Every Lipschitz Function is Uniformly Continuous Proof
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- Опубликовано: 12 мар 2020
- In this video I go through the proof that every Lipschitz function is uniformly continuous. I hope this video helps someone who is studying mathematical analysis/advanced calculus.
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Thank you:)
I'm non-English speaker, and your lessons is really understandable!!! keep your good work up
Thank you!
The way you write on the board has so much swag. Nice video, love your channel.
Haha thanks
I really needed a playlist with proofs like this, to practice being more through with my proofs in my first semester.
Could you kindly upload more videos just like these? I always wanted to learn mathematical analysis on a graduate level. Measure theory in particular 😁. In any case - thank you!!!
Yes more coming😄
@@TheMathSorcerer please make more real analysis video cover all rudin topics :D
Your channel is a treasure trove.
Thank you for your video! I wasn’t able to find any source that was able to explain this concept as well as you do
you are welcome!!
Thank you this was actually super clear
I love this channel. P.S you've grown so much this year!
👍
This video is very helpful. Thank you sir 👍🏻.
Man you really helped me out! Our teacher didn't show us this proof. Thank you!
You're very welcome!
This video was awesome. Thank you very much.
you are welcome!
simple and accurate. thanks a lot
You are welcome!
thank you sir
Thank you!
You're welcome!
Thank you so much. I must say I don't understand anything not the function nor the proof and this is something related to me for sure, can I know what is going on here and what video should I watch to understand this function?
Can you upload the video of peano
Thank you . Your videos are really helpful for me. Can we talk sir....
Professional way of writing on board
hehe thank you:)
Thank you for the very informative video! But what I don't quite understand is, how does one decide, that when M > 0, you choose delta as δ= ϵ/M?
M also has to be either bigger or equal to M, right?
Lip. condition is like here’s all the work for you in your epsilon delta proof done for you, and tells you recall you’ve got control over the size of |x - y| as always lol
Lol ya
I think M can be negative if x=y for all x,y in D (i.e. D is a singleton set), since then we will have |f(x)-f(y)| = |f(x) - f(x)| = 0 and M|x-y| = M|x-x| = 0 for all x,y in D. Therefore, |f(x)-f(y)| 0.
Thank you , but in assuming it is lipschitz is assumed, ok but how do you know that absolute value of x-y is less than delta?
Oh sorry, I see it now
Amazing :)
Could you upload a video proving there exists an M for every differentiable and continuous
function such that |f(x)-f(y)|
thank you!
A function f : [a, b] → R is said to satisfy a Lipschitz condition with Lipschitz constant L on [a, b]
if, for every x, y ∈ [a, b], we have |f (x) − f (y)| ≤ L|x − y|.
a. Show that if f satisfies a Lipschitz condition with Lipschitz constant L on an interval [a, b], then
f ∈ C[a, b].
b. Show that if f has a derivative that is bounded on [a, b] by L, then f satisfies a Lipschitz condition
with Lipschitz constant L on [a, b].
c. Give an example of a function that is continuous on a closed interval but does not satisfy a
Lipschitz condition on the interval.
it's clear, it is clear, it is clear, why do u solve a clear problem ??? Actually, u repeated the clear thing :) same as books
hahaha yeah, eveyrone says it's clear, and, it actually is, but still, worth to show the details! You'd be surprised:D
All I think of is rugrats when I think of lipshitz 😆