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Continuity on infinity

Proving uniform continuity on an intervalContinuity of a constant functioncontinuity and uniform continuty of $1/x$Prove that the Rational function $fleft(xright)=frac{pleft(xright)}{qleft(xright)}$ is uniformly continuousProof of uniformly continuous for nth root of x on [0,+∞)Given a continuous function with an asymptote, prove that the function is uniformly continuous.How to prove that $arg: S^1setminus{-1}to (-pi,pi)$ is continuous? (With restrictions.)$f$ be real uniformly continuous function on the bounded set $E$ of $Bbb R$ then $f$ is bounded on $E$“Trick” to finding Delta in proving continuity/uniform continuity?Real Analysis - Uniform continuity

1

$begingroup$

We have $f: [a.b] rightarrow mathbb{R}$ continuous, and
$$c_n = sup{c in [a,b] : |f(x) - f(c_{n-1})| < epsilon text{ for any } x in [c_{n-1},c]}$$
with $c_1 = a$. In a previous exercise we saw that $c_n$ goes to $b$, so we have those intervals $[c_{n-1},c]$ covering the whole of $[a,b]$.

What I have been tasked to do is to prove, using this, that a continuous function on a closed and bounded interval is uniformly continuous. I have some clue about how to go about this, yet I cannot figure out how to prove this is the case if for some $x$ and $delta$, $x-delta$ is on one of the intervals and $x+delta$ is on another.

real-analysis sequences-and-series analysis continuity uniform-continuity

share|cite|improve this question

edited Mar 12 at 15:59

David Gravanita DGDASMASTER

asked Mar 12 at 15:28

David Gravanita DGDASMASTERDavid Gravanita DGDASMASTER

525

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Instead of "converges uniformly", did you intend to write "is uniformly continuous"?
  $endgroup$
  – jawheele
  Mar 12 at 15:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did, thanks for the catch!
  $endgroup$
  – David Gravanita DGDASMASTER
  Mar 12 at 15:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And did you show in the previous exercise that $c_n$ goes to $b$ in finitely many steps? Or only that $lim_{n to infty} c_n =b$?
  $endgroup$
  – jawheele
  Mar 12 at 16:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did see that it goes in finitely many steps
  $endgroup$
  – David Gravanita DGDASMASTER
  Mar 12 at 16:09
  

  
  
  
  

add a comment | 

1

$begingroup$

We have $f: [a.b] rightarrow mathbb{R}$ continuous, and
$$c_n = sup{c in [a,b] : |f(x) - f(c_{n-1})| < epsilon text{ for any } x in [c_{n-1},c]}$$
with $c_1 = a$. In a previous exercise we saw that $c_n$ goes to $b$, so we have those intervals $[c_{n-1},c]$ covering the whole of $[a,b]$.

What I have been tasked to do is to prove, using this, that a continuous function on a closed and bounded interval is uniformly continuous. I have some clue about how to go about this, yet I cannot figure out how to prove this is the case if for some $x$ and $delta$, $x-delta$ is on one of the intervals and $x+delta$ is on another.

real-analysis sequences-and-series analysis continuity uniform-continuity

share|cite|improve this question

edited Mar 12 at 15:59

David Gravanita DGDASMASTER

asked Mar 12 at 15:28

David Gravanita DGDASMASTERDavid Gravanita DGDASMASTER

525

$endgroup$

• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  Instead of "converges uniformly", did you intend to write "is uniformly continuous"?
  $endgroup$
  – jawheele
  Mar 12 at 15:40
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did, thanks for the catch!
  $endgroup$
  – David Gravanita DGDASMASTER
  Mar 12 at 15:59
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  And did you show in the previous exercise that $c_n$ goes to $b$ in finitely many steps? Or only that $lim_{n to infty} c_n =b$?
  $endgroup$
  – jawheele
  Mar 12 at 16:02
  

  
  
  
  


• 
  
  
  

  
  

  
  
  
  
  
  

  
  $begingroup$
  I did see that it goes in finitely many steps
  $endgroup$
  – David Gravanita DGDASMASTER
  Mar 12 at 16:09
  

  
  
  
  

add a comment | 

$begingroup$

We have $f: [a.b] rightarrow mathbb{R}$ continuous, and
$$c_n = sup{c in [a,b] : |f(x) - f(c_{n-1})| < epsilon text{ for any } x in [c_{n-1},c]}$$
with $c_1 = a$. In a previous exercise we saw that $c_n$ goes to $b$, so we have those intervals $[c_{n-1},c]$ covering the whole of $[a,b]$.

