Real Analysis | Uniform Convergence and Continuity

[@lvlanson]

Very nicely explained, thank you Michael!

[@alphalunamare]

Very Lucid! I am retired now, it's a long time since I studied Mathematics but I find your presentation so easy to follow, especially the way you explain the what goes on between the lines of the proof, something you rarely find in either books or the average lecturer. - well in those day's at least. They should treble your Salary :-)

[@pandas896]

Hey , after having mathematics are you not doing anything. I mean having so much knowledge doesn't yields anything?

[@alphalunamare]

@@pandas896 It's like Poetry, being able to appreciate it doesn't mean you can write it. I do those Japanese problems when I wake up at night, you know, think it through for fun. Career wise you don't get to use Mathematics much unless you get tenure somewhere and you most likely end up in the job market. I made the mistake of becoming a Software Engineer, in the early days it was well respected but then 'C' came along and, to be honest, the industry entered into a dark age of chaos and cheap ill thought out solutions for the lowest cost. There was no room for Mathematics. In the 21st Century, the average Software Engineer has the same deal as a Coal Miner in The Appalachians .. certainly not a career path to be recommended to anyone with a glint of talent. I survived the industry and not rekindle my interests by channels such as this. I might go up in the attic and pull out my notes on The Prime Number Theorem, The Reimann Zeta function and all that.complex analysis .. now that is beautiful. I suppose the bottom line is that the question of yield depends mightily upon the field :-)

[@alphalunamare]

@@pandas896 I am really pissed. I wrote a long response but youtube dumped it in its infinite wisdom. Suffice to say that yield has nothing to do with field. And screw String Theory ... The Reimann Zeta Function was made for The Prime Number Theorem not nonsense. Life when complex is so beautiful :-)

[@JM-ty6uq]

@@alphalunamare Hey Brian, thanks for your insight. I hope this is not in bad taste: Can you share what made you choose to become a software engineer all those years ago, rather than pursuing further math education?

[@alphalunamare]

@@JM-ty6uq Pragmatism. I'd had a sort of meltdown and screwed up my personal life as well as my PhD I'd had a job offer a couple of years prior and so I took them up on it. I spent the next few years fixing things and then Children came along and I needed the job even more :-) I had an MSc in Computing Mathematics so it wasn't a total wipe out :-)

[@kevin326520]

Another good video. One small problem, the "

[@alphalunamare]

Not true ..... it is fashionable in some books to use

[@jonaskoelker]

Also, we can take your observation and generalize it: If there exists epsilon > 0 such that some quantity < epsilon then there exists epsilon' > 0 such that some quantity = epsilon e.g. epsilon' = epsilon. If there exists epsilon > 0 such that some quantity 0 we have some quantity < epsilon then for every epsilon' > 0 we have some quantity

[@user-nd7vb7th8t]

Agree!

[@alexmason6839]

I really love your channel, it is midnight by the way.🙂

[@franciscomackenney7664]

it's 2 am

[@arvindsrinivasan424]

Real analysis for the win

[@Evan-ne5bu]

Hello Micheal, i really like this playlist, and I have a question: will talk about (maybe not going to deep) measure theory?

[@User-gt1lu]

I would like that too. Maybe just a short introduction to sigma algebras

[@CM63_France]

Hi, For fun: 12 "let's go ahead and", including 3 "so let's go ahead and do that", and 1 "so now let's go ahead and do that", 1 "let's may be go ahead and write that down", 1 "I want to go ahead and", 1 "I can go ahead and", 6 "great", 1 "now the thing I want to do", 1 "now what we are going to do", 1 "now I want to notice".

[@TheMauror22]

I missed the real analysis videos. Great video though!!

[@sinqobilebandile6558]

yesss

[@heberfarnsworth]

Question. When you use continuity of f_N at a you assume that the epsilon doesn't depend on N. Is that a problem?

[@jonaskoelker]

So something struck me: Pointwise convergence says for all epsilon > 0 for all x exists N in IN such that [insert conclusion(x, N, epsilon)] Uniform convergence says for all epsilon > 0 exists N in IN such that for all x: [insert conclusion(x, N, epsilon)] Another way of saying this is: View N as a function of epsilon and x, N = g(epsilon, x). Fix a particular epsilon and project out g_epsilon(x). In pointwise convergence, g_epsilon can be any function of x. In uniform convergence, g_epsilon must be a constant function of x. Viewed this way, a question naturally arises: the requirements are at two extremes. What if we impose much more intermediate requirements on g_epsilon, like monotonicity, continuity, differentiability? Oh wait g_epsilon: IR -> IN. How about this: for every x and every epsilon > 0 there exists delta > 0 and N in IN such that for every y if |y-x| < delta then [insert conclusion(y, N, epsilon)] I feel tempted to call this local uniformity: uniform convergence lets us conclude conclusion(y, N, epsilon) given conclusion(x, N, epsilon) for _every_ y. Locally uniform convergence lets us conclude conclusion(y, N, epsilon) only when y is sufficiently close to x. I think local uniformity is sufficient for many purposes, since many of the properties we want to prove-chiefly continuity and differentiability-are local in some way (e.g. pointwise). I suspect most proofs that assume uniform convergence become proofs built on local uniformity if you let delta' = min(delta1, delta2) as an additional proof step. Does anyone know if this concept has been explored? It should be easier to prove than global uniformity since global implies local, and thus more function sequences should be locally uniform, and thus we can conclude continuity, differentiability etc. about a greater number of functions. Has my idea been tried and found wanting?

