Banach Spaces - Lec02 - Frederic Schuller

This is from a series of lectures - "Lectures on Quantum Theory" delivered by Dr.Frederic P Schuller

[sardanapale]

As a mathematics professor I must admit I love seeing my colleague wrestling to prove that a Cauchy sequence is necessarily bounded (or getting totally confused on how to prove that the bounded operators form a Banach space).
It happens to the best of us hahaha! Feels good to see that.
These are not easy subjects to teach.
That being said, I am loving these lectures and the ones he made on General Relativity. He is a very good teacher.

[ahmedrefaat]

I was struggling with mathematical physics 8 years ago and I wish I wish I have seen these lectures, it is so great. Thank you.

[ark]

This is the quantum mechanics class I wish I took.

[shinuito]

For anyone interested, when he's showing A is bounded, as in the limit of a cauchy sequence of bounded operators is bounded, he should've used completes of the real numbers. ||A_n|| is a cauchy sequence of real numbers by the reverse triangle inequality, and that allows you to replace lim ||A_n|| by a finite real number.

You can also find this as a theorem in Reed and Simon volume 1 ~Thm III.2 depending on the edition.

[SD]

Someone knows how it's possible to obtain the problem sheet?

[MathematicsMadeSimple]

This was fantastic, I like the way the topic was introduced.

[millerfour]

8:50, 35:00, 39:15, 42:50, 47:30, 49:40, 54:53, 1:05:45, 1:18:27, 1:38:37, 1:44:13, 1:47:54

[sanketthakkar]

how can we find problem sheet for this lecture?

[neelmodi]

At 1:19:00 on the blackboard, I don't see why the boxed expression being less than epsilon for every f implies that the supremum is less than epsilon. Surely it is possible that epsilon could be equal to the supremum? This is probably nitpicky since we could just start by taking the epsilon to be eps/2 or something and then that is less than epsilon, but just want to confirm that to be absolutely concise one should do this.

[stipepavic]

thxxx alot, finaly found your channel, great lecturer with deep math, subbed!!

[ActionPhysics]

The dual vector space exists in linear algebra without any restriction to being bounded. I don't see why the restriction to bounded linear map also named as dual. Are these 2 notions different ?

[michaellewis]

Use a sequence to show the operator is unbounded 31:00

[anirbanmukherjee]

A Banach space is not really a generalization of a Hilbert space. The norm in the Hilbert space follows from the inner product. So yes, the Banach space is more general in that it admits norms that don’t correspond to an inner product. But it also entirely lacks the structure of the inner product.

[donoyonkai]

Very good videos...I'll be aiming an A+ this time...Arigato Sensei!!!

[andrea.dibiagio]

Oh man I wish I found these when I was studying for my masters degree.

[GaryPansey]

A Banach Space is a complete normed space (every Cauchy sequence converges). A Norm has 4 properties: >=0; =0 only when the function being normed is the 0 vector; scalar multiplication holdes; and; the norm of a sum = the sum of the norms.

[giannisniper]

to show that the operator is not bounded, can you say for all M in R you can find a function in C1 such that sup(...)>M?

[reinerwilhelms-tricarico]

What does he mean with ADDU and CADI ? I forgot but can't find it anywhere defined.

[junfan]

1:22:45 if the {fn} converges to f in V, isn't V a Banach space instead of just being a normed space?

[themenace]

@Frederic Schuller, could you provide the problem sheet, please?