Functional Analysis 17 | Arzelà-Ascoli Theorem

[@xwyl]

Hard to believe I've made it through half of this series. Unlike the thick books, The Bright Side shows me there is light and hope.

[@brightsideofmaths]

Thank you very much :)

[@randomname7013]

Thanks so much for such Intuitive formulation, there are only few channels like your who focus on raising the intuition level of concept than rather blathering about definitions👌🤗

[@Domzies]

I think I am most thankful for showing me that parts of my understanding of functional analisis were very incomplete and shaky. And ofcourse helping to fix those parts.

[@FraserIland]

Dear The Bright Side of Mathematics thank you so much for you beautiful series!
Nevertheless in my opinion you made a lesser error at time 10:31 because the mean value theorem needs "=" sign.
Only at time 10:43 you must apply "

[@brightsideofmaths]

You are absolutely right: You get an equality sign in the mean value theorem. Of course, it is not an error at all but I should have clarified that. Thanks!

[@tsshamoo2376]

Its here sooner than I thought itd be :)

[@nguyenvietdung7588]

Thanks a lot

[@davidescobar7726]

Thanks man... Great video.

[@felipegomabrockmann2740]

Mega Video!! Thank you so much!!

[@ahmedamr5265]

I love your videos!
Q: Is compactness guaranteed for a closed and bounded set in a finite-dimensional normed vector space because we can do basis isomorphism into R^(dim(X)) and apply the Bolzano-Weierstrass theorem there?

[@brightsideofmaths]

Exactly! :)

[@ahmedamr5265]

@@brightsideofmaths thanks to you I can now understand such things 🙏

[@MLeconometrics]

Thank you so much for the amazing videos!!! They are extremely helpful :)
I have one question on the first example (a) you gave: it says that "A compact iff A closed + bounded" for finite dimensional normed space, but how does this apply to [0,1] interval in a metric space (R, discrete metric) you discussed in the last video? You explained that this is not compact, but it seems like it is closed, bounded, and finite dimensional.

[@brightsideofmaths]

Thanks a lot! The answer is simple: we talk about vector spaces here!

[@MLeconometrics]

​@@brightsideofmaths Understood! Thank you very much :)

[@qiaohuizhou6960]

Hi, thank you so much for your video! If it's not because of these videos I have no idea how many misconceptions I have towards maths! Just for clarification, making sure I am not getting any ideas wrong. Does example (a) implies that for a set of a continuous function on [0,1], we may still find a sequence of such function that doesn't converge according to the definition of ||·||, but does converge if we make further restrictions on the continuous function that we choose?

[@brightsideofmaths]

For example (a) you really should have the definition of equicontinuity in mind. This is reason why we choose the sequence of functions.

[@paramanandadas1319]

At 10:12, I mean in example (b), why do we need the continuity of derivative?

[@brightsideofmaths]

We don't need it but it's easier to formulate by seeing A as a subset of C^1.

[@lenguyenbach7794]

Purrfect!!!!

[@nguyenvietdung7588]

Hege

[@danielparra6902]

I love you

[@MathPhysicsEngineering]

What software do you use for the annotations?

[@brightsideofmaths]

Xournal

[@randomname7013]

Sir why arzelo ascoli theorem is not completed?

[@brightsideofmaths]

Completed?

[@randomname7013]

@@brightsideofmaths Sorry, I thought it's incomplete🤗

[@sorundegil4993]

ich dachte (C[0,1],||•||) mit supremunsnorm ist kein Banachraum

[@brightsideofmaths]

But it is.

[@denklem40]

@@brightsideofmaths achso ja stimmt ich hatte an C^1[0,1] gedacht. C ist die Menge der stetigen Funktionen. C^1 hingegen 1 mal stetig diffbar. Ok dann passt das danke! :)