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00:00
@BalarkaSen I understand the definition but I cannot fully comprehend the intuition or a visual approach (if there is one).
it's not so much the standards as the ability of the teachers to meet the standards. they spell it out in a more or less normal way. but the follow-through is where we fall behind.
one of my more controversial opinions is that the united states rested very heavily upon talented women not having other career opportunities to populate its teachers for many generations. when women could do other stuff a lot of them did do other stuff and now what. this is unorthodox because it implies that teaching is not a calling but a field where people respond to incentives.
you have to give talented workers something or they will go elsewhere unless encumbered by other burdens.
i dunno.
not to malign the many talented people who stick with teaching, but most people will not martyr themselves to a field.
Bob
Bob
I think high school teachers have to do a lot of non-teaching activies.
very good point. high school teachers have to do so much stuff that isn't teaching that there just isn't time for teaching.
Bob
Bob
I know high school teachers are afraid that if they do something in their personal life, such as get drunk, it can hurt them professionally.
I think that is very wrong.
from what I have seen, there is a drop in standards in college
one of my friend's husbands is a high school teacher and he brings food to school every day because his students do not have enough to eat.
00:10
@leslietownes I would state that teaching is definitely a vocation. Not everyone should teach, even if one is able to.
Bob
Bob
but I have not seen all that much
I had a friend ( he has died ) who taught adjunct for many years
One semester he had bad students so he gave all Ds and Fs.
He was not asked back.
AMDG i agree that it is not like the field i work in. you have to believe in it. but it's bad if the system is premised upon people believing in it. maybe give them good wages too. although so many people would do it for free.
Also, I haven't read the whole conversation, but yes, US schooling is hardly school and is very poor quality. I went through it. This is even in consideration of my own personal experience which was more of the bottom 10% experience of the bell curve. There are many issues with the system, and my experience was poor because I did not care about grades, but the schol system did.
@leslietownes Bruh, let me tell you about teacher's salaries and how they get paid, or at least how it was at the time I was in school...
i do remember resenting my high school english teacher complaining about his salary when it was more than my parents made. it certainly isn't a functional relationship where more dollars equals more quality.
Bob
Bob
from what I have seen, tenure teachers are well paid
and do not work very hard
non-tenure teachers are not well paid and do work hard
00:14
How much they earn is contingent upon the performance of the students. The issue with that is that that means if a student (such as me) performs poorly for lack of incentive to get good grades, or if a student simply can't understand the material, the teacher is punished for something that he or she cannot possibly have any control over.
Bob
Bob
I think tenured teachers should be paid on a lower scale
@Bob I'm glad that you aren't in charge, then.
it is a patchwork. different K-12 schools have wildly different pay structures and opportunities for advancement. this is a big problem in US education, there are like 15000 different systems, not one system.
This is a terrible system, and it penalizes the teachers for mistakes that they aren't even making in the first place on top of their already poor salaries.
very hard to generalize.
the fact that someone somewhere is getting away with something doesn't mean that everybody is.
00:16
i think there is a lack of clear syllabus & consistent (statewide, at least) testing. rubbish 1,000 page back breaking books don't help.
california has no uniformity whatsoever. you can have different schools in the same city teaching wildly different curricula.
@leslietownes That's the issue you get with allowing everyone's subjective ideas to reign supreme instead of imposing a single system, even a terrible one, upon everyone and then taking honest feedback to improve it.
Bob
Bob
I am signing off now
teaching to the test is better than what i have observed.
it all nominally meets state standards but it is a goofy way to run something.
Bob
Bob
00:17
have a nice evening
cheers, bob.
palindromic bob
So, that's the other thing that I still do not understand: state standards.
a friend of mine is from korea where he has given me the belief that essentially the whole curriculum is nationalized, so everyone is reading the same textbook. that at least focuses the conversation on one textbook.
Everyone in "school" gets this uniform learning agenda that does not promote specialization or learning, but instead incentivizes optimization of grades.
00:19
when i was in iowa there was some guy lobbying the legislature to adjust standards so that his book would be the only thing that satisfied them, and i think they went with it. as did a number of other flyover states.
preposterous.
So ultimately, the student, ironically, learns (I know, shocking). Namely, he learns to find the path of least resistance to barely passing and pleasing the teachers and/or his parents.
i think the common core standards make a lot of sense but there is no population of teachers to implement them. there is a population of people who mainly are qualified only in conveying vague ed school soup.
