Mathematics - 2021-01-28 (page 1 of 4)

BigSocks

00:15

@robjohn for what? wealth kutch or watchet hulk?

user2103480

00:47

"Let $H$ be a separable real Hilbert space with inner product $\langle \, , \,\rangle_H$ and $H^\ast$ its dual. Let $V$ be a reflexive Banach space, such that $V \subset H$ continuously and densely. Then for its dual space $V^\ast$ it follows that $H^\ast \subset V^\ast$ continuously and densely."

The embedding $H^\ast \hookrightarrow V^\ast$ is given by the obvious restriction of $\langle h, \cdot \rangle$ to elements of $V$, and we proved that it's continuous, but I don't see how to prove density. Anybody have a clue?

Thorgott

01:10

what does continuously mean here

subspace inclusions are always continuous

user2103480

V has it's own inner product

it's a subspace set-theoretically but has a different norm etc

Thorgott

what a horrible set-up

Unknown x

7

Q: Confused about proof that diameter of a closure of a set is the same as the diameter of the set.

Definition Let $E$ be a nonempty subset of a metric space $X$, and let $S$ be the set of all real numbers of the form $d(p,q)$, with $p,q \in E$. The supremum of $S$ is called the diameter of $E$. Theorem If $\bar{E}$ is the closure of the set $E$ in a metric space $X$, then $ \text{diam} \ \bar...

user2103480

Rigged Hilbert space

In mathematics, a rigged Hilbert space (Gelfand triple, nested Hilbert space, equipped Hilbert space) is a construction designed to link the distribution and square-integrable aspects of functional analysis. Such spaces were introduced to study spectral theory in the broad sense. They bring together the 'bound state' (eigenvector) and 'continuous spectrum', in one place. == Motivation == A function such as the canonical homomorphism of the real line into the complex plane x ↦ e i x...

enjoy

Unknown x

How do I prove the infinite case? Is it enough to say if diam(Y)=$\infty$.

Thorgott

01:14

oh yeah, this is some horrible stuff physicists do, right?

V is only Banach, so Riesz rep won't help, meh

user2103480

@Thorgott PDE theorists in general

Unknown x

$Y\subset of \overline Y$

user2103480

@Thorgott I tried riesz anyways for the hilbert space and didnt manage

This gets even worse directly afterwards

Unknown x

diam(Y)=$\infty <diam(\overline Y)\implies diam(\overline Y)=\infty$

Is my argument true for infinite case?

Thorgott

yes, your argument works

but the argument in the question you linked also works for the infinite case

Unknown x

01:17

Thank you

user2103480

We somehow force operators such as the laplacian to be continuous operators into the dual, i.e. $\Delta \colon H^1(D) \rightarrow H^1(D)'$

and we define PDEs as ODEs with values in hilbert spaces with differential operators being "continuous"

and then we throw a stochastic integral over some operator-valued stochastic process onto that

I'm very much lost in layers of abstractions there

But somehow satisfied that this also happens in hard analysis lol

Thorgott

this set-up is ugly

the obvious diagram with $V,H,V^{\ast},H^{\ast}$ need not commute

only analysts would do this

user2103480

I think this should not commute

you mean that the inclusion of $V$ into $V^\ast$ is equal to $V \hookrightarrow H \simeq H^\ast \hookrightarrow V^\ast$?

Thorgott

yeah, it needn't be

but I want it to

user2103480

The right map is "designed" not to be an isomorphism

Thorgott

01:27

even the set-up on Wikipedia is nicer than this

user2103480

Since riesz would give an isomorphism in the hilbert space case

Thorgott

even they don't give the subspace a different inner product

user2103480

friendship with base case ended

hilbert space case is now my best friend

Thorgott

this is the worst thing since immersed submanifolds

user2103480

This is beautiful

Note that since H is a separable inclusion into $V^\ast$ this is a separable as well and so $V$ is separable

beautiful argument, very straightforward

looking forward to learning these concise and intuitive theorems and their proofs

... by heart, since it's an oral exam

remembering the sheer number of conditions should be worth an A

incidentally, that was basically how it worked with my last oral exam with the same professor, in stochastic filtering theory

