@AlexPetzke Can you be more specific about what you are planning to learn?

I recently posted an answer to a 7 year old question, anyone wanna give it a look because it was quite lengthy and I want to know whether it meets the criteria for good answer

Without sounding spammy

Consider $w=(1,0,0)$, $\alpha=\{(1,0,0),(0,1,0),(0,0,1)\}$ and $\beta=\{(1,0,0),(0,2,0),(0,0,3)\}$

Yes, I see this example helps me to understand better what I am trying to ask. Now if we suppose that $w$ isn't in $\alpha$ or $\beta$?

consider $2w$

I see, thanks

:)

@AlessandroCodenotti what's the use of fixed point property in continuum theory?

Please do not call people out by name.

@XanderHenderson you mean username?

@Jakobian Are you confused about the meaning of the statement that I made, or are you just being pedantic? Don't. Call out. Other. Users.

Focus on the content, not the people.

What do you mean, how else am I supposed to call them

@XanderHenderson alright. If you're deleting everything then you probably should delete the link too. Thanks

@Jakobian Again, the point is that it is fine to call out bad content. It is not okay to attack other users.

@XanderHenderson No I was just confused. Because you said "by name".

So I thought the problem is the name

I wouldn't say I was being pedantic, I just understood it too literally

I see your point and admit my fault

@SoumikMukherjee I'm fine with introductory group theory and at the moment I'm casually working through some group theory qualifying exam problems, if that gives any sense of the level. My plan is to dig into group theory, especially local analysis. More than anything it would be fun to connect to someone else teaching themselves math because I'm pretty isolated from the math community (beyond MSE).

@AlexPetzke Oh okay, interestingly I am currently reading a book called Global Analysis. But I am also willing to learn more group theory.

I'm working the following exercise. Prove directly that if $$K_n(s)=\begin{cases}n &\text{if }|s|\leq1/(2n)\\ 0 &\text{if }|s|>1/(2n),\end{cases}$$ and $f$ is a continuous function at the origin, $$\lim_{n\to\infty}\int_\mathbb{R}K_n(s)f(s)ds=f(0).$$ For starters, I'm wondering whether this exercise actually means for $f$ to be continuous in a neighborhood of the origin, because otherwise the integrand will never be (Riemann) integrable. Do you agree?

The exercise is from Fourier Analysis and its Applications by Vretblad.

I mean... are we working with Riemann integrals?

Yes.

Then assume $f$ is Riemann integrable near the origin

right, then the exercise makes sense

Your concern is a valid one

but maybe not necessarily we want to fix it by assuming that $f$ is continuous in a neighbourhood of $0$, thats pretty strong

@Jakobian I've been given a solution which I do not quite understand. Maybe you can take a look at it. It assumes I think the function to be simply continuous at $0$, not near $0$.

By definition:

$$\tfrac1n \inf_{|x| \leq \tfrac 1{2n}} f(x) \leq \int_{-1/2n}^{1/2n} f(x)\,dx \leq \tfrac1n \sup_{|x| \leq \tfrac 1{2n}} f(x).$$

And by continuity of $f$ at $0$:

$$\lim_{n \to \infty} \inf_{|x| \leq \tfrac 1{2n}} f(x) = f(0) = \lim_{n \to \infty} \sup_{|x| \leq \tfrac 1{2n}} f(x).$$

well... uh... I guess its valid but, yeah

Is that the correct definition of continuity?

well its not definition

is it some proposition? THose are not limsups and liminfs, right?

I'd just write $\int K_n(s)f(s)ds - f(0) = \int_{-\frac{1}{2n}}^{\frac{1}{2n}} n(f(s)-f(0))ds$

@psie lets not focus on some person's attempt at a solution

So now $|\int K_n(s)f(s)ds - f(0)| \leq \int_{-1/(2n)}^{1/(2n)} n\cdot |f(s)-f(0)|ds \leq \varepsilon $ where $n$ is chosen to be small enough so that $|f(s)-f(0)|\leq \varepsilon$ for $|s|\leq 1/(2n)$

you can just pick an epsilon

@psie technically they are

$\limsup_{x\to 0}$ and $\liminf_{x\to 0}$

its not something I'd write tbh

@Jakobian ok, do you mean to choose $n$ large enough?

@psie yeah. $n$ to be large enough so that $1/(2n)$ is small enough

ok 👍

so it seems like the requirement that $f$ be continuous at $0$ is totally fine

i.e. we do not require continuity near $0$

yeah just at the point

you require integrability, for Riemann integrals this requirement might suck to force

but for Lebesgue integrals, if $f$ is assumed to be measurable, then since it's bounded near $0$, it will be Lebesgue integrable near that point

@Jakobian This still confuses me though, because the expression $$\lim_{n\to\infty}\int_\mathbb{R}K_n(s)f(s)ds=f(0),$$ is saying that for large enough $n$, the integral of $n\cdot f(s)$ over $[-1/(2n),1/(2n)]$ is close to $f(0)$, right? So somehow, if we're working with Riemann integrals, we need to require $n\cdot f(s)$ to be Riemann integrable over some small neighborhood around $0$.

