Hyperbolic Filling, L5: quasisymmetry
10:34
Hyperbolic Filling, L4: uniformization
14:10
Hyperbolic Filling, L2: Examples
15:39
19 сағат бұрын
Hyperbolic Filling of Metric Spaces
0:58
21 сағат бұрын
Pansu Derivative
14:23
3 ай бұрын
14. Stoke’s Theorem Example
8:57
13. Stoke’s Theorem
15:13
4 ай бұрын
6. Green’s Theorem
37:44
4 ай бұрын
Пікірлер
@zhikunli475
@zhikunli475 22 күн бұрын
I love you prof, great video
@YoungMeasures
@YoungMeasures 22 күн бұрын
Wow, thanks!
@rheat7109
@rheat7109 Ай бұрын
Nice explanation! Thank you!
@YoungMeasures
@YoungMeasures Ай бұрын
Thanks! Glad you liked it.
@TolulopeOriowo
@TolulopeOriowo 2 ай бұрын
Good work from you! We came across your course on Udemy and it's such an outstanding course with so much opportunity of gaining a stronger feet amidst other courses in your niche. Are you open to for us to explore the path to the success of your course together?
@heartpiecegaming8932
@heartpiecegaming8932 2 ай бұрын
I think the answer is a no. If you take the topologists sine curve, together with adjoining the two end points of the sine curve via a different route, then the resulting set is rectifiably connected but has infinite diameter (with respect to the rectifiable curve metric you introduced).
@BehnamEsmayli
@BehnamEsmayli 2 ай бұрын
Wonderful! That will do.
@bagalo
@bagalo 3 ай бұрын
My guess is to look at a closed, bounded subset of R^2 with an inward cusp. Like the closure of the bounded component of a standard cardioid. There can be no bilipschitz map of this set to the same set but with the length metric.
@BehnamEsmayli
@BehnamEsmayli 3 ай бұрын
Yes. But is the length distance not compact? For example, is it not true that every sequence has a convergent subsequence?
@YoungMeasures
@YoungMeasures 3 ай бұрын
kzbin.info/aero/PLleGBpoKCrJn21-tCrNRk6JAyh4gwtgks&si=8KfKhmytMwOjSjLd (Professor Le Donne's playlist on Sub-Riemannian geometry)
@YoungMeasures
@YoungMeasures 3 ай бұрын
kzbin.info/www/bejne/hIjTeGh8bt2LkKc
@YoungMeasures
@YoungMeasures 3 ай бұрын
kzbin.info/www/bejne/fYrdpHuaiJWgd7Msi=j8cK4e-Gjdufsdfg&t=1325
@sushobhandas7752
@sushobhandas7752 4 ай бұрын
Why in 24:17 that inequality holds? Do you assume that \lambda_1<\lambda_2<\lambda_3?
@YoungMeasures
@YoungMeasures 4 ай бұрын
Yes. Lambda is the percentage of "surviving" intervals. We keep increasing them in order to preserve as much as we can. In limit, you remove almost nothing, so lambda is 50% -- remember there are two copies of size lambda. So, in limit, you basically remove nothing. The denominators and log work out to give that inequality.
@sushobhandas7752
@sushobhandas7752 4 ай бұрын
@@YoungMeasures thanks.
@sushobhandas7752
@sushobhandas7752 4 ай бұрын
@@YoungMeasures have you made some video on uniform rectifiability? I have been referred to your channel long back by some professors in Finland. Your content is very good. Keep doing this.
@YoungMeasures
@YoungMeasures 4 ай бұрын
@@sushobhandas7752 I have no videos on that particular topic. I would look up Jonas Azzam’s channel for them.
@sushobhandas7752
@sushobhandas7752 4 ай бұрын
@@YoungMeasures can you share the link of that video or channel here.
@jorgebecerril2361
@jorgebecerril2361 4 ай бұрын
Very nice tools, I will follow your advise on creating a website! Another tool I like is research rabbit. It's similar to mathscinet where you can look for paper abstracts and related research but free, and you also have a nice graph view of how papers connect with each other so you can easily find related work.
@Jorge-c7d
@Jorge-c7d 4 ай бұрын
Very nice exersice! I have one question tho, when you refer to equicontinuity of the family F you are actually referring to uniform equicontinuity, is that right? I ask because, apparently, the continuity modulus delta you choose for the family F holds for every pair of elements in the domain, or maybe im missing something?
