Functional Analysis 23 | Dual Space - Example

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This is my video series about Functional Analysis where we start with metric spaces, talk about operators and spectral theory, and end with the famous Spectral Theorem. I hope that it will help everyone who wants to learn about it.

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00:00 Introduction
00:31 Example
03:13 Proof

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I hope that this helps students, pupils and others. Have fun!

(This explanation fits to lectures for students in their first and second year of study: Mathematics for physicists, Mathematics for the natural science, Mathematics for engineers and so on)

[davidescobar]

Mann, you're on Thanks for your content :'p

[trondsaue]

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.

[sergiohuaman]

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.

[StratosFair]

Very very nice, I will have to come back to this one again in the future to fully absorb it I think

[sergiohuaman]

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!

[arturo]

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!

[arcstur]

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?

[王雅舒-ri]

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.

[trondsaue]

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 ?

[bumdeedum]

Thank you so much! Extremely well explained!

[circlesofMATH]

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.

[elsa]

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?

[vinithakumari.v]

Sir, can u suggest a topic in functional analysis for my phd research

[saqarkhaleefah]

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

[vanrltv]

I dont understand the step at 8:35, isnt inequality supposed to be something like ||y(x)|| <= ||y|| ||x||? But you dont have the ||y(x)|| at the start you just have y(x). I feel like am missing something.

[Dr.kcMishra]

What is the difference between Banach Space and Hilbert Space?. Thank you

[TheWombatGuru]

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?

[Dr.kcMishra]

Do functionals and operators have same sense?

[xwyl]

What is T here? Is it just a sign to tells us that x now is for calculating a sum?

[mathiasbarreto]

Why do we require the map T to be bounded?