I never fully understand the math content that your channel covers, but I always appreciate your calm voice and outlook on school and life. Keep it up!
@Latronibus
3 ай бұрын
My sloppy idea: the goal is to say the oscillation is less than K m(I) for some TBD constant K. Split I into E_1 and F_1, the contribution to the oscillation from E_1 is less than m(E_1). At this stage, try to invoke uniform integrability of |f-f_I| with epsilon=K m(I)-m(E_1) and get a delta; if m(F_1)
@soyoltoi
3 ай бұрын
BMO: Who wants to -play video games- study Fourier analysis?
@Jaeghead
3 ай бұрын
I never fully understand your calm voice and outlook on school and life that your channel covers, but I always appreciate the math content. Keep it up!
@kobemop
3 ай бұрын
I first saw Bounded Mean Oscillations in Fourier/Harmonic Analysis. Grafakos in his textbook already sort of goes over this.
@ashveet420
3 ай бұрын
My favorite KZitem channel ❤❤
@torgeirHD03
2 ай бұрын
My only idea is to do some kind of Calderone Zygmund decomposition for |f-f_I| at height 1, then you can bound the average values to 2 over each dyadic cube/interval meaning BMO is at most 4 on I\E.
@chris-tt6yp
3 ай бұрын
Do u use Feynman study technique when studying? Great channnel btw
@ojalajim
3 ай бұрын
Hello, my brother. I hope you are doing well! I am looking forward to seeing your new post soon.
@De2Venner
3 ай бұрын
If you can somehow show that the set of all E's is compact over the integration I, that should justify changing the integral over I into a sum of integrals over different intervals E, which should be easier to show a bound for. Maybe you can show it's compact from the fact that its measure is bounded from below of the original interval?
@johnny-wm4uo
3 ай бұрын
Funnily enough I first heard of bmo in numerics of pdes
@matt61956q
3 ай бұрын
I first hear of BMO in Adventure Time :)
@Latronibus
3 ай бұрын
I am curious how this problem actually turned out. My attempt from earlier (you can still see it in another comment) seems to hit a snag because I need to bound the number of times that we have to cut before the integral of |f-f_I| over the remaining subdomain is less than K m(I) for some K dependent on f but not on I. That is because each "good set" contributes potentially as much as alpha m(I) to the integral, even as they get smaller and smaller.
@oo_rf_oo8824
3 ай бұрын
I remember javier duoandikoetxea's Fourier analysis has a chapter that talks about BMO, and it might be of your interest to take a look at it.
@oo_rf_oo8824
3 ай бұрын
Another thing, such E is presence in every Lebesgue measurable set X, due to the way that Lebesgue outter measure is constructed. So maybe we can overestimate the integral by something akin to geometric series with some coefficient.
@sdgiohsoidghsoidfhsoigh
3 ай бұрын
wow, what a last name
@oscarspolander318
3 ай бұрын
Do you have anything to say about the order in which you recommend one studies the Princeton Analysis quartet by Stein and Shakarchi? I currently own a copy of Real Analyisis and consider getting the other three.
@PhDVlog777
3 ай бұрын
I don’t the order matters too much, but if you want to study it the way I did, recommend 1) real analysis 2) complex analysis 3) Fourier analysis 4) functional analysis.
@oscarspolander318
3 ай бұрын
@@PhDVlog777 Brilliant! Then I can get going right away :) And keep doing the great work that you do man, I am rooting for you and you should be proud of what you have done and what you do in Mathematics! Wishing you well from Stockholm, Sweden
@Polaris12295
3 ай бұрын
Just curious. You know a lot of very complex and difficult mathematics. Have you found that you have forgotten some of the basic mathematics like algebra? Just curious.
@willclark8946
3 ай бұрын
what??
@PhDVlog777
3 ай бұрын
Absolutely lol. I need to review my forgotten calculus 3 skills, I need it now more than ever
@abcdefghij50
3 ай бұрын
hint: forget about the expression and try proving that the optimal alpha is actually 1
@johnny-wm4uo
3 ай бұрын
the value of alpha depends on the function f and is only 1 in some uninterresting cases
@abcdefghij50
3 ай бұрын
@@johnny-wm4uowhat i was thinking is if you are given an alpha>0 that you can actually make alpha bigger as long as it is less than 1. Since I/E is measurable it can be covered by a countable set of disjoint intervals of measure m(I/E) + epsilon. Now do the same for each of these intervals and take the union of all such E’. Their total measure that is contained in I is at least aplha*(m(I/E)+epsilon) - epsilon >= alpha*((1-alpha)*m(I)+epsilon)- epsilon. So by sending epsilon to zero you can increase alpha by alpha(1-alpha)
@abcdefghij50
3 ай бұрын
Typo it should be =< instead of >=
@Latronibus
3 ай бұрын
@@abcdefghij50 Doing that, you pick up more pointwise error each time you cut the bad set, and you need to take into account how much that additional error piles up. Your version as stated is obviously wrong, since it's saying everything's within 1 of the global average, which is obviously not true for something like f(x)=x on [0,10]. If you follow the basic idea of iterating, I now think you hit a snag. As I said in my own comment, you have |f-f_I|
@abcdefghij50
3 ай бұрын
First of all i don't get the counterexample since this is an if question not if and only if. I kinda get what you are trying to do but let's just forget about BMO stuff for a minute since my argument is purely measure theoretic. Suppose we have a measurable set E with property that m(I intersect E)>= alpha*m(I) for all intervals I and some alpha>0. I basically want to prove that E is equal to R a.e. Why would someone want to prove that? Well if E cuts intervals with some alpha and alpha!=1 or 0 than E obviously can't be measurable since E can't be aproximated well with the outer measure. Now you do what I did. You fix an interval I and pick E that cuts I with coef. alpha and take E'= E intersect . I further proved since I/E' is measurable that for arbitrary epsilon you can prove that E intersect I has measure alpha+alpha(1-alpha+epsilon)-epsilon times m(I). Now if alpha < 1 you get m(E')
@drjohnespy
3 ай бұрын
For fun, Find a 1:1(f) from the R*R into the R (R = Reals, sorry no math symbols)
@Basedgwad
3 ай бұрын
very sexy book
@Dmitriy-qu6hv
2 ай бұрын
kindergarden, second month, hardly a grad. student level.
@jrgen7903
3 ай бұрын
sixth
@replex8889
3 ай бұрын
maybe second
@knight4689
3 ай бұрын
Third
@GreyWilliams2718
3 ай бұрын
Maybe first
@GreyWilliams2718
3 ай бұрын
That aside, thanks for your video as always. I like your stuff a lot. You talk about math, and you are honest about your struggles. You are like a friend I wish I had irl.
@aryansaxena4978
3 ай бұрын
Hi struggling grad student, is there any way to contact you via mail or some social media, i had some doubts related to PhD admissions in the US, I'd really appreciate your help
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