$ is valid for all $bf x$.

I wonder if the following is true:

Question: Can we always construct an extension of $barf$ of $f$, such that $barf(bf x) = f(bf x)$, for $bf x in U$, and $barf$ uniformly differentiable on $mathbbR^n$?

derivatives continuity uniform-continuity

share|cite|improve this question

edited Jul 19 at 0:12

asked Jul 18 at 21:10

Guillaume F.

351211

share|cite|improve this question

share|cite|improve this question

edited Jul 19 at 0:12

edited Jul 19 at 0:12

edited Jul 19 at 0:12

asked Jul 18 at 21:10

Guillaume F.

351211

asked Jul 18 at 21:10

Guillaume F.

351211

asked Jul 18 at 21:10

Guillaume F.

351211

351211

add a comment | 

add a comment | 

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Consider $n=m=1$, $U=(-1,1)$, $f(x)=sqrtx+1$.

Obviously $f$ is differentiable on $U$, but not at $x=-1$.

In fact, the tangent there would become vertical, or equivalently:

$f'(x)$ would increase beyond any bound.

--- rk

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edited Jul 18 at 21:50

answered Jul 18 at 21:38

Dr. Richard Klitzing

7586

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your $f$ is differentiable on $U$, but not uniformly differentiable on $U$.
  – Guillaume F.
  Jul 19 at 0:13
  

  
  
  
  

add a comment | 

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1 Answer
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1 Answer
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active

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oldest

votes

up vote
0
down vote

Consider $n=m=1$, $U=(-1,1)$, $f(x)=sqrtx+1$.

Obviously $f$ is differentiable on $U$, but not at $x=-1$.

In fact, the tangent there would become vertical, or equivalently:

$f'(x)$ would increase beyond any bound.

--- rk

share|cite|improve this answer

edited Jul 18 at 21:50

answered Jul 18 at 21:38

Dr. Richard Klitzing

7586

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your $f$ is differentiable on $U$, but not uniformly differentiable on $U$.
  – Guillaume F.
  Jul 19 at 0:13
  

  
  
  
  

add a comment | 

up vote
0
down vote

Consider $n=m=1$, $U=(-1,1)$, $f(x)=sqrtx+1$.

Obviously $f$ is differentiable on $U$, but not at $x=-1$.

In fact, the tangent there would become vertical, or equivalently:

$f'(x)$ would increase beyond any bound.

--- rk

share|cite|improve this answer

edited Jul 18 at 21:50

answered Jul 18 at 21:38

Dr. Richard Klitzing

7586

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your $f$ is differentiable on $U$, but not uniformly differentiable on $U$.
  – Guillaume F.
  Jul 19 at 0:13
  

  
  
  
  

add a comment | 

up vote
0
down vote

up vote
0
down vote

Consider $n=m=1$, $U=(-1,1)$, $f(x)=sqrtx+1$.

Obviously $f$ is differentiable on $U$, but not at $x=-1$.

In fact, the tangent there would become vertical, or equivalently:

$f'(x)$ would increase beyond any bound.

--- rk

share|cite|improve this answer

edited Jul 18 at 21:50

answered Jul 18 at 21:38

Dr. Richard Klitzing

7586

Consider $n=m=1$, $U=(-1,1)$, $f(x)=sqrtx+1$.

Obviously $f$ is differentiable on $U$, but not at $x=-1$.

In fact, the tangent there would become vertical, or equivalently:

$f'(x)$ would increase beyond any bound.

--- rk

share|cite|improve this answer

edited Jul 18 at 21:50

answered Jul 18 at 21:38

Dr. Richard Klitzing

7586

share|cite|improve this answer

share|cite|improve this answer

edited Jul 18 at 21:50

edited Jul 18 at 21:50

edited Jul 18 at 21:50

answered Jul 18 at 21:38

Dr. Richard Klitzing

7586

answered Jul 18 at 21:38

Dr. Richard Klitzing

7586

answered Jul 18 at 21:38

Dr. Richard Klitzing

7586

7586

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your $f$ is differentiable on $U$, but not uniformly differentiable on $U$.
  – Guillaume F.
  Jul 19 at 0:13
  

  
  
  
  

add a comment | 

• 
  
  
  

  
  

  
  
  
  
  
  

  
  Your $f$ is differentiable on $U$, but not uniformly differentiable on $U$.
  – Guillaume F.
  Jul 19 at 0:13
  

  
  
  
  

Your $f$ is differentiable on $U$, but not uniformly differentiable on $U$.
– Guillaume F.
Jul 19 at 0:13

Your $f$ is differentiable on $U$, but not uniformly differentiable on $U$.
– Guillaume F.
Jul 19 at 0:13

add a comment | 

 

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