Mathematics - 2018-05-19 (page 1 of 2)

Ted Shifrin

12:00 AM

But the multiplication rule doesn't affect our argument, does it?

Jon

@feynhat sorry to sound redundant but then is $L(x)$ V and $x$ is W right? In the graph I provided earlier.

Sha Vuklia

I think it doesn't, because we have $\{(n,e):n\in N\}$ and $\{(e,h):h\in G\}$, which boils down to the regular product

but in a sense where they are disjoint (plus e)

it's as if we've made them disjoint, is that what you were trying to say before, Ted?

Leaky Nun

@ShaVuklia "they have trivial intersection" are the words you're looking for

Sha Vuklia

I was just being lazy actually:p

GettingNifty

https://www.youtube.com/watch?v=oBRXMD2m8MU

This is stuff I posted the day it came out and this is 20th time my posts have been on the radio.

Sha Vuklia

12:03 AM

but thanks

GettingNifty

and I render adventadors

brake lights son

ÍgjøgnumMeg

At the risk of invoking the wrath of @Ted again, what was it about the equation $(n^\prime)^{-1}n = h^\prime h^{-1}$ that meant that $(n^\prime)^{-1}n = h^\prime h^{-1} = e$?

Ted Shifrin

Nah, @ÍgjøgnumMeg, you're right that I didn't complain enough.

ÍgjøgnumMeg

aw

kms

Leaky Nun

@ShaVuklia (h,e) isn't even a thing

Sha Vuklia

12:10 AM

I'll start over

oh oops, I realise we actually get $(n,e)(e,h)=(n,h)$. right, then it follows trivially that there is a unique decomposition

Leaky Nun

@ShaVuklia lol

Ted Shifrin

Once you write ordered pairs, @Sha, it's like the product of sets, as I said earlier.

anakhronizein

HEY TED.

Sorry about the other night!

Sha Vuklia

@TedShifrin yes but, in the case of a product of sets, we need disjoint subgroups right, for the property to hold that I was initially asking about

so we can't really use that as an argument (unless you were not even trying to use that as an argument, but were just pointing it out)

Ted Shifrin

Disjoint, again, meaning trivial intersection. But you have that as well for the semi-direct case. The only thing that changes is the rule for $(nh)(n'h')$.

Sha Vuklia

12:16 AM

oh yea, we have a trivial intersection indeed

and $(nh)(n'h')$ isn't relevant to the argument, only $(n,e)(e,h)$, which behaves like the direct product

Ted Shifrin

There you go :)

Sha Vuklia

btw Ted

I'm thinking of splitting my third year in two years

because I can't keep up with all the math electives in one year, and I was also thinking of joining some social stuff at my uni, and doing a minor in computer science. kind of to broaden my horizon I guess

Ted Shifrin

Slowing down isn't a bad thing, if you can afford the time and money. (I guess things aren't that bad in the EU.)

Sha Vuklia

yea we are very lucky

Sha Vuklia

12:39 AM

guys, I hope I don't come off as too lazy, but I have another question; my book says that $hnh^{-1}=\tau(h)(n)$ for each $h\in H,n\in N$. here $\tau\colon H\to\operatorname{Aut}(N)$ is the automorphism that defines the semi-direct product. It seems to me then that $\tau$ is then automatically an inner automorphism. but err, any hints on why this should be the case?

Ted Shifrin

It's inner if $N$ and $H$ live inside some group to start with. But abstractly, it's not, if you have an action of $H$ on $N$ by that rule.

Oh, but it won't make sense literally unless that $hnh^{-1}$ is computed in some larger group.

Sha Vuklia

hm, but inner just means $\phi_a(x)=axa^{-1}$ right?, where $x\in G$, for some group $G$ (so no larger groups involved per se?)

Ted Shifrin

That definition is for $a,x\in G$.

Sha Vuklia

oh

Ted Shifrin

But if all you have is $N$, what sense does $hnh^{-1}$ make?

Sha Vuklia

12:43 AM

all we can work with then is the semi-direct product

Ted Shifrin

Huh? I'm saying that you need $H$ and $N$ sitting inside some $G$ where that conjugation makes sense.

Sha Vuklia

oh like that

right, so that is assumed, is what you're saying

Ted Shifrin

So this is why I was trying to get you to think that way an hour ago :P

Sha Vuklia

right, well that went over my head:p

Ted Shifrin

I thought you were taller than I am.

