a neat fact about uniform continuity

Uniform Continuity and Derivatives
In this video, I present a really neat test for uniform continuity, which has to do with derivatives. Check out this video to find out what it is!
Uniform Continuity: • What is Uniform Contin...
Mean Value Theorem: • Mean Value Theorem Proof
Continuity Playlist: • Limits and Continuity
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[@saminzamil3023]

Just wanted to tell you that I love your videos. Keep it up Sir❣️

[@drpeyam]

Thank you!!!

[@dylank6191]

Continuously differentiable functions with bounded derivatives even are Lipschitz continous, as seen in your proof if we just stop at |f(x) - f(y)|

[@lucaug10]

I was thinking the same thing! Good to know I wasn't wrong hahahah!

[@arturcostasteiner9735]

The derivative doesn't need to be continuous. All that's required is it's bounded.

[@yoav613]

What agrand finale to the series!!👌

[@ethandole2218]

I was wondering about this- good to know that it is indeed true!

[@jsusss]

Muchas gracias! 😊

[@rasad5285]

Thankyou so much i was struggling with this topic ❤

[@Peter_1986]

I don't think that it is possible to be any more charming than Dr Peyam.

[@deeptochatterjee532]

I'm guessing this is specifically for connected intervals then for the MVT to apply

[@drpeyam]

I think so

[@arturcostasteiner9735]

Every interval is connected

[@deeptochatterjee532]

@@arturcostasteiner9735 Yeah I guess I should have said connected subsets, which maybe by definition is an interval

[@curiosityzero2151]

I love analysis videos from Dr Peyam

[@santhoshhb3397]

It's very awsome

[@dgrandlapinblanc]

Ok. Thanks.

[@Ghjsinshshbdgdbd]

This is so neat thank you for not waiting and releasing early😫💦

[@sumittete2804]

Hello Sir...If a function is uniformly continuous on a closed interval, could we refine the definition of uniform continuity by replacing the condition |x-y| < δ and |f(x) - f(y)| < ε with |x-y| ≤ δ implying |f(x) - f(y)| ≤ ε ?

[@user-wu8yq1rb9t]

Hello Dear Dr Peyam. I have a request. If it's possible, please make a video (actually another one, from a to z) about Cauchy integral and Residue theorem. Thank you

[@drpeyam]

I think I already have one

[@user-wu8yq1rb9t]

@@drpeyam Yeah you have and I watched it. But I think it's just an example; please if it's possible think about making a new video. And also I have another suggestions. Please make some videos about Group Theory and somehow explain to us one of the groups (like SO(5), SO(10) ...). Thanks

[@user-wu8yq1rb9t]

@@drpeyam Another suggestions: As an Iranian, it would be nice if you make a video about *Khayyam Pascal's triangle*

[@dwaipayansharma2282]

Can we use this theorem to prove that f(x) = (√x)sinx is not UC on R(since its derivative is not bounded on R) Thanks in advance.

[@e-learningtutor1351]

Sir can you please explain about PROBABILITY DENSITY THEORM I was searching for this n iam not getting any proper video about it anywhere. So, can you please help me out sir!! Thanks sir,i like the way you lecture

[@drpeyam]

Never heard of it

[@e-learningtutor1351]

@@drpeyam even I didn't sir But when I made a video on probability in my super 5 series,one viewer commented about it..and it was a month ago or so I was searching for it from so long ,iam not even finding related concepts of that..so I have asked you It basically says that,we can find probability even without knowing the total observations.. I was shocked after knowing this..so I have asked you sir

[@yassinemohamed2235]

Mr the thoerem is not applied to the interval [a,b) .because the conditions of continous should have at the interval [a,b] but not [a,b) .else you remplace this interval by [a,oo)

[@drpeyam]

It works for [a,b) as well I think, and also for [a,oo)

[@sudinbhattarai3964]

Your fan from NEPAL

[@reeeeeplease1178]

I think the statement is even stronger, my memory might play tricks on me tho (was HW exercise): Let f be cont. on [a,b] and diff. on (a,b). Then i) and ii) are equivalent: i) f' is bounded ii) f is UC on [a,b] *As I said, I'm not sure if it's correct* Should be intuitively correct tho; if f is UC then f can't grow "infinitely fast" (referring to def of UC), so f' should always be bounded

[@reeeeeplease1178]

Apparently, we need f to be Lipschitz in order for the equivalence to hold: |f(x)-f(y)| < L |x-y| Now "solve" for L L > | ... | for all x,y in (a,b) Since f is diff., we know |f'(x)| = | lim (...)| = lim | ... | < L Where *

[@jamesshelton3827]

Any possibility you could do some videos on showing holder continuous functions?

[@drpeyam]

Check out the playlist

[@Wooflays]

ASU! Hope it's working out for you!

[@drpeyam]

Actually I’m at TAMU now haha

[@tomoki-v6o]

تحياتنا

[@MathAdam]

For some reason, my profanity filter blocked the thumbnail?

[@drpeyam]

LOL, how? 😂😂😂

[@MathAdam]

@@drpeyam My attempt at childish humour. Look at the letters in the thumbnail and read them aloud like a 9-year-old boy would. Not very sophisticated humour, I know. Please don't block me. :D

[@drpeyam]

Oooooh I get it now 😂

[@lorenzojulian7080]

Fachero reyy

[@tricky778]

f' =>UC that