@user193319: If the limit exists, lim inf, lim, and lim sup are all equal.

But in general, $\liminf x_n \le \limsup x_n$, and equality holds iff the limit exists.

Hi chat

Hi Eric ... You dun recovered from your quarter?

Partially

Problem: Let $\{f_n\}$ be a sequence of measurable functions that converge pointwise to $f$ on $E$. Let $M \ge 0$ be such that $\int_E f_n \le M$ for all $n$. Show that $\int_E f \le M$. Proof: Since $\int_E f_n \le M$ for all $n$, it must be that $\lim_{n \to \infty} \int_E f_n \le M$. But Fatou's lemma says $\int_E f \le \lim \inf \int_E f_n$. Putting these inequalities together, $\int_E f \le \lim \inf \int_E f_n \le \lim_{n \to \infty} f_n \le M$.

Well, you have a few more days for the rest of the recovery, Eric

I did p good so I'm kind of finally starting to actually calm down now that grades are out

@TedShifrin So the argument I just posted is not valid because $\lim_{n \to \infty} \int_E f_n$ might not exist?

You really need to learn not to stress over those, Eric.

If you have the money...

It's hard not to I think

@user193319: You have not justified the claim that $\lim \int f_n$ exists!.

Depends on the person, Eric. I never did.

@TedShifrin I know. That's what I asked. But is there any guarantee that it will exist?

It's kind of hard not to compulsively check when it's all up on the internet

Oh, but you don't need it. All you need from Fatou is that $\liminf \int f_n \le M$.

Well, I grew up in the dark ages, Eric, where we waited for grades in the mail — unless profs posted them on the wall.

Morning @TedShifrin

heya Faust

I got called for jury duty the other day and that was another fountain of stress

@TedShifrin think your alittle young for the dark ages.

Since they kind of gave me the run around despite not being able to return to Florida for a long time

@EricSilva at least your not in the hospital

wait, Eric, where are you registered?

Back in Florida

oh

anyone know the proof behind extreme value theorem?

Yes.

But no way I could afford a plane ticket back there while paying my rent and dealing with school

Good argument to register where you're in school.

@Ted didn't want to cause of the election

Certainly you want to be a resident for grad school.

But I'm mot confident I can type it out with this much morphine.

just Send it back by mail saying your not in the country

Is it accurate to say that the maximum value is defined as: $\exists d \in [a,b] : \forall x \in [a,b], ~f(x) \leq f(d)$ ?

You thought you'd make more difference in FL than in IL, Eric?

I mean Florida matters more so yes

@user525966: $f(d)$ would be the max value, yes.

But then when I negate it it almost seems to say the same thing

Lol liberals at work

$\forall d \in [a,b] : \exists x \in [a,b], ~f(x) > f(d)$

That's correct, @user525966.

if i switch the for-all and the there-exists part there it looks almost identical to the earlier piece

$\exists x \in [a,b] : \forall d \in [a,b] , ~f(d) < f(x)$

That's how negation works, @user525966. Quantifiers switch and things get negated.

All I did there now was change order in my negation and it looks almost just like the original

@Zee conservative voters also have more impact in Florida so I don't understand the content of your comment

You screwed up an inequality there, @user525966, one of the times.

10/10

hi Balarka

The liberal part was the fact that academics are liberal

Hi @Ted

I just wanted to drop that meme and scurry off

Which I shall do now

Meme drive by

For the most part that's explained by the fact that conservatives tend to be quite ignorant and stupid in this day and age, Zee. Perhaps I'll put you on ignore.

Come on , I didn’t insult anybody

I don't think that is precisely the reason academics are skewed liberal

@user525966: Your second thing is just nonsense.

People who need other people to survive lean liberal , academics need society to support them so they value social support and that’s the essence of being a liberal

fixed?

@user525966: You can't randomly rearrange phrases and quantifiers.

No. Wrong.

which line is wrong?

2nd or 3rd or both?

The last line. The first negation is correct.

there is a difference whether the for-all or exists comes first?

Absolutely.

Did you not take a basic intro to higher math course?

@user525966 a big one

hi demonic @Alessandro :)

@TedShifrin can you help me with some elliptic integration problem?

