I want to prove $T(\varphi)=\sum_k \varphi^{(k)}(1/k)$ is a distribution on $(0,\infty)$.

First I'm trying to prove that the series converges. If the support is some $[a,b]$ then there is only finitely many terms. But what if support is $(0,a]$? ($(0,a]$ is compact in $(0,\infty)$ right?). Then it's an infinite series.

How to go from here?

Is there any relation between $\varphi^{k}(0)$ and $\varphi^{k}(1/k)$?

no (0,a] is not compact

its always finite by what you said, compactly supported smooth functions have support contained in [a,b] for some 0<a<b<+inf

then proving its a distribution is pretty much immediate by definition, since phi_n -> phi iff phi_n,phi are all supported in some [a,b], and all derivatives of phi_n converge uniformly to corresponding derivatives of phi on [a,b]

that was definitely my main question, namely, what function space are these "distributions" acting on. the answer will focus on that, and maybe follow trivially from that, unless your source is making very unusual choices.

$\mathcal D(0,\infty)$

i think there arent that many possibilities

usually its C_c(0,+infinity) viewed as a LF-space

well i'd certainly hope so, my only point was, the answer falls out of whatever it's defined to be.

i.e. as an inductive limit of C_c(Omega_k) for some exhausting sequence of compacta Omega_k, where each subspace has the usual topology

yeah

I was thinking that $(0,a]$ is closed in $(0,\infty)$ and bounded so by Heine Borel... I guess, here Heine Borel can't be applied

@PNDas compactness is intrinsic, it has nothing to do with any embedding into a larger topological space

find in your reference material the definition of \mathcal D (0, \infty). that is the key thing here. not so much whether (0, a] is compact, because it isn't. there's no "compact in [subspace]"

the "closed" in heine borel refers to the topology of R, not a subspace topology. that's ,aybe a subtle point and a good one to appreciate, but it's separate from the point here.

So, this distribution has no finite order.

yeah it has infinite order

yes, that sounds right.

Thank you.

The range of a real indefinite quadratic form on \Bbb {R}^n is all of \Bbb{R}.

Quadratic forms are continuous.

So I think the Darboux property is the best choice.

Indefinite implies Q(x) >0 for some x and Q(x) <0 for some x

Is there any method to check whether a quadratic form represents 0 non trivially?

Positive definite and negative definite doesn't represent 0 non trivially.

An indefinite form represents 0 as Q(x) >0 for some x and Q(x) <0 for some x implies Q(x) =0 for some x.

But we don't know x is trivial or not.

As an answer to this link chat.stackexchange.com/transcript/message/63554673#63554673), I will post this link arxiv.org/pdf/1911.07458.pdf as an answer I found on Arxiv.

This is also found here math.stackexchange.com/questions/4698750/…

Hey there. I tried to prove that whenever the graph is compact, the function is continuous: If $(x_n, f(x_n))$ is any sequence in $\mathcal G$ that converges to some $q \in X \times Y$, we have $q=(p,f(p)) \in \mathcal G$ for some $p \in E$, since compact subsets of metric spaces are closed, but then we have $\lim_{x \to p} f(x) = f(p)$, which proves continuity. I feel suspicious of this argument, is it correct?

Don't want to look at a solution, for sure, so just asking about my approach.

Sorry, I misspelled: it is not $E$, it is $X$; didn't notice, it was typed in a short form from my solution.

Happy Monday y'all

@noballpointpen Why (x_n, f(x_n)) converges?

@user977780 we picked any sequence which converges. Do you want me to look at the possibility that we cannot have any convergent sequence at all?

We could use Cauchy sequence here then, as in compacts it converges anyway.

I mean, considering the most trivial example for non-compact graph of $f(x) = -1$ (for $x \leq 0$) and $f(x)=1$ (for $x>0$), we have a limit point $(0,1)$ that is not in the graph set, and it is obvious that the function is not continuous. This was my idea to approach the problem.

A question from the final I gave last week:

> Complete the definition of the Intermediate Value Property: A function $f$ is said to have the intermediate value property on an interval $[a,b]$ if for every $M$ between $f(a)$ and $f(b)$, there exists some $c \in (a,b)$ such that _______.

One student's answer: $f(ck)$.

@noballpointpen How you choose a cauchy sequence?

@XanderHenderson :( f(ck) =M for k=1

@user977780 so my proof with such a random pick is flawed from the beginning then?

@noballpointpen Consider $f : [0,1] \to \mathbb{R}$ defined by $$ f(x) = \begin{cases} 0 & \text{if $x<1/2$, } \\ 1/2 & \text{if $x = 1/2$, and} \\ 1 & \text{if $x > 1/2$.} \end{cases}$$

The graph is compact, no?