Thorgott

01:35

@user2103480 mood

user2103480

the funny thing is that this stuff is still analytically easier than probability theory III which I skipped

Thorgott

tfw I haven't done any serious analysis in over a year

user2103480

comes back fast after some trouble

and least you don't forget as much analysis as I forget algebra in the span of mere months

Ted Shifrin

01:52

@Thor at least two remarks I want to mark with LOL

user2103480

@Thorgott uhh isn't the setup worse

they explicitly mention that $\Phi$ is a topological vector space with continuous inclusion map. Having just a different hilbert space is nice

Thorgott

well, you can restrict the inner product of $H$ to $\Phi$ at least

instead of putting a different inner product on there

user2103480

But... that's... not the point

that would be way too nice

Thorgott

yeah..

user2103480

with this setup you can consider consider continuous maps on a compact space as continuously embedded into, say, $L^2$

and if you want a hilbert space, you can take sobolev spaces included in $L^2$

Thorgott

02:05

but why

user2103480

Although the space would probably need some conditions for continuous maps being dense in $L^2$ so uh just take some nice subset of $\Bbb R^d$

@Thorgott it gives us some kinds of solutions to PDEs uhh I'm not deep into the phenomenology yet

Mike Miller

02:29

I'm not reading anything that came before this but I can answer PDE questions if you ask me again

user2103480

Do gelfand triples count as PDE to you?

Mike Miller

What's that

user2103480

Doesn't matter for the question I have, since it's about steps in the definition. Since you don't want to read anything above I quote myself

"Let $H$ be a separable real Hilbert space with inner product $\langle \, , \,\rangle_H$ and $H^\ast$ its dual. Let $V$ be a reflexive Banach space, such that $V \subset H$ continuously and densely. Then for its dual space $V^\ast$ it follows that $H^\ast \subset V^\ast$ continuously and densely."

The embedding $H^\ast \hookrightarrow V^\ast$ is given by the obvious restriction of $\langle h, \cdot \rangle$ to elements of $V$, and we proved that it's continuous, but I don't see how to prove density. Anybody have a clue?

(The gelfand triple is $(V, H, V^\ast)$)

Ted Shifrin

02:44

Doesn't look like PDE, but maybe Sobolev embedding fits somehow.

user2103480

Hm, not sure

It's a kind of approach to pde. The context of this definition is proving the following theorem in that setup

For $V = H^1((0,1))$ and $H = L^2((0,1))$ we can take $A$ as the laplacian by "extending" it to a map from $H^1((0,1))$ to its dual

going to sleep now, bye bye

Mike Miller

03:26

Sure this is a standardish setup I guess you're working with parabolic equations

@user2103480 Thank you for respecting my demands

@user2103480 If $H^*$ is not dense in $V^*$ then $V = V^{}$ is not dense in $H = H^{}$ by Hahn Banach, yes?

Contrapositive

polite proofs

Could someone have a look at some early Spivak proofs of mine, just to make sure I understand the style that he wants for these very 'basic' proofs?

Dr. Shifrin!

Ted Shifrin

03:56

Oh oh ...

Give a proof.

polite proofs

Okay, I'll go with (v) from problem 1. Proving that $x^n - y^n = (x - y)(x^{n-1} + x^{n-2}y \dots + xy^{n-1}).$

Let us start with $(x - y)(x^{n-1} + x^{n-2}y \dots + xy^{n-1})$. By (P9), we have that $x(x^{n-1} + x^{n-2}y \dots + xy^{n-1})+(-y)(x^{n-1} + x^{n-2}y \dots + xy^{n-1}) = (x^{n} + x^{n-1}y \dots + xy^{n-1}) - ((x^{n-1}y + x^{n-2}y^2 \dots + xy^{n})$. Then, by P3, we see that $(x-y)(x^{n-1} + x^{n-2}y \dots + xy^{n-1}) = x^n - y^n$

I think these early proofs may sometimes be more tricky than later proofs, since you have to be very careful

Ted Shifrin

Oh, i told you to skip the first few chapters.

polite proofs

You did?