Sha Vuklia

12:45 AM

lol:p

but just to be sure: so they actually assume there is a larger group. I almost can't believe it. why wouldn't my book mention that

but even if I assume that

Ted Shifrin

They probably did.

Sha Vuklia

I feel like I should still show that it holds?

Ted Shifrin

Show what holds?

Sha Vuklia

that we have $hnh^{-1}=\tau(h)(n)$

Ted Shifrin

That's the definition.

You should write the equation reversed.

Sha Vuklia

12:48 AM

they just said that $\tau$ maps to an automorphism, not an inner automorphism, and which one

it's in Dutch, so I can only translate here

ohh

no wait

maybe I've seen something before about $\operatorname{Inn}(G)$

Ted Shifrin

An inner automorphism of $N$ would be conjugation by an element of $N$, not by an element outside of $N$.

It doesn't. Unless you mean inner automorphism of the large group (that's what I said before). It won't be an inner automorphism of $N$ itself.

Sha Vuklia

right, I didn't realise that the image of $\tau$ is $\operatorname{Inn}(N)$ (assuming the larger group thing)

Ted Shifrin

It's still not Inn($N$).

Right. It would be an inner automorphism of $G$, but only considering the action on $N$.

loch

yes you have your usual real points and pairs of complex-conjugate points (although i don't think people usually identify them with the upper half plane though..) - and in general over an arbitrary field $k$ your closed points would consist of galois orbits of points in $\bar{k}$ under $Gal(\bar{k}/k)$

but yes that's why things being algebraically closed makes life much easier !

Sha Vuklia

well.. I'm afraid I don't see it:( but I also don't want to hold you up too much!

Ted Shifrin

12:55 AM

You need to do concrete examples, @Sha.

Leaky Nun

@loch nice

Sha Vuklia

alright, I'll read through an example that's given right after this in my book

Ted Shifrin

And work out details!

loch

1:06 AM

@LeakyNun if you want to see a lot of examples on how to visualise schemes then the book 'geometry of schemes' is pretty good (although i've never gone through much of it)

Leaky Nun

thanks for your recommendation

are the circle and the ellipse isomorphic as affine varieties?

loch

yes

mercio

o. .o

Leaky Nun

why?

@mercio que?

eisenbud :o

the modular guy

loch

well a circle is cut out by $x^2+y^2 = 1$ right, and an ellipse is cut out by $(x/a)^2 + (y/b)^2 = 1$ (up to scaling and translation i guess) , so you get from one to the other by something like $x\mapsto x/a, y\mapsto y/b$

Leaky Nun

1:11 AM

oh ok

what is the isomorphism class of the circle?

mercio

isomorphism of what ?

Leaky Nun

isomorphism class, as in everything that is isomorphic to it

loch

if we are thinking of them as projective varieties (take its projective closure), then they are all isomorphic to $\mathbb{P}^1$

Leaky Nun

how can an affine variety be isomorphic to a projective space?

loch

oh so i took the projective closure (because this is something i know off the top of my head haha i have to think a little bit if im just looking at the affine variety - although it shouldn't be hard)

i.e. i was looking at $X^2+Y^2=Z^2$ in $\mathbb{P}^2$
(but if you're not interested in this then it's fine - although imo this is pretty cool)

Leaky Nun

1:18 AM

I see

mercio

what is an affine variety

Leaky Nun

@mercio a closed subset of an affine space

under the zariski topology

Balarka Sen

There's lots of real affine varieties biregular to the circle. Just take $x^2 + y^2 = 1$ and consider a polynomial map $f : \Bbb R^2 \to \Bbb R^2$ such that $f$ has polynomial inverse. Then take $f_1(x, y)^2 + f_2(x, y)^2 = 1$. That's biregular to the circle.

$f = (f_1, f_2)$

mercio

and what is an isomorphism of affine varieties

Balarka Sen

Biregular morphisms

Leaky Nun

1:22 AM

@BalarkaSen but not many polynomial maps have polynomial inverse?

Balarka Sen

@LeakyNun In the multivariable world, they do, no?

mercio

check out the Cremona group

Balarka Sen

The essential condition is that the Jacobian of $f$ is a (nonzero) constant polynomial.

mercio

oh ?

Balarka Sen

In fact the corollary is exactly what the Jacobian conjecture states.

