Real Analysis | Intro to uniform continuity.
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- Опубликовано: 9 сен 2020
- We introduce the notion of uniform continuity and give some motivating examples and calculuations.
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Really liked how you highlighted the order of quantifiers, it makes things very clear and intuitive in my opinion.
Thanks for clarifying the difference. Basically if it's uniformly continuous, Delta should work for all a. If it's only continuous, for all a, there's a Delta that works.
It was really great that you explained the subtle yet massive difference based on quantifiers. At first glance the definitions look almost the same.
I was almost scared we were going to stop in a place that was not a good place to stop....
At 10:31 that inequality is true only if a>=0. Generally, you need to multiply by 3|a|, not by 3a, because you are not sure of the sign of a. So, the correct inequality is 3a|a| - 3|a|
exactly. I was about to comment about it.
true but then you get in trouble when you add the inequalities.
When you write het inequations for a>0 and a
A super helpful and simple video !!
Hey Michael great video! Would you consider doing a playlist on questions from previous years' Preliminary Exams for Masters/PhD? I know this would be greatly appreciated by students trying to prepare for Masters Exams this upcoming year.
Here's my (not very rigorous) attempt at proving x³ is not uniformly continuous: Fix ε = 1; then if x³ is uniformly continuous then ∃δ > 0 such that |x − a| < δ ⇒ |x³ − a³| < 1. Now pick x = a + δ/2, so clearly |x − a| < δ is satisfied. Now x³ − a³ = 3a²(δ/2) + 3a(δ/2)² + (δ/2)^3 > 3a(δ/2)(a + δ/2) > 3(δ/2)a². But when a = 1/√δ, this is equal to 3(δ/2)/δ = 3/2 > 1, i.e. |x³ − a³| ≮ 1. Thus no choice of δ satisfies the criteria and therefore x³ is not uniformly continuous.
why'd you say not very rigorous?
Thank you very much.
thank you so much ,great video :)
What a legend are you
I am from India your teaching style ossm sir 😊
Thanks for clarifying vedio
14:13 Almost forgot to say the line 😛
11:15 why does 3a^2 +3|a|+1 need the absolute value on the a in 3a?
So functions which are concave downward and bounded below are uniformly continuous? And functions which are concave upward and bounded above are uniformly continuous? This is just like an intuitive hunch that i have without any rigorous proof behind my statement. Is it true? Can someone give me an example where my statement is false.
I don't remember uniform continuity over R being particularly useful. Uniform continuity over bounded intervals on the other hand is far more important if I'm not mistaken. For instance x^3 is not uniformly conituous over R, but is uniformly continuous over any bounded interval, so it behaves "nicely". On the other hand any continuous function that is not uniformly continuous over a bound interval (I'm sure you'll introduce exemples later, I will not spoil there) has a far more "interesting" behaviour.
@VeryEvilPettingZoo totally agree. And that's why I don't see uniform continuity over R as particularly useful, as not being uniformly continuous there does not give you much info.
How can u multiply 3a without sure about the sign of a..if a is negetive then the order of the inequality don't remain same..
Called it! Okay, how long to general topology?
I don't think you need all of this to start reading Munkres' book
12:45 - I'm constantly confused by this in such proofs, can anyone shed some light? Why do we require the min argument here and how do we handle it when we reverse the calculations in the proof? Can we just ignore the |x-a|
The easiest way to think about it is that it is a cheeky way to give ourselves two inequalities. Since we are going to assume |x-a| < delta , if delta = min(a,b) then |x-a| < a AND |x-a| < b.
You might be confused because we imagine epsilon to be small and so delta should be smaller than 1 anyways. However, epsilon can be ANY positive real number, even if the force of the definition is strongest when epsilon is small.
@Garry Cotton we need |x-a|
And as always, we are interested in the case where the epsilons and the Deltas are very small. Thus, the assumption |x-a|
Thanks for the replies guys.
I totally get why we need delta in terms for epsilon, it’s the constant 1 that confuses me. It seems arbitrary? Presumably I’m just missing the connection.
Edit: I think I get it now, it’s because of the assumption that delta was equal to 1 in the scratch work. Thanks for helping me clear it up.
@@garrycotton7094 it might help you to unterstand what happens when we choose an Delta which is greater or equal to 1. :)
Isn't uniform continuity just continuity + bounded derivative over the domain? Are there any cases where these two conditions are not necessary and sufficient to prove uniform continuity?
bounded derivative over the domain + continuity imples uniform continuity, but a founction can be uniformly continuous without being derivable (a simple exemple would be |x| )
@@Falanwe Doesn't |x| still have a bounded derivative over all domains? While the derivative does not exist at 0, the derivative never approaches infinity. Thus the range of the derivative would be bounded by [-1,1]
@@jrkirby93 you need to be differentiable to have a bounded derivative
@@Falanwe Perhaps the term I meant is "not-unbounded"?
@@jrkirby93sqrt(x) is uniformly continuous over its domain (Sorry, I have no idea how to write radicals in those comments), but its derivative is unbounded where it's defined (everywhere except at 0). as it tends towars infinity when you approach 0.
Even worse; the Weierstrass function is uniformaly continuous but differentiable nowhere!
So your two condnitions are sufficient to prove uniform continuity, but absolutely not necesary.
10:03 how can he just multiply it by 3a? What if 3a is negative?
If a|3a²+3a| for all a
🔥🔥🔥
'Ello Brofessor Penn
11:09 I think it should be
3a²+3|a| ≦ |x²+ax+a²|, |x-a|·|x²+ax+a²| < ε
→ |x-a| < ε/(3a²+3|a|)
I like to just whoop out the |x| < min{| -1 + a | , |1 + a |} := M kind of things and throw them all over the place lol. Not elegant but it's easy and thoughtless to do
Hi,
I was wondering why you needed consulting your notes to say the ending sentence 😛
Our good place to stop colleague is asleep
wooow
14:15
He confused me