Ted Shifrin

Yup, I did. The first few chapters have issues. To make sense of $\dots$ takes summation or induction, which come later. I trust you to go to later chapters.

You should read, of course, but you know this stuff.

There are a few interesting problems, but move on.

I can look up a few, maybe, that you should ponder.

polite proofs

Okay, I'll go to the end of chapter 3 to do the exercises?

Yes, please

Semiclassical

04:08

setting aside English issues, am I missing the point with what this person is saying? i feel like i'd just be repeating myself right now: math.stackexchange.com/questions/4002718/…

Ted Shifrin

OK, problems like Chapter 1: #6, 12(v), and using the triangle inequality (and being sure not to do backwards proofs) will be important. You can do 20, 21, 22, 23 if you want to know keys in Chapter 5. Chapter 1 #7 and Chapter 2 #22 are interesting if you've never done them. Chapter 2 #27, 28 are good puzzler problems, but not relevant to calculus.

Semiclassical

8 hours ago, by Ted Shifrin

I'm getting meaner and meaner to the people who post homework/exam questions with zero effort.

agree

polite proofs

Thanks, I'll do those then

Semiclassical

there's the meta post which says "don't just vote to close on people's first posts" but...hard to feel like they deserve more of a response than that when they post a problem verbatim

Ted Shifrin

@Semiclassic You agree that I'm meaner?

Semiclassical

04:10

hmm

agree that they deserve it

Ted Shifrin

Yup.

I don't vote to close, but I ask for effort. If I see none later, I may vote to close.

Semiclassical

that's probably better, yeah

polite proofs

Wait, can I use induction for Chap. 1., #6?

I saw that induction is introduced later, so that's why I am asking

Semiclassical

but, to quote from the above (cleaning up some of the English):

copper.hat

That is a reasonable approach Ted. A lot of folks like to shoot first.

Lots of wind & rain here atm.

Ted Shifrin

04:14

@polite That's a problem with the beginning of Spivak. It's hard to sort it out with infinite rigor, but I like the book regardless.

polite proofs

So is that a yes?

Ted Shifrin

Yes.

polite proofs

Thanks

Semiclassical

"Given an $n$-by-$n$ matrix $A$ of order n, its characteristic polynomial is of degree $n$ and so it will have $n$ roots (fundamental theorem of algebra). Let this root be $\lambda_1$ and its corresponding eigenvector be $e_1$. Now consider $K=(\text{Span}\{e_1\})^\perp$. Now again we have an $(n−1)$-degree characteristic polynomial. So again (by above) we have an eigenvalue and corresponding eigenvector..."

Ted Shifrin

Hmm

We need invariance ...

Semiclassical

04:16

The question I asked in response was "in what sense is the lower-degree polynomial the characteristic polynomial of a matrix?"

as in, characteristic polynomial of what matrix

it's not the characteristic polynomial of $K$, at least not naively. if you make a matrix whose column space is the orthogonal complement of $e_1$, then that'll be $n$-by-$(n-1)$ and so not have a characteristic polynomial

Ted Shifrin

You need that hyperplane to be invariant. E.g., a symmetric map.

Semiclassical

Well, this is in the context of Hermitian or real symmetric matrices. So I think that can be granted.

copper.hat

@politeproofs always a good idea to look for simplifications before starting: if $y=0$ the equality is immediate, otherwise you can divide across by $y^n$ and then the problem reduces to showing $t^n-1 = (t-1)(t^{n-1}+\cdots + 1)$ which follows from the geometric series formula.

Semiclassical

(maybe)

polite proofs

Let me do a fun proof: We wish to show that $\forall n \in \mathbb{Z}, \forall (x,y) \in \{ (x,y) \in \mathbb{R}^2 : 0 \le x < y \}, x^n < y^n.$

Suppose that, on the contrary, there is some positive integer $m$ for which this is not true. By the well-ordering principle, which we can take as an axiom, we know there is a least positive integer $m$ for which it is not true. Then, $x^{k} < y^{k}$ for $1 \le k < m$. Then $m = k + 1$, and so notice that $y^m = y^{k+1} = y^k y > x^k y > x^{k+1},$ which is a contradiction.