@XanderHenderson the points $(1/2, 0)$ and $(1/2, 1)$ are limit points of the graph that are not in the set, so the set is not closed and therefore not compact.

@noballpointpen Oh, derp.

I had little closed dots in my head.

I'm an idiot.

Nevermind.

I'm tired of grading. :/

:'(

and it is affecting my brain.

It's ok

ARG! They are doing so badly!

I HATE asynchronous classes. :(

Oh, my jeezus! I have a student who can't even identify a point in quadrant 3. :(

That was supposed to be a gimmie!

:'(

@noballpointpen To show f is continuous, start with a sequence (x_n) that converges to x. Then show that f(x_n) \to f(x)

@XanderHenderson Hang in there captain, we're almost there.

Whelp... I just found a typo that not a single student noticed. And it really screws up the problem. :(

@noballpointpen I have found this.

@user977780 thanks for the link.

@XanderHenderson Bonus points?

can $A \subseteq B$ be interpreted as a homomorphism if both $A$ and $B$ are rings?

i'd like to apply unique homomorphism results to $\subseteq$ if this is the case

@shintuku yes, the identity mapping

No, the inclusion. If $A$ is actually a subring of $B$ .

right, $A$ is a actually a subring of $B$

alrighto, thanks!

@TedShifrin Heh.

Ugh... this semester was a slaughter. Out of 27 students that started in my precalc class, only 5 successfully completed the class.

Granted, I was down to 10 by the end of week 2 (I think that a lot of folk decided against asynchronous instruction very early on), but... yikes.

That's rough but it feels like that's becoming par across the board for math requisites for non-math majors.

I run an R class and it looks like I'll only have about 6 passing, maybe 7 out of a starting 20 class.

Okay... grades are due tomorrow.... I've been here since 4 this morning, and it is now 10:30. Three of four classes are done. I think that I will go home now, and deal with calculus tomorrow. There are only two students who took that final, so it should be easy (?).

Also, I had to submit a report for academic misconduct.

That upset me. I hate doing that kind of thing.

No one enjoys that, not in any era.

Yar... I'm going home.

Yup, I always felt like it was a slap in the face to me, although the students didn’t look at it that way.

Ok. I took a break and returned to the problem. We assume $x_n \to x$, and (to a contra.) $f(x_n) \nrightarrow f(x)$. We can find a subseq. with $(x_{n_i}, f(x_{n_i})) \to (x,y)$ ($y \neq f(x)$), and then $(x,y) \in \mathcal G$, which assumes $y=f(x)$, so $\perp$. This?

Yes, seems like a valid solution. I scrolled down the post given to me above, there is the same solution below the topological one.

How to find the Euler-Lagrange's equation for the functional $I[u]=\int_{B}(1+|\nabla{u}|^2) ^{\frac{1}{2}}$ ?

Just follow the usual procedures?

In case of I[y]=\int_{a}^{b} F(x, y, y') dx, Euler-Lagrange's equation is \partial{F}\partial{y} -\frac{d}{dx} [{\partial{ F}}{\partial{y'}]=0

My fun math problem has an answer and I can't tell if I should be excited about its simplicity or disappointed.

@user977780 Yes, and so?

I just realized that that solution is a little different; but the idea is the same, so I still think mine is correct. Could anyone then tell me if I missed some critical detail in my proof?

What are your hypotheses, exactly?

@TedShifrin is the question to me or the person with Euler-Lagrange's?

You

@TedShifrin $X$, $Y$ are metric spaces and $f$ maps $X$ to $Y$. We have that $X$ is compact and the graph $\mathcal G$ of $f$ is compact. We want to show that $f$ is continuous.

Now write your proof :)

Already did above.

@noballpointpen As I have told earlier, why the sequence (x_n, f(x_n)) converge?

He had a convergent subsequence.

Yes.

@user977780 you seem to address my old proof. Look above, I did another.

But we don’t know $Y$ is compact. So how did you get the convergent subsequence?

The graph is compact and we have a sequence $(x_n, f(x_n))$ in it.

We know $f(x_n)$ is not converging to $f(x)$. Now what?

This is the 🔑 step.

@noballpointpen This is the same proof given in the mentioned post.

I don’t see the point of the contradiction. It just muddies it.

Suggest me any reference on Euler-Lagrange's equation for multivariable function.

@user977780 you told me that I should start with $x_n \to x$ and prove that $f(x_n) \to f(x)$, and I did exactly this. Why not?

Maybe I don't see something obvious...

