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12:01 AM
if you start with initial condition $v_1$ then the solution is $e^t v_1$. If you start with $v_2$ it is $t e^t v_1 + e^t v_2$ and if you start with $v_3$ the solution is $e^{-3t} v_3$. To find the exponential, you need to find solutions to $x'=Ax$ for the initial conditions [1 0 0]', [0 1 0]', [0 0 1]', so you need to express these in terms of the $v_k$.
 
@copper.hat ok, thank you 🙏
 
🥰 Ted's here
 
i suspect he is aware of that
 
Not in emoticon for
format
 
12:08 AM
format?
 
yes
:D
ascii
😀😀
unicode
You have to format your text right or all day you're looking at plain ascii
causes you to think you're in the 80's
All you have to do is hit Windows + .
Windows + Shift + s and Windows + v are my power tools
 
X4J
12:36 AM
@TedShifrin Yes I did
 
12:52 AM
OK, so what question do you have?
 
X4J
I am not sure about the solution provided there, more specifically, I am not sure why we can assume $f$ is given by its Jordan form without the loss of generality and what are the invarient subspaces which compound the direct sum
 
You can write a linear map as a matrix with respect to any basis you want. So you pick the basis that gives Jordan form. I told you days ago to wirk on $h=1$ and $h=2$ explicitly first.
 
X4J
I've used some examples that but I struggle to see what I can understand from it
Thing is if I pick a specific basis, why it is still generalized for any basis
 
1:29 AM
The chain basis or whatever you were calling it is precisely how you build Jordan form.
Equations the linear map satisfies, rank, nullity are all the same no matter what basis you use.
 
X4J
I see, I will take some more examples
 
there really are very few ml/ai sort of convex questions in the last while, maybe its going away
 
Not much interesting diff geo either.
 
2:29 AM
3
A: Ring homomorphisms between $C([0,1], \mathbb{R})$ and $C(\text{Cantor set},\mathbb{R})$

HallaSurvivorGelfand Duality says that the category of Compact Hausdorff Spaces $\mathsf{CompHaus}$ is antiequivalent to the full subcategory of rings of the form $C(X)$ for a compact hausdorff space $X$, $\mathsf{C^*Alg}$. Everything I'm about to say is well treated in chapter IV.4 of Johnstone's Stone Space...

This answers the question but the exam where it was asked, won't require the knowledge of Category theory, so is there an easier approach to answer the question?
 
@SoumikMukherjee Consider the surjection from the Cantor set to $[0, 1]$
$$C\ni \sum \frac{x_n}{3^n}\mapsto \sum \frac{x_n/2}{2^n}\in [0, 1]$$
 
Okay so if we call this map $h$ then for any $f$ in $C([0,1])$ we get a corresponding map $g \in C(K)$ such that $g=f\circ h$, so there is a surjection from $C(K)$ to $C([0,1])$, hence an injection in the opposite direction
 
2:45 AM
No
The map $f\mapsto f\circ h$ is an injection from $C([0, 1])$ to $C(K)$
In fact a ring homomorphism
 
@Jakobian Oh okay
Thank you
 
@copper Maybe you like this one?
 
3:35 AM
@TedShifrin Interesting, thanks. Not sure I have any insights, though
 
X4J
Ted, I think the idea is that the number of chains in the chain basis equals the length of each chain and since the number of the chains in such a basis is always the same then so the length
 
Why is the number the length? Surely the dimension comes in somewhere.
 
 
3 hours later…
6:35 AM
what is the meaning of "ꟻ" in math? (is it even used?)
 
$\unicode{xa7fb}$ I don't know of any use.
 
 
6 hours later…
12:11 PM
Does anyone know some references that can point to piecewise-quadratic approximation for a function of two variables $f(x,y)$ defined on a square $[0,1] \times [0,1]$? I know that piecewise-constant approximation is to subdivide the square into smaller squares and to replace $f(x,y)$ by its value in the center of a square (for example).
For piecewise-linear approximation by splitting each small square along the main diagonal we can subdivivide it into two triangles and the replaces $f(x,y)$ by a plane passing through the points that $f$ takes at the corner of each triangle. But I have no idea how the approximation procedure goes for the case of quadratic approximation for two variables.
 
