Functional Analysis 18 | Compact Operators

[@Jooolse]

9:02 By the way, it is not necessary for A to be bounded: it is suffisent to show A is point-wise bounded to apply Ascoli Theorem.
i.e. {g(x)/g in A} bounded for each x in [0,1]

[@TavinCole]

Where is this particular formulation of Arzela-Ascoli written up?

[@zachchairez4568]

10 minutes with you >> 1 week in class

[@felipegomabrockmann2740]

I am totally agree

[@jonasw4791]

great video! the series is amazing 😊
at 9:20 I dont understand why bounded and equicontinuity holds for the closure as well?

[@brightsideofmaths]

Indeed, this is something you could try to prove. Maybe start with bounded and then check equicontinuity.

[@Jooolse]

​@@brightsideofmaths When applying the Ascoli Theorem, we don't need to worry about the closure: if A is equicontinuous and pointwise bounded, then A is relatively compact = its closure is compact.

[@brightsideofmaths]

@@Jooolse Thanks. Still something one needs to prove :)

[@Jooolse]

​@@brightsideofmaths Hmm to prove what? I just stated the conditions of the Ascoli Theorem! The conditions are on A and you get the compactness of its closure Ā :)
A is relatively compact (i.e. Ā is compact) A is equicontinuous and pointwise bounded.

[@brightsideofmaths]

@@Jooolse But that the closure is also equicontinuous one has to prove, right?

[@AadityaVicramSaraf]

I'm couldn't really understand what happened at 3:53 for a particular f in C[0,1], how and why are you defining Tf at s with another function k? What is the purpose of this k, and how did it come, in the first place?

[@brightsideofmaths]

Thanks for the question! We define an operator by using a kernel function k. So for a given k, we get an operator T by the chosen definition.

[@aleksherstyuk8319]

This might come in a later video, but what can we say about compact operators? Why would it be nice to know that an operator is compact?

[@brightsideofmaths]

This will eventually come, yes! However, compact operators have some nice properties, a simple spectral theorem and other things. Therefore, if you know that an operator is compact, you know immediately a lot of things. We will discuss this :)

[@marcuslaurel5758]

There’s a cool theorem called the Leray-Schauder fixed point theorem, which requires a compact operator in one of its hypotheses. It’s a theorem that can be used to prove the existence of solutions for certain types of nonlinear PDE, which is pretty neat!

[@StratosFair]

How to show that the closure of an equicontinuous set of functions is equicontinuous though ?

[@brightsideofmaths]

Very good question! This is indeed not totally clear. It is helpful to try to prove it :)

[@StratosFair]

@@brightsideofmaths challenge accepted !

[@ativjoshi1049]

4:30 Is T_k a distribution?

[@brightsideofmaths]

Not quite. A distribution has numbers as the codomain.

[@kellyramsay4896]

What software do you use to write the math? Also love your videos thanks for these!

[@brightsideofmaths]

Xournal and many thanks :)

[@spin7765]

Excellent video!

[@brightsideofmaths]

Thanks :)

[@Jooolse]

8:40 uniformly -> uniformely -> uniformly 👀

[@brightsideofmaths]

Yeah, I corrected the typo.

[@Dr.kcMishra]

Great... Thank you

[@lissiecoutts9808]

So helpful! Thank you :)

[@ferry7185]

can I ask for verification. if the operator is bounded, we can also say that image of B_1(0) of T also bounded?

[@brightsideofmaths]

Yes, indeed :)

[@trondsaue7860]

As usual, great ! Will there be anything about Hilbert-Schmidt operators ?

[@brightsideofmaths]

Thank you very much! I like Hilbert-Schmidt operators and can definitely make a video about them!

[@conradmorris8881]

thank you!

[@brightsideofmaths]

You're welcome!

[@JamieLynnSelman]

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