This is a proof for #6 without using induction

I hope it works, since I didn't think too hard about it

copper.hat

04:23

@politeproofs surely you can use the $x^n-y^n=...$ result to show it directly?

polite proofs

@copper.hat No?

copper.hat

why not?

polite proofs

Or at least I don't follow how you mean

Semiclassical

$x^n-y^n=(x-y)(x^{n-1}+...+y^{n-1})$?

123

Hello Guys..

polite proofs

04:25

@Semiclassical Did you typo?

Semiclassical

no, i lazied

fixed typo

copper.hat

if $x^n-y^n = (y-x)(\text{ stuff from above })$ then $x^n,y^n$ have the same order relation as $x,y$.

Ted Shifrin

That was Spivak’s intention, yes.

polite proofs

What about my proof?

It avoids induction, which is nice

Ted Shifrin

No it doesn't.

polite proofs

04:26

:(

copper.hat

you mean i have to think :-)

Ted Shifrin

Well ordering is equivalent.

polite proofs

Induction uses well ordering, but Induction is a theorem that we have to prove

Ted Shifrin

They are logically equivalent.

Semiclassical

if it's equivalent, then it hardly matters that you used one as a definition and not the other

polite proofs

04:27

Dr. Shifrin, you are so hard to impress!

copper.hat

proof is fine, but induction in disguise.

Ted Shifrin

Yup. Just ask Semiclassic.

copper.hat

transfinite induction will never be natural for me.

Semiclassical

(Hardly matters in terms of the logical substance, i mean. can matter in how quick it is to write the proof)

copper.hat

the best proofs (imo) are those that make the result 'obvious'.

that way i can remember them

Semiclassical

04:29

it's more restricted but i like visual 'proofs' when they're available

copper.hat

i have a preference for visual too.

polite proofs

I must be the only person in the world that dislikes visual proofs. :)

copper.hat

proof by picture was what my advisor called it

Semiclassical

like the visual demonstration that odd integer sequences sum to squares

polite proofs

Only because I have a bad history of them coming back to haunt me later

copper.hat

04:30

i think rudin dislikes visual.

polite proofs

I heard there was not a single visual in Hoffman & Kunze

Semiclassical

i mean, they're definitely very limited in what they can accomplish

polite proofs

And I don't mean proofs

Not a single picture, period.

Which, for linear algebra, is an interesting approach to say the least.

copper.hat

rockafellar is another lad who eschews pictures

which is odd, because convex is just a step away from geometry

Ted Shifrin

Rudin has no single picture. Even my algebra book has hundreds of figures.

polite proofs

04:33

I like pictures, sometimes. i.e. illustrations of $\sup$ and $\inf$ were helpful when I first learnt about them

Ted Shifrin

Even if you're not a visual learner, more people are and you should learn to draw and understand them.

copper.hat

well, sometimes a picture gives a fast way of understanding what a paragraph of words takes to elucidate.

Semiclassical

it's just another part of the tool kit

Ted Shifrin

Pictures alone do not a proof make, but usually they give informative intuition. Not alwsys.

copper.hat

apparently pontryagin developed his maximum principle when blind.

i can hardly follow it with sight.

right tool for the right job.

Semiclassical

04:35

Euler was basically blind for the last 17 years of his life and he was still productive but

1) he had scribes

and 2) he was Euler

Ted Shifrin

The proof for turning the sphere inside out (with all the models in 1010 Evans) was figured out by Morin, a blind mathematician.

copper.hat

amazing really.

i like those models.

Ted Shifrin

They were stolen.

Semiclassical

oof

Ted Shifrin

I guess Smale proved it. Blind Morin visualized it,

copper.hat

04:37

i found smale incomprehensible.

i mean his lectures.

Ted Shifrin

He's a great character, super smart. Not a great teacher.

Balarka Sen

It's quite interesting that the visual idea was already in Smale's proof.

copper.hat

i stumbled on a lot of such folks.