12:47 PM
I have a question. Consider $g(x)=\frac1{\sqrt{x^2+\sqrt x}}$ and the operator $T:L^p(]0,+\infty[)\to \Bbb C$ such that $Tf=\int_0^{+\infty}f(x)g(x)dx$. For which $p\in[1,+\infty]$ is $T$ continuous? First of all, I found that $g\in L^q(]0,+\infty[) \iff q \in ]1,4[$. I consider applying Holder inequality when $1/p+1/q=1$, so $T$ is continuous for all $p\in ]1/4,1[$. I don't know how to procede in general
 
1:12 PM
@TedShifrin i guess whoever does the hiring then sort of tries to subjectively evaluate that each individual professor has good enough standards for the department's calculus course, or something of the sort?
maybe there were some informal, office-politics-ish mechanisms to slap professors on the wrists if they didn't keep up or were too harsh?
 
1:42 PM
Let $f$ be a differentiable function on R such that $|f'(x)|>1|$ for all x. Then preimage of compact set is compact. True/ false?
 
1:53 PM
@SoumikMukherjee why is there an additional $|$
anyway, if $|f'(x)| > 1$ then such function will be strictly increasing or strictly decreasing so it has a continuous inverse
In fact $|f'(x)| > 0$ is enough
I guess why they want $|f'(x)| > 1$ is so that $f$ is a bijection too
because of things like $\arctan$
But yeah, $|f'(x)| \geq r > 0$ for some fixed $r$ will imply that $f$ a diffeomorphism of $\mathbb{R}$ with itself
this discussion has been really one-sided
 
2:14 PM
@Jakobian I was typing on the mobile and was on road
@Jakobian Thanks
 
2:51 PM
There were some interesting problems in the exam today. I will start typing those once I reach home
 
3:19 PM
@Ted Hello!
 
3:34 PM
yesterday, by Sine of the Time
@ThomasFinley in $B_1$ there is $(0,1,0)$ whereas $e_3=(0,1,1)$
@SineoftheTime Oh, I get it now, it was a typo. $e_3$ is indeed (0,1,1) and in $B_1$ it should be $(0,1,1)$ instead of the previously written typo, $(0,1,0).$
I have fixed the typo, but is my solution after the change, still valid?
Here is the link of the post( in case you have trouble finding it):
2
Q: Find the change of basis matrix $P$ (say) for the change from the basis $B=\{(1,0,0),(0,1,0),(0,0,1)\}$ to $B_1=\{ (1,1,0),(0,0,1),(0,1,1)\}.$

Thomas FinleyFind the change of basis matrix $P$ (say) for the change from the basis $B=\{(1,0,0),(0,1,0),(0,0,1)\}$ to $B_1=\{ (1,1,0),(0,0,1),(0,1,1)\}.$ I am familiar with change of coordinate matrix. My definition of change of coordinate matrix is: Let $V$ be a finite dimensional vector space such that, $...

 
3:59 PM
@ThomasFinley It's still valid and I answered it assuming $e_3=(0,1,1)$
 
4:25 PM
I have a doubt regarding this ex about Laurent series: find the expansion of $f(z)=\frac1{(z-1)(z-2)}$ in the punctured disk $\{0<|z-2|<1\}$. My try: $f(z)=\frac1{z-2}-\frac1{z-2+1}=\frac1{z-2}-\sum_{k\ge 0}(-1)^k(z-2)^k$. Is this correct?
Or alternatively $\frac1{z-2}\sum_{k\ge0}(-1)^k(z-2)^k$
 
4:39 PM
@TedShifrin Sure. But where do you draw the line between "I am teaching something controversial" and "I am teaching something which doesn't meet the standards"? (Whose standards, by the way?)
In Arizona, the legislature has essentially told us that our standards cannot include "critical race theory". Academic freedom should protect the rights of faculty to teach that topic. But the standards say...
In general, academic freedom is interpreted fairly broadly, and so you don't often see highly coordinated courses in the US.
I'm not saying that this is good (or bad)---it just... is.
 
4:51 PM
In conclusion, to solve the Collatz conjecture as true or false is to determine whether or not $x\equiv \lfloor\frac{4^n}{3}\rfloor \pmod 2$. For convenience, it is also self-evident that the rule reduces to the unconditional "extract odd part of number, then do $3x+1$".
But someone more interested in that can go figure out the rest. As for me, well, presumably mapping XOR or any one other bitwise operation to elementary wholewise operations would solve that and many other discrete mathematical problems.
 