And instead of raising standards and reorganizing and restructuring the US school system, in order to look better, the schools then instead lower their standards such that the schools as individuals have the appearance of a high pass-rate, when in reality, they've played themselves into their own system as an inevitable conclusion of its clockwork: the school now learns to find the path of least resistance to making students pass.
@leslietownes I apologize in advance for stating this in Mathematics chatroom: none of what I learned in my HS math schools I cared for nor retained because I saw no need for them. Of those things that I have need of now, I never had a chance of learning back then. I am now having to "waste" my time learning these things on my own. The school system is nothing more than something to waste years of your life teaching you things you likely will never need. That's what we have right now.
no need for apology. i think a lot of california public schools' duty amounts to teaching people how to be and stay poor and ignored and not listened to.
so, you have found fertile ground for your opinions.
for more on my radical political views, listen to my podcast, Leslie Speaks.
I've never gone to uni/college (and I never will because the internet allows one to teach himself for the cost of internet), but of those whom I know of, their degrees, I observe, are wasted; are left with thousands of dollars of debt for said schooling; and never put those skills to good use because something better came up, or it wasn't really what they wanted, etc.
00:30
@Bob Clearly you’re the expert.
@leslietownes Mostly apologizing that I basically had no respect for math, and if no one else, to myself because math is a useful tool and model for describing reality, and I, like many others who care more about knowledge than a piece of paper, have been cheated out of a proper education.
i got nothing decent until 12th grade and after that i had to pay for it, so i get the vibe.
I would love to learn more math, but it's so slow to try to find material on my own. It'd be faster if I just had like an advisor in every field that I'm interested in so I could bounce ideas between me and him and thereby learn faster and obtain a better understanding than slowly figuring it out on my own through contemplation which by nature is a slow and methodical process.
@vitamind Tell me the definition
Well, cheers to you guys. I think I shall poor myself a glass of chareau aloe liqueur and find some way to enjoy the rest of this Sunday evening.
00:39
that doesn't sound enjoyable
What do you mean? lol
i'm texting with my best friend about the best recipe for something to throw on toasted baguette chunks. all is right with the world.
The liqueur actually tastes quite good, though it's a bit too sweet for my liking.
idk what's chareau aloe liqueur
Ah nice, I wouldn't have expected it to be sweet from the name idk
nuanced drinks are not my forte.
00:40
A liqueur made with aloe vera juice, cucumber, eau de vie, lemon peel, muskmelon, spearmint, sugar, and water.
Same, I stick to beer of varying brightness
I'll chug some water; cheers
I like to try various kinds of things. Try my palette and see what I like. So far, I'm quite happy with Tassajara Pinot Noir.
if i have a cocktail, which is rare these days, it's vodka plus something. and the something doesn't have 20 herbals in it.
If there were nothing else to drink in the world, I would be content with this Pinot Noir.
00:42
Korn + peach iced tea
tassajara had a very good vegetarian cookbook, if it's the same collective. i don't know where the trademark rights are.
if I had to have a cocktail it's going to be two different COVID vaccines at once
I always prefer something herbal or flavorful but not sweet. This here is great: hellacocktail.co/pages/bitters-soda#dry-aromatic-bitters-soda
they were some hippies in the monterey area. some of the recipes are disgusting but the good ones are very good.
00:43
do hops count as herbs?
definitely not.
then I steer clear of herbs
I've taken a liking to a particular set of icelandic beers. Nice taste.
as a northern californian i cannot object to 'hella cocktail' although i have used the word hella non ironically exactly zero times.
00:44
I also enjoy stouts, and the German black beer I had yesterday was quite excellent.
Helles is my beer of choice
I think Icelandic bear just means bear but 5 gallons of it
Beer not bear
thats how they stay warm
that is a different helle
tmw beer autocorrects to bear
00:45
Icelandic bear is... oof
@BalarkaSen Here?: en.wikipedia.org/wiki/Pseudoconvexity (Levi pseudoconvex suffices in my case)
I prefer icelandic bear how I like my steak: cooked, rare, and dead.
Emphasis on the dead part.
@vitamind When I ask you to tell me the definition I do not mean send me the Wikipedia page :)
do you do sloppy steaks?
What're those?
00:47
it's steak where you pour a glass of water on it and it gets all over the place.
Should I copy paste it??
Sounds more appealing than soggy biscuit
@vitamind Be my guest. I'm just going to ask you what every word means in what you quote, and in the process the meaning of plurisubharmonicity would be clear.