Was too poor, self-studying what I missed

@Zee this is definitely reductive

Aha.

Depends on the problem, @Cows.

Can someone help me with an elliptic integration problem?

@TedShifrin perfect

@AlessandroCodenotti What's the difference?

I don't mean to be obtuse I just don't see the difference

@TedShifrin Let me type it out

how So @eric

Try the sentence for every $d$ there is an $x>d$, @user525966.

Is that the same as saying there is an $x$ so that for all $d$ we have $x>d$ ??

Oh hell no.

Hi @Ted

@TedShifrin hehehehe

@user525966, in the first case, we can pick a new x depending on the d each time. In the second case, we need an x that works for all d.

@TedShifrin can you point me to some resources to help me compute this?

Is there a logical/Boolean equivalent to that relationship?

I see the difference now but is there a formalization?

@Cows. No. But didn't you literally have a mess over the square root of the precise same mess?

it almost seems like a bunch of a implies b statements

@user525966: You need to do some thinking and not have it be formal.

@TedShifrin yes, it was horrible looking stuff

Ya lots of math is informal

@user525966 $\forall x\exists y(y<x)$ is true, $\exists x\forall y(y<x)$ is false, do you see why?

@AlessandroCodenotti take $x=\infty$ for the second.

But $u/\sqrt u = \sqrt u$, @Cows.

@TedShifrin yeah I can't even find good easy to grasp resources about this stuff

yes

0celo, you're not helping.

@0celo7 I'm working in $\Bbb R$ <.<

boo :)

@0celo7 please help me with this

help with what

@0celo7 I want to solve an elliptic integral. The root of a degree 4 polynomial.

I don't know what that means

@0celo7 let me type an example I am working on

I know what a parametric elliptic integrand is.

@0celo7 $$ \int dt {(2t^4 - 6 t^3 + 3 t^2 - 12 t -3)^\frac{1}{2}} $$

Why would anyone be able to do that integral

Okay so if the correct negation is $\lnot P = \forall d \in [a,b], \exists x \in [a,b]: ~f(x) > f(d)$ is the general idea to show that this contradicts continuity of $f$ over $[a,b]$?

@0celo7 lol

why would you want to

@0celo7 it's a long story, but I need to.

@user525966 what are you trying to do?

Prove the extreme value theorem

Mathematica will give it to you in terms of standard elliptic functions.

@0celo7 here is what I have done so far

@user525966: You need some real mathematics. Compactness or the least upper bound property.

@user525966 Do you know that a bounded sequence has a limit?

And that $[a,b]$ is closed?

That's not valid @0celo.

Bounded and strictly increasing

Subsequence, of course.

I would need to know the formal terms to know what these things technically mean I suppose

Or just increasing actually

well, correct statements help someone who's just starting out.

@0celo7 I have rewritten it in rational form, and I was hoping to factor the denominator, but it turns out I have to factor a quartic. I tried doing synthetic division, but I suspect the roots are complex. The goal was to see if I could do partial fractions and express as a sum of $\int$

@Cows I see no reason to do that integral. Why are you trying to do it?

To me, extreme value theorem is "obvious" and yet so much in analysis is about proving things we take for granted so I don't want to get too informal and just assert something as true without proof

@TedShifrin You're right.

@user525966 Well what do you know?

@0celo7 well it resulted from a very trivial math thing I was trying to compute, for a project

I figure that we need to show $\lnot P = \forall d \in [a,b], \exists x \in [a,b]: ~f(x) > f(d)$ is false

@user525966 I don't think this is provable by pure logic.

no but I mean it guides the next steps

It's not an easy theorem compared to how early one meets it. It usually takes more to prove it than what the students know the first time they see it.

this is saying, correct me if I'm wrong, that if there is no maximum, then for every single f(d) in our interval (there are infinitely many of them), there is an even bigger value of f(x) somewhere which must also come from our interval technically. This alone feels like a contradiction to me but I'm not sure if this is enough to "show it"

Take a calculus of variations approach. Consider a sequence $(x_n)\subset [a,b]$ such that $f(x_n)\to \sup f$. Get a convergent subsequence. Then use continuity.