5:23 PM
uhhh
 
5:37 PM
?
 
6:01 PM
@XanderHenderson sorry I hadn't seen this! huh, this proceeded on the basis of a personal initiative?
I guess it's also significantly related to the resources an administration has? or the relative seniority of the professorship at any particular institution
@XanderHenderson heh, that's something else I hadn't thought about. giving the capacity to enforce standards also means giving the capacity to enforce political agendas of the sort, maybe
 
@shintuku No? Not really?
 
i don't know i'm clueless heheh
 
At my masters institution, we typically had around 800 students taking precalculus every semester. Rather than put them into three or four very large lecture sections, they were divided into sections of 25-30 students. Each of these sections was typically taught by a graduate student TA or an adjunct lecturer.
Each instructor was given broad autonomy to teach the class however the hell they wanted, but all of the students took a common final.
This is the most coordinated form of instruction I have ever taken part in.
Typically, the assumption is that if an instructor is hired, they are the content expert, and they know how to teach the material. It is rare for there to be a lot of interference from a department chair or, g-d forbid, an administrator.
At my current institution, every class we teach is housed in a database. The database contains course forms which outline "learning outcomes" and "course topics". As long as I teach to that course form, I am given latitude to do what I think is best for my students.
We do have some institution-wide practices to ensure that students are meeting those objectives, but this is mostly in the form of our department making reports to the vice president for learning and student services on an annual basis (which then gets put into our accreditation report).
In other news, my daughter actually seems kind of excited to celebrate a traditional Jewish Christmas in two weeks. Minor victories.
 
6:17 PM
I see! thanks a lot for the perspective, really insightful and interesting to learn about the behind-the-scenes part. I hadn't expected administrators to be such that we'd ask forgiveness for even considering the possibility of them intervening in your work heheh
 
@shintuku In my opinion, if the academy is working correctly, administrators should almost never be telling instructors what to do. There are extreme circumstances in which an administrator might need to step in, but (again, in my opinion) instructors are the experts, and should given latitude to do what they believe is best.
 
right that makes a lot of sense
 
 
1 hour later…
7:55 PM
@SoumikMukherjee f' follows IVT (Darboux property) so either $f'(x)>1$ for all x or $f'(x)<-1$ for all x, i.e., f is strictly monotonic. $f$ is invertible and inverse is also continuous. So true.
 
I am wondering if anyone knows how to split bezier curves based on its matrix, rather than using De Casteljau algorithm
I came across this when I was reading "primer on bezier curves" ch 11. Which I didn't understand.
 
@XanderHenderson A traditional Jewish Christmas? You light different branches of the tree every night for 8 nights? Christmas Gelt for 8 days?
@Xander @shin There are departmental administrators, too. Department head, assistant/associate head (which I was for 8 years). And then course coordinators for certain courses. As associate head I certainly adjudicated a number of grade complaints, and often the instructor was not in the right.
@Xander I agree with you in principle on course content issues; however, academic freedom does not give someone the prerogative to cover only half the calculus syllabus.
Our curriculum committee wrote up, as part of the syllabus, a guide to what levels of competency grades of ABCD should have for first semester calculus. (Precalculus was tested already on computers.) We didn't go on to second semester because it was first semester that was often terminal/gateway.
 
To add more context. I don't understand under what assumption the author multiply t by z here
 
@Serilena It's not that we're ignoring you, but I'm not sure we have any people here who know what you're talking about. I know I do not.
 
@TedShifrin was my wording that bad? I assume you know bezier curves. (I mean they are an interesting mathematical objects so I made the assumptions people where knows them)
 
8:07 PM
Do not blithely make assumptions.
 
Alright.
 
Numerical analysts, of course, are completely familiar with this topic. But as a differential and complex algebraic geometer, I never once worked with them (other than using Adobe Illustrator).
 
@TedShifrin Chinese food and a movie. :)
 
Oh, of course. I've done that on Christmas eve a few times with a friend in ATL. We cooked a fancy dinner on the day.
 
2 hours ago, by Xander Henderson
At my current institution, every class we teach is housed in a database. The database contains course forms which outline "learning outcomes" and "course topics". As long as I teach to that course form, I am given latitude to do what I think is best for my students.
 
8:13 PM
My dad gave in eventually to my younger sister and got a Christmas tree (which was duly decorated) instead of just doing Chanukah. That was during my high school and college days, I suppose.
 
