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Functional Analysis 20 | Minkowski Inequality

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The Bright Side of Mathematics

The Bright Side of Mathematics

Күн бұрын

Пікірлер: 26
@Anteater23
@Anteater23 4 жыл бұрын
You teach everything at just the right time
@reptilewithsadhumaneyes
@reptilewithsadhumaneyes 3 жыл бұрын
Seconded
@StratosFair
@StratosFair 2 жыл бұрын
I mean, there are no wrong times to learn some mathematics :)
@psyspin
@psyspin 4 жыл бұрын
As usual, excellent content!
@Jay-ms1dv
@Jay-ms1dv Жыл бұрын
The step (**) turns the Lp norm into a well-designed L1 nom is genius... I cannot imagine how the inventor finds this trick.. Nice video as usual :-).
@brightsideofmaths
@brightsideofmaths Жыл бұрын
Yes, it is a nice trick :D
@mnada72
@mnada72 2 жыл бұрын
Thank you for these interesting lessons, for me this part of mathematics is completely new. For that I watch the lessons to get a course view on the subject. Can you suggest a prerequisite for getting the most of this topic when I revisit the topic in the future. Also can you tell me which software you use for writing on screen. Thank you
@brightsideofmaths
@brightsideofmaths 2 жыл бұрын
Glad you like them! You can watch my Real Analysis series for the foundations. I use Xournal :)
@arturo3511
@arturo3511 Жыл бұрын
If we apply Holder's inequality to tristar (***), then it reads, SUM (aj+cj)bj =< ||a+c||_p * ||b||_p'. However I don't see how we are allowed to say this if we divide it into two sums: ||a||_p * ||b||_p' + ||c||_p * ||b||_p' . If I understand we are using Holder's inequality two times instead on the right side of 3:36? THanks:)
@brightsideofmaths
@brightsideofmaths Жыл бұрын
Exactly, in (tristar) we apply Hölder two times!
@babuinayat4623
@babuinayat4623 2 жыл бұрын
Please make a video on application on Mankowski inequality, u r teaching like a pro, Mashallah
@dibeos
@dibeos 4 жыл бұрын
In the red star inequality, why did you have to write the infinite sum as a finite sum with the limit when n goes to infinity? In the end (6:50) you took the limit again. Why couldn’t you work all the proof dealing with infinite sums?
@brightsideofmaths
@brightsideofmaths 4 жыл бұрын
I want to use Hölder's inequality and have proved this for a fixed (finite) n in the previous video. Whenever you have an infinite sum, you always need to check if it really exists. By working with the finite sums, you don't have this problem and only have to check the existence of the limit in the end.
@xwyl
@xwyl 2 жыл бұрын
This is mind-blowing. Is there an easier way to prove it?
@angelmendez-rivera351
@angelmendez-rivera351 3 жыл бұрын
Wait, did you actually mean to say that you would generalize this to abstract *measure* spaces, or did you mean abstract *metric* spaces? You said the former in the video, but you probably meant the latter, seeing that it makes way more sense 😅
@brightsideofmaths
@brightsideofmaths 3 жыл бұрын
Abstract *measure* spaces, indeed :)
@metalore
@metalore 2 жыл бұрын
In case you don't know what a measure space is, he has a playlist called "Measure Theory" that explains what a measure space is.
@angelmendez-rivera351
@angelmendez-rivera351 2 жыл бұрын
@@metalore I know what a measure space is, but thank you for the suggestion.
@AnjaliSheoran-hf7yv
@AnjaliSheoran-hf7yv Жыл бұрын
@@brightsideofmaths will you please explain the sufficient and necessary condition for the equality in minkowski inequality?? I read it from our reference book (G. de. Barra). I don't get the part of necessary condition that is a|f|=b|f+g|^(p-1) same for g. Will you help me with this?
@AnjaliSheoran-hf7yv
@AnjaliSheoran-hf7yv Жыл бұрын
For p=1, equality holds iff we have almost everywhere either f(x)•g(x)=0 or sgn f(x)=sgn g(x). For p>1, iff sf=tg a.e. where s and t are not both zero.
@shaymahamad1896
@shaymahamad1896 4 жыл бұрын
Sorry l have same question about this lesson
@brightsideofmaths
@brightsideofmaths 4 жыл бұрын
Just ask :)
@abhishekparihar7866
@abhishekparihar7866 5 ай бұрын
Why are you write so small world that i can't even read
@brightsideofmaths
@brightsideofmaths 5 ай бұрын
Which parts are not readable?
@abublahinocuckbloho4539
@abublahinocuckbloho4539 Жыл бұрын
this is too convoluted, there should be a way to prove the inequality without having to redefine terms that are already redefined terms. hard to track what terms mean what
@brightsideofmaths
@brightsideofmaths Жыл бұрын
I redefined the term to make it less convoluted :)
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