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Different definition of continuity

1

$begingroup$

Condition: $f:Itomathbb R$ is continuous. For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ such that $$ sum_{k} |y_{k} - x_{k}| < delta$$
implies
$$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon.$$

Is this condition a necessary or sufficient condition of absolution continuity? Note that the order of the logic identifiers has changed.

A function $f: I to mathbb{R}$ is absolutely continuous on an interval $I$ if for every $epsilon > 0$ there is a $delta > 0$ such that whenever a finite sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$ satisfies
$$ sum_{k} |y_{k} - x_{k}| < delta$$
then
$$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon$$

real-analysis calculus limits analysis continuity

share|cite|improve this question

edited Mar 26 at 1:47

High GPA

asked Mar 25 at 8:19

High GPAHigh GPA

956422

$endgroup$

add a comment | 

1

$begingroup$

Condition: $f:Itomathbb R$ is continuous. For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ such that $$ sum_{k} |y_{k} - x_{k}| < delta$$
implies
$$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon.$$

Is this condition a necessary or sufficient condition of absolution continuity? Note that the order of the logic identifiers has changed.

A function $f: I to mathbb{R}$ is absolutely continuous on an interval $I$ if for every $epsilon > 0$ there is a $delta > 0$ such that whenever a finite sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$ satisfies
$$ sum_{k} |y_{k} - x_{k}| < delta$$
then
$$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon$$

real-analysis calculus limits analysis continuity

share|cite|improve this question

edited Mar 26 at 1:47

High GPA

asked Mar 25 at 8:19

High GPAHigh GPA

956422

$endgroup$

add a comment | 

$begingroup$

Condition: $f:Itomathbb R$ is continuous. For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ such that $$ sum_{k} |y_{k} - x_{k}| < delta$$
implies
$$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon.$$

Is this condition a necessary or sufficient condition of absolution continuity? Note that the order of the logic identifiers has changed.

A function $f: I to mathbb{R}$ is absolutely continuous on an interval $I$ if for every $epsilon > 0$ there is a $delta > 0$ such that whenever a finite sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$ satisfies
$$ sum_{k} |y_{k} - x_{k}| < delta$$
then
$$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon$$

real-analysis calculus limits analysis continuity

share|cite|improve this question

edited Mar 26 at 1:47

High GPA

asked Mar 25 at 8:19

High GPAHigh GPA

956422

$endgroup$

share|cite|improve this question

edited Mar 26 at 1:47

High GPA

asked Mar 25 at 8:19

High GPAHigh GPA

956422

share|cite|improve this question

edited Mar 26 at 1:47

High GPA

asked Mar 25 at 8:19

High GPAHigh GPA

956422

asked Mar 25 at 8:19

High GPAHigh GPA

956422

asked Mar 25 at 8:19

High GPAHigh GPA

956422

2 Answers
2

active

oldest

votes

1

$begingroup$

Given any function $f:Itomathbb R$ (not necessarily continuous), the condition: For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ such that $$ sum_{k} |y_{k} - x_{k}| < delta$$
implies
$$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon$$
is trivially true.

Proof: Given any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, just choose $delta = frac{1}{2} sum_{k} |y_{k} - x_{k}|$. Then the condition $ sum_{k} |y_{k} - x_{k}| < delta$ will be false and so the implication "$ sum_{k} |y_{k} - x_{k}| < delta$
implies
$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon$" will be trivially true.