@Koro Sure. This is more or less my argument. Just no one asked me for it
I was expecting at least a question
 
@Xander I saw that. We of course were mandated to have that garbage in our syllabus. It is basically meaningless crap.
@Jakobian It seemed obvious to me :P
 
@Jakobian happens to everyone sometimes :).
 
@leslie This is leek patrol.
 
elsewhere often I expect at least a follow up question but I receive only 'OK'.
 
8:15 PM
I just bought wonderful cilantro, baby eggplants, arugula, radishes ...
Sometimes what you say is actually correct and clear, @Koro.
 
@TedShifrin this a bit counter intuitively to me, aren't curves are somewhat very connected to geometry? (the main reason I assumed they be very interesting to most people is that you can define a tangent line without limits)
 
Rare occasions, I realize.
 
at least that's some improvement :).
 
Bezier curves are important for numerical computations on computers, not particularly in theoretical circumstances.
 
Hmm that is a very good point.
 
8:16 PM
@TedShifrin Maybe the systems you are used to are less robust than ours? I don't think that our course forms are "meaningless crap"...
 
how nice! even pinging feels so calm now. I didn't feel a thing on my ears when both of you pinged.
 
I was imagining Soumik would ask me why I'm assuming $f$ is $C^1$ and I'd go reveal Darboux property of $f'$ as my triumph card
 
I disabled that ear blast sound.
 
To create a course, we must write learning outcomes and various other things for the college and university curriculum committees to approve the course. But in point of fact, they are sufficiently vague as to be meaningless crap in the global scheme of asking about course quality and grading standards.
Yes, of course every Calc I student shall learn to find derivatives with the rules, shall learn to do related rates and applied max/min problems, blah blah blah.
 
@Jakobian of course you mean trump card.
 
8:19 PM
@TedShifrin The course forms are not really meant to say anything about quality or grading standards---the instructors are supposed to be the experts, and are trusted to to the job in a fair and meaningful manner.
 
but whatever
 
Or did your department exercise significantly more oversight over the faculty who taught those courses?
 
@Koro I don't know, triumph because I'm triumphing when its revealed
 
It may also be noted that the number of courses at a college or university with such a standardized and universal syllabus is diminishingly small (a couple of math classes, some intro science classes, and composition?).
 
8:21 PM
In my experience, that trust is not warranted in many circumstances. But I'm thinking of large universities with lots of part-time instructors who are not supervised at all. Honestly, I had issues with many colleagues. As I mentioned the other day, one of my coauthors was a notoriously generous teacher (who also didn't return exams until the day before the final exam); typically students who got at best C's from me got A's from him.
 
Most college level classes are more specialized.
 
Is it a mistake here that $L^\infty(X,\Omega,\mu)$ is claimed to have an identity?
 
No.
I think the function that is $1$ at every point is bounded a.e.
 
but not integrable
with respect to a non-finite measure
which $\mu$ could be
 
It's measurable, not integrable. Check your definitions.
 
8:22 PM
@TedShifrin Yeah, I didn't make any value judgement. I was just trying to explain how things actually work at most places. Whether or not we should trust the folk who are hired to teach the classes, we do (from an institutional perspective).
Though, again, I think that it is common to have common finals for intro level classes like calculus and precal.
 
@TedShifrin yes, so it's not in $L^\infty(X,\Omega,\mu)$
 
You have a bizarre definition of $L^\infty$.
 
Not universal, but not unusual by any stretch.
 
You're saying it has to be $L^1$?
I agree with that, @Xander, although our faculty rebelled at such. We mandated that the instructors and grad students had to use the uniform finals, though.
 
o wait
 
8:24 PM
It has to be measurable, Sha. As I said, please go look up actual definitions.
You are badly confused.
 
lol I'm not
I just expected that integrability was part of the definition
I realise now it's not
 
@TedShifrin If I had my druthers, I would insist upon common finals for any course over a certain level of enrollment (e.g. at my PhD institution, precalc had 600+ every semester---there should be a common final).
 
"just expected"? Seriously. Then $L^\infty$ is just bounded $L^1$ functions? Hrumph.
@Xander We got around that by having all exams, including the final, for precalc done on computers.
 
In practice, there was a common final, because one ladder faculty was responsible for teaching all of those students (typically in three sections of 200 students each, with recitation sections run by TAs...).
 