@BehnamEsmayli
@BehnamEsmayli 4 ай бұрын
Yes yes. I did not know not uniform equicontinuity, i.e., pointwise equicontinuity, was a thing until this comment! LoL!
@YoungMeasures
@YoungMeasures 4 ай бұрын
A second lecture will complete the discussion here.
@YoungMeasures
@YoungMeasures 4 ай бұрын
Apologies for the type in definition of the Hausdorff measure. It is lim of inf of sums not "sup". There is equivalent definition as sup of inf of sums, which I had initially prepared, but when switching to limit I erased the wrong item :(. Obviously that supremum is always infinite, except if your space is discrete or something, LoL!
@yahya1031
@yahya1031 4 ай бұрын
I was just scrolling and i found this video hopefully i will get motivated to study more math
@nin10dorox
@nin10dorox 5 ай бұрын
Where can I find a proof of the theorem that is used as a starting point?
@YoungMeasures
@YoungMeasures 5 ай бұрын
It is called CARATHEODORY^S CRITERION. It is Theorem 5 in first edition of Measure Theory and Fine Properties of Functions, by C. Evans and F. Gariepy. Other advanced books must have a proof too.
@nin10dorox
@nin10dorox 5 ай бұрын
@@YoungMeasures Thank you!
@bagalo
@bagalo 6 ай бұрын
Very nice video. Thanks for the illuminating examples. Did you construct them yourself? Is there a reference where notions in GMT are explained using examples?
@YoungMeasures
@YoungMeasures 6 ай бұрын
Thanks. Any GMT book will do. Examples? I do not know of one doing examples as explicit. Maybe they toss them as exercises. I did come up with them for the lecture but to be fair, they are easy to come up with.
@YoungMeasures
@YoungMeasures 6 ай бұрын
I have a video on different books for GMT. Might help to check out.
@Top_Maths
@Top_Maths 7 ай бұрын
Congratulations! Looking forward to more
@TraderZeta
@TraderZeta 7 ай бұрын
I would love to see more on writing papers. Things like citations and general conventions. Also, 10/10 channel. Love this HQ math content.
@YoungMeasures
@YoungMeasures 7 ай бұрын
Noted!
@YoungMeasures
@YoungMeasures 7 ай бұрын
If we use absolutely continuous curves or Lipschitz curves in the infimum in the definition of d_cc, then the triangle inequality is indeed obvious, as well. But with C1 curves not so -- there is technical details with concatenation of curves.
@YoungMeasures
@YoungMeasures 7 ай бұрын
!!LIVE!! On Sunday May 26, 11:00am EST (USA & Canada) time. Can’t wait to see you there! Any questions I can ponder on already?
@erwin55226
@erwin55226 7 ай бұрын
Hi, can you please suggest some material for Heisenberg group.
@YoungMeasures
@YoungMeasures 7 ай бұрын
The standard book on the topic is: Capogna, Luca and Danielli, Donatella and Pauls, Scott D. and Tyson, Jeremy T., An introduction to the Heisenberg group and the sub-Riemannian isoperimetric problem. But depending on you need and focus there may be other (possibly non-book) options. If you give more details I might find better references --Young measures
@YoungMeasures
@YoungMeasures 8 ай бұрын
kzbin.info/www/bejne/jHaWiomJac6SetEsi=lcwMHUmva8y--MKQ&t=871
@TraderZeta
@TraderZeta 8 ай бұрын
Great videos, very clear explanations. Looking forward to poincare inequalities.
@akashpradhan6039
@akashpradhan6039 9 ай бұрын
Hello sir! Can you please suggest books related to metric measure space
@YoungMeasures
@YoungMeasures 9 ай бұрын
Lectures on Analysis on Metric Spaces by Juha Heinonen Sobolev Spaces on Metric Measure Spaces An Approach based on Upper Gradients Juha Heinonen Pekka Koskela Nageswari Shanmugalingam Jeremy T. Tyson Topics on Analysis in Metric spaces, Luigi Ambrosio, Paolo Tilli. First published in 2004, approximatly 130 pp. Easy to read? Yes. Exercises? Yes. Nice ones! Focuses on: Lipschitz curves/maps into metric spaces, geodesic problem, Sobolev spaces on metric spaces
@akashpradhan6039
@akashpradhan6039 9 ай бұрын
@@YoungMeasures Thank you very much sir for suggesting these books. Appreciated!