@psie we do require that

well, we require that for $f$ because who cares

$n\cdot f$ or $f$

so the exercise is missing something

yes we've already discussed that

I'm just repeating myself

@psie just to understand the solution: what is this saying? That the limit of the integral $\int_{-1/2n}^{1/2n} f(x)\,dx$ is $0$? What does that say?

@psie $\int K_n(s)f(s)ds\to f(0)$

@psie but here we have $\frac{1}{n}$ both in the lower and upper bound, which tends to $0$ as $n\to\infty$...

so $\int_{-1/2n}^{1/2n} f(x)\,dx$ is squeezed between $0$, right?

$\int K_n(s)f(s)ds$ the same as $\int_{-1/2n}^{1/2n} f(s)ds$?

no

then what is it

right, how does $\int_{-1/2n}^{1/2n} f(x)\,dx\to 0\implies \int K_n(s)f(s)ds\to f(0)$?

why are you claiming things no one said

because you're confused I guess

No, no one is saying that

answer my question, what is $\int K_n(s)f(s)ds$

it is $\int_{-1/2n}^{1/2n} nf(x)\,dx$.

and what is it in terms of $\int_{-1/2n}^{1/2n} f(s)ds$

$n$ times that

right

@psie so what is this inequality saying

in terms of $K_n$

aha! that $\int_{-1/2n}^{1/2n} nf(x)\,dx$ is squeezed between $f(0)$?

that its bounded from above and below by sequences that converge to $f(0)$

no one is saying that $\int_{-1/2n}^{1/2n}f(s)ds\to 0$ implies anything, this conclusion is too weak for our purpose

ok

its the other way around, $\int K_n(s)f(s)ds\to f(0)$ implies $\int_{-1/2n}^{1/2n}f(s)ds \to 0$, if anything

ok, happy new year to you!

here it's raining :)

sure, happy new year

I'm not sure why this was downvoted. Would anyone like to suggest improvements, please?

Could someone point me to some references to understand the material in this question? math.stackexchange.com/q/4834535/1146260

Happy New year!!

I’m on mobile but Happy new year everyone

In this year I learned that the Riemann zeta function for can be thought of as a sum of discrete orbits of a vector field which can then be continued to the entire complex plane aside from a pole

Suppose $f$ is piecewise continuous and bounded on $\mathbb R$, and $\lim\limits_{t\to 0^-}f(t)=1$ and $\lim\limits_{t\to 0^+}f(t)=3$. Then I need to find $$\lim _{n\to \infty }\frac{n}{2}\int _{\mathbb{R}}^{ }e^{-n\left|t\right|}f\left(t\right)dt.$$

My initial attempt was to split up the integral around $0$, so to get rid of the absolute value in the exponent, and then use partial integration, but I realized that I do not have an expression for $f'(t)$. Any ideas on how I can tackle this problem?

I don't know if it helps, but $\frac{n}{2}e^{-n|t|}$ is a positive summability kernel.

could you answer the question if the limits were 0 and 0, respectively?

@leslietownes yes, because then it would be continuous at 0, and hence we'd get $f(0)=0$ I think, right?

I'm referring to a theorem in my book when I say "we'd get $f(0)=0$..."

@psie Letting $u = tn$ we have $\lim_n (1/2) \int e^{-|u|}f(u/n)du$

I believe this helps

Because now you can just use Lebesgue's dominated convergence theorem

You have $\lim_n (1/2)\int ... = (1/2)\int_{-\infty}^0 e^{-|u|}du + (1/2)\int_0^\infty 3e^{-|u|}du$

$2$?

yes

the Lebesgue dominated convergence theorem seems pretty advanced. Is there any alternative route?

its pretty basic

its just that Riemann integrals are outdated

I'm not going to come up with other method, but I can check your solution

psie: so if you let g be the function that is 3 for x >= 0 and 1 for x < 0 and consider the corresponding limit of f - g, where f satisfies your hypotheses, you get 0. can you compute the limit for g?

hmm, I'm not sure I understand. The limit of g will be the same as the limit of f, will it not?

The limit of the integral with $g$ instead of $f$.

aha

let me think

The standard way to approach these problems (not needing to know about Lebesgue integrals) is to show that as $n\to\infty$ the integral is more and more concentrated in a small interval $[-\delta,\delta]$ and then use the $\delta$-$\epsilon$ definition of (almost) continuity.