@ShaVuklia f(x)=1 is essentially bounded.
so f is in L^oo.
 
8:26 PM
@TedShifrin Yikes. Doesn't that make it difficult to ask non-multiple choice questions?
 
no?
 
UGA has not gone the large lecture route at all. When I got there in 1981, I volunteered to do some large calculus lectures, but no one was interested.
 
Or for students to demonstrate process?
@TedShifrin Lucky you. Large lectures are terrible. It's one of the reasons that I am not super disappointed to have not gotten a tenure track position at a four-year college.
 
the set f^{-1}(1,oo) is of measure 0 , where f(x)=1 for all x.
 
No multiple choice. It's all very sophisticated software ... has been for years and years. Now there are various platforms for precalc and calc that use the software. Publishers have been doing that for 20 years.
 
8:27 PM
so the constant function x\mapsto 1 is in L^oo.
 
@Xander In my lack of humility, I knew that I would do a better job in a large lecture than many of the faculty/instructors were doing in small classes. I still stand by that assessment. But it would have opened the floodgates to people who would not do a good job.
 
@Koro yea I assumed integrability was part of the definition
hence why I thought the constant function 1 wouldn't work necessarily
but that was wrong
 
@TedShifrin Heh.
 
I mean you don't assume that even with $L^2$ right
 
As I said, it is important not to "think" but to know basic definitions.
This is not like you, Sha. :)
Jakobian, as often is the case, makes an important logical point.
 
8:29 PM
I don't think that I would do a very good job teaching a large lecture class. I think that I am a competent, but not excellent, lecturer. I am better at curriculum design on the front end, and in small group or one-on-one instruction. I don't like lecturing at large groups of students.
(This is actually pretty well borne out by my teaching evaluations---those students who come to office hours, or hang out after class, or deal with me over email really like me; those who don't, don't).
 
I did 350-person lectures of multivariable at MIT back in the day, and I believe I did an excellent job. But it was a lot of work, making sure that all the recitation instructors got notes for what I did and telling them what things I wanted to make sure they covered; and I had to go visit every recitation at least once or twice each semester. Not a typical postdoc thing, but I wanted to do it.
(And they knew me and my teaching skills from my undergraduate days there.)
But, yeah, in my career, I've always insisted that office hours are essential. I often had 10-15 kids in each office hour. The hallway got busy. That's too many.
 
large lecture can be very good if the lecturer is 'choice' and the people running discussion sections are capable of following instructions.
at berkeley the weak link was often the second part.
 
If $X\in M_d(\Bbb{C})$ is a matrix with eigenvalue $\lambda$. Let $v\in \Bbb{C}^d$ be the corresponding eigenvector. Let $\{u_1,...,u_d\}$ be an orthonormal basis of $\Bbb{C}^d$. Why is then $\{v,u_2,...,u_d\}$ also an orthonormal basis?
 
Yes. And often the first, too.
Although they had some superb lecturers, too.
In my day.
 
@TedShifrin Good for you. I've never seen anyone else do this kind of work.
@leslietownes At UCR, too.
 
8:34 PM
I had the honor, as I've mentioned here before, of having Gil Strang do a recitation for me in the multivariable course my second year. It was specified for students who were recommended by their first-semester recitation instructors as needing a slower class with more individual attention.
 
user: it's not, necessarily? think of d = 2 and X the identity matrix and {u_1, u_2} the "standard" orthonormal basis and v any unit vector that is not on the axes.
user: there might be missing hypotheses on the eigenspace for X corresponding to lambda and/or the relation of the basis {u_1, ..., u_d} to that eigenspace.
 
Not to mention that no one said $v$ was a unit vector. Go back to square one.
 
@TedShifrin Wow! I've always found Strang to be a phenomenal lecturer. I want to be him when I grow up.
 
@TedShifrin I mean I can chose $v$ s.t. $\|v\|=1$
 
I adore Strang, but I find some of his lectures and book-writing too meandering and not clear enough. But, yeah, he was great.
 
8:36 PM
@TedShifrin How did you deal with your office hour when there are a lot of people waiting?
 
@user123234 You said the corresponding eigenvector. That is not cool. Yes, you can choose a unit vector.
 
@TedShifrin I consider those things to be features, not bugs, but I can see how that style is not everyone's cup of tea.
 
Everyone adores Strang.
 