So this conditionis not sufficient for absolute continuity or even continuity.

share|cite|improve this answer

answered Mar 28 at 10:31

RamiroRamiro

7,36421535

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I also saw this after you gave the detailed explanation in the other question
  $endgroup$
  – High GPA
  Mar 28 at 22:25
  

  
  
  
  

add a comment | 

2

$begingroup$

Yes, they are equivalent. Suppose you choose $delta$ according to the usual definition of absolute continuity with $epsilon$ repalced by $epsilon /2$. If $(a_k.b_k)$ is a disjoint sequence of interval with total length less than $delta$ then $sumlimits_{k=1}^{N} |f(b_k)-f(a_k)| < epsilon /2$ for each $N$. Let $N to infty$ to complete the proof.

share|cite|improve this answer

answered Mar 25 at 8:24

Kavi Rama MurthyKavi Rama Murthy

75.9k53270

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Many thanks for your teaching! So you proved that the definition "$forallepsilonexistsdelta(forall text{finite subintervals we have} (sum|y_k-x_k|<delta Rightarrow sum|f(y_k)-f(x_k)|<epsilon))$" is equivalent to "$forallepsilonexistsdelta(forall text{countable subintervals we have} (sum|y_k-x_k|<delta Rightarrow sum|f(y_k)-f(x_k)|<epsilon))$". However, my first condition means "$forall text{countable subintervals}(forallepsilonexistsdelta text{we have} (sum|y_k-x_k|<delta Rightarrow sum|f(y_k)-f(x_k)|<epsilon))$". Not sure my understanding is correct, though.
  $endgroup$
  – High GPA
  Mar 25 at 22:05
  
  
  

  
  
  
  

add a comment | 

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1

$begingroup$

Given any function $f:Itomathbb R$ (not necessarily continuous), the condition: For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ such that $$ sum_{k} |y_{k} - x_{k}| < delta$$
implies
$$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon$$
is trivially true.

Proof: Given any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, just choose $delta = frac{1}{2} sum_{k} |y_{k} - x_{k}|$. Then the condition $ sum_{k} |y_{k} - x_{k}| < delta$ will be false and so the implication "$ sum_{k} |y_{k} - x_{k}| < delta$
implies
$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon$" will be trivially true.

So this conditionis not sufficient for absolute continuity or even continuity.

share|cite|improve this answer

answered Mar 28 at 10:31

RamiroRamiro

7,36421535

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I also saw this after you gave the detailed explanation in the other question
  $endgroup$
  – High GPA
  Mar 28 at 22:25
  

  
  
  
  

add a comment | 

1

$begingroup$

Given any function $f:Itomathbb R$ (not necessarily continuous), the condition: For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ such that $$ sum_{k} |y_{k} - x_{k}| < delta$$
implies
$$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon$$
is trivially true.

Proof: Given any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, just choose $delta = frac{1}{2} sum_{k} |y_{k} - x_{k}|$. Then the condition $ sum_{k} |y_{k} - x_{k}| < delta$ will be false and so the implication "$ sum_{k} |y_{k} - x_{k}| < delta$
implies
$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon$" will be trivially true.

So this conditionis not sufficient for absolute continuity or even continuity.

share|cite|improve this answer

answered Mar 28 at 10:31

RamiroRamiro

7,36421535

$endgroup$

• 
  
  
  

  
  1
  

  
  
  
  
  
  

  
  $begingroup$
  Yes, I also saw this after you gave the detailed explanation in the other question
  $endgroup$
  – High GPA
  Mar 28 at 22:25
  

  
  
  
  

add a comment | 

$begingroup$

Given any function $f:Itomathbb R$ (not necessarily continuous), the condition: For any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, we have $forallepsilonexistsdelta$ such that $$ sum_{k} |y_{k} - x_{k}| < delta$$
implies
$$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon$$
is trivially true.

Proof: Given any countable sequence of pairwise disjoint sub-intervals $(x_k, y_k)$ of $I$, just choose $delta = frac{1}{2} sum_{k} |y_{k} - x_{k}|$. Then the condition $ sum_{k} |y_{k} - x_{k}| < delta$ will be false and so the implication "$ sum_{k} |y_{k} - x_{k}| < delta$
implies
$sum_{k} |f(y_{k}) - f(x_{k})| < epsilon$" will be trivially true.

