Every Lipschitz Function is Uniformly Continuous Proof

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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[MoHaMmADAlmotaref]

I'm non-English speaker, and your lessons is really understandable!!! keep your good work up

[TheOskro]

The way you write on the board has so much swag. Nice video, love your channel.

[rfLm]

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!!!

[paperstars]

I really needed a playlist with proofs like this, to practice being more through with my proofs in my first semester.

[blackmathematician]

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

[Sentuyashi]

Man you really helped me out! Our teacher didn't show us this proof. Thank you!

[datsmydab-minecraft-and-mo]

I love this channel. P.S you've grown so much this year!

[epiccheeseburgercat]

Thank you this was actually super clear

[ARYANSAXENA-qjii]

Sequential criteria of uniform continuity:
If (x_n) and (y_n) are two equivalent sequences i.e. lim (x_n-y_n)=0, then (fx_n) and (fy_n) are eqivalent.

Lipschitz implies UC proof

Let (x_n) and (y_n) be equivalent. |f(x_n)-f(y_n)|<=M |x_n-y_n| (by assumption). Taking limit on both sides, we have that f(x_n) and f(y_n) are equivalent.

[tomascampo]

This video was awesome. Thank you very much.

[theannakata]

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?

[laisathasneem]

This video is very helpful. Thank you sir 👍🏻.

[jameyatesmauriat]

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?

[caziobay]

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)| <= M|x-y| for all x, y in D. But this will still imply uniform continuity since In this case we can choose delta to be any positive value, like 1. so for all x, y in D s.t |x-y|<1 (which is all values in D) we have |x-y| = |x - x| = |0| = 0 < 1 -> |f(x)-f(y)| = |f(x)-f(x)| = |0| = 0 < epsilon for all epsilon > 0.

[rupamsharma]

Thank you . Your videos are really helpful for me. Can we talk sir....

[learngermanwithvanessa]

M also has to be either bigger or equal to M, right?

[coreymonsta]

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

[SW]

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?

[علاءعبدالامير-هو]

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.

[davidgarmroudi]

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