Room for Liad and Daniel Fischer

Daniel Fischer

15:39

?

Liad

hi :)

Daniel Fischer

Hi. What's up?

Liad

Im good and you?

Daniel Fischer

So-so.

Liad

what's wrong?

Daniel Fischer

15:50

The weather outside is dreadful.

Liad

where are you from?

Daniel Fischer

Europe

Liad

So i guess you dont like the winter :P

i dont like it either .. but i guess in europe the winter is much colder

Daniel Fischer

I don't like the eternal rain.

Liad

yea it can make you sad

do you like measure theory ? ^^

Daniel Fischer

15:55

I'm fairly impartial on that. Some nice things, others not so nice.

Liad

i think i just solved the question i wanted to ask you lol

Daniel Fischer

Cool. What was it about?

Liad

i think your "impartial" is better than some very good students

$f_n , f \in L2$ and $lim f_n = f$ a.e

$||f_n|| \le M$

first i need to show that $f\in L\ ^ 1$

the second part is to show $||f_n- f||_1 \to 0$

the space is finite $\mu(X) \lt \infty$

so we get that $L2 \subset L1$

so the first part is easy

Daniel Fischer

Yeah, that's a crucial assumption. Would be terribly wrong for infinite measure.

Liad

i found a mistake in my solution to the other part :/

i thought using Egorff thm.

you familiar with it?

Daniel Fischer

15:59

Almost everywhere convergence gives almost uniform convergence when the measure is finite.

Liad

its supposed to be $||f_n||_2 \le M$

yea right so we have $A $ with $\mu (A \ ^ c) \lt \epsilon$ and $||f_n -f||_1 \to 0$ on $A$

Daniel Fischer

Yes. And on $A^c$, Cauchy-Schwarz helps.

Liad

now im stuck ^^

Cauchy-Schwarz?

Daniel Fischer

Famous inequality.

Liad

yea i know it

do you mean to use Holder's inequality?

Daniel Fischer

16:02

$$\int_X \lvert f_n - f\rvert \cdot \chi_{A^c}\,d\mu$$

Well, Cauchy-Schwarz is the special case $p = q = 2$ of Hölder.

Liad

yea right. wanted to write it :P

so $||fg||_1 \le ||f||_2||g||_2$

you suggest taking $f_n-f$ and $x_{A \ ^ c}$ ?

Daniel Fischer

Sounds like a plan.

Liad

so we get $||f_n-f||_1 \le ||f_n-f||_2 \epsilon$

on $A \ ^ c$

Daniel Fischer

There's a square root missing.

Liad

i feel like $||f_n-f||_2 $ should tend to $0$ \

Daniel Fischer

16:07

Need not be. But we don't need that. It's bounded.

Liad

right because $||f||_2 \le M$ also

right ? ^^

and you are right it should be $\sqrt{\epsilon}$

Daniel Fischer

Yes. But even if it could be larger, we still would have $\lVert f_n - f\rVert_2 \leqslant \lVert f_n\rVert_2 + \lVert f\rVert_2 \leqslant M + \lVert f\rVert_2$.

Liad

right , because it's in $L2$ then $||f||_2$ would be finite

i think it would be bounded by M because of dominated convergance

Daniel Fischer

Do we have a dominating function? Not necessarily.

Liad

$M$

wait no.

Daniel Fischer

16:13

But we have weak convergence, and that implies $\lVert f\rVert_2 \leqslant M$.

Liad

im talking no sense

dont we need the convergance to be monotone to get that?

and the functions to be positive

Daniel Fischer

Monotone convergence would immediately imply $L^2$ and $L^1$ convergence. If we had that, Egorov would be unnecessary.

Liad

how did you conclude $||f||_2 \le M$ ?

Daniel Fischer

Weak convergence. The closed balls are weakly closed, and $f_n$ converges weakly to $f$.

But, as said above, we don't need that, a bound independent of $n$ is enough.

Liad

right ok

i have an exam on Thursday :P

Daniel Fischer

16:22

Good luck.

Liad

Thanks. i think measure theory is the best course so far that i took

do you have a favorite field? i mean measure theory / complex analysis /algebra/ .. ?

Daniel Fischer

Complex analysis, point set topology, topological vector spaces, analytic number theory.

Liad

I have an exam on complex analysis course in 2 weeks ^^

Can i have one more question ?

Daniel Fischer

You could try.

