Thanks so much! Glad you found it useful

You're welcome Pedro, glad you found it useful

Thank you Mansi, it's a nice proof, Arrow/Hahn later used the same idea for existence of general equilibrium in their book

Glad you found it useful Alexandra

This is done at 2:47 (both domain and codomain are the set of real numbers)

Found one: f(x) = (x^2)/2 +0.1 in [0,1]. It's definetely a pseudo-contraction, since if we apply it to any x1, x2, in [0,1], the distance between them gets contracted. However, its not a contraction because as x->1, the slope tends towards 1, so we can't find a λ<1 such that we'll always have |f(x1)-f(x2)|< λ|x1-x2|. Make x2=1, and try to pick a λ. You can't, because for whatever λ you pick, you can always choose an x1 that's closer to x2, and then that λ won't work anymore, and you'll have to pick a bigger lambda. Hence, its not a contraction.

Also: someone has posted a quick question on your lecture on StackExchange, I have attempted to give a formal answer, feel free to verify it ! UHC is a niche concept maybe the questioner is struggling with it, it's posted just a month ago. math.stackexchange.com/questions/3737960/continuity-of-correspondence-hemi-continuity/3799418#3799418

Gosh, I don't know if I have the time or stamina for more, but happy to know you found these useful.

Hit the space bar to pause, go at your own pace, rewind as needed (and remember these are free of charge, unlike the books)

You're welcome David, happy to help

Thanks David, glad you found it useful.

Yes of course, but please provide students with a link to the playlist.

You're welcome, glad to be of help

Matthew, it is true that the image of x_2 minus delta will contain points outside of the image of x_2 as you say, but these points will still be in an epsilon neighborhood of the image of x_2, for any given epsilon, provided that delta is small enough.

Ah, perfect. I missed the epsilon argument in the definition. Thanks for the prompt reply.

Thank you very much!

You're very welcome Iris

Maybe the green arrow at 9:30 will help. For any sequence (x) converging to x1 from the right, you can find a sequence (y) such that each y is in the image of the corresponding x, but the limit of the (y) sequence is not in the image of x1. Note that the image of x1 is a single point.

Yes, that's correct, thanks for spotting the typo.

thanks!

but maybe gamma star could be single value, for example, at the point x4 in this video(06-1) kzbin.info/www/bejne/hXvJq32eqKhmiMU. Also in the lucas and stokey text book, pages 57, exercise 3.11, it says, 'show that if gamma is singled valued and uhc, then it is continuous' which is pretty straightforward. But the point is, maybe gamma or gamma star could be single valued..

Yes, Gamma^* could be single valued, but it is still a subset of Gamma, not an element. It has to be defined this way to allow for the general possibility that the optimization problem has multiple solutions.

Rajiv Sethi thanks professor Rajiv. I think i get it. Gamma star is generally a subset of gamma, and if it is singled valued, then gamma start is still a subset if gamma but with one element. I guess that is it...right?

Yes, exactly, it's a singleton in that case

Gamma^* is a subset of Gamma, not an element. For any given theta, there can be multiple solutions to the optimization problem, even though the maximized value of the objective function is unique. See segment 07-2 kzbin.info/www/bejne/l5nJqpKKmMyjhbs at 9:26 for an example.

ok...I get it if the assumption of multiple solutions is made. Thank you. But, in the video here kzbin.info/www/bejne/l5nJqpKKmMyjhbs, though you showed that there could be multiple solutions, both of the examples you provided failed either lhc or uhc, which kind of makes gamma star not continuous, right? so if gamma star is not continuous, can it still be a solution to the problem? In other words, I want to ask if there could be multiple solutions and at the same time, both lhc and uhc would be satisfied? Thanks.