[@monku1521]

Doesnt fn(x) = x^n converge uniformly to f where f maps [0,1) to 0 and 1 to 1, but still isn’t continuous?

[@hybmnzz2658]

Not uniformly

[@learnmatheasily7112]

Q.1. Prove or give a counter example that uniform continuity preserves convergence of sequences. Q. 2. Prove or give a counter example that continuity preserves the length of an interval. Q. 3. Let 𝑓(𝑥) = [𝑥 + 1] − 𝑥 − 1, ∀𝑥 ∈ 𝑅 Where [𝑥] represents the smallest integer ≥ 𝑥. Determine all points of discontinuity and continuity of the function. Q. (4). Consider 𝑓(𝑥) = sin(𝑥) , 𝑥 ∈ [ −𝜋 4 , 5𝜋 4 ] (10) (a). Determine whether Rolle’s theorem or the Lagrange’s Mean value theorem is applicable here. Explain (b). Apply the theorem that is applicable and find all values of c that satisfy the statement. Also draw a rough sketch of what you get in this discussion. (you can use calculator to compute values of inverse trigonometric functions) Q. 5. Prove or give a counter example that differentiable function over an interval is bounded. Q. 6. Consider 𝑓(𝑥) = 𝑥 3 3 − 3𝑥 2 2 + 2𝑥 + 1. Use derivatives to separate the regions where function is increasing or decreasing. Q. 7. Let 𝑓(𝑥) be derivable over [𝑎, 𝑏] such that 𝑓 ′ (𝑎) = 10 and 𝑓 ′ (𝑏) = 20 . Prove that there exist 𝑐 ∈ (𝑎, 𝑏) such that 𝑓 ′ (𝑐) = 15.

[@learnmatheasily7112]

please solve these..

[@kurax9115]

I really Like these Videos. Im so bad when it comes to creating ideas for proofs although i already knew how to do the first two its amazing to see such content

[@laurensiusfabianussteven6518]

is it true that the proof for continuity of uniform convergent sequence fn(x) can be extended to if and only if statement?

[@andrewkepert923]

It sort of depends which bit you want to make the iff bit match to. ... but in most interpretations of your question the answer is no. E.g. it is possible to have a sequence of continuous functions that convergent to a continuous functions, but the convergence is not uniform. Eg f_n(x)=arctan(x+n) converges pointwise to the continuous function g(x)=π/2, but it is not uniform convergence. It is also possible to have a sequence of functions that are discontinuous at a functions that converges uniformly to a function that is continuous at a. More generally a sequence of nowhere continuous functions that converges uniformly to a continuous function. E.g. if f is the characteristic function of the rationals, let f_n(x)=f(x)/n. Then f_n -> 0 uniformly. The only sort of converse is that for any f continuous there is a sequence of continuous functions {f_n} that converges uniformly to f. E.g. f_n(x)=f(x). So that’s too trivial to mention.

[@laurensiusfabianussteven6518]

@@andrewkepert923 thanks for the answer, appreciate it!

[@rektator]

This essentially shows that the metric space (C(X,Y), e), of continuous functions from a compact space X to a complete metric space (Y,d), equipped with the supremum metric e [e(f,g) := sup_{x\in X}d(f(x),g(x)) for f,g\in C(X,Y)) defines a complete metric space. The completeness of C(X,Y) is important for example in the proof of Existence and Uniqueness theorem of first order differential equations. The completeness of C(X,Y) makes it possible to use Banach fixed point theorem to find solutions to the differential equations when the underlying vector field of the differential equation is Lipschitz continuous.

[@alphalunamare]

So what?

[@ethannguyen2754]

Oh don’t mind me. I’m just waiting for him to get demonetized for saying “left” and “right.”

[@goodplacetostop2973]

18:31

[@shivanggoyal7280]

Homeworkkkk

[@ashimchakraborty2908]

Can someone please solve the problem?? A sequence a_n is defined by a_1=1, a_2=12, a_3= 20, and a_(n+3) = 2a_(n+2) + 2a_(n+1) − a_n, n > 0. Prove that, for every n, the integer 1 + 4 a_n a_(n+1) is a square.

[@ashimchakraborty2908]

@@alphalunamare I edited the question, it was a mistake

[@alphalunamare]

keep on editing.

[@ashimchakraborty2908]

@@alphalunamare could not get you what are you trying to say?

[@alphalunamare]

@@ashimchakraborty2908 More editing is required ... it is not clear what is being asked to be proven.

[@ashimchakraborty2908]

@@alphalunamare look at the question carefully and you will see every term of the sequence is integer and it is asked to prove (1+ 4 a(n).a(n+1)) is perfect square for every n€lN.

[@djvalentedochp]

epic