amazing to see how all the previous contents consolidate into higher mathematics.
@sergiohuaman60843 жыл бұрын
second time here and 100% understood. thanks!
@minglee5164 Жыл бұрын
The topic that I do not understand in linear algebra.
@siharsanan-3794 Жыл бұрын
Thanks alot for you videos,, can you please make some videos on nuclear maps in C* algebra?
@filipbecanovic5072 жыл бұрын
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?
@brightsideofmaths2 жыл бұрын
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.
@ahmedamr52655 ай бұрын
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?
@brightsideofmaths5 ай бұрын
Yeah, maybe you should call the two spaces isometrically anti-isomorphic.
@jonasw47912 жыл бұрын
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?
@zhizhongpu8937 Жыл бұрын
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.
@KaiseruSoze3 жыл бұрын
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?
@brightsideofmaths3 жыл бұрын
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.
@Dr.kcMishra3 жыл бұрын
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...
@ХаньцинВан10 ай бұрын
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.
@csirnetiitjammathematics78883 жыл бұрын
Pls upload some problems of functional analysis
@brightsideofmaths3 жыл бұрын
With solutions?
@rajshreejoshi97652 ай бұрын
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❤❤
@brightsideofmaths2 ай бұрын
Thank you very much :)
@Dr.kcMishra3 жыл бұрын
Great. At 0:56 what is F? Thank you
@JonixMaroni3 жыл бұрын
Probably the accompanying field.
@brightsideofmaths3 жыл бұрын
F is R or C. Sorry, I have used this a lot in previous videos that I totally forgot to explain it again here :)
@Dr.kcMishra3 жыл бұрын
@@brightsideofmaths thanks a lot.
@Dr.kcMishra3 жыл бұрын
@@JonixMaroni thanks
@salvatoregiordano1273 жыл бұрын
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
@brightsideofmaths3 жыл бұрын
Do you have a timestamp for this?
@salvatoregiordano1273 жыл бұрын
@@brightsideofmaths yes, sorry, 3:55
@brightsideofmaths3 жыл бұрын
@@salvatoregiordano127 Thanks! Since x is fixed in this equation, the norm of x is a finite number. That is all we need there.
@salvatoregiordano1273 жыл бұрын
@@brightsideofmaths I see. Thank you so much. Your videos are the best!
@brightsideofmaths3 жыл бұрын
@@salvatoregiordano127 Thank you very much :)
@ahmedamr52656 ай бұрын
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?
@brightsideofmaths6 ай бұрын
Very good question. However, also easy to answer: for each element x in X, we get a linear operator
@ahmedamr52656 ай бұрын
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
@haiangang95613 жыл бұрын
Why is there an ||x||_X on almost every equation? I don't see the reason why.
@brightsideofmaths3 жыл бұрын
I don't understand the question, sorry.
@angelmendez-rivera3513 жыл бұрын
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.
@sieni2212 жыл бұрын
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.
@Seskulinear_47 Жыл бұрын
Why C tilde is finite not infinete?
@brightsideofmaths Жыл бұрын
Because C was finite as a norm.
@Kevin-rt5fo2 жыл бұрын
why do we need to show l is bounded?
@abublahinocuckbloho4539 Жыл бұрын
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
@mathiasbarreto96333 жыл бұрын
What was the definition of F? (I mean the codomain of all functions in the dual space of X)
@brightsideofmaths3 жыл бұрын
Either R or C, depending if you are working in a real or a complex vector space. It stands for "field".
@mathiasbarreto96333 жыл бұрын
@@brightsideofmaths Thanks! Greetings from Paraguay :D
@brightsideofmaths3 жыл бұрын
@@mathiasbarreto9633 You are welcome!
@hardywhite58713 жыл бұрын
Wrong at 2:37. The isometric is not guaranteed. Hence only Ismorphosim, no isometric. Please change that.
@brightsideofmaths3 жыл бұрын
You are joking, right? I showed almost the whole proof of this in Part 15.
@angelmendez-rivera3513 жыл бұрын
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.