@marclcf
@marclcf 10 ай бұрын
Great video but your corollary is wrong... The condition is Hs(A)>0 (not Hs(A) finite)!
@YoungMeasures
@YoungMeasures 10 ай бұрын
Yes. I think you are right! The least confusing way to state it is to assume both finite and positive. I will either cut it out or add clarification. Thanks.
@sarojchhatoi1307
@sarojchhatoi1307 11 ай бұрын
thanks for the nice presentation! Are you aware of the generalization to the case where the domain of measurable function U(x) is unbounded?
@YoungMeasures
@YoungMeasures 11 ай бұрын
Unfortunately, I am not aware of any references. But was boundedness used in any essential way?
@AhmedAbubakr1
@AhmedAbubakr1 11 ай бұрын
I want to send you my thesis about area formula, I want your opinion
@YoungMeasures
@YoungMeasures 11 ай бұрын
Sure. Please do.
@AhmedAbubakr1
@AhmedAbubakr1 11 ай бұрын
I don't know why the comment with the link is being deleted ... Could I send it by any other mean?
@AhmedAbubakr1
@AhmedAbubakr1 11 ай бұрын
​@@YoungMeasures I have sent it via your email in the description of the channel
@YoungMeasures
@YoungMeasures 11 ай бұрын
Of course. On channel descriptions you can find my emails.
@AhmedAbubakr1
@AhmedAbubakr1 11 ай бұрын
@@YoungMeasures I sent it to your gmail. Thank you
@StratosFair
@StratosFair Жыл бұрын
Nice :)
@loscerdospablomateo
@loscerdospablomateo Жыл бұрын
Spent hours trying to understand my professor notes, you saved me!
@chrisstewart1492
@chrisstewart1492 Жыл бұрын
Love your content, sir! They are really helping me a lot. Could you recommend some textbooks that can help me to better follow your videos? I have only done real analysis before, so a lot of definitions mentioned in your videos are totally unfamiliar, things like rectifiable paths and so on. That'll be much appreciated, thanks!
@YoungMeasures
@YoungMeasures Жыл бұрын
Thanks for the feedback! I have covered multiple topics, so if you tell me which playlist or topic you like most and also with what math you need to learn, I can find some titles for you.
@jayprich
@jayprich Жыл бұрын
Thank you I like the concreteness, I'm trying to get a feel for Lie algebra and Lie group from category theory that's rather abstract. Do you have any tips about how to imagine this exponential map itself intuitively in H1?
@YoungMeasures
@YoungMeasures Жыл бұрын
Around min 23 I explain that we identify G with g via exp map. This then means that in our special setting the exponential map is the identity! It is initially a bit difficult to live with this but you can go over the process to convince yourself that it is true! The real reason is Campbell Hausdorff formula.
@NikitaEvseev
@NikitaEvseev Жыл бұрын
Hi! Thank you for that video, I liked it very much. Then It would be interesting for me the further introduction to Newtonian-Sobolev spaces. Also I wander if it is possible to go from Newtonian-Sobolev space back to Sobolev space on a Riemannian manifold ?
@YoungMeasures
@YoungMeasures Жыл бұрын
Absolutely. Up to precision with a.e. representations, the two classes should coincide. How easy it is to prove depends on how you define the Sobolev functions between manifolds. The Newtonian-Sobolev spaces are also defined for maps into Banack space targets as well. I will refer to the book by Jeremy T. Tyson, Nageswari Shanmugalingam, Juha Heinonen, Pekka Koskela.
@mamahuhu_one
@mamahuhu_one Жыл бұрын
Very interesting videos, thank you! Is there a video talking about tangent measures?
@YoungMeasures
@YoungMeasures Жыл бұрын
I refer to Azzam’s channel for these. He has fantastic videos on GMT. kzbin.info/aero/PLp0TNqYe2DSEA8r9xYJnJCibS1pN-NQq3&si=M7aYBvhygsEaGkwt
@luisguamushig9122
@luisguamushig9122 Жыл бұрын
help me please, i want to know covering lemma on doubling metric space, i can find in the metric space, but i cant find in the doubling metric space, can you help me ? pleaaaase im mathematic
@YoungMeasures
@YoungMeasures Жыл бұрын
I do not understand the question itself. Do you want a proof of the 5r covering lemma in metric setting?