Liad

$\mu ,v$ finite measures. i need to prove that there is disjoint measurable sets $X = A \cup B$ s.t $\mu \bot v$ on $\Sigma \cap A$ and $\mu <<v << \mu$ on $\Sigma \cap B$

i have a partial solution.

we have that either $\mu \bot v$ or there is $\epsilon \gt 0$ and $A \in \Sigma$ s.t $v(A) \le \epsilon \mu(A)$

if $\mu \bot v$ we are done

if not , we are not ^^

Daniel Fischer

16:36

Have you looked at the Radon-Nykodim derivatives with respect to $\mu + v$?

Liad

wow you are fast :P

hm. let me try it

do you mean that we have $f,g$ s.t $d\mu = fd(\mu+v)$ and $dv = g d(\mu+v)$ ?

Daniel Fischer

Yes.

Liad

do i need to use the $\epsilon $ from my partial solution?

and the $A$ of course

Daniel Fischer

You certainly don't need to. Whether it would be useful, I don't know.

Liad

alright . we have $\int_A f-g d(\mu+v) = \mu(A) -v(A)$ for all $A \in \Sigma$

any chance for a bigger hint ?

i dont see how to get those $A,B$

Daniel Fischer

16:46

When, on some $M \subset X$, you want to have $\mu \ll v$, what conditions on $f$ and $g$ does that impose?

Liad

hmm

if $g=0$ a.e then $f=0$ a.e ?

or $f-g=0$ a.e ?

Daniel Fischer

a.e. with respect to which measure? But note that $f + g = 1$ ($\mu + v$-a.e.).

Liad

right.

im dog asks for a walk ^^ i will think about your hints while walking with him. can i write to you like in an half hour what i've got?

Daniel Fischer

I'll be away for dinner then, but you can, I'll read when I return.

Liad

17:46

@DanielFischer you there?

Daniel Fischer

18:19

@Liad Not then.

Liad

first i thought taking $A=\{x: f(x) >g(x)\}$

and $B=A \ ^ c$

but now i think that the set $B= \{x: f(x) + g(x) = 1\}$

would be better

not sure if im right, im just checking it

its not working either

Daniel Fischer

Not quite. On a set where $\mu \ll v$, you can write $d\mu = h\,dv$. What does that tell you about $f$ and $g$?

Liad

$fd(\mu+v) = hdv$ ?

i really dont know where to take this..

Daniel Fischer

And $dv = g\,d(\mu + v)$, so …

Liad

f=hg

Daniel Fischer

18:28

And since $f + g = 1$ (a.e.) …

Liad

(h+1)g = 1 a.e?

is there something obvious im missing?

Daniel Fischer

$f = h(1-f)$, whence $h = f/(1-f)$ ($= f/g$)

Liad

alright

btw if we work with positive measure, do we know that $f$ is positive?

Daniel Fischer

Nonnegative. And I was assuming we speak of positive measures. We'd need a couple of absolute values and some modification if we're speaking about complex or signed measures.

Liad

yea we are working with positive measures, just wanted to know

alright where we going with $h=f/g$ ?

Daniel Fischer

18:37

When does that break?

Liad

when $g=0$ ?

Daniel Fischer

Right.

Liad

so $B = \{x : g(x) \ne 0 , f(x) \ne 0\}$ ?

Daniel Fischer

Yes. We could add any subset of the set where both vanish, but that only complicates things.

Liad

im writing it down to see if i understand it

so on $A$ we have that $\mu=v=0$ correct?

Daniel Fischer

18:46

No. That may be, but usually isn't. Consider two point masses at different points.

Liad

but on $A$ the functions vanishes

so $\mu(E) = \int_E 0 d(\mu+v)$

Daniel Fischer

At every point of $A$ one of them (possibly both) vanishes.

Liad

for all $E\in \Sigma\cap A$

i thought $A=\{x : f(x) = g(x) = 0\}$ , oops

so we can take $C = \{x : f(x) = 0\}$

$C \subset A$

and then $C , A-C$ would show $\mu \bot v$

on $\Sigma \cap A$

Daniel Fischer

Yes.

Liad

alright part 1 is done ^^

we saw that if $\mu <<v$ then the $h$ you wrote will be $f/g$

but we need to prove the converse dont we? that $\mu <<v $ on $\Sigma \cap B$

and $v << \mu$

Daniel Fischer

18:52

So suppose $N \subset B$ and $v(N) = 0$. Then how do you see that $\mu(N) = 0$?