So this conditionis not sufficient for absolute continuity or even continuity.

share|cite|improve this answer

answered Mar 28 at 10:31

RamiroRamiro

7,36421535

$endgroup$

share|cite|improve this answer

answered Mar 28 at 10:31

RamiroRamiro

7,36421535

answered Mar 28 at 10:31

RamiroRamiro

7,36421535

answered Mar 28 at 10:31

RamiroRamiro

7,36421535

2

$begingroup$

Yes, they are equivalent. Suppose you choose $delta$ according to the usual definition of absolute continuity with $epsilon$ repalced by $epsilon /2$. If $(a_k.b_k)$ is a disjoint sequence of interval with total length less than $delta$ then $sumlimits_{k=1}^{N} |f(b_k)-f(a_k)| < epsilon /2$ for each $N$. Let $N to infty$ to complete the proof.

share|cite|improve this answer

answered Mar 25 at 8:24

Kavi Rama MurthyKavi Rama Murthy

75.9k53270

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Many thanks for your teaching! So you proved that the definition "$forallepsilonexistsdelta(forall text{finite subintervals we have} (sum|y_k-x_k|<delta Rightarrow sum|f(y_k)-f(x_k)|<epsilon))$" is equivalent to "$forallepsilonexistsdelta(forall text{countable subintervals we have} (sum|y_k-x_k|<delta Rightarrow sum|f(y_k)-f(x_k)|<epsilon))$". However, my first condition means "$forall text{countable subintervals}(forallepsilonexistsdelta text{we have} (sum|y_k-x_k|<delta Rightarrow sum|f(y_k)-f(x_k)|<epsilon))$". Not sure my understanding is correct, though.
  $endgroup$
  – High GPA
  Mar 25 at 22:05
  
  
  

  
  
  
  

add a comment | 

2

$begingroup$

Yes, they are equivalent. Suppose you choose $delta$ according to the usual definition of absolute continuity with $epsilon$ repalced by $epsilon /2$. If $(a_k.b_k)$ is a disjoint sequence of interval with total length less than $delta$ then $sumlimits_{k=1}^{N} |f(b_k)-f(a_k)| < epsilon /2$ for each $N$. Let $N to infty$ to complete the proof.

share|cite|improve this answer

answered Mar 25 at 8:24

Kavi Rama MurthyKavi Rama Murthy

75.9k53270

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Many thanks for your teaching! So you proved that the definition "$forallepsilonexistsdelta(forall text{finite subintervals we have} (sum|y_k-x_k|<delta Rightarrow sum|f(y_k)-f(x_k)|<epsilon))$" is equivalent to "$forallepsilonexistsdelta(forall text{countable subintervals we have} (sum|y_k-x_k|<delta Rightarrow sum|f(y_k)-f(x_k)|<epsilon))$". However, my first condition means "$forall text{countable subintervals}(forallepsilonexistsdelta text{we have} (sum|y_k-x_k|<delta Rightarrow sum|f(y_k)-f(x_k)|<epsilon))$". Not sure my understanding is correct, though.
  $endgroup$
  – High GPA
  Mar 25 at 22:05
  
  
  

  
  
  
  

add a comment | 

$begingroup$

Yes, they are equivalent. Suppose you choose $delta$ according to the usual definition of absolute continuity with $epsilon$ repalced by $epsilon /2$. If $(a_k.b_k)$ is a disjoint sequence of interval with total length less than $delta$ then $sumlimits_{k=1}^{N} |f(b_k)-f(a_k)| < epsilon /2$ for each $N$. Let $N to infty$ to complete the proof.

share|cite|improve this answer

answered Mar 25 at 8:24

Kavi Rama MurthyKavi Rama Murthy

75.9k53270

$endgroup$

share|cite|improve this answer

answered Mar 25 at 8:24

Kavi Rama MurthyKavi Rama Murthy

75.9k53270

answered Mar 25 at 8:24

Kavi Rama MurthyKavi Rama Murthy

75.9k53270

answered Mar 25 at 8:24

Kavi Rama MurthyKavi Rama Murthy

75.9k53270