What I have been tasked to do is to prove, using this, that a continuous function on a closed and bounded interval is uniformly continuous. I have some clue about how to go about this, yet I cannot figure out how to prove this is the case if for some $x$ and $delta$, $x-delta$ is on one of the intervals and $x+delta$ is on another.

real-analysis sequences-and-series analysis continuity uniform-continuity

share|cite|improve this question

edited Mar 12 at 15:59

David Gravanita DGDASMASTER

asked Mar 12 at 15:28

David Gravanita DGDASMASTERDavid Gravanita DGDASMASTER

525

$endgroup$

share|cite|improve this question

edited Mar 12 at 15:59

David Gravanita DGDASMASTER

asked Mar 12 at 15:28

David Gravanita DGDASMASTERDavid Gravanita DGDASMASTER

525

share|cite|improve this question

edited Mar 12 at 15:59

David Gravanita DGDASMASTER

asked Mar 12 at 15:28

David Gravanita DGDASMASTERDavid Gravanita DGDASMASTER

525

asked Mar 12 at 15:28

David Gravanita DGDASMASTERDavid Gravanita DGDASMASTER

525

asked Mar 12 at 15:28

David Gravanita DGDASMASTERDavid Gravanita DGDASMASTER

525

1 Answer
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1

$begingroup$

You've shown that for any given $epsilon>0$, you have a corresponding partition ${c_n}_{n=1}^{N}$ of the interval $[a,b]$ such that $c_1=a,$ $ c_N=b$, and for each $1 leq n < N$, $forall x in [c_n,c_{n+1})$ you have $|f(x)-f(c_n)| < epsilon$. Note this implies also that $|f(c_{n+1})-f(c_n)| leq epsilon$.

Fix an $epsilon>0$, and consider the above sequence ${c_n}$ corresponding to $frac{epsilon}{3}$. Choose
$$delta= min_{1 leq n < N}{c_{n+1}-c_{n}}$$
Then for any $x,y in [a,b]$ such that $|y-x| < delta$, suppose WLOG that $x leq y$ and consider that $exists$ $1 leq n < N$ such that $x in [c_n,c_{n+1}]$, and we either have $y leq c_{n+1}$ or $y > c_{n+1}$.

In the former case, $y in [c_n,c_{n+1}]$, so
$$|f(y)-f(x)| leq |f(y)-f(c_n)|+|f(x)-f(c_n)| leq frac{epsilon}{3} + frac{epsilon}{3} < epsilon$$
By construction of the partition.

In the latter case, $y > c_{n+1} geq x$ (note $n leq N-2$ for this case to occur), so
$$y-c_{n+1} leq y-x < delta leq c_{n+2}-c_{n+1}$$
and hence $y < c_{n+2}$, i.e. $y in [c_{n+1},c_{n+2})$, and we have
$$|f(y)-f(x)| leq |f(y)-f(c_{n+1})| + |f(c_{n+1})-f(c_n)| + |f(x)-f(c_n)| < frac{epsilon}{3} + frac{epsilon}{3} + frac{epsilon}{3} = epsilon$$
again by construction of the partition, showing that $f$ is uniformly continuous on $[a,b]$.

share|cite|improve this answer

answered Mar 12 at 16:55

jawheelejawheele

30219

$endgroup$

add a comment | 

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1

$begingroup$

You've shown that for any given $epsilon>0$, you have a corresponding partition ${c_n}_{n=1}^{N}$ of the interval $[a,b]$ such that $c_1=a,$ $ c_N=b$, and for each $1 leq n < N$, $forall x in [c_n,c_{n+1})$ you have $|f(x)-f(c_n)| < epsilon$. Note this implies also that $|f(c_{n+1})-f(c_n)| leq epsilon$.

Fix an $epsilon>0$, and consider the above sequence ${c_n}$ corresponding to $frac{epsilon}{3}$. Choose
$$delta= min_{1 leq n < N}{c_{n+1}-c_{n}}$$
Then for any $x,y in [a,b]$ such that $|y-x| < delta$, suppose WLOG that $x leq y$ and consider that $exists$ $1 leq n < N$ such that $x in [c_n,c_{n+1}]$, and we either have $y leq c_{n+1}$ or $y > c_{n+1}$.

In the former case, $y in [c_n,c_{n+1}]$, so
$$|f(y)-f(x)| leq |f(y)-f(c_n)|+|f(x)-f(c_n)| leq frac{epsilon}{3} + frac{epsilon}{3} < epsilon$$
By construction of the partition.