@TedShifrin but why does this then tell me that $\{v,u_2,...,u_d\}$ also an orthonormal basis
 
I find his linear algebra lectures less than ideal, to my taste. I prefer my own :P But that's not surprising. His presentation in linear algebra is also not good for a proof-oriented course, but he's not trying to be.
@user123234 Of course it does not, as leslie already said. Why should $v$ replace $u_1$ and not $u_5$? And why is it orthogonal to the others?
 
8:39 PM
okey but does there exists an orthonormal basis where I replace one of the $u_i$ with $v$?
because this arises in a solution of a homework and I don't get why
2c) is the exercise
 
You are not doing the right thing, @user123234. You start with a unit eigenvector $v$ and extend it to an orthonormal basis. Think.
 
@TedShifrin Yeah, I don't disagree.
 
ah so I start with one and say that I can extend it to an orthonormal basis
which can be always done by finding $d-1$ vectors which are pairwise orthogonal with length $1$ right?
 
mm, that description is too general, and too unrelated to the thing you started with.
 
I don't get what you mean. why does this not work?
 
8:48 PM
a more standard narrative would be "extend v to any basis whatsoever" [there are matrix recipes for doing this if you really want to go into specifics], and then do gram-schmidt, which will not change the first vector if you started with a unit vector.
because you're not saying how to find those orthogonal vectors orthogonal to the first one you started with.
 
ah okey but I mean I only need to know that it's possible to do this
so for this exercise
 
agreed, but you also need to say it in a way that indicates truth. e.g. "it is possible to find u_2, ..., u_d such that {v, u_2, ..., u_d} is an orthogonal basis" is true. "just grab d-1 vectors {u_2, ..., u_d} which are pairwise orthogonal and with length 1, then {v, u_2, ..., u_d} is an orthognal basis" is not true.
 
okey I see, thanks!
 
so if you include narrative about how to do it [which maybe you don't], that narrative should not suggest that any choice of u_2, ..., u_d would work.
 
sure!
 
8:50 PM
The buzz word you should know is Gram-Schmidt.
 
Yes I remember it now. Thats the method how to "orthogonalize" a basis
 
@user123234 are you aware that the Frobenius norm is submultiplicative? If so, there is an easy proof of (c).
 
9:24 PM
@user123234 An easy characterisation of the Frobenius norm is $\|A\|_F^2 = \sum_k \|Ae_k\|_2^2$, where $e_k$ are the unit vectors. This is (c) for the unit vectors. Now apply the earlier results to conclude.
 
@copper.hat the only place I heard about Frobenius norm is my friend that studied robotics
 
@Jakobian In some ways, it is the most natural norm for matrices.
 
I was considering matrices recently, I had to prove that compact-open topology on $M_n(\mathbb{R})$ coincides with standard topology
for an argument I was making
That a continuous function $X\to M_n(\mathbb{R})$ corresponds to continuous map $X\times\mathbb{R}^n\to\mathbb{R}^n$ where for each fixed $x\in X$ the latter map is linear
 
9:44 PM
@Jakobian As copper said, it's the most natural norm ... although I am fond of the operator norm, it is much harder to compute. If you write a matrix as a long vector, then this is just the usual $\ell^2$ norm.
 
In what ways is it the most natural one
 
For the reason I just gave. The disadvantage is that it depends on an inner product structure if you want to define it for a linear map and not just a matrix.
Howdy @robjohn!
 
Hey, @Ted!
 
How are things there? It was very windy and cold here yesterday, but it is better now.
 
9:57 PM
Another question about Conway's book... He proves this form of the Riesz representation theorem by invoking results from either of two books, that actually work with $C_{00}(X)$ (functions vanishing in a neighbourhood of infinity) instead of with $C_0(X)$ (functions vanishing at infinity). I also looked in my go-to measure theory book, where the result was simply proven for $X$ compact. Is the statement here correct, and if so, does anyone know a reference?
 
I mean if its just the standard one for $\mathbb{R}^{n^2}$ then why mention its the most natural? Can you elaborate the point?
 
@user85795 Very emphatic.
 
Indeed!
:D
 
wait, actually the statement of the Riesz representation thm in those two books is a bit different
 
Beautiful sunny day today, @robjohn.
 
10:00 PM
maybe there is a variation where they do work with $C_0(X)$
 
Functions vanishing in a neighborhood of infinity? That's an awkward way of saying that they have compact support.
 
Yeah...
 