It’s the same. Just put in all the variables and component functions. You can see a brief discussion in section 4 of chapter 3 of my diff geo text (linked in my profile).

@TedShifrin Awesome :)

@noballpointpen See the second answer of this

@user977780 I already did and after wondered if the fact about projection compactness was crucial. See, we have a sequence $x_n \to x$, and since the function is well-defined, we have a sequence $(x_n, f(x_n))$ in the graph. I don't say anything about its convergence. I just use the fact, after this, that the graph is compact.

I have to go. Will be back in a hour or so and read your answer.

@noballpointpen So here's a question for you. If $(x_n,y_n)$ is a sequence in $X\times Y$, $x_n\to x$, and $y_{n_i}\to y$. When can you say that $(x_n,y_n)\to (x,y)$?

Is anyone else using the latest Firefox version and getting rendering issues with Mathjax: i.imgur.com/DCoQi2L.png ? I copy/pasted this code here mathb.in/75316

I have no issues with Firefox (on a Mac).

I did quit using Safari for other reasons.

Why is 2|a||b| < |a|^2 + |b|^2 ? for real a,b

False

er, for integers

sorry

something something binomial

You could have equality? This is just the fact that the square of a real number is nonnegative.

yea

Hm, italics seem not working.

noballpoint: the step (5), specifically where you say "where y \neq f(x)," could use slightly more justification. implicitly, what's (maybe) going on is from the fact that f(x_n) does not converge to f(x), you're finding some subsequence where f(x_n) stays more than some epsilon away from f(x) the whole time in the subsequence, and then a subsequence of that is convergent to something, and that something can't be f(x). all true, but could use explanation.

at the moment, it kinda sounds like you could take any subsequence of f(x_n) and obtain that desired property, which is not true. that isn't what you're saying, but without saying more about how the subsequence is chosen, you're relying on the reader to fill that gap. (the issue is that f(x_n) can fail to converge to f(x) without every subsequences of f(x_n) also failing to converge to f(x).)

@leslietownes It comes from the fact that a sequence converges in a metric space if and only if every its subsequences converge to the same point.

also in (6) the last bit isn't so much "by the definition of mathcal G" but by the assumption that mathcal G is assumed to compact, and hence closed.

@leslietownes compactness is used where I mention that $(x,y) \in \mathcal G$. The graph is defined to consist of all the points $(x, f(x))$, why not by definition, then?

@noballpointpen there's some subtlety there. "if z_n is a sequence in a metric space that does not converge to z, then there is w different from z in the metric space and a subsequence of z_n that converges to w" is not generally true in every metric space. (take R to be the metric space, z_n the sequence that is 0 when n is even, and n when n is odd. this sequence does not converge to 0, but there is no subsequence of z_n that converges to anything else.)

The set is compact.

here there's this compactness thing that's helping out, but again, if someone just skims (5), they don't expressly see how it's being used, other than "by compactness of G."

all i'm saying is that you could add more detail to more clearly connect the hypotheses to the conclusions.

Ok. Then I appreciate your help. I sometimes see this to be too pedantic... maybe, I am making myself problems in the future by this belief...

there's always some matter of opinion or taste in these things, but, i think everyone would agree that a proof (whatever that is) is slightly more than a list of true assertions. while none of the stuff you are doing above is particularly controversial, it's also maybe appearing in a context where convincing a reader that you know which dots you are connecting is as important, or more important, than connecting the dots.

You seem to be toxic and unable to answer this question - because I feel like I don't need to explain my live in order to understand the question.

And yes I'm asking because I don't know math or the answer.

here, that background characterization of convergent sequences, is maybe that thing. roughly, "if z_n doesn't converge to z, then there is a subsequence of it that converges to something else" is not quite what you get from negating "z_n converges iff every subsequence is convergent to the same thing." it's what you get if you additionally assume compactness, which rules out the possibility of having subsequences with no convergent subsequences.

@leslietownes I also would like to ask about this. Does your claim state that I need the fact that the y-projection set must be compact from $\mathcal G$ (which is true)? I thought I could use the fact of graph compactness only. It is indeed subtle!

noball: most of what i was saying above was specific to the structure of your proof, i think some of the things that appear to be needed in that argument are not needed in general. see e.g. moishe kohan's answer in math.stackexchange.com/questions/231797/…

Thank you.

if you just want to assume compactness of the graph, you seem to be able to get the full result if you assume only that X is hausdorff (which is automatic for metric spaces).

i note that moishe's answer assumes facts like "continuous bijection from compact space to hausdorff space has continuous inverse" to be known, which i might not assume in any context where i was being asked to prove that :D

@AnArrayOfFunctions This seems uncalled for. As far as I can tell, leslie's previous comment was certainly not directed at you.

ted: it seems to be a reaction to a series of comments on the question on the main site. none of whom, of course, are present here.