In the latter case, $y > c_{n+1} geq x$ (note $n leq N-2$ for this case to occur), so
$$y-c_{n+1} leq y-x < delta leq c_{n+2}-c_{n+1}$$
and hence $y < c_{n+2}$, i.e. $y in [c_{n+1},c_{n+2})$, and we have
$$|f(y)-f(x)| leq |f(y)-f(c_{n+1})| + |f(c_{n+1})-f(c_n)| + |f(x)-f(c_n)| < frac{epsilon}{3} + frac{epsilon}{3} + frac{epsilon}{3} = epsilon$$
again by construction of the partition, showing that $f$ is uniformly continuous on $[a,b]$.

share|cite|improve this answer

answered Mar 12 at 16:55

jawheelejawheele

30219

$endgroup$

add a comment | 

1

$begingroup$

You've shown that for any given $epsilon>0$, you have a corresponding partition ${c_n}_{n=1}^{N}$ of the interval $[a,b]$ such that $c_1=a,$ $ c_N=b$, and for each $1 leq n < N$, $forall x in [c_n,c_{n+1})$ you have $|f(x)-f(c_n)| < epsilon$. Note this implies also that $|f(c_{n+1})-f(c_n)| leq epsilon$.

Fix an $epsilon>0$, and consider the above sequence ${c_n}$ corresponding to $frac{epsilon}{3}$. Choose
$$delta= min_{1 leq n < N}{c_{n+1}-c_{n}}$$
Then for any $x,y in [a,b]$ such that $|y-x| < delta$, suppose WLOG that $x leq y$ and consider that $exists$ $1 leq n < N$ such that $x in [c_n,c_{n+1}]$, and we either have $y leq c_{n+1}$ or $y > c_{n+1}$.

In the former case, $y in [c_n,c_{n+1}]$, so
$$|f(y)-f(x)| leq |f(y)-f(c_n)|+|f(x)-f(c_n)| leq frac{epsilon}{3} + frac{epsilon}{3} < epsilon$$
By construction of the partition.

In the latter case, $y > c_{n+1} geq x$ (note $n leq N-2$ for this case to occur), so
$$y-c_{n+1} leq y-x < delta leq c_{n+2}-c_{n+1}$$
and hence $y < c_{n+2}$, i.e. $y in [c_{n+1},c_{n+2})$, and we have
$$|f(y)-f(x)| leq |f(y)-f(c_{n+1})| + |f(c_{n+1})-f(c_n)| + |f(x)-f(c_n)| < frac{epsilon}{3} + frac{epsilon}{3} + frac{epsilon}{3} = epsilon$$
again by construction of the partition, showing that $f$ is uniformly continuous on $[a,b]$.

share|cite|improve this answer

answered Mar 12 at 16:55

jawheelejawheele

30219

$endgroup$

add a comment | 

$begingroup$

You've shown that for any given $epsilon>0$, you have a corresponding partition ${c_n}_{n=1}^{N}$ of the interval $[a,b]$ such that $c_1=a,$ $ c_N=b$, and for each $1 leq n < N$, $forall x in [c_n,c_{n+1})$ you have $|f(x)-f(c_n)| < epsilon$. Note this implies also that $|f(c_{n+1})-f(c_n)| leq epsilon$.

Fix an $epsilon>0$, and consider the above sequence ${c_n}$ corresponding to $frac{epsilon}{3}$. Choose
$$delta= min_{1 leq n < N}{c_{n+1}-c_{n}}$$
Then for any $x,y in [a,b]$ such that $|y-x| < delta$, suppose WLOG that $x leq y$ and consider that $exists$ $1 leq n < N$ such that $x in [c_n,c_{n+1}]$, and we either have $y leq c_{n+1}$ or $y > c_{n+1}$.

In the former case, $y in [c_n,c_{n+1}]$, so
$$|f(y)-f(x)| leq |f(y)-f(c_n)|+|f(x)-f(c_n)| leq frac{epsilon}{3} + frac{epsilon}{3} < epsilon$$
By construction of the partition.

In the latter case, $y > c_{n+1} geq x$ (note $n leq N-2$ for this case to occur), so
$$y-c_{n+1} leq y-x < delta leq c_{n+2}-c_{n+1}$$
and hence $y < c_{n+2}$, i.e. $y in [c_{n+1},c_{n+2})$, and we have
$$|f(y)-f(x)| leq |f(y)-f(c_{n+1})| + |f(c_{n+1})-f(c_n)| + |f(x)-f(c_n)| < frac{epsilon}{3} + frac{epsilon}{3} + frac{epsilon}{3} = epsilon$$
again by construction of the partition, showing that $f$ is uniformly continuous on $[a,b]$.

share|cite|improve this answer

answered Mar 12 at 16:55

jawheelejawheele

30219

$endgroup$

share|cite|improve this answer

answered Mar 12 at 16:55

jawheelejawheele

30219

answered Mar 12 at 16:55

jawheelejawheele

30219

answered Mar 12 at 16:55

jawheelejawheele

30219