Correction: German dark beer, not black beer. "Weihenstephaner".
@leslietownes lol
What does meaning mean for you in this case?
00:50
i've never done sloppy steaks on my own account but i will do them if my company is picking up the tab.
Tonight, I've also got another of the German ones. This one's a lager.
Which lager?
@vitamind That seems too philosophical for my tastes, we should converse about mathematics. Tell me what a plurisubharmonic function is.
@EdwardEvans I found the icelandic ale that I like. einstokbeer.com/our-brews/arctic-lager and their white ale are quite nice.
00:53
:)
I only really know german beers
@EdwardEvans Oh right. Yeah it's the same Weihenstephaner brand. "Original Premium".
Ah okey nice
I think we had that in the UK
one of the only beers widely available in the UK that doesn't taste like piss
LOL
I honestly don't think I could ever try the "cheap" american beers we have here like budweiser and busch. They're probably fine, but I'm content with what I've tried so far.
I tend to like trying exotic things and seeing what my palette lands on after spinning the wheel.
bud has a few OK beers. strong emphasis on OK.
01:01
Honestly, it just sounds like the Great Value of beers. You got your glass bottle Irish stout sitting on the top row of high-class bleachers and then you got yer buds and yer miller lites in the splash zone in the baseball stadium. Maybe something fun to use for target practice. I mean, when I had the opportunity to see NASCAR races from the bleachers, these beers littered the seats in abundance, not just one here or there.
I was not asking for a philosophical answer. I can write the definitions here but that's not what I am looking for. My question is: Is there some kind of image I can have when thinking of pseudoconvex domains?
Yes, first you have to tell me the definition so that I can tell you how to start thinking about them
Go on, tell me the definition of plurisubharmonicity you understand.
(Again [ctrl+v]): Let $X$ be a complex manifold. An upper semi-continuous function
$f\colon X\to {\mathbb {R} }\cup \{-\infty \}}f \colon X \to {\mathbb{R}} \cup \{ - \infty \}$
is said to be plurisubharmonic if and only if for any holomorphic map $\varphi \colon \Delta \to X}\varphi\colon\Delta\to X the function $f\circ \varphi \colon \Delta \to {\mathbb {R} }\cup \{-\infty \}}f\circ\varphi \colon \Delta \to {\mathbb{R}} \cup \{ - \infty \}$ is subharmonic, where $\Delta \subset {\mathbb {C} }}\Delta\subset{\mathbb{C}}$ denotes the unit disk.
What's the easiest example of a complex manifold?
"more/many + under + quality of harmony [ in the context of mathematics]" is what I get out of that word that I've never seen before.
Latin prefixes. Familiar root.
Let's see how close I was to the definition...
01:10
The crowning result of the several complex variables course I took in 1973 (!) was that pseudoconvex is equivalent to domain of holomorphy.
@vitamind Good. What about arbitrary dimension?
i'm not sure that 1973 was even a time.
Simplest example of a complex n-manifold
unit ball?
01:12
@TedShifrin The converse is the famous Levi problem I believe.
@vitamind Unit ball where?
you had it and then deleted it @vitamind
Ok well C^n
converse of equivalence? :P
@TedShifrin Ah. Domain of holomorphy implies pseudoconvex is the Levi direction I believe, I meant.
@vitamind Good. Can you specialize the above definition to tell me what a pseudoconvex function f : C^n -> R is?
I guess the other direction should be relatively easy. I've totally forgotten this stuff and no longer possess Hörmander.
01:16
the USPS lost my copies of hormander.
Maybe Tromp should have appointed you to run the USPS.
I thankfully have yet to lose something to a postal service. I've had things ship abroad and locally and they've all managed to be delivered.
I'm probably going to get that happening seldomly if I should suddenly find myself regularly ordering palettes.
he should have.
You hate it almost as much as DeJoy.
if i could zip them all up in a bag and spirit them away i would. i think that's non-actionable.
01:22
Well, the one friend I have who is a mail carrier works his butt totally off.
"Them" might be referring to other parties.
@vitamind This shouldn't take as much time. You just replace wherever there's X by C^n. Let me do it for you: f : C^n -> R is plurisubharmonic if for every holomorphic map g : D -> C^n, the composition f o g : D -> R is subharmonic.
Agreed?
the fact that nobody knows what 'them' means is part of what makes it non actionable.
i don't mean my post-person.
