Functional Analysis 6 | Norms and Banach Spaces

Рет қаралды 80,354

Күн бұрын

Пікірлер: 40

@wdacademia2329 2 жыл бұрын

A quick remark: it may seem rather ad-hoc to restrict ourselves to R and C as the only two options for the field. Or in the definition of a metric space, having the metric be a function into R (why not some other ordered topological field). It turns out though, that when we consider all desired properties for our analysis we get nice categoricity results. For example, up to isomorphism, R is the unique complete ordered field as well as the unique complete Archimedean field (order completeness in the former and metric completeness in the latter). Another result is that R and C are the only connected, locally compact, topological fields. So these restrictions are not so ad-hoc after all.

@samtux762 10 ай бұрын

Why should we care about this space? Euclidian or hilbert space seem to bo the same job.

@jordanmatin8498 3 жыл бұрын

Hello, You are doing a very good job! Thank you for strengthening my intuiton while remaining rigorous :)

@zoedesvl4131 4 жыл бұрын

I'd like to add some generalization here, many of which will be learned in the future I think. There is a hidden property of this metric d induced by norm: d is translate invariant. To be precise, we naturally have d(x+z,y+z)=||(x+z)-(y+z)||=||x-y||=d(x,y) for all z in X. But is norm always a thing? By Axiom of Choice, any vector space is normable (i.e., we are able to define a norm whatsoever), but some norm results in abnormal structure, in which case we don't want to admit the existence of that norm. (If you are interested, learn this theorem: A topological vector space X is normable if and only if its origin has a convex bounded neighborhood.) To solve this problem, mathematicians introduced a more generalized topological vector space: F-space. A topological vector space X is called a F-space if it has a complete translate invariant metric (i.e. d(x,y)=d(x+z,y+z) for all z in X). So the blue box at the end will be updated in the future: norm be 'upgraded' to translate invariant metric. By the way this channel is great! I always prefer recommend this channel to people learning math over many others. It's important to keep the seriousness of mathematics and this channel deals with it nicely.

@brightsideofmaths 4 жыл бұрын

Thank you for your generalisations and the recommendations :) I will cover Fréchet-spaces maybe in the end of this series. I really like them but I have the feeling that it is easier first to do a lot of functional analysis with Banach spaces before going into this direction.

@rin-or3no 4 жыл бұрын

Thanks. I know its hard to make videos so fast but I enjoy this. Waiting for the next one. (One video each day will be great)

@PunmasterSTP 2 жыл бұрын

Norms and Banach spaces? More like "Now these videos grace us" with tremendous amounts of knowledge and understanding. Thank you so much for making and uploading all of them!

@kristiantorres1080 3 жыл бұрын

Amazing content, very easy to grasp. Thank you! Subscribed!

@parianhatami 8 ай бұрын

ALL OF YOUR VIDEOS ARE AWESOME!

@ishaangoud3180 2 жыл бұрын

Does a Banach Spaced form an Abelian Group(+) under addition . Like a vector space?

@brightsideofmaths 2 жыл бұрын

Yes!

@ishaangoud3180 2 жыл бұрын

@@brightsideofmaths Thank you!

@pebotin 4 жыл бұрын

Thanks for uploading..very much appreciated..😊😊

@raycopper9229 4 жыл бұрын

Great intro video for Functional Analysis, easy to digest.

@brightsideofmaths 4 жыл бұрын

Glad you like it. We will do harder stuff later :)

@raycopper9229 4 жыл бұрын

@@brightsideofmaths Looking forward to it! XD

@rakshithasp1279 4 жыл бұрын

Sir please explain metric completion theorem and its uniqueness please do explain sir I'll be waiting sir

@brightsideofmaths 4 жыл бұрын

Coming soon :)

@rakshithasp1279 4 жыл бұрын

@@brightsideofmaths I will be waiting sir thank you for replying sir

@ROni_ROmio 4 жыл бұрын

thanks for ur efforts,,,, and ur term,,, very useful

@Independent_Man3 4 жыл бұрын

Difference between Hilbert space and Banach space ?? What if the norm is induced by an inner product and X is complete with respect to this norm? Is the following true? Banach space = complete normed vector space Hilbert space = complete inner product space Further, what if the vector space is infinite dimensional?

@brightsideofmaths 4 жыл бұрын

Yes, both things are true. The dimension does not play a role in this definition, a priori.

@davidaugustyn9234 11 ай бұрын

What do you need to understand this course

@vipilvijay7116 11 ай бұрын

Set Theory Some notions from Analysis Linear Algebra Basic notions from Abstract Algebra & Topology are very helpful and most importantly Mathematical Maturity.