JacobsonRadical

It is recommended and nearly required to write some of your thoughts and attempts. And perhaps provide more context.

证明基本是一样的

那个HW2， 你可以看stein，下个pdf然后直接搜索dense这个词就能找到

我先去做饭了

我也不知道是不是完全正确但是目前看起来没问题

哈哈哈哈

my WeChat is 9176800596

since the proof is over [0,1], even if we use (a,b] the thing will be a little bit complicated. Since the total space is compact.

are you Chinese?

yeah

Let us add in WeChat?

Are you a Chinese student?

I guess you are a student at Courant and perhaps taking course with Varadhan?

thank you so much

you are correct I believe

I got it

okay

let me take a look

I went to out to throw away trash..

sorry for replying late

hello

I got it, so $\partial_{t}F=\partial_{t} u(xe^{-t},1-e^{-2t})=u_{t}(xe^{-t}, 1-e^{-2t})\cdot 2e^{-2t}$, so at $t=0$, we have the limit is $2u_{t}(x,0)|_{t=0}$? but what is this thing? I am really sorry for these many confusions...

I got this, but then how do I compute my limit? I understand $F(x,t)=u(xe^{-t},1-e^{-2t})$, so that $$\lim_{t\searrow 0}\dfrac{F(x,t)-f(x)}{t}=\lim_{t\searrow 0}\dfrac{u(xe^{-t},1-e^{-2t})-f(x)}{t}.....$$ I still don't know how to compute it...

so in your notation, $$\lim_{t\searrow 0}\dfrac{u(x,t)-u(x)}{t}$$ is the infinitesimal generator of a brownian motion? what should I do to involve the $u(xe^{-t},\sqrt{1-e^{-2t}})$?

my fiancee finishes her task, good

need to sleep. Have a wonderful day!

it seems like undergraduates are the same all over the world

hahahah

hard to modify..

bad habit from my undergraduate years

yeah

I need to stay up late hahahaha

but it is just not me anymore

I don't know. I tried to live a healthier life

ahah yes

by working on my Homework :(

i just be with her

so

my fiancee is working overnight, she has a heavy task

I am 3:29 am

wow

so we have time difference right? you are still morning or afternoon?

yeah..

$f(x):=e^{ix}$ and $f_{k}(x)=e^{ikx}$

due to this counter example

well so I said $f_{k}(x)$ does not necessarily have such a limit

and asked does $f_{k}(x)$ have a limit in $L^{1}$

and define $f_{k}(x):=f(kx)$

the question gives a general $f\in L^{1}(\mathbb{S}^{1})$

yes