Astonishing that in my 50 years or so as a mathematician, I've never before encountered this result.
I prefer CS to AM/GM, of course.
 
@TedShifrin I think its related to uncertainty principle
 
How big is a neighborhood of infinity...
 
10:06 PM
sha as a side note, there is a lot of variation in how people state "the" riesz representation theorem. almost to the point where it's not that good of a name to use. :)
 
I think thats how I got to know it, someone mentioned its related to Heisenberg principle from physics
 
yea I am aware, but in this case I'm interested in $C_0(X)^*$, and I've found three books that actually rather seem to work with $C_c(X)^*$ (replacing $C_{00}(X)$ by $C_c(X)$, because that's less awkward :))
 
lots of choices in whether you have a limiting assumption like X being compact, or sigma-compact, or second countable (easiest mode), or some very general thing.
 
the assumption is that $X$ is locally compact Hausdorff
 
it basically means that many references' "riesz representation theorem" or their proofs are not interchangeable with one another.
 
10:08 PM
this lemma I've seen proven for $C_c(X)$ instead of $C_0(X)$ in the two books referenced, and also in my third source
 
$C_0(X)\cong C(\alpha X)$ where $\alpha X$ is the one-point compactification of $X$
so, without checking carefully, the argument for compact space should translate to LCH space
 
related to my semi rant with koro's radon measures.
Dec 4 at 23:19, by leslie townes
very late to all of this, koro, but for purposes of working across multiple references, jakobian is right that the definition in those notes [and apparently not given in folland] is the standard one. although it is very wise to pay extra attention to definitional details within references, because there is a good amount of variation in how different authors present that material.
 
@Jakobian Well, the uncertainty principle (as I taught it years ago) is just CS, intermixing position and momentum spaces.
 
@Jakobian ah!
that might solve it
wait...
wouldn't $C_0(X)$ be identified with a subspace of $C(\alpha X)$?
 
I think why they're not using $C_c(X)$ is that its not a Banach space unless $C_c(X) = C_0(X)$
 
10:11 PM
namely, the subspace consisting of functions that vanish at infinity
 
@ShaVuklia yeah good point
 
So mod out by a maximal ideal.
 
$f\in C(\alpha X)$ such that $f(\infty) = 0$
 
@TedShifrin It is similar here. 70°
 
I did my usual Sunday trek to the farmers market 25 miles north. Always fun.
 
10:15 PM
@Jakobian Wouldn't really matter here right, because we are considering the dual anyways?
 
@TedShifrin not walking, I hope.
 
Though I guess we still have $C_0(X)^*\cong C(X_\infty)^*/C_0(X)^\perp$
Maybe unicity is still preserved if we restrict the corresponding Radon measure $\mu\in M(X_\infty)$ to $X$
which would boil down to showing that:
if for all $f\in C_0(X)$: $\int_X f d\mu_1=\int_X f d\mu_2$, then $\mu_1=\mu_2$
on compact sets they coincide
which implies that $\mu_1=\mu_2$ because they are Radon
(I should check that this restriction works as I expect it to work)
 
10:45 PM
It seems to work out, since $C_0(X)$ is a closed subspace of $C(X_\infty)$, and the restriction of a $\sigma$-finite Radon measure to a Borel measurable subspace is again a Radon measure
and the Radon measures from the Riesz representation theorem are finite, so things should work out
hm, something looks wrong, because this seems to imply that $C_0(X)^*\cong C(X_\infty)^*$ (here I mean the canonical map), which shouldn't be true, since $C_0(X)^\perp\neq 0$, by Hahn Banach
oh maybe this goes wrong: if $\mu_1\vert_X=\mu_2\vert_X$, then we don't necessarily have $\mu_1=\mu_2$ (on $X_\infty$)
if that is correct, then we don't have $C_0(X)^*\cong C(X_\infty)^*$, so there is no contradiction
yea, that is it. we can perfectly well have $\mu_1(\{\infty\})\neq\mu_2(\{\infty\})$ while $\mu_1\vert_X=\mu_1\vert_X$. Simply start with a measure on $X$, and then define $\mu\{\infty\}$, which we can do freely
great, then I consider it solved
 
11:05 PM
good you were able to handle it
(I was too lazy to step in and try to guide you)
 
it was probably even better this way! (active vs passive learning) regardless, I am happy you were here and mentioned the one-point compactification
 

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