Could you please explain why at 1:46 we are able to take the supremum of f to be 1 without loss of generality? Aren't we looking for the supremum over all f in X?
Hi, I don't quite get why at 1:55 we can simply choose f such that the ||f|| =1 without changing the equality of the sup? Could you please explain that a bit?
This is a scaling argument. If the norm of f is not 1, you can consider the vector f/||f|| instead. Just try to write that down and you will see that it works :)
@@brightsideofmaths thank you so much for your explanation! I think I get what you meant here. Just to make sure I didn't get the idea wrong, you mean I can simply choose a vector f=f/||f||, so by taking the norm the denominator will go to 1 anyway.
At 0:32, Can u please clarify how the supreme norm of a continuous function is defined? Is this defined in one of the earlier video of this series? Thank you.
@@brightsideofmaths Many thanks for quick reply. Afraid that it is not easy to locate a definition in the real analysis series. I would guess, it is probably defined as \sup(|f(D)|) where D is the demain of f?
Wonderful and informative video. I had a query. Can you kindly explain how, while defining the operator norm of ||Tg||=norm of ouput by norm of input (1:27 onwards) is shown as absolute of output by norm of input. i.e shdnt it be ||Tg|| = { ||Tg(f)|| / ||f|| } but the video shows.... ||Tg|| = { |Tg(f)| / ||f|| }. Can you kindly clarify and correct me where I am going wrong? Thank you soo much
i don't see how you can assume there is no loss in generality when choosing the function h to be the complex conjugate of g. how do you recreate any other function from h and your lower bound remains true?
We don't need the generality in the second step. We just want to find one function that satisfies the other inequality. Together with the general first step, we have shown the equality then :)
![Functional Analysis 15 | Riesz Representation Theorem [dark version]](http://i.ytimg.com/vi/OVvugoEt2ZQ/mqdefault.jpg)
![Functional Analysis 15 | Riesz Representation Theorem [dark version]](/img/tr.png)
Functional analysis is really fun! Thank you for making these videos!
Thank you for watching :)
Very nice course. You really make it easier for people majoring in engineering to understand functional analysis! Have supported on steady~
Thank you very much :)
All your explanation and demonstration are extremely rigorous, I love it 😊😊!
Thanks so much! 😊 Also thanks for your support!
Could you please explain why at 1:46 we are able to take the supremum of f to be 1 without loss of generality? Aren't we looking for the supremum over all f in X?
Yeah, you just pull 1/norm(f) into the absolut value and then, by linearity, into the function T_g.
@@brightsideofmaths ahh I see thanks!
But h is not continous, so Th shouldn’t be defined, right? Wouldn‘t you habe to argue with a sequence of continous functions that converges to h?
Why shouldn't h be not continuous?
These videos are amazing.
Thank you :)
Hi, I don't quite get why at 1:55 we can simply choose f such that the ||f|| =1 without changing the equality of the sup? Could you please explain that a bit?
This is a scaling argument. If the norm of f is not 1, you can consider the vector f/||f|| instead. Just try to write that down and you will see that it works :)
@@brightsideofmaths thank you so much for your explanation! I think I get what you meant here. Just to make sure I didn't get the idea wrong, you mean I can simply choose a vector f=f/||f||, so by taking the norm the denominator will go to 1 anyway.
At 0:32, Can u please clarify how the supreme norm of a continuous function is defined? Is this defined in one of the earlier video of this series? Thank you.
It's defined in my real analysis course :)
@@brightsideofmaths Many thanks for quick reply. Afraid that it is not easy to locate a definition in the real analysis series. I would guess, it is probably defined as \sup(|f(D)|) where D is the demain of f?
Teacher, why is the absolute value of f(x) less than the norm of f at 2:39 in the video?
That is the definition of the supremum norm :)
Wow ,these videos are amazing and very helpful , keep it up 👍
Thanks a lot :)
Wonderful and informative video.
I had a query. Can you kindly explain how, while defining the operator norm of ||Tg||=norm of ouput by norm of input (1:27 onwards) is shown as absolute of output by norm of input. i.e shdnt it be ||Tg|| = { ||Tg(f)|| / ||f|| } but the video shows.... ||Tg|| = { |Tg(f)| / ||f|| }. Can you kindly clarify and correct me where I am going wrong? Thank you soo much
Thank you so much :) The absolut value is the norm of the corresponding space here.
i don't see how you can assume there is no loss in generality when choosing the function h to be the complex conjugate of g. how do you recreate any other function from h and your lower bound remains true?
We don't need the generality in the second step. We just want to find one function that satisfies the other inequality. Together with the general first step, we have shown the equality then :)
@@brightsideofmaths ah yes you are right ^^
Great video!
Isn't Tg(f) an inner product over X?
It depends how you precisely define it. For complex vector spaces, it's not :D
why did you include the condition of g having no zeroes?
See calculation at 3:40
@@brightsideofmaths so am i correct in assuming that it is not necessary to show the inequality, and only useful to show that equality is possible?
@@rohankapoor8289 Yes, we show equality :)
Very helpful!! Thanks for this video!
This time construction of h(t) is a bit easier to understand: to make h(t) as big as possible to maximize the integral.
Please Auto english subtitle on!!!
Sadly, they don't work so well...