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1:42 AM
Very weird question for me...seems simple: Take $r = \min_{0 \leq j \leq m} r_j$. Am I missing something?
EE18: not that i can tell. note, for purposes of quibbling with exposition, that it would have been nicer prose for him to put "for some r > 0, one has B(a,r) subset..." instead of putting the quantifier after
in the general spirit of "it's nice to first introduce what a symbol represents before you make other statements about it"
2:12 AM
I agree, I hate when they switch up the order
Just a weird question...their questions are usually not one liners like this
Are (1) and (2) necessary here? Isn't a ring automorphism in particular a field auotmorphism when between two fields?
2:53 AM
i don't know what you mean by "necessary," those are properties of complex conjugation, they just, are that way. did you mean to paste something else?
if you're thinking of replacing conjugation with a more general map f, you can probably deduce f(0) = 0 from other usual assumed properties of homomorphisms, but f(1) = 1 is something you usually need to add in
(e.g. the zero map is both additive and multiplicative)
also, FYI, wolfram is written with, uh, minimal care for consistency within itself, i would not expect it to be minimal in presentation of anything
minimality in the sense of "can this be deduced from some subset of the things i am assuming" or worse "can this hypothesis be deduced from shorter list of assumed things" is kind of its own rabbit hole and not worth spending too much self-study time on
@ThomasFinley not sure what you mean by motivation
But yes, that theorem is useful
@leslietownes What I mean is that (1) and (2) are not needed in the definition of field automorphism right (I looked around and more and am now clear on that after a main site answer)
I think $f(1) = 1$ follows from $f(a) = f(a1) = f(a)f(1)$ so that by uniqueness of the element obeying $xf(a) = f(a)$ we have $1 = f(1)$
EE18: that condition only requires f(1) to 'act like a multiplicative identity' on the range of f. think of the zero map
i'm going to get out the warning sign of "please don't talk about 'rings without unity'" in case i need to tap it
@leslietownes Am not sure I understand what you mean Leslie? But just so I am clear on what I'm saying the claim is merely that f(0) = 0 and f(1) = 1 obtain for any ring automorphism between fields (where ring automorphism is defined merely as preserving multiplication and addition)
May I shamelessly ask for an examination of whether the following two questions are useful: first one I think no, it just seems to be basic verification but maybe this isomorphism is interesting? Second one vaguely reminds me of part of the proof of FTA from Hubbards' book
@EE18 automorphism also assuming surjectivity there (which you would need to do)?
i guess you could do that, but nobody does that. they usually define an automorphism to be a kind of homomorphism, and mapping 1 to 1 is baked into the notion of homomorphism for rings
even if you could deduce it from an assumption that you had a surjective additive and multiplicative map
3:10 AM
It does (automorphism = bijective homomorphism), but I don't quite see why that's relevant? Ohhhh or are you saying that I have to specify nonzero ring homorphism for the claim to hold
I think I see what you mean. This is getting into the waters of different definitions for ring homomorphism etc
OK, I won't worry too much here
EE18: you cannot deduce f(1) = 1 from "f(1) f(x) = f(x) for all x"
all the latter tells you is that f(1) y = y for all y in the range of f, where you'd want that for all y in the codomain of f
Just going back through my notes here to see what I was thinking, one moment. Sorry for the confusion Leslie
and the zero map is a great example of an additive and multiplicative map from any ring to any other ring that does not send 1 to 1
@EE18 the exercise is just a calculation. i guess if you weren't familiar with the "picture" of the multiplication of complex numbers, you could derive it from this picture (e.g. it shows that 'multiplication by i' in the complex algebra side is 'rotation by pi/2' in the 'linear map of R^2 side', which is most of what you need to start up that picture)
This is what I was thinking of with my earlier comment (as applied to the multiplicative group obtained from a field by excluding the 0 element)
the exercise with the estimate about |p(z)| seems kind of pointless outside of some wider context (as you mention, some proofs of the FTA, and other things about polynomials, involve this sort of estimate)
3:18 AM
@leslietownes This is interesting. I guess I see that by the $a = 0, b = 1$ case. Thanks for this, will include it in my notes but then ignore the actual work...I've done enough boring quesitons
EE18: if you remember the polar decomposition of a linear operator from a linear algebra class, and the polar form of a complex number, those two notions coincide in that picture :)
with the same lack of uniqueness in both sides
I am too far from my last LA text and too early in Hoffman Kunze to appreciate that sadly :(
there's more you can do with polar decompositions of operators on larger dimensions, where you can kinda write them like matrices like that
halmos had a good paper or part of a paper on it
i'm looking for it and can't find it, so maybe its in one of his books
or maybe it was all just a beautiful dream
Halmos tends to induce such dreams. THE MAN
3 hours later…
6:39 AM
Q: Given an inconsistent overdeterminate system AX=b where $A\in M_{m×n}(R)$ and $b\in R^m$ with rank A=n. Find the least square approx. solution of AX=b