But Leslie is showing you a different approach, which you should understand.

i should have said 'the corresponding limit for f - g' instead of 'the corresponding limit of f - g'

So, have the Munchkins made their New Year's wishes?

the general idea being to consider the limit as a function of "f" and look at properties of that function

ted: one of them just yelled for a new diaper. does that count?

I should have resolved to say resolutions, instead.

Are you the duly appointed diaperer?

i was that time. he's resolved to yell more.

And "adult" Munchkin?

she's still asking for presents. i'm not sure she's clear on the concept of a new year

Boy, she probably wouldn't even be happy with the eight days of Chanukah.

@leslietownes ok, so we'd get $$\lim_{n\to\infty} \left(\frac{n}{2}\int_{-\infty}^0 e^{nu}du + \frac{n}{2}\int_0^\infty 3e^{-nu}du\right)=\frac12+\frac32=2$$ But can we simply replace $f$ with $g$ (as you specified it) in the integral?

psie: i don't know what other stuff you might mean by "simply replace f with g" so i will rephrase. if f was your function satisfying the original hypotheses, then f = (f - g) + g, so if you were to multiply both sides by your kernel [as a function of n] and integrate, then you've shown (by a theorem in the case of f - g, by evaluation in the case of g) that the RHS tends to a limit as a function of n

ok, thank you, this makes more and more sense

taking one step back, the vibe is that because you have one of those kernel dealies, the value of the limit as n goes to infinity of integral kernel(n,t) f(t) dt [when it exists] should only depend on the limiting behavior of f at 0, so at a vibe level, yes, for purposes of computing the limit, you can replace the otherwise unknown f(t) satisfying those hypotheses at 0, with a simpler function that has the same behavior at 0 where you can actually evaluate the integrals

or, if the question is "can we simply replace f with g," at the vibe level the answer is yes, and at the theorem level there's your theorem about what the limit is for continuous functions at 0, and the linearity of the integral, and linearity of limits, etc.

👍

so in general if you had L and R as left and right hand limits, the integrals would converge to (L+R)/2, the average of the two best candidates for a value of the function at the point. sort of generalization of the formula for continuous functions

you see this sort of behavior even with things that might not quite be positive summability kernels. maybe that is the next section of your text :)

waits impatiently for the Gibbs phenomenon

why would $dg_x = df_x$?

I don't see a reason why $x\in H^m$

oh wait

$\pi(x) = 0$ means precisely $x\in \partial H^m$

that was a little dumb

I have a basic question (from one of my earlier problems). If a function $f(x)$ is continuous at $x=0$ and Riemann integrable in a neighborhood around the origin, does this imply that the function is continuous in a neighborhood around the origin? I know Riemann-integrable over $[a,b]$ does not imply continuity on $[a,b]$, but here we have the extra condition that $f$ is continuous at $0$. I can't think of a counterexample.

@psie no

Every Riemann integrable function has lots of continuity points, so every Riemann integrable function satisfies this for some point

In measure theory terms we say that the set of points where the Riemann integrable function is discontinuous is of measure zero

this doesn't show anything since I haven't gave you a counter-example, but Thomae's function is such a counter-example

Its Riemann integrable, continuous at every irrational number, and discontinuous at every rational

interesting 👍

@Jakobian by satisfies this I mean, uh, it has a continuity point

I'm not sure what I was trying to say

Sanity check: consider $f(o,x) \mapsto g(x)$ where we sum over a discrete and infinite set $o\in O$ to obtain $g(x)$. Am I correct to conclude that this mapping is linear?

because i know that we can sorta do this in the cont. case with integral transforms..

and those guys are linear always

what do you mean?

the statement of sanity check makes little sense

$$ f(o,x) \mapsto \sum_{o\in O}f(o,x)$$

boom

still makes no sense

why

you have $o$ as a bound variable and as a variable you index over

perhaps you meant $$f\mapsto \sum_{o\in O} f(o, \cdot)?$$

oh yeah that is what I meant...sorry

For every $f$ and $x$, is the amount of $o$ such that $f(o, x) \neq 0$ finite?

at $x=1/k$ the sum blows up but elsewhere converges

for say $k=e$

what does that mean

it blows up?

blows up meaning is not finite

positive infinity?

yeah

is $f$ a non-negative function then?

yeah

how can we talk about linearity then

Well anyway since I already know its positive real numbers, I'm able to say that $\sum_{o\in O} f(o, x) + \sum_{o\in O} g(o, x) = \sum_{o\in O} (f(o, x)+g(o, x))$ for all $x$

and similarly, $a\cdot \sum_{o\in O} f(o, x) = \sum_{o\in O} a\cdot f(o, x)$

since $f$ are non-negative functions, we can't quite talk about linearity, but you can say the function is closed under addition and multiplication by non-negative numbers

With the convention that $0\cdot \infty = 0 $, of course

gotcha, yeah that makes sense now.