What is the difference between uniformly continuous and Lipschitz continuous? I understand that Lipschitz is more strict, but I don't see the difference.

Question

What is the difference between uniformly continuous and Lipschitz continuous?  I understand that Lipschitz is more strict, but I don't see the difference.

fibonaccichick666 · Answer

@zzr0ck3r Do you mind?

fibonaccichick666 · Answer

so I know for Lipschitz we find some K that makes it so we don't depend on epsilon, but is that really it?

zzr0ck3r · Answer

Do you understand the difference in uniform convergence and regular convergence?

fibonaccichick666 · Answer

convergence no... that is chapter 6

zzr0ck3r · Answer

err I meant continunity

fibonaccichick666 · Answer

Uniform convergence means that delta is the same for all values, so like for f(x)=2x, delta =2, right?

fibonaccichick666 · Answer

@amberpruitt thank you, but I'm sorry, I don't understand your reasoning/explanation due to the use of gradient.

zzr0ck3r · Answer

it is basically all about the order of the quantifiers

its not that its one delta... its always one delta

zzr0ck3r · Answer

its the fact that delta cant depend on c

zzr0ck3r · Answer

so our example before with x^2 would not work, but the one with 3x would

fibonaccichick666 · Answer

ok, so delta=epsilon/2 would work?

zzr0ck3r · Answer

note that in one we had delta less than something with c in it, that was the linear one

zzr0ck3r · Answer

yes

zzr0ck3r · Answer

we choose delta before we choose "c"

zzr0ck3r · Answer

so it cant depend on it

fibonaccichick666 · Answer

ok and then for Lipschitz, how do we eliminate epsilon?  What is the process?

zzr0ck3r · Answer

ps @amberpruitt was just posting from here.... https://www.physicsforums.com/threads/lipschitz-vs-uniform-continuity.702500/

zzr0ck3r · Answer

I cant find a good picture....

fibonaccichick666 · Answer

oh

zzr0ck3r · Answer

lipschitz is just $|x-x_0|< c|x-x_0|$ for some c right?

fibonaccichick666 · Answer

right

fibonaccichick666 · Answer

so like 3x?

zzr0ck3r · Answer

err the function... yeah

fibonaccichick666 · Answer

or no

zzr0ck3r · Answer

|f(x) - f(x_0)| < c|x-x0|

zzr0ck3r · Answer

so I would think about the right hand side as a secant line

fibonaccichick666 · Answer

ok

fibonaccichick666 · Answer

so if delta=epsilon/3 then would c=1/3?

zzr0ck3r · Answer

yes

zzr0ck3r · Answer

for 3x

zzr0ck3r · Answer

good luck with x^2

fibonaccichick666 · Answer

I'll do it on [0,1] :P

zzr0ck3r · Answer

:)

fibonaccichick666 · Answer

no... I don't have that memorized at all...

zzr0ck3r · Answer

what?

fibonaccichick666 · Answer

when x^2 is unif cont.

zzr0ck3r · Answer

I don't think it is.

fibonaccichick666 · Answer

ok so, in the end, for uniform, we have one delta that works for every c, and for Lipschitz, we have a delta such that it does not depend on anything but epsilon.

fibonaccichick666 · Answer

and yea, I know it is.   It's one of the few things I have memorized due to the frequency it comes up

zzr0ck3r · Answer

I don't think there are any delta epsilon in lippelletz

fibonaccichick666 · Answer

lol, not technically, but that is just how I am understanding it

zzr0ck3r · Answer

$\exists   \ C \ \forall u,v \ \ \ |f(u)-f(v)|\le C|u-v|
 
$

fibonaccichick666 · Answer

right

zzr0ck3r · Answer

I see

fibonaccichick666 · Answer

but for me to understand that, when I don't know it is lippellets, i need that

zzr0ck3r · Answer

its all forcing things under inequality signs

fibonaccichick666 · Answer

right

fibonaccichick666 · Answer

ok so I think this myth is confirmed to the point that I understand.  Thank you again :)
I really have no idea what I'd do without you.

zzr0ck3r · Answer

npz

think of lipschitz like this

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