So the geometric issue is to relate subharmonicity to convexity
I would mean every one associated with the "GOP."
that's what i meant too, but that's just between us. not public.
Certainly not. We wouldn't want to be bounced.
01:25
or pepper sprayed.
Ok, obviously, thought there is more to simplify.
@vitamind The geometry comes from what the boundary of the domain looks like.
We're getting there. For n = 1, what does this say about the relationship between subharmonic vs plurisubharmonic?
duct tapes mouth closed and leaves it to Balarka
01:29
a @Balarka: I have literally heard from Mike precisely once since his big event. I assume you still speak with him.
Yeah, he's been extremely busy with teaching and writing two papers at once.
I'll let you know in case he wants to send you a word
*let him know
Plurisub is sub
Cool. Believe it or not, I miss interacting with him, but I'm glad he's good busy.
@vitamind Excellent! I'm going to ask you to tell me a definition of a C^2 subharmonic function f : C -> R now (sorry for asking you definitions everytime, I promise this is leading to something)
nothing prurient to add unfortunately.
01:48
Isn't it exactly the same but with the nice plus condition that f is C^2?
Since obviously {upper semi continuous} \in C^2
Well sure. But C^2 subharmonic has an equivalent description that makes it clear why it is called "sub harmonic"
I mean your plurisubharmonic definition depends on knowing what subharmonic means. You haven't told me that.
Can I copy paste a link?
No. Tell me if you know, or say if you don't. Either is fine.
I can tell you if you're not familiar
Ooh. Well obviously I'm not familiar haha, I mean I understand the definition after reading it but I cannot tell you from my mind after looking at it for a minute.
OK I understand. Do you know what a harmonic function f : C -> R is off the top of your head?
01:57
Yes, if the laplacian of f exists and is zero.
Excellent. So Del f = 0. Subharmonicity is making this equality an inequality.
Del f >= 0. That's all. That's what a subharmonic function is.
oh all right got it
C^2 is just to ensure the Laplacian exists as you said.
Can you write Laplacian of a function f : C > R in terms of the z, zbar coordinates instead of x, y coordinates (where it is Del f = d^2f/dx^2 + d^2f/dy^2)
That is to say using Wirtinger derivatives
kind of reminds me of cauchy riemann equations
They're all related. We can discuss how after this discussion ends.
02:05
wonders how vitamind got to plurisubharmonic, which is pretty sophisticated stuff
sorry it's already after 4am but I'll try to think
We can continue later. I was thinking of getting a nap as well :)
Time zones.
Good night, all! :)
Good night Ted!
Good night @Ted
Wirtinger derivative wrt z times wirtinger wrt conj z times 4?
02:13
Correct. Very good!
Not quite, $\partial^2/\partial z\partial\bar z$ — not the product.
Watch out.
smacks Balarka
Oh, I thought by times he meant composition of operators
Ted is correct of course. It's a mixed partial of order 2.
Not a product of partial derivatives.
I wasn't sure, so I pounced. Now good night.
Yes, as you should have. Night!
@TedShifrin Yes sorry
02:16
It's OK. Just wanted to make sure you don't misunderstand.
Let me round the discussion off for today by saying @vitamind has proved f : C -> R is subharmonic if d^2f/dzdzbar >= 0, and a plurisubharmonic function f : C^n -> R is one which restricts to a subharmonic function on every holomorphic curve (= a copy of C) in C^n
Next time we'll see what this means in terms of the multivariable Hessian and mean curvature.
Good night. I'm off to another round of nap
Well if that's what it takes to visualize pseudoconvexity... Good night!
@vitamind In $\mathbb{C}^n$, pretty much
$\frac{\partial^2}{\partial z\partial\bar{z}}$ is the complex version of the Laplacian
what is the purpose of the Wirtinger derivative (as opposed to the usual)?
02:37
@copper.hat hadn't really seen them called that before, but they look like the normal $\frac{\partial}{\partial z}$ and $\frac{\partial}{\partial\bar{z}}$, are they not?
02:59
@copper very powerful. Cauchy-Riemann is the statement $\partial f/\partial\bar z=0$.
So Cauchy is immediate from Green/Stokes with nothing more.
03:17
monica?
some very storied SE thing. people had 'free monica' in their usernames. i'm unsure of what it was about and don't know if there is a link for it.
also in a state of near ignorance.