If $f : \Bbb A_k^n \to \Bbb A_k^n$ is a regular function (so polynomial on each component), and $Df$ is a nonzero constant, $f$ is a biregular function (so has an inverse which is a regular function)

$k$ is an arbitrary field of char 0

That's the big conjecture

loch

1:27 AM

TIL

anyway in any case over $\mathbb{C}$, the variety $x^2+y^2=1$ is isomorphic to $\mathbb{C}$ minus a point i think

Leaky Nun

Rudin managed to define the winding number without using "argument" :o

I'm thoroughly impressed

Balarka Sen

What's bad about argument

(except if it's argoment, in which case it's invalid)

Leaky Nun

well Rudin himself said that this is the advantage of his definition

> One virtue of the preceding proof is that it establishes the main
properties of the index without any reference to the (multiple-valued) argument
of a complex number.

"index" = "winding number"

Balarka Sen

How does he do it? Using $\gamma'(t)/\gamma(t) dt$? That's just $d \log \gamma(t)$ written without having to worry about complex logarithms.

Leaky Nun

sure, I know that, but still

loch

1:31 AM

One way to see it is, knowing that $\mathbb{P}^1$ is isomorphic to the projective closure, then removing a point of the projective closure is isomorphic to $\mathbb{A}^1$. The projective closure of $x^2+y^2=1$ is given by $X^2+Y^2=Z^2$ in $\mathbb{P}^2_k$, and where you see that you have points at infinity given by $[1,i,0]$ and $[-1,i,0]$ (this corresponds to the fact that this is a degree $2$ hypersurface, which intersects a line at two points).

So removing two point in the isomorphism shows that the affine guy $x^2+y^2=1$ is isomorphic to $\mathbb{A}^1_k$ minus a point.

mercio

yup

@loch can you guess, if you take a real ellipse and look at its tangents of slope $i$, where are the real points where they intersect ?

loch

no

maybe i can if i think about it

Leaky Nun

real ellipse doesn't have any tangent of slope i

Balarka Sen

Yeah you can write $(x + iy)(x - iy) = 1$, change coordinates to make it $zw = 1$. That's just the hyperbola, which projects to $\Bbb A^1_{\Bbb C} - \{(0, 0)\}$

Leaky Nun

@BalarkaSen you win

loch

1:34 AM

yeah it shouldve been obvious!

Leaky Nun

@loch hmm

Balarka Sen

@loch Haha it wasn't to me until you wrote down the algebraic map to $k[t, 1/t]$

mercio

I meant tangents of slope $i$ and $-i$

(if you only pick tangents of slope $i$ they only intersect at infinity, which is boring, and at a complex point at that)

Balarka Sen

you and your complex slopes man

mercio

:s

Balarka Sen

1:38 AM

slipper slopes should be simple to slide along, not complex!

(I'm not a rapper)

mercio

if you take a point inside a circle you can draw the two complex tangents to the circle going through it, then draw the line going through those two points and it wil be a real line

and then if you take a point on that line on the axis of symmetry of the whole figure, and draw the two tangents going through there, and then draw the line between the two tangents points, it will again be a real line, and it will go through the original point inside the circle

Faust

Morning

loch

anyway i actually don't happen to know a lot about classifying e.g. affine curves (off the top of my head) because people usually just jump to projective curves - probably because to any affine curve you can take its projective closure..

for projective curves though it turns out that genus 0 curves are all isomorphic to $\mathbb{P}^1$ (at least over an alg. closed field, need to check vakil if this is true for all fields..) - for plane curves you have the degree genus formula $g=\frac{1}{2}(d-1)(d-2)$, so this tells you that any conic is isomorphic to $\mathbb{P}^1$

on the other hand if you care about isomorphism classes then you're getting into this whole moduli space of curves thing which is also very interesting

Leaky Nun

Let $\Omega \subseteq \Bbb C$ be an open set, and $\Delta \subseteq \Omega$ be a 2-simplex, $p \in \Omega \setminus \Delta$ be a point, $f:\Omega \to \Bbb C$ be continuous in $\Omega$ and differentiable in $\Omega \setminus \{p\}$. Then, $\displaystyle \oint_{\partial \Delta} f(z) = 0$.

Rudin's proof: bloody barycentric subdivision

@BalarkaSen

Balarka Sen

Indeed.

Leaky Nun

1:44 AM

I'm thoroughly impressed

Balarka Sen

Goursat's theorem is proved using iterative subdivision

Leaky Nun

I wish I had read this book a year ago

this so dank

Balarka Sen

If you assume $f$ is $C^1$ this technicality can be dispensed with in the process of proving Cauchy's theorem

Just Gauss-Green theorem kicks in

Or shall I say... Newton-Leibniz-Gauss-Green-Ostrogradsky-Stokes' theorem

Leaky Nun

don't you need Cauchy's theorem to prove that differnetiable implies C1?

