Thanks for making these! Would you mind clarifying what you mean by taking the maximum of the operator norm at 6:13 in the context of proving boundedness for the first N operators of the sequence?
The first N operators are bounded by definition. Hence there is a maximal operator norm. Together with the inequality for the infinitely many operators, we have a constant that holds for all n.
Yes, in finite-dimensions the dual is not so interesting. This is simply the case because in finite-dimensions, we always are in a Hilbert space in some sense.
Carl bender(a physicist and mathematician) says that when we have non hermitian operators (matrices) in that case not all it's Eigen basis are orthogonal in given Hilbert Space but we have to use the dual of that Hilbert Space where all Eigen basis will be orthogonal...I don't know how to visualise this...
Question: Shouldn't the Riesz representation theorem result in X' being associated to X via an antilinear map? (as opposed to a linear one) If so, is this still called isometric isomorphism?
Great video again! One question though at 3:50 :(1 / ||x||) * |Ln(x)-Lm(x)| < ||Ln - Lm|| is ture because |Ln(x)|-Lm(x)| is in F and therefor the norm of it is the absolute value?
There are four equivalent definitions of ||f||. One definition is ||f|| = sup |f(x)| / ||x|| for all none-zero vector x. Then ||f|| >= sup |f(x)| / ||x|| for all none-zero vector x.
8:40, why for all x, you can take the supremium. It is possible that for different x, there exists different n s.t. the inequality holds 1/|x|*|ln(x) - lm(x)| < e.
I hope our educator teaching like that. They are just focused on a marks on every theorem. But no one tell a concept behind a theorem. My concepts are clear now. Thank you. love from India❤❤
Excellent Videos! I have a question, please. You define an epsilon'=epsilon * norm(x), but in order for this to make sense, norm(x) needs to be bounded. How do we know norm(x) is bounded? Thanks
Great video, thanks! At 2:07, I understand that Riez representation theorem states injectivity from X' to X. But what about the surjectivity required for bijectivity?
Thank you! Two more questions: You say that in this case "X' is essentially X again". 1) Since the set X' itself is only composed of inner products, X' is not exactly X. Am I correct? 2) Is X' an inner product space? Does it make any sense to take the inner product of a set whose elements themselves are inner products? Sorry if I'm overly confused here... @@brightsideofmaths
Because the norm being used in those specific instances is specific to the normed space (X, ||·||), so the norm map has to be indexed by the corresponding set X. It is completely necessary to do this.
Could you make some video of hilbert space application to the theory of fourier series because my functional analysis course has bunch of fourier series too.
in order to satisfy the riesz representation theorem the operator must be bounded. There was a previous video in which it was shown bounded operators are continuous operators. this continuity is key to doing any sort of analysis on a given space
The isometry is guaranteed by the theorem, and it was actually proven during the video on the theorem. There is no need to change anything in the video. What does need to change is the fact that you are not paying sufficient attention.
amazing to see how all the previous contents consolidate into higher mathematics.
second time here and 100% understood. thanks!
The topic that I do not understand in linear algebra.
Thanks alot for you videos,, can you please make some videos on nuclear maps in C* algebra?
Thanks for making these! Would you mind clarifying what you mean by taking the maximum of the operator norm at 6:13 in the context of proving boundedness for the first N operators of the sequence?
The first N operators are bounded by definition. Hence there is a maximal operator norm. Together with the inequality for the infinitely many operators, we have a constant that holds for all n.
If you have a dual position space (e.g., Euclidean) where it's dual is a velocity space, intuition says the metric dual should be the same. No?
Yes, in finite-dimensions the dual is not so interesting. This is simply the case because in finite-dimensions, we always are in a Hilbert space in some sense.
Carl bender(a physicist and mathematician) says that when we have non hermitian operators (matrices) in that case not all it's Eigen basis are orthogonal in given Hilbert Space but we have to use the dual of that Hilbert Space where all Eigen basis will be orthogonal...I don't know how to visualise this...
Question: Shouldn't the Riesz representation theorem result in X' being associated to X via an antilinear map? (as opposed to a linear one)
If so, is this still called isometric isomorphism?
Yeah, maybe you should call the two spaces isometrically anti-isomorphic.
Great video again!
One question though at 3:50 :(1 / ||x||) * |Ln(x)-Lm(x)| < ||Ln - Lm|| is ture because |Ln(x)|-Lm(x)| is in F and therefor the norm of it is the absolute value?
There are four equivalent definitions of ||f||. One definition is ||f|| = sup |f(x)| / ||x|| for all none-zero vector x. Then ||f|| >= sup |f(x)| / ||x|| for all none-zero vector x.
8:40, why for all x, you can take the supremium. It is possible that for different x, there exists different n s.t. the inequality holds 1/|x|*|ln(x) - lm(x)| < e.
Pls upload some problems of functional analysis
With solutions?
I hope our educator teaching like that.
They are just focused on a marks on every theorem.
But no one tell a concept behind a theorem.
My concepts are clear now.
Thank you.
love from India❤❤
Thank you very much :)
Great. At 0:56 what is F? Thank you
Probably the accompanying field.
F is R or C. Sorry, I have used this a lot in previous videos that I totally forgot to explain it again here :)
@@brightsideofmaths thanks a lot.
@@JonixMaroni thanks
Excellent Videos! I have a question, please. You define an epsilon'=epsilon * norm(x), but in order for this to make sense, norm(x) needs to be bounded. How do we know norm(x) is bounded? Thanks
Do you have a timestamp for this?
@@brightsideofmaths yes, sorry, 3:55
@@salvatoregiordano127 Thanks! Since x is fixed in this equation, the norm of x is a finite number. That is all we need there.
@@brightsideofmaths I see. Thank you so much. Your videos are the best!
@@salvatoregiordano127 Thank you very much :)
Great video, thanks!
At 2:07, I understand that Riez representation theorem states injectivity from X' to X. But what about the surjectivity required for bijectivity?
Very good question. However, also easy to answer: for each element x in X, we get a linear operator
Thank you!
Two more questions: You say that in this case "X' is essentially X again".
1) Since the set X' itself is only composed of inner products, X' is not exactly X. Am I correct?
2) Is X' an inner product space? Does it make any sense to take the inner product of a set whose elements themselves are inner products?
Sorry if I'm overly confused here...
@@brightsideofmaths
Why is there an ||x||_X on almost every equation? I don't see the reason why.
I don't understand the question, sorry.
Because the norm being used in those specific instances is specific to the normed space (X, ||·||), so the norm map has to be indexed by the corresponding set X. It is completely necessary to do this.
Could you make some video of hilbert space application to the theory of fourier series because my functional analysis course has bunch of fourier series too.
Why C tilde is finite not infinete?
Because C was finite as a norm.
why do we need to show l is bounded?
in order to satisfy the riesz representation theorem the operator must be bounded. There was a previous video in which it was shown bounded operators are continuous operators. this continuity is key to doing any sort of analysis on a given space
What was the definition of F? (I mean the codomain of all functions in the dual space of X)
Either R or C, depending if you are working in a real or a complex vector space. It stands for "field".
@@brightsideofmaths Thanks! Greetings from Paraguay :D
@@mathiasbarreto9633 You are welcome!
Wrong at 2:37. The isometric is not guaranteed. Hence only Ismorphosim, no isometric. Please change that.
You are joking, right? I showed almost the whole proof of this in Part 15.
The isometry is guaranteed by the theorem, and it was actually proven during the video on the theorem. There is no need to change anything in the video. What does need to change is the fact that you are not paying sufficient attention.