I see that using them results in simpler statements of formulae, but I was wondering if there was something deeper underlying this?
i was thinking black dress
(of course)
TL;DR Monica Cellio was unceremoniously fired by SE from I believe it was six positions over a supposed refusal to adhere to a CoC, which was not even in effect at the time, which was really a misunderstanding between people arguing/discussing in TL at the time.
not my own black dress, of course. i am a very angry dresser.
@hyper-neutrino She was an SE employee?
just a volunteer mod
03:21
i only wear a black dress when i am going undercover and trying to seduce a multimillionaire.
there's a timeline someone made here: stackexchange-timeline.webflow.io
Deeper? Yes, the real analytic $f(z,\bar z)$ doesn’t depend on $\bar z$.
that is curious
thanks @hyper-neutrino.
no worries. you could probably find more asking someone who actually knows more about it or digging through MSE (though there are many reasons why one should not do the latter)
the curious was a complex remark
03:24
Seemed real simple to me.
seemed that way to me too, but perhaps that was imaginary
We’re talking about something else!
i do not have an iota of understanding
Why not?
03:40
just a play on $i$ :-)
(even though the $i$ is not actually iota.)
I have always found z, zbar coordinates mystifying. But they're the natural coordinates on C, x and y are coordinates of C as a real 2 dimensional object. z = x + iy does not suffice as a single coordinate on C because you lose the anti-holomorphic part. z, zbar are jointly sufficient.
@copper.hat you got me.
The Wirtinger derivatives are natural because if f : C -> R is a smooth function df = (df/dx)dx + (df/dy)dy = (df/dz)dz + (df/dzbar)dzbar
They're the coefficients of df when written in terms of the induced basis {dz, dzbar} on smooth 1-forms of C
i see that, but i guess i was hoping for something more geometric.
i'm very demanding tonight.
It is analysis, after all.
03:44
true
maybe it is like compactness, it becomes obvious with time :-)
Is conformality geometric? Then you need holo or anti-holo.
So you can get that out of this pretty directly.
i need to spend more time looking at this.
Distracts you from algebra.
03:50
i'm still working through your book :-). i'm slow.
Do you hate it? ;)
I can tell you for a complex manifold M, the d/dz, d/dzbar coordinates are coordinates on TM o_R C really. But I'm not sure if you want me to take you in that rabbithole.
i like it so far.
@BalarkaSen my FIDE math rating is far lower than yours...
@copper.hat Think of $df = A dz + B d\bar z$ and pull back the Euclidean metric. When is it a multiple of the Euclidean metric?
Lol I'd probably be in trouble if FIDE was putting out ratings for mathematicians
03:54
i was interviewing a potential consultant the other day and his firm's representative mentioned his h-index, which was better than my h-index of 3. the guy said 'you have an h-index?' and i said, exactly.
all citation based measures are noise to me but it was a funny moment.
that leads naturally into one's obama or trump number.
you're the only one who has one of those, i think.
ugh, unless i have one through HLS somehow.
:-) the most unlikely of my friends has a very low trump number
the ironies of life
i know a trump judge. life truly is a mystery.
How to recognize an American in a conversation
Step 1: hear "Trump" or "Obama"
03:56
the mask?
Indeed … so much for complex geometry.
sry, it was a distraction while i play with my pullback
when i lived near HLS i used to give tourists guidance to the place where obama had a well circulated photo taken outside of austin hall. the children loved seeing themselves photographed where obama had been. i was fine with that. my extreme liberal views are incompatible with obama's incrementalism but i did it anyway.
I hope vitamind is enjoying and not getting lost in plurisubharmonicity.
Note the usual metric is something like $dz\otimes d\bar z$ symmetrized.
So what is $df \otimes d\bar f$?
03:59
i'd see people wandering around and ask if they wanted to go to the spot. they all did.
Nice one, Ted.
I like this explanation
Never thought about it before until copper coerced me.
glad something came out of it, i am struggling to understand the one form $d \bar{z}$ on $\mathbb{C}$.
It is $\overline{dz}$.
It is dx - i dy to mortals who have to resort to R
04:03
i a little more comfortable with real things :-)
I'm rolling some virtual eyes at you now copper
there is a long line :-)
I have only a pair unlike Ted
Just see that $dz\otimes d\bar z$ symmetrized is $dx\otimes dx + dy\otimes dy$.
the eyes scream
04:05
Bring me ice cream.
Are you aware of what a Weinstein manifold is, Ted?