Balarka Sen

That's why I said if you assume $f$ is $C^1$, this can be dispensed with.

Not a-priori

Leaky Nun

1:47 AM

proof that the metric topology on $\Bbb C$ is the product topology on $\Bbb R^2$

Balarka Sen

LOL

mercio

lol

that's nifty

Balarka Sen

Ideally there should be a bigger square encapsulating the whole thing

Well, rectangle, whatever

Leaky Nun

that's the boundary of picture :P

Balarka Sen

lmao

Fair!

@loch That's very weird. What's the canonical embedding here? The one coming from Kodaira embedding?

Leaky Nun

1:52 AM

@loch right, $\Bbb R$ is all BS, let's just do everything in $\Bbb C$

and forget about the metric topology, just put the zariski topology there

Balarka Sen

Holy shit

Gehenna - First Spell (Full Album)

►

Antonios-Alexandros Robotis

whats dat @BalarkaSen

Balarka Sen

It's black metal with lots of groovy organs and keyboards

Growling vocals over folk-sounding tunes which could have come from Ulver, "Bergtatt"

Pretty damn cool

Antonios-Alexandros Robotis

oh that is pretty cool lol

Leaky Nun

18 mins ago, by Leaky Nun

Let $\Omega \subseteq \Bbb C$ be an open set, and $\Delta \subseteq \Omega$ be a 2-simplex, $p \in \Omega \setminus \Delta$ be a point, $f:\Omega \to \Bbb C$ be continuous in $\Omega$ and differentiable in $\Omega \setminus \{p\}$. Then, $\displaystyle \oint_{\partial \Delta} f(z) = 0$.

@BalarkaSen now let $p \in \partial \partial \Delta$. Then, cut your bloody triangle to three pieces, two of which has zero integral, and the other one of which approaches $p$

then bloody finish it by ML since $f$ is continuous

Balarka Sen

2:03 AM

That's how the proof proceeds, yeah

Leaky Nun

and then let $p \in \Delta^o$. trisect your triangle wrt $p$, using the result for $\partial \partial \Delta$

loch

@BalarkaSen hm i'm not familair with the kodaira embedding - but here when i say canonical embedding i mean im taking the induced map to projective space given by its canonical divisor (equivalently the map to projective space induced by your canonical sheaf = cotangent sheaf = cotangent bundle (which is a line bundle) on your curve)

Balarka Sen

I think that's exactly how one Kodaira embeds. If your variety $X$ admits a certain line bundle $L/X$ choose a basis of sections $s_0, \cdots, s_n$ of this and define $f : X \to \Bbb P^n$ by $f(x) = (s_0(x) : s_1(x) : \cdots : s_n(x))$.

I suppose that's what you mean by "induced", with $L$ being the cotangent bundle in this case?

I'mma knock off to bed now

loch

yeah

Balarka Sen

Keep the memes up

Leaky Nun

2:15 AM

So $\displaystyle f(z) = \frac 1 {2 \pi i} \oint_\gamma \frac {f(\xi) - f(z)} {\xi - z} \mathrm d\xi$ as long as $z \notin \gamma$?

Faust

5:31 AM

It's too quiet

Daminark

shouts

Anyway hey @Faust, @Alessandro, and @Eric!

Lol for some reason this chat sorta shows me who's online 5 minutes ago and then doesn't update when people leave

Faust

O.o

most people just kinda idle off

they dont exactly "leave"

dont microwave pie

skull

did you guys hear about 0celo7?

Faust

no

he ok>

Daminark

Banned for a year apparently

Faust

5:43 AM

wtf

what did he do

Daminark

I'll let skull take the floor on that since I've got no clue

Faust

i can't imagine what he coulda done to manage a year ban

Daminark

Also @Faust did microwaving pie not go well for you?

Faust

hey mike

@Daminark freacking terrible turned it into mushy baby food

Daminark

Rip

Faust

5:44 AM

i ate it anyway but was gross

Daminark

Hey @Mike

Faust

was too lazy to try and microwave other food

hmm seems mikes not who i thought he was

skull

here

Faust

seems pretty harsh

he can be a bit of a jackass but i still like him

anyone seen that depressed suicidal guy that was off his meds lately?

i been really sick so havent been on much just want to make sure hes ok

skull

i haven't seen him for awhile

Faust

5:51 AM

hmm ok

skull

(months?)

