Equicontinuity related proof

[robertdeniro]

Homework Statement

let A represent the family of linear functions (ie. mx+b) on [0, 1] with distinct slopes

show that if A is equicontinuous, then the absolute value of the slopes are bounded, that is |m|<=C, for all functions in A

Homework Equations

The Attempt at a Solution

equicontinuous means for given e>0, can find &>0 such that |f(x)-f(y)|<e for |x-y|<&

|f(x)-f(y)|=|m(x-y)|<e

if i divide through by |(x-y)| then I am in trouble, since x and y could be arbitrary close

any help?

 

[Dick]

Why would you want to divide through by |x-y|? Divide both sides of the epsilon equation by |m|. Now how should epsilon be related to delta?

 

[robertdeniro]

oh i was trying to isolate |m|

so if i do what u said i would get |x-y|<e/|m|, but we know |x-y|<&...

not sure what to do from here

 

[Dick]

oh i was trying to isolate |m|

so if i do what u said i would get |x-y|<e/|m|, but we know |x-y|<&...

not sure what to do from here

What's the relation between epsilon and delta you get from that for that single function with slope m?

 

[robertdeniro]

What's the relation between epsilon and delta you get from that for that single function with slope m?

sorry, i must be missing something really obvious here

i have |x-y|<e/|m| and |x-y|<&

but i cannot see a relationship between e and &

 

[Dick]

sorry, i must be missing something really obvious here

i have |x-y|<e/|m| and |x-y|<&

but i cannot see a relationship between e and &

The definition of continuity says you are supposed to be able to choose a delta such that if |x-y|<delta then |f(x)-f(y)|<epsilon. Remember? You have that |f(x)-f(y)|<|m||x-y| which is then <|m|*delta, right? To make sure that's less than epsilon, you want to choose a delta such that |m|*delta<epsilon. Right, right? If you want to skip to the chase and equicontinuity, if you want to pick a single delta for all members of your family, what happens if |m| is unbounded? Can you pick such delta?

 

[robertdeniro]

The definition of continuity says you are supposed to be able to choose a delta such that if |x-y|<delta then |f(x)-f(y)|<epsilon. Remember? You have that |f(x)-f(y)|<|m||x-y| which is then <|m|*delta, right? To make sure that's less than epsilon, you want to choose a delta such that |m|*delta<epsilon. Right, right? If you want to skip to the chase and equicontinuity, if you want to pick a single delta for all members of your family, what happens if |m| is unbounded? Can you pick such delta?

ohhhhh...

if |m| is unbounded then i can't pick such delta because |m|*delta would also be unbounded

right? right?

 

[Dick]

ohhhhh...

if |m| is unbounded then i can't pick such delta because |m|*delta would also be unbounded

right? right?

Basically, right, right. If |m| is unbounded then given a fixed epsilon, the only delta that would satisfy |m|*delta<epsilon for all m is delta=0. But you want a delta>0.