Ted Shifrin

I actually forget, a Balarka.

copper.hat

i am never quite sure of the accepted protocols in a chat room. do people say hi or is that considered noisy?

not that accepted protocol ever limited me...

dc3rd

04:40

Hi @copper.hat :)

Semiclassical

@TedShifrin in truth, i wonder how much being blind is an obstacle to visualizing things which can't actually be visualized in everyday 3D space

Balarka Sen

If you were to read Smale's proof from a geometric point of view, he wants to break S^2 as upper + lower hemisphere, and turn the hemispheres inside out while keeping something fixed along a neighborhood of the equator.

copper.hat

@dc3rd hi :-)

Balarka Sen

This requires making a neighborhood of the equator flexible by twisting it around a lot

This was the main geometric idea that was exploited later by sphere eversion visualizers

copper.hat

must have incredible visio-spatial development

Ted Shifrin

04:42

TBH, I never grokked it.

@Semiclassic But talent to communicate the pictures to sighted persons.

Semiclassical

@TedShifrin yeah

Balarka Sen

The Morin-Francis pictures are excellent. It's surprisingly complicated to see everything that is going on at once.

Ted Shifrin

Charles Pugh built the (now stolen) models.

Balarka Sen

Here and here.

Ted Shifrin

I'm so not a topologiet.

copper.hat

04:46

he was my quals advisor :-)

Balarka Sen

@TedShifrin Using chicken fence, right?

Ted Shifrin

Chicken wire, yeah. There is a movie somewhere.

Balarka Sen

Yup, I have seen it

copper.hat

pugh wrote a proof of the rademacher theorem for me, i have it somewhere :-)

Ted Shifrin

Pugh would have been my adviser if Chern hadn't gotten stuck.

copper.hat

04:48

really?

Ted Shifrin

Oh, I wrote notes for his course with that. But long gone.

copper.hat

i think pugh would have been more fun :-)

Ted Shifrin

Yeah, I loved his dynamical systems course my first year. We became good food/wine buddies.

copper.hat

i liked him because he was very down to earth.

he advised some of my irish friends as well.

Ted Shifrin

Chern was a sweet and generous man. His wife liked me because I cooked gourmet food for them :)

copper.hat

04:50

:-)

polite proofs

Is this the very short proof of the reverse $\triangle$ inequality? $|x - y| + |y| \ge |x - y + y|$ by $\triangle$ inequality, then that $= |x| \Rightarrow |x - y| + |y| \ge |x| \iff |x-y| \ge |x| - |y|$?

Ted Shifrin

Yup. Good!

copper.hat

which shows that the absolute value is Lipschitz with rank 1.

polite proofs

(Only took.... 20 minutes!!)

Balarka Sen

Lipschitzness of the distance function is one of my favorite facts in mathematics.

copper.hat

04:53

you need a solid foundation. a lot of people (even rather famous folks) do not have solid foundations.

Ted Shifrin

All over my books I have sequential problems trying to train people to use what has been established and not to go back and start over. @Balarka will so testify.

polite proofs

Was that directed at me?

copper.hat

I like that Lipschitz functions are differentiable ae.

@politeproofs yes :-).

Balarka Sen

It's a cool fact but I feel that I rarely use it

polite proofs

Oh trust me I know. That's why I am so obsessive when I don't understand a single sentence that I read in a book

It might even be a side comment, but as soon as I don't understand, I stress over it :(

copper.hat

04:55

@politeproofs i understand, but don't let that force you to learn sequentially.

Ted Shifrin

You overdo that, polite. I realize you can't help it, but work on it.

copper.hat

trust yourself, its ok to have a little forward debt :-)

dc3rd

It's the aspiration to perfection. I think all us budding mathematicians have it when looking up to people like the senior folks in this room........ :)

copper.hat

unfortunately i am neither budding nor a mathematician...

dc3rd

you're too humble Copper.....from the help you've provided me in the past you are quite the mathematician.....you've graduated beyond budding.

you're in the arena

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$Mathematics$ Mathematics