Great video as usual! I will check 20 and 21 again just to make sure the concepts sink in... I saw you are adding some new videos here on spectral theory... thanks for that!
Could you explain that how do you proof the injective (12:15)? I don't understand why we set Tx to 0 and y to e^k, and then how we get the injective from this.
Okay, I try to explain. Injectivity means that from Tx = 0 follows x = 0. So we assume Tx = 0. Then we apply the e_k vectors and get (Tx)(e_k) = 0. This implies x_k = 0 for all k. Therefore x = 0, which is what we wanted to show.
So great videos, thank you so much! The dual space is isometric isomorphic to ℓ^{p'}. When p=2, then ℓ^{p'} = ℓ^p and then we have a Hilbert space and Riesz representation theorem?
I got a little confused by your notation. You introduce the definition T: l^p'(N) -> (l^p(N))^'. Should I read this as "T is a linear map from the infinite sequence space l^p' to the dual of the infinite sequence space l^p ? It is the parenthesis with a prime to the right that confused me.
So, we take a sequence from lp' and a linear functional from (lp)' and form a mapping that sends it to some field (which is the codomain of all linear functionals on (lp)') , that it is. If i understood everything correctly. Which can be seen for p=2 as a riesz rep theorem , that it is for each functional and its x we identify it with some x' such . that it we measure it projection.
I love your videos. So, does identifying the dual space up to an isometry imply that we can find other mappings than the angle brackets with the Lq space? Maybe by choosing another base than the e_k vectors?
I see that well defined comes often in these proofs. From what I found well defined mean: "if g=h implies f(g) = f(h)" now how does showing that the operator (Tx) is linear and bounded shows that the initial operator (T) is well defined? Thankss!
@@brightsideofmaths Ah okay, that makes sense. Could you please just elaborate on the notation. When you say |y(x)| does this mean just absolute value or norm? Are these two actually same in this case? Kinda confused when working with functionals.
I allow myself two more questions: 1) In the definition of (Tx)(y) appears an infinite sum, but when you apply the triangle inequality for absolute values you instead use a finite sum and will afterwards take the limit to infinity. You do talk about continuity of the absolute value, but I need to fresh up on this. Is this discussed in a previous video ? 2) At the end of the video you want to show that T is injective. You point out that if Tx is zero, then each component of x has to be zero. I have learned about injectivity in a different manner, namely that f(x1) = f(x2) implies x1 = x2. Is this equivalent ?
I have the same confusion as your question 2. [Edit: linearity gives us that (Tx=Tx' implies x=x') (Tx=0 implies x=0)] Additionally, I don't understand why T needs to be injective :/
@@brightsideofmaths yes that I know, so the notation with is familiar to me. But you’re saying that the (Tx)(y): definition is the same as the inner product in l2 but you don’t have the conjugation inside the sum after you define (Tx)(y) so I thought you were missing it there right?
Hmmm. How close is being isometric isomorphic to being equal? For which properties (e.g. convergence, boundedness, compactness (?)) is looking at a space V that is isometric isomorphic to W enough to show that W itself has these properties?
In used part 21 to explain isomorphisms of Banach spaces: If two Banach spaces are isometric isomorphic, they act completely the same as Banach spaces.
@@brightsideofmaths But then doesn't condition (5) imply (3)? If f is an isometry then the for all x, y, such that their distance is < e, we have that: norm(f(x)-f(y)) = norm(x-y) < e. Therefore isometry implies continuity (and boundness). Please correct me if I am wrong
@@brightsideofmaths but what exactly does T do? Your videos are always so profound that I have to watch at least 3 times to grasp the concepts and proof, and many more hours of thinking and watching other channels' videos to understand your logic. But they're still good when I want to know what to learn. Sometimes your videos are really hard to understand, despite that I have a PhD in EE and am currently a researcher.
@@xwyl T sends sequences to functionals. This all is not an easy concept. Maybe, it helps if you look at an example sequence such that you can see what T does.
![Functional Analysis 25 | Hahn-Banach Theorem [dark version]](/img/n.gif)
Mann, you're on fire..... Thanks for your content :'p
Greetings from Peru by the way. You make the pandemic go on easier by making us forget about it for some minutes. God bless you.
Very very nice, I will have to come back to this one again in the future to fully absorb it I think
Great video as usual! I will check 20 and 21 again just to make sure the concepts sink in... I saw you are adding some new videos here on spectral theory... thanks for that!
Thank you so much! Extremely well explained!
Could you explain that how do you proof the injective (12:15)? I don't understand why we set Tx to 0 and y to e^k, and then how we get the injective from this.
Okay, I try to explain. Injectivity means that from Tx = 0 follows x = 0. So we assume Tx = 0. Then we apply the e_k vectors and get (Tx)(e_k) = 0. This implies x_k = 0 for all k. Therefore x = 0, which is what we wanted to show.
@@brightsideofmaths great! I get it by remembering the case of nullspace in matrices, if nullspace is 0 the operator is invertible for the domain.
Sir, can u suggest a topic in functional analysis for my phd research
So great videos, thank you so much! The dual space is isometric isomorphic to ℓ^{p'}. When p=2, then ℓ^{p'} = ℓ^p and then we have a Hilbert space and Riesz representation theorem?
Yes, that is indeed correct :)
I got a little confused by your notation. You introduce the definition T: l^p'(N) -> (l^p(N))^'. Should I read this as "T is a linear map from the infinite sequence space l^p' to the dual of the infinite sequence space l^p ? It is the parenthesis with a prime to the right that confused me.