Thomas FinleySuppose $A$ is a real matrix of order $m\times n$ with $m>n,b\in\Bbb R^m$ be such that the over determined system of linear equations $AX=b$ is inconsistent and $\text{rank} (A)= n.$ Let $W$ be the column space of $A$ and $b_0$ be the orthogonal projection of $b$ onto $W$ ($\Bbb R^m,\Bbb R^n$ are...

Can someone please help me with this?
@SoumikMukherjee Can you please take a look at this? :D
2 hours later…
8:33 AM
@ThomasFinley I'll see it later
2 hours later…
10:20 AM
Let's consider a Cartesian monometric orthogonal reference system in space S. Let's consider the following plane:
$$\pi:4x+2y+3z=0$$After choosing a point $P$ not belonging to the plane $\pi$, then:
**$(1)$ determine a line $r$ passing through the point $P$ and parallel to the plane $\pi$.**
I stopped on this thing
I can choose infinitely many points $P$ not belonging to the plane $\pi$. To simplify calculations, let's take $P=(1,0,0)$, since $4 \cdot 1 + 2 \cdot 0 + 3 \cdot 0\neq 0$, and thus $P \not\in \pi$.

There are infinitely many lines passing through point $P$ and parallel to the plane $\pi$. Let's write the parametric equation of the line:
x = x_0 + \lambda \rho \\
y = y_0 + \mu \rho \\
z = z_0 + \nu \rho
\end{cases}$$ Imposing passage through point $P$:
x = 1 + \lambda \rho \\
my doubt is : the equation admits infinite solutions, can I choose any values ​​for the 3 variables?
in the solution it says: choose the values ​​$ \lambda = 1, \mu = −2$ so the equation becomes $\nu = 0$
Is there a reason why these values ​​are chosen? Have they been obtained?
10:51 AM
I think I can choose any value because there are $\infty$ solutions, but could I also choose the values ​​of $\mu$ and $\nu$ and obtain $\lambda$?
x = 1 + \rho \\
y = -2\rho \\
z = 0
if for example I had chosen $\mu = -2$ and $\nu = 3$ then the parametric equation of the line would have been:
x = 1 -\frac{5}{4}\rho \\
y = -2\rho \\
z = 3\rho
Is it wrong to do this?
11:41 AM
@EE18 your bound is the wrong way. the sequence is convergent, so it is bounded.
2 hours later…
1:33 PM
@SoumikMukherjee Ok...
2 hours later…
3:56 PM
@Thorgott I still don't quite see why my original instinct was wrong, but this is the solution I did following Jakobian: chat.stackexchange.com/transcript/message/65708364#65708364
4:06 PM
@Thorgott its not in the wrong way
purpose of the bound is to bound $|x^{1/q}-x_n^{1/q}|$ by $C\cdot |x-x_n|$ where $C$ is some constant
when you use the formula $x^n-y^n = (x-y)(x^{n-1}+...+y^{n-1})$ or $x-y = \frac{x^n-y^n}{x^{n-1