Liad

is it legal to say $d(\mu+v) = d\mu /f$ ? ^^

wait that's not giving something useful.

Daniel Fischer

More conventional to write that $\frac{1}{f}\,d(\mu +v)$.

That's of course not valid everywhere, but on $B$ it's fine.

Liad

but it make no sense to me. i see the equation $d\mu = fd(\mu+v)$ as :

$\mu(M) = \int_M f d(\mu+v)$

so how can we write $d(\mu+v) = 1/f d\mu$ ? what does it mean?

the same? $(\mu+v)(E) = \int_E 1/f d\mu$ ?

Daniel Fischer

It means $$(\mu + v)(M) = \int_M \frac{1}{f}\,d\mu$$

Liad

ok that fixes things to me

wasn't sure about it

so $v(E) = 0$

implies $\int_E g/f d\mu = 0$

does this implies $\mu(E) $ = 0?

Daniel Fischer

19:14

$v(E) = 0$ implies $\int_E \frac{1}{g}\,dv = 0$.

Liad

you sure its $1/g$ ?

Daniel Fischer

We could take any function. In view of what we want, $f/g$ is the more direct route, but $1/g$ is great too.

Liad

$0 = v(E) = \int_E g d(\mu+v) = \int_E g/f d\mu$

Daniel Fischer

Yeah, that works too. We want $\mu(E) = 0$. Writing $\mu(E) = \int_E w\,dv$ yields that immediately. Writing it as $\int_E g/f\,d\mu$, we need to say that $g/f$ is strictly positive a.e. on $E$, hence $E$ must be a $\mu$-null set.

Liad

it is possitive because g and f both are positive

Daniel Fischer

19:21

Yes.

Liad

im not sure why there is something that bugs me with this.

we have $\int_X f d\mu = 0$ iff $f=0 $ a.e

Daniel Fischer

Using that $f \geqslant 0$. Yes. What bugs you?

Liad

we got $\int_E g/f d\mu = 0$

so $g/f = 0$ a.e , and this is not true

Daniel Fischer

It is, because $E$ is a null set.

Liad

so its a claim , if $f\gt 0$ on $E$ then $\int_E f = 0$ iff $\mu(E) = 0$ right?

Daniel Fischer

19:29

You need to write $\int_E f\,d\mu$ there. Yes, the integral of a strictly positive function is zero if and only if the domain of integration is a null set.

Liad

thank you very much

is it alright if i stay in this room and if i get stuck on something to consult with you?

you really helped me with that question ..

Liad

20:38

@DanielFischer im trying to prove the claim you wrote. if $\int_E f d\mu =0$ , and $f>0$ .is there any nice way to show $\mu(E)= 0$? i cant say $f>c$ for some $c >0$..

Daniel Fischer

Consider $\int_E n\cdot f\,d\mu$ for $n \in \mathbb{N}$.

Liad

hmm

Daniel Fischer

MCT

Liad

its going to infinity O_o

so $\int_E \infty d\mu =0$

i got what you meant?

Daniel Fischer

Yep

Liad

20:49

i must find a question that it will take more than a second to solve ^^

i think i found it

Riemann hypothesis

In mathematics, the Riemann hypothesis is a conjecture that the Riemann zeta function has its zeros only at the negative even integers and complex numbers with real part 1/2. It was proposed by Bernhard Riemann (1859), after whom it is named. The name is also used for some closely related analogues, such as the Riemann hypothesis for curves over finite fields. The Riemann hypothesis implies results about the distribution of prime numbers. Along with suitable generalizations, some mathematicians consider it the most important unresolved problem in pure mathematics (Bombieri 2000). The Rieman...

can you point me in the right direction? ^^

Daniel Fischer

Look at the growth of the Mertens function.

Liad

lol. no idea what that is

im going to sleep now, thanks a lot for the help today.
can i stay in this chat and maybe ask you some questions tomorrow if i'll have some? i will try my best to minimize the amount of questions ^^

Daniel Fischer

Mertens function

In number theory, the Mertens function is defined for all positive integers n as M ( n ) = ∑ k = 1 n μ ( k ) {\displaystyle M(n)=\sum _{k=1}^{n}\mu (k)} where μ(k) is the Möbius function. The function is named in honour of Franz Mertens. This definition can be extended to positive real numbers as follows: M ( x ) = ...

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$Mathematics$ Room for Liad and Daniel Fischer