Uniform continuity

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Homework Statement:

Prove that if f Uniform continuity in R, then for any two series xn and yn limyn-xn=0 so limf(xn)-f(yn)=0

Relevant Equations:

Uniform continuity
There are two parts to the question Let's start with part :)
I understand the definition of Uniform continuity And I think I'm in the right direction for the solution but I'm not sure of the formal wording.
So be it ε>0
Given that yn limyn-xn=0 so For all ε>0 , ∃N∈ℕ so that For all N<n , |yn-xn|<ε so ∃δ>0 so that |yn-xn|<δ And because f is Uniform continuity
|f(yn)-f(xn)|<ε.
I know my formulation is not good but I can't formalize it nicely
 

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  • #2
Math_QED
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Homework Statement:: Prove that if f Uniform continuity in R, then for any two series xn and yn limyn-xn=0 so limf(xn)-f(yn)=0
Homework Equations:: Uniform continuity

There are two parts to the question Let's start with part :)
I understand the definition of Uniform continuity And I think I'm in the right direction for the solution but I'm not sure of the formal wording.
So be it ε>0
Given that yn limyn-xn=0 so For all ε>0 , ∃N∈ℕ so that For all N<n , |yn-xn|<ε so ∃δ>0 so that |yn-xn|<δ And because f is Uniform continuity
|f(yn)-f(xn)|<ε.
I know my formulation is not good but I can't formalize it nicely
The correct term is sequences. Not series.

I think you got the right idea but it contains some mistakes.

Let ##\epsilon>0##. Choose ##\delta>0## as in the definition of uniform continuity.

Choose ##N## such that ##|x_n−y_n|<\delta ## if ##n\geq N##.

Can you conclude?
 
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  • #3
PeroK
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Homework Statement:: Prove that if f Uniform continuity in R, then for any two series xn and yn limyn-xn=0 so limf(xn)-f(yn)=0
Homework Equations:: Uniform continuity

There are two parts to the question Let's start with part :)
I understand the definition of Uniform continuity And I think I'm in the right direction for the solution but I'm not sure of the formal wording.
So be it ε>0
Given that yn limyn-xn=0 so For all ε>0 , ∃N∈ℕ so that For all N<n , |yn-xn|<ε so ∃δ>0 so that |yn-xn|<δ And because f is Uniform continuity
|f(yn)-f(xn)|<ε.
I know my formulation is not good but I can't formalize it nicely
You must be much clearer about how you get ##N, \delta, \epsilon##.

Hint: try working backwards from the way you did it:

Let ##\epsilon > 0##

How do you get ##|f(y_n) - f(x_n)| < \epsilon##?

Answer: if ##x_n, y_n## are close enough together. This gives you ##\delta## (depends on ##\epsilon##).

How do you get ##x_n, y_n## close enough together?

Answer: by taking ##n## large enough. This gives you ##N##, which depends on ##\delta##.

Can you formalise that strategy?
 
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  • #4
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You must be much clearer about how you get ##N, \delta, \epsilon##.

Hint: try working backwards from the way you did it:

Let ##\epsilon > 0##

How do you get ##|f(y_n) - f(x_n)| < \epsilon##?

Answer: if ##x_n, y_n## are close enough together. This gives you ##\delta## (depends on ##\epsilon##).

How do you get ##x_n, y_n## close enough together?

Answer: by taking ##n## large enough. This gives you ##N##, which depends on ##\delta##.

Can you formalise that strategy?
I understand what you mean I tried for an hour now to translate my formulation into English.
But my English is not good and it turns out really not good.
I'm not sure how I write the passage from that if |xn−yn|<δ so |f(yn)−f(xn)|<ϵ
I'd love if you could help me with that.
 
  • #5
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The correct term is sequences. Not series.

I think you got the right idea but it contains some mistakes.

Let ##\epsilon>0##. Choose ##\delta>0## as in the definition of uniform continuity.

Choose ##N## such that ##|x_n−y_n|<\delta ## if ##n\geq N##.

Can you conclude?
The correct term is sequences. Not series . Thanks I will use it
Can you please help me with a formal wording: /
 
  • #6
Math_QED
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The correct term is sequences. Not series . Thanks I will use it
Can you please help me with a formal wording: /
Let ##\epsilon > 0##. Because ##f## is uniformly continuous, we may choose ##\delta > 0## such that for all ##x## with ##|x-y|< \delta## we have ##|f(x)-f(y)|< \epsilon##. Because ##\lim_n (x_n-y_n)= 0##, there is ##N## such that ##|x_n-y_n|< \delta## whenever ##n \geq N##. Thus, combining the two previous statements, we get ##|f(x_n)-f(y_n)|< \epsilon## whenever ##n \geq N##. Thus we have proven that for any ##\epsilon>0##, there is ##N## such that ##|f(x_n)-f(y_n)|< \epsilon## whenever ##n \geq N##. This is exactly the definition of ##\lim_n (f(x_n)-f(y_n)) = 0## and we are done. ##\quad \square##
 
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  • #7
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Let ##\epsilon > 0##. Because ##f## is uniformly continuous, we may choose ##\delta > 0## such that for all ##x## with ##|x-y|< \delta## we have ##|f(x)-f(y)|< \epsilon##. Because ##\lim_n (x_n-y_n)= 0##, there is ##N## such that ##|x_n-y_n|< \delta## whenever ##n \geq N##. Thus, combining the two previous statements, we get ##|f(x_n)-f(y_n)|< \epsilon## whenever ##n \geq N##. Thus we have proven that for any ##\epsilon>0##, there is ##N## such that ##|f(x_n)-f(y_n)|< \epsilon## whenever ##n \geq N##. This is exactly the definition of ##\lim_n (f(x_n)-f(y_n)) = 0## and we are done. ##\quad \square##
Thanks so much for your understanding.
This stuff is new to me and I need to get used to a good formulation
 

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