I know the person well. Something symplectic …
alan who?
i was being deliberately quiet
Oh, I didn't realize Weinstein was recent. Yeah, it is something like a noncompact manifold with an exact symplectic form preserved by flow of a vector field X such that Flow^t_X* w = e^t w
04:09
he's responsible for my lowest grade in graduate school. you, balarka, have brought up some stuff.
Another Berkeley faculty, student of Chern.
@leslietownes Ahaha oops
and the berkeley crowd assembles.
I’m sure the grade was well-deserved.
he had a take-home exam and basically everyone in the class cheated by working together except for me. i saw them working on it in a mexican restaurant.
i don't hold it against weinstein.
04:11
la fiesta?
Euclid.
What course was it
the elements of a good meal
it was on center. i don't think it's a mexican restaurant anymore.
04:12
214 i think
craving a tamale now
Grad manifolds @Balarka
the block between campus and the BART station. half of my class was in there.
i'm not annoyed at them for not inviting me because i would not have gone.
change your username to groucho
04:14
groucho is one of my main comedic inspirations.
I wonder who gets that, copper.
my twisted mind
i think irish learned doublespeak during british rule
a survival tactic for a nation who could not shut up
(surprised)
no offence to the british intended.
i had about 60 minutes of conversation with my grandfather and 30 minutes was him telling me never to tell anything to the police.
please be nice to my daughter
I asked because I was reminded during conversations with vitamind that pseudoconvex domains are Weinstein. There's an exact symplectic form and the antiderivative of the form keeps growing larger and larger as you flow by the gradient of the plurisubharmonic exhaustion function
04:17
you have to wonder what went on, he was saying this to an 8 year old boy. stuff is ingrained very early.
There's some results by Eliashberg et al which go the other way. The phenomenon is known as "from Stein to Weinstein to Stein"
i have some similar embedded rules.
Interesting @Balarka
i own the domain etaliter
just btw
the page was created in order to clear a bump in refinancing while not a full time employee
will try to shut up now
before we're all arrested. thank you.
04:23
Finish computing to understand my deep explanation!
I guess the way to go about it is to think of a domain G in C^n with compact boundary dG which is a (totally) real codimension 1 submanifold of C^n, and dG = f^-1(0). f^-1(c) for various c > 0 foliate a collar neighborhood of dG in G by hypersurfaces, the area of f^-1(c) is O(e^(1/c)) as c -> 0
Hello, I am trying to prove ($g:Y\rightarrow Z$ continuous iff $gf:X\rightarrow Z$ continuous) implies $f:X\rightarrow Y$ is an identification map. Why are we allowed to take $Z=Y$?
Not quite area. Hard to say. I'll think later
Larry — if that suffices to prove the result, why not?
Because it seems there may be cases when $Z\neq Y$ where it still holds, so those cases won't be covered
04:28
You’re not thinking about the logic here.
$Z$ doesn’t appear in what you’re concluding.
You should know proofs where the hypothesis has a “for all,” but to prove some specific thing you only need one case, not all.
Wouldn't it be for all such functions g and gf, including where Z is not equal to Y?
You’re a broken record.
I don't understand still
Do the proof as they suggested. Then ask your question.
I have the solution of the proof and am trying to understand the solution
04:37
Do this proof. Suppose $f$ is an injective linear map. Prove that $f(x)=0 \implies x=0$.
04:54
Existence of maxima (local) and minima (local)of $$f(x)=\begin{cases}x\sin\frac 1x: x\ne 0\\ 0: x=0
\end {cases}$$
is to be shown in arbitrary neighborhood of $0$.

Clearly, $f$ is differentiable on $\mathbb R\setminus \{0\}$ and not differentiable at $0$.

We take any $\epsilon \gt 0$ and consider the interval $(-\epsilon, \epsilon)$. $f$ is an even function so let's consider only the half neighborhood $(0,\epsilon)$.
If $f$ attains maxima/minima at an interior point $c$ of the open interval $(0,\epsilon)$, then by Fermat's theorem, we must have $f'(c)=0\implies \sin (\frac 1c)-\frac 1c\co
@TedShifrin I think I got it, could you check my reasoning please? ($g:Y\rightarrow Z$ cts iff $gf:X\rightarrow Z$ cts) implies ($g:Y\rightarrow Y$ cts iff $gf:X\rightarrow Y$ cts) implies $f:X\rightarrow Y$ identification map. Since the first implication is obvious, and the second one is the proof itself, it works out.

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