Strange way to ask for help.

noballpoint: :D we could discuss all day whether it is good pedagogy or bad pedagogy for rudin to include a bunch of general topology without ever telling his audience that it is general topology and not specific to metric spaces.

My intuition says that if there's a random coin flip and I always guess "heads" or I always guess "tails" or I guess "heads" or "tails" "at random," then in the long run my probabilities of being right must all be the same.

Prof. Shifrin, what do you think about my clarification of the argument we debated a bit a few hours ago?

Oh, I stopped paying attention; sorry. :)

I personally like the non-metric space proof for someone who knows the topology, as then it just depends on the projection being a closed map.

uh oh, somebody woke ted up

Yeah, it seems nicer for me too.

I'm always a fan of any appearance of the tube lemma in topology :D

i also like the non metric space proof. sequence-based proofs sometimes capture the 'intuition' of what's happening at the cost of considering cases, or passages to subsequences, or other choices, that take up space from a proof writing point of view but do not reflect logically necessary parts of the argument.

i like them better when you need them, e.g. when you can only get the thing that you want to reason about as a limit of some subsequence, and it turns out that not every subsequence will do.

Also, regarding the step (5), again. I understood that the thing I should have pointed is the compactness of y-projection; the question is not about the step, but of a more general setting: I didn't even bother to prove y-projection compactness (even beyond my posted steps, but I admit that I probably knew the result somewhere in the mind intuitively), could it be the case, in an attempt to prove any other statement, that the thing that seems to come naturally and logically, end up false?

You talked about topology, I didn't study it now beyond Rudin's second chapter.

I mean, I should have proved a result that is "in-between" and overstepped it without justification thinking that just from compactness of $\mathcal G$ I would have a required subsequence, which I thought out from the actual theorems and actual results.

that subsequence selection stuff is definitely using compactness, although whether you choose to locate it in the graph or the projection of the graph is kinda up to you. (in terms of the sequence argument, the idea is that for any f(x_m) you're talking about, there's also the point (x_m, f(x_m)) in some compact space that it's a part of, and you can use compactness of that space to pull out subsequences of f(x_m), whether or not you explicitly think of this in terms of the projection map.)

and yeah, switching back and forth between these perspectives is almost automatic in this context, but might not be in a more complicated setting, where i could see someone goofing up.

the difference is that here, once you think of filling in the detail, it's pretty clear how to add more information to do that.

Ok

If I have a double sum, am I allowed to swap the indices of the sums? The comment on my question here math.stackexchange.com/q/4699898/965485 suggests it

you are allowed to swap the order of the sums, although you may need care to adjust the bounds so that you still get what you want. e.g. if the bounds of the inner sum depend on something involved in the outer sum, you cannot just interchange the two sigmas, bounds and all; the thing won't make sense.

but if the bounds are just fixed numbers you can do that.

if you've met changing the order of integration in multivariable calculus, the attention paid to bounds is a lot like that.

Convergence in one order does not guarantee convergence in the other …

we need Fubini-like things.

oh, i was also assuming finite sums.

I cannot begin to read that. I leave it to anti-ChatJax leslie.

yes, this is fine. note that if the bounds were more complex you could not simply interchange n with m in "fixing" the bounds in the reversed order. it's helpful here that everything being summed is nonnegative. this avoids any issues that might otherwise arise from the order of summation

Sorry Ted, I know its bad.

i love it. it looks fine to me. :D

The "m != n" condition didn't mess up the swapping?

why would it? the set {(m,n) in Z^2: n neq m} is the same thing as {(m,n) in Z^2: m neq n}. my only observation was, if the condition on the inner sum is more complicated, you may have to pay close attention at this point to be sure that whatever you use as the bounds actually do describe the same set.

so for example it's not as simple as swapping m with n, although if you happened to do that in this case, you do get the same set.

ah ah

Hi :) Quick question: Is there a name for a group $G$ for which the torsion subgroup ${\rm Tor}(G)$ is nontrivial?

There's torsion groups, but they're when $G={\rm Tor}(G)$.

I can't seem to find anything online.

Up until a few hours ago, I thought torsion groups referred to any groups with nontrivial torsion subgroups.

I suppose "not torsion-free" works . . .

Yeah . . . That makes sense. There's not much point defining $\lnot A$ when you already have $A$.

I guess I answered my own question. Thank you nonetheless :)

WE HELPED

Feeling really sluggish this morning

Probably shouldn't have played video games all night