Faust

yeah me 2

hopefully he just deleted his account again and will come back ok at some point

skull

yup, he's been doing that for years

i talked him into going on meds

(if we're talking about the same guy)

Faust

yeah i heard he was going to try going back on meds last time i talked to him

skull

didn't you mention that you're taking Japanese this year?

Faust

5:56 AM

yeah im taking an intensive course right now

2.5 hrs a day 5 days a week

and a joke math class

skull

why that language?

Faust

mm i wanted to learn german or japanese cause im intrested in both and need a language credit

but german wasnt offered

japanese was :P

skull

What about French or Russian?

Faust

i hate french i know a bit of russain

russains not a pretty language but its fairly straight forward

skull

hmmm, also more math related

grad school etc

Faust

6:00 AM

yeah i wanna goto grad school

got a gpa of 8.5/9 for the last few semesters 9 in the last two so i should be able to get in

skull

cool

Faust

was told needa gpa of 8

to get into pure math

7 for applied or discreete

not sur ehow accurate that is but meh

ima try anyway

skull

don't you have to write the GRE?

Faust

nah im canadian

i dont even know what a gre is

well i do

but not really

skull

6:56 AM

in The Skunk Works, Mar 26 '15 at 0:36, by infinitesimal simplicio

A mathematician, a physicist and a engineer are asked by a student what the meaning of $$\int \frac{1}{dx}$$ is.

The mathematician says it is meaningless.

The physicist ponders it for a moment and wonders if there is some way to give it meaning.

The engineer says, "Hmmmm, I used to know how to do that."

2

Alessandro Codenotti

Hi @Dami

I called it, tell 0celo7 about Washington DC if you get in contact with him @Balarka @Dami

tatan

7:12 AM

If y=(x^2+2x+a)/(x^2+4x+3a) is a surjective mapping, what is the range of 'a' ? How do I proceed ?

Leaky Nun

@tatan so for every y, the discriminant of x >= 0

tatan

so, I arrange it into a quadratic x^2(y-1)+x(4y-2)+(3ya-a)... right?

Now should I make D>=0?

Leaky Nun

yes

Secret

@skull So how exactly the engineer abuse that notation, i.e. what is the rule they use when interpret that integral?

tatan

@LeakyNun So, I have got a quadratic in y and a... now ?

Leaky Nun

7:22 AM

@tatan so that quadratic is always >= 0

what can you say about its discriminant?

tatan

No idea... sorry

Leaky Nun

how many roots can our new quadratic have?

(hint: what happens between two distinct roots?)

tatan

y should belong to the set of all real numbers as the mapping is from domain of func. ->All real numbers...

Leaky Nun

so what's your quadratic?

tatan

(16-12a)y^2+(16a-16)y+(4a+4)>=0

Leaky Nun

7:28 AM

right

tatan

yup

Leaky Nun

so let g(y) = (16-12a)y^2+(16a-16)y+(4a+4)

tatan

Okay.

Leaky Nun

if g has two distinct roots, what would be the sign of g between the roots?

tatan

maybe +ve or -ve depending on whether its an upward or downward opening parabola

Leaky Nun

7:29 AM

well but g is always >= 0

skull

@Secret perhaps, infinitesimal simplicio would know :P

Leaky Nun

so which one can it be?

Faust

one

Secret

@skull uh that just tag you, oh well, we need a time machine...

past selves are a tricky business...

skull

:D

Secret

7:32 AM

I have a special notebook that allows me to keep track of most of my past selves, so like the time machine of a mac, I can revert to my past selve's thinking and thus understood from their point of view if necessary

tatan

If g is always >=0 then how can it have 2 distinct roots? If it has two distinct roots then the region between the roots should be of opposite sign than the other part of the parabola... isn't it? @LeakyNun

Leaky Nun

@tatan right, so g can't have 2 distinct roots

what does this tell us about the dsicriminant of g in terms of y?

tatan

It should be negative... right?

Secret

hmm...

$$\int \frac{1}{dx} = \int \frac{1}{1 - (1 - dx)} = \int 1 + (1 - dx) + (1 - dx)^2 + \cdots = \int \frac{1}{dx} dx + (1 - dx) + (1 - dx)^2 + \cdots$$

tatan

It does not have any real roots... so D (of g)<0... this would give the range of y... @LeakyNun

Faust

7:35 AM

thats not entirely true

Leaky Nun

@tatan well it can also have one repeated root

Yeah... at 0

so D_g <= 0

yep

and this would give you the range of a, not y

tatan

7:37 AM

Typo there... I meant 'a'...