@felipesants8936
@felipesants8936 Жыл бұрын
VERY GOOD VIDEO. THANK YOU SIR
@professorno9080
@professorno9080 Жыл бұрын
great video, thank you!
@gojo-kunanimes902
@gojo-kunanimes902 Жыл бұрын
Congratulations on the class, very good. I have some questions, I hope you can help me. A Carnot distance was defined in H^1, do you know if at this distance H^1 is a complete space? Then you gave a "norm", Koranyi norm, in fact from your expression it is not homogeneous, but do you know if it is possible to induce a norm in H^1 that makes it Banach?
@YoungMeasures
@YoungMeasures Жыл бұрын
It is a complete metric space yes. For example because convergence with respect to carnot distance is equivalent to convergence as if in usual R^3. Do you mean if there is a norm on H^1 that is, say, bilipschitz to the Koranyi, and as a 3D vector space over R, H^1 becomes a Bsnach space? I have to say no, but dont have a simple readon why. Maybe the scaling would provide a contradiction. Because scaling that is consistent with Carnot geometry is different from the usual scalar product.
@gojo-kunanimes902
@gojo-kunanimes902 Жыл бұрын
@@YoungMeasures That's right, the only thing we don't have for the Koranyi norm is the property ||ax|| = |a|.||x|| and to define Banach space, it is assumed that a norm satisfies such properties. My question is precisely whether I could either define a norm in H^1 that makes it Banach, or whether the definition of Banach space (but maintaining the known properties) can be relaxed to meet this case in which the norm is not homogeneous. Maybe this isn't trivial, or doesn't make sense, as I haven't found anything about it.
@kz1662
@kz1662 Жыл бұрын
nicely explained! keep making videos please
@jacopofontanesi
@jacopofontanesi Жыл бұрын
oh thank you finally, i didn't find it elsewhere
@fanalysis6734
@fanalysis6734 Жыл бұрын
Can you please give a page number with the full theorem referenced at the end of the video, about "if the projection has hausdorff measure zero then E is purely unrectifiable"? Glanced at simons and it isn't obvious which theorem is being referenced. Thank you this is a helpful video.
@YoungMeasures
@YoungMeasures Жыл бұрын
Will do soon.
@YoungMeasures
@YoungMeasures Жыл бұрын
Ok, Simon has some notes online, which is basically a newer version of his book's material. There in Remark 3.4 it is stated that: Fix an (orthogonal) basis of R^(n+m). If projection of a set S onto any of the hyperplanes given by span of n-many of these basis vectors has zero Hausdorff-n measure then S is purely H^n-unrectifiable.
@YoungMeasures
@YoungMeasures Жыл бұрын
If you know or can find a reference for that fact that in any rectifiable set with positive measure you can find a set with positive measure that is bi-Lip to a subset of Euclidean space with biLip constant as close to 1 as we wish, then this should not be difficult to see: Working locally, the projection of a part of the set onto a suitable tangent space (a hyperplane) must have positive measure. Because the set and its projection onto the tangent are (working on even smaller neighborhoods) extremely nearby, the projection of the set to the hyperplanes that are in general position with regard to that tangent plane, is pretty much the same projecting the set onto tangent plane first then projecting onto that hyperplane. Since projection onto the hyperplane does not vanish the area, we are left with a projection of S of positive area. (If you want to draw a picture, first draw a 2D manifold'ish set in 3D, i.e. S, then a tanget plane to it, the tangent plane projects nontrivially to one of xy, or xz or yz planes. )
@ARBB1
@ARBB1 Жыл бұрын
Thanks for the video
@asdasds4679
@asdasds4679 Жыл бұрын
Accoridng to wikipedia, its the Chebyshev ineq that you are talking about at 13:16 , but in the book Concentration inequalities by Boucheron, Lugosi & Massart they call it Markovs ineq in pag 19 Also note that in Boucheron et al, they use Mf(x) as the Median of the function and not the Maximal 🤷
@YoungMeasures
@YoungMeasures Жыл бұрын
Thank you for very clear references.