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?
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.
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:)
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
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 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?
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.
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
![Functional Analysis 25 | Hahn-Banach Theorem [dark version]](/img/n.gif)
You teach everything at just the right time
Seconded
I mean, there are no wrong times to learn some mathematics :)
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?
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.
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 :-).
Yes, it is a nice trick :D
As usual, excellent content!
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:)
Exactly, in (tristar) we apply Hölder two times!
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
Glad you like them!
You can watch my Real Analysis series for the foundations.
I use Xournal :)
This is mind-blowing. Is there an easier way to prove it?
Please make a video on application on Mankowski inequality, u r teaching like a pro, Mashallah
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 😅
Abstract *measure* spaces, indeed :)
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.
@@metalore I know what a measure space is, but thank you for the suggestion.
@@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?
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.
Sorry l have same question about this lesson
Just ask :)
Why are you write so small world that i can't even read
Which parts are not readable?
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
I redefined the term to make it less convoluted :)