Yes, that is how you should read it. I used parentheses such that you would not miss the prime :)
So, we take a sequence from lp' and a linear functional from (lp)' and form a mapping that sends it to some field (which is the codomain of all linear functionals on (lp)') , that it is. If i understood everything correctly. Which can be seen for p=2 as a riesz rep theorem , that it is for each functional and its x we identify it with some x' such . that it we measure it projection.
I love your videos. So, does identifying the dual space up to an isometry imply that we can find other mappings than the angle brackets with the Lq space? Maybe by choosing another base than the e_k vectors?
Thanks! Of course, you could also find other isometries if they could be helpful for some problems.
Hi! why is functional analysis called "functional analysis"?
I see that well defined comes often in these proofs. From what I found well defined mean: "if g=h implies f(g) = f(h)" now how does showing that the operator (Tx) is linear and bounded shows that the initial operator (T) is well defined? Thankss!
Well-defined just means that the definition makes sense :)
I dont understand the step at 8:35 , isnt inequality supposed to be something like ||y(x)||
Yes, we have |y(x)| = y(x) because everything is positive here.
@@brightsideofmaths Ah okay, that makes sense. Could you please just elaborate on the notation. When you say |y(x)| does this mean just absolute value or norm? Are these two actually same in this case? Kinda confused when working with functionals.
@@vanrltv Yes, they are the same in this case: the norm on R is the absolute value.
@@brightsideofmaths Thank you for the quick response on a 2 year old video!!
@@vanrltv Oh wow. It's already 2 years!?
I allow myself two more questions: 1) In the definition of (Tx)(y) appears an infinite sum, but when you apply the triangle inequality for absolute values you instead use a finite sum and will afterwards take the limit to infinity. You do talk about continuity of the absolute value, but I need to fresh up on this. Is this discussed in a previous video ? 2) At the end of the video you want to show that T is injective. You point out that if Tx is zero, then each component of x has to be zero. I have learned about injectivity in a different manner, namely that f(x1) = f(x2) implies x1 = x2. Is this equivalent ?
I have the same confusion as your question 2. [Edit: linearity gives us that (Tx=Tx' implies x=x') (Tx=0 implies x=0)]
Additionally, I don't understand why T needs to be injective :/
if you’re missing the conjugation in the first definition (Tx)(y): how is it the same thing as the notation? I’m confused by what you mean there
In the inner product there is a complex conjugation in the first argument :)
@@brightsideofmaths yes that I know, so the notation with is familiar to me. But you’re saying that the (Tx)(y): definition is the same as the inner product in l2 but you don’t have the conjugation inside the sum after you define (Tx)(y) so I thought you were missing it there right?
Yes, that is the reason that I put the conjugation in as well. Then you have it two times there :)
@@brightsideofmaths but if you go to that part in the video it is NOT there in the summation, is that a mistake perhaps?
@@saqarkhaleefah6159 It's the definition (Tx)(y). There is not complex conjugation there.
Hmmm. How close is being isometric isomorphic to being equal? For which properties (e.g. convergence, boundedness, compactness (?)) is looking at a space V that is isometric isomorphic to W enough to show that W itself has these properties?
Do you maybe have a simple application of dual spaces? An example that without the notion of dual spaces would be very hard to prove and/or grasp?
In used part 21 to explain isomorphisms of Banach spaces: If two Banach spaces are isometric isomorphic, they act completely the same as Banach spaces.
Please make va video on dual of Hilbert Space also ...with examples.. thank you
@@Dr.kcMishra Do you mean this? th-cam.com/video/rKiy6wEiQIk/w-d-xo.html
@@brightsideofmaths yes..
What is the difference between Banach Space and Hilbert Space?. Thank you
A Banach space is only equipped with a norm or distance, whereas a Hilbert space also has an inner product
Thank you
Do functionals and operators have same sense?
Functionals are a special kind of operators.
@@brightsideofmaths thank you
Why do we require the map T to be bounded?
If T is not bounded, it can't be an isometry, right?
@@brightsideofmaths But then doesn't condition (5) imply (3)? If f is an isometry then the for all x, y, such that their distance is < e, we have that: norm(f(x)-f(y)) = norm(x-y) < e. Therefore isometry implies continuity (and boundness). Please correct me if I am wrong
@@mathiasbarreto9633 Yes, (5) implies (3). But for our proof it is easier first to prove (3). Then one can use (3) and prove (5). See 11:53.
What is T here? Is it just a sign to tells us that x now is for calculating a sum?
T is the operator :)
@@brightsideofmaths but what exactly does T do?
Your videos are always so profound that I have to watch at least 3 times to grasp the concepts and proof, and many more hours of thinking and watching other channels' videos to understand your logic. But they're still good when I want to know what to learn.
Sometimes your videos are really hard to understand, despite that I have a PhD in EE and am currently a researcher.
@@xwyl T sends sequences to functionals. This all is not an easy concept. Maybe, it helps if you look at an example sequence such that you can see what T does.
Have watched this 6 times, now I understand the proofs, but the ideas are still elusive to me. I need not just proofs, but also purpose of every step.
@@xwyl I am glad that you could understand the proof!
Apparently functional analysis is part of my linear algebra exam, lucky me
Very good! So you learn a lot of concepts in one course :)