Secret

grrr... I need an abstract integration, one that has no dependence whatsoever to the measure chosen. I guess that is always defined for constant functions...?

$$\int 1 d \mu = ?$$

tatan

@LeakyNun You are such a great teacher... thanks ...<3

Secret

Differentiation of integrals

In mathematics, the problem of differentiation of integrals is that of determining under what circumstances the mean value integral of a suitable function on a small neighbourhood of a point approximates the value of the function at that point. More formally, given a space X with a measure μ and a metric d, one asks for what functions f : X → R does lim r → 0 1 μ ( ...

skull

best of luck to JDH Hamkins

Secret

hmmmmmm...

Recall previously we have:

Secret

8:00 AM

Mar 8 at 0:22, by GFauxPas

you mean to say that $\int \phantom{x}^2 d \lambda$ isnt good notation? :P

Thus $$\int d\lambda^{d\lambda}$$ becomes:

Let $dk = d\lambda^{d\lambda} = e^{D \ln (I+(D-I))} \lambda$

Harry Evans

If say $g_k, h_k \in L^2 (I)$, I'm interested to know if $K : I \times I \rightarrow \mathbb{R}$ defined by
$$K(x,y) = \sum_{n \in \mathbb{N}} g_n(x)h_n(y)$$
is in $L^2 (I \times I).$

Secret

$= e^{D ((D-I)\lambda - \frac{1}{2}(D-I)^2 \lambda +\frac{1}{2}(D-I)^3 \lambda + \cdots)} = e^{D(D-I)\lambda - \frac{1}{2}D(D-I)^2 \lambda +\frac{1}{2}D(D-I)^3 \lambda + \cdots}$

Secret

how about $g_k = 1$ and $h_k = \frac{1}{\sqrt{k}}$, that should blow up?

Secret

$$\int d\lambda^{d\lambda} = \int e^{D(D-I)\lambda - \frac{1}{2}D(D-I)^2 \lambda +\frac{1}{2}D(D-I)^3 \lambda + \cdots}$$ which gives some differential operator

We can check that this is consistent with intuition as $\int d \lambda = \int D (\lambda) = \lambda$ by first fundemental theorem of calculus

Now back to:

$$\int \frac{1}{dx}$$

We can expand this as follows"

Harry Evans

8:16 AM

If I check whether
$$\int_{I \times I} \left| g_n(x) h_n (y) \right|^2 dx dy < \infty$$

Does that work?

Secret

Yeah, an easy approach to the counterexamples will be something such that $g_k, h_k$ does not blow up in $L^2 (I)$, but does in their product.

and harmonic series is the easiest thing I can think of that can blow up

14

Q: Product of two Lebesgue integrable functions not Lebesgue integrable

I have a homework problem that says; Give Borel functions $f,g: \mathbb{R} \to \mathbb{R}$ that are Lebesgue integrable, but are such that $fg$ is not Lebesgue integrable. I saw this page too: Product of two Lebesgue integrable functions, but the question does not mention boundedness. I also a...

real-analysis

Here's an in depth discussion that highlighted the counterexample I used

Secret

$$\int \frac{1}{dx} = \int \frac{1}{1-(1-dx)} = \int I + (I-D)x + (I-D)^2 x + (I-D)^3x + \cdots = \int () + \int x() + x() + \int x() - \int Dx () - \int xD() + Dx () + \cdots$$

where $D$ is understood to act on anything to its right until the next + or - sign

and thus, we have an integrodifferential operator

More generally, provided it is well defined (details to be worked out) a generic expression

$$\int f(dx)$$

corresponds to some integrodifferential operator

The next question we can ask is whether the following result:

Mar 25 at 10:12, by Secret

$$\int^{\int^{\int}}f(x) dx = 0$$

holds for the more general case:

$$\int^{\int^{\int}} f(dx) = \textbf{0}$$

where the RHS is the zero map

Harry Evans

8:32 AM

I see, so it is possible that $K$ is not in $L^2 (I \times I)$. I was just thinking that is $K$ is, then by a theorem we used in class, I could show that $T : L^2(I) \rightarrow L^2(I)$ defined by
$$Tf(x) = \int_I K(x,y)f(y)dy$$
is compact.

Leaky Nun

10:27 AM

@BalarkaSen hmm, I don't like how Rudin uses Fourier to prove Liouville and maximum modulus...