Distributions 2 | Test Functions

[@MikeMagTech]

Thank you. Your clear, detailed, and direct explanations make complex subjects easily accessible. As you know, in math explanations are everything. So many teachers lack the ability, or the willingness, to explain subjects clearly, and often omit critical details.

[@brightsideofmaths]

Glad it was helpful! :) And thanks for your support!

[@brightsideofmaths]

Thanks for the quick response in the comments. I had some confusing typos in this video, which are now repaired :) Thank you very much!

[@brazni]

Hi really like your videos, I just wanted to say in your playlist on distributions you seemed to have added part 4 twice and forgotten part 3. cheers

[@ysun0]

Thanks for continuing to make videos! They are great!

[@aquamanGR]

Thanks for this! I just completed your very nice series on Measure theory - how about a few videos on the basics of Martingales! (in case you are looking for ideas on what to discuss next). Thanks again!

[@mattetor6726]

Yes that would be awesome. I want to learn more about stocastic claculus at some point :)

[@morbidmanatee5550]

Interesting. Look forward to the next video!

[@mohamedaymanerrahmouni]

Thanks for your efforts! we appreciate your work =)

[@intensivemathematics5943]

Very useful video! Thanks! My research is related to distribution and I learn a lot from your videos!

[@brightsideofmaths]

Thank you :)

[@MagicBoterham]

At 6:23 "because we want to have a closed subset of R^n, we have to add the closure above" then a line is drawn above, does this overline indicate that the subset is closed?

[@brightsideofmaths]

It denotes the closure. I have video in my manifolds series about this.

[@MagicBoterham]

@@brightsideofmaths Thanks for the quick reply, I like the infinite scrolling whiteboard a lot and the explanations are really clear. In the example b at 5:16 you have `0 for ||x||≥1and exp(1/(1-||x||²)) for ||x||1 and exp(1/(1-||x||²)) for ||x||≤1` to make the support closed?

[@brightsideofmaths]

@@MagicBoterham You are not allowed to put 1 into exp(1/(1-||x||²)) :)

[@MagicBoterham]

​@@brightsideofmaths Good point. From Wikipedia: some sets are neither open nor closed, for instance the half-open interval [0,1) in the real numbers. en.wikipedia.org/wiki/Closed_set#Examples_of_closed_sets From the topological definition of a closed set "In a topological space, a closed set can be defined as a set which contains all its limit points." I would say that the function has an open support, as its limit point, 1, is not contained in the set.

[@brightsideofmaths]

@@MagicBoterham By definition, the support is always closed. That says the line above the set :)

[@apostoloskountouris5144]

Hello! thank you for these excellent videos, they make the topic of distributions a lot more accessible ! Just a quick question... for me it is not self-evident that we can have a function infinite differentiable and 0 outside some bounded interval. I cannot see how it can be smooth if it becomes 0 (so to speak instantaneously) at some point. The existence of test functions seems counter-intuitive ): Do you have an idea where one should look to get some more insight on this ? Where could I find some examples of test functions other that exp ( -1 / 1 - x^2) ?

[@brightsideofmaths]

I will make a whole video about the construction of such functions.

[@MrOvipare]

You define D(Phi) as the space of test functions, but later you use D^alpha as an kind of "generalized differential operator". Is this a valid interpretation?

[@brightsideofmaths]

Of course, these are different Ds. D^alpha is a differential operator and the D(R^n) is the space of test functions.

[@weirdo-jw9kc]

Just to be clear Infinitely differentiable functions with compact support forms sobolev space right. I have read about this somewhere.

[@HPTopoG]

The compactly supported smooth functions form a small (meager) SUBSPACE of the Sobolev spaces. Sobolev spaces themselves are defined by the notion of a “weak” derivative i.e. whether you can use the integration by parts formula on a function and get an integrable function back out as the weak derivative.

[@jacobvandijk6525]

Oops, didn't expect this one on KZbin ;-)

[@张博南]

I wonder if here the compact means just the bold and closed?

[@beastgames1234]

Yes, in R^n compactness is equivalent to closed and bounded

[@blankino-1824]

This looks like an awesome video, but it got me a bit confused. What are the prerequisite topics I would have to know in order to fully understand this? Thanks in advance!

[@brightsideofmaths]

You need a good knowledge of real analysis and maybe some things about partial differential equations.

[@blankino-1824]

@@brightsideofmaths Thank you! I'll be back in a few years then lol

[@jesusvaldes9957]

Awesome video!! any book that you recommend for this topic?

[@mattetor6726]

Iva started to work on Guide to distibution theory and fourier transform by Strichartz

[@edwardhartz1029]

Would φ=1 be a valid test function?

[@brightsideofmaths]

No, it does not have a compact support on R^n.

[@rahmatkhan3982]

your lectures very helpful..can you share with me some lectures notes or any other material related to Test function.Generalized functions.?

[@antoniodeoliveiranginamaub2845]

you did not explain well why zero is a test function, the smoothness is pretty obvious, but about having a compact support i did not get it, you did not explain that

[@StratosFair]

Its support is empty so it is a compact set (closed and bounded)

[@hangmingsheng4825]

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[@scollyer.tuition]

I note that you are requiring the space of test functions to be C^\infty, but you're not requiring them to be analytic. Is it generally true that a test function need not be analytic?

[@brightsideofmaths]

Yes, the functions need not to be analytic.

[@scollyer.tuition]

@@brightsideofmaths I guess that is because you want to define the test functions in a piecewise way, and analyticity is thus likely to fail in general? (as in the classic f(x)=e^{-1/x} x> 0, f(x)=0 x

[@brightsideofmaths]

@@scollyer.tuition Yes, analyticity will fail.