Uniform Continuity (Example 4): Not Uniformly Continuous

Showing a function is not uniformly continuous
In this video, I give another example of a function that's not uniformly continuous. Compare this with Example 1 in the description below.
What is Uniform Continuity? • What is Uniform Contin...
Example 1 • Uniform Continuity (Ex...
Example 2 • Uniform Continuity (Ex...
Example 3 • Uniform Continuity (Ex...
Example 4 • Uniform Continuity (Ex...
Continuity Playlist: • Limits and Continuity
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[@e-learningtutor1351]

Amazing presentation and excellent explanation sir
Thank you

[@dr.rahulgupta7573]

Excellent presentation. vow !!

[@fernandobezerrademenezes1622]

Congratulations, my dear. Amazing video. Hard math made simple and easy to understand!

[@flowers9652]

Great proof!🙌🏻🙌🏻🙌🏻🙌🏻

[@TrueVasember]

Amazing video, watching from india for mid semester tests..

[@ZipplyZane]

This one is the interesting one to me. I don't remember the classic proof of continuity of f(x)=x² resulting in a delta that was defined in terms of x.

[@saiharivatsalboyanapalli2344]

Super Video

[@aneeshsrinivas9088]

epsilon=1 and x=1/delta, and y=1/delta+0.5*delta would work right? y^2-x^2=(x+0.5*delta)^2-x^2=x^2+2*0.5*delta*x+0.25delta^2-x^2=delta*1/delta+0.25*delta^2=1+0.25*delta^2, if delta>0 then 0.25*delta^2 is also >0 and thus 1+0.25*delta^2>1.

[@Bedoroski]

A dedicated explanation as always. But what do you mean by: let Epsilon be announced, cuz i'm not a native speaker

[@SontoshKumarSahani]

Hi Can you give the proof/Disproof for
(a) The product of two uniformly continuous functions on IR is also uniformly continuous.
(b) The product of two uniformly continuous functions on [0; 1] is also uniformly continuous.

[@Tletna]

Thank you for the video. I wanted to know if there is a different way of describing what 'uniformly continuous' vs 'not uniformly continuous' actually means that doesn't involve deltas and epsilons. What is the usefulness of making such a distinction (both within math and outside of math)?

[@Hi_Brien]

There's stuff we can do on functions that are uniformly continuous that we can't do on functions that aren't. The things we can do with it are very important.
I'm sure you've learned this in 2 years.

[@shadaanhassan5936]

Yes you can use the concept of Sequential Criteria for Uniform Continuity to show whether a function is U.C or not. By taking any two arbitrary sequence and showing there difference converging to 0 but there difference of function value limit does not converge to 0

[@Avinashkumar-uc8qd]

Sir do we need to fix the delta to get an Epsilon Or delta can be arbitrary....
Because in this example at 10:50 you wrote let delta be arbitrary but in previous example you wrote let delta be given( like delta is fixed)....

[@drpeyam]

We fix epsilon and then get delta, as in step 2

[@user-fy5tn7sy3t]

Yes it’s super video but need other watching and reversion
Thank u for video

[@xfilesz]

Is there a more formal way of saying let epsilon be "whatever" in proofs?

[@drpeyam]

Let epsilon be TBA

[@xfilesz]

@@drpeyam What exactly does that mean? It is fixed?

[@svetlanarusnakova5640]

let epsilon be given

[@saptarshisahoo5075]

What is the negation of "a implies b"?

[@wannabehuman.production]

A and -B

[@lacasadeacero]

7:27 why did u erase the abs?

[@drpeyam]

x and y are positive

[@user-wu8yq1rb9t]

Hello Dear Dr. Peyam.
Please let me ask you one (maybe personal) question.
Are you Iranian (specially because of your name)?
Thanks

[@drpeyam]

Yes

[@user-wu8yq1rb9t]

@@drpeyam
Thanks for your attention Dear Peyam.
I'm Iranian too.
I like you before, but now I love you!
Big Fan.

[@OndrejPopp]

Ok Tx dr Peyam! Ok, now let me think 🤔.. If you take the derivative you get 2x, so... on the positive interval 0 -> ♾️ the slope is increasing and depends on x and so you will always find x + delta that outruns Epsilon. And there it is.. I think 🤔💭

[@drpeyam]

I mean intuitively yes, but there are counterexamples to your statement I think

[@arturcostasteiner9735]

Formalizing a bit more
your resoning we can do as follows:
Fix d > 0 and let |y - x| = d. By the mean value theorem, there is s between x and y such that |f(y) - f(x)| = 2|s| |y - x| = 2s d. Letting |x| --> oo, then |y| --> oo, |s| --> oo and therefore |f(y) - f(x)| --> oo. So, given eps > 0, no matter what d you choose, there'll always be x and y with |y - x| eps. Hence, f is not u.c. on R. The same reasoning shows every polynimial od degree > = 2 is not u.c. on R.

[@OndrejPopp]

@@arturcostasteiner9735 Ok tx Artur. I was just preparing my dinner and then I realized that actually that is what you see in your mind 😃 how x + delta outruns epsilon! But then I should have said that f(x + delta) - f(x) outruns epsilon. So thank you for the full formal statement 😃

[@OndrejPopp]

@@drpeyam Counterexamples? Don't think so Dr Peyam. Because either x^2 is not uniformly continuous like you have proven by yourself, or it is which would be your counter example but you can't have it both ways... unless you misunderstood what I meant how x + delta outruns epsilon, but that's not what I meant I meant f(x + delta) - f(x).. Sorry about that! 😃

[@desertpointshacks6299]

Don’t know if this is what you meant but the classic counterexample is f(x)=sqrt(x) which has an unbounded derivative but yet is still uniformly continuous. If you know a function is not uniformly continuous then its derivative will be unbounded.

[@Sonyakumari-jv5do]

I wonder a lot finally get solution here .

[@purim_sakamoto]

関係ないけど、√εってあまり見ないから可愛い😋

[@yashj1072]

ma man

[@David-km2ie]

I would have taken |x-y|=delta/2 then |x+y|>=2epsilon/delta.
So take x=epsilon/delta and y=epsilon/delta+delta/2 and you are good
Fun video though to get it exactly epsilon big

[@user-dj4vf1yw2m]

ایرانی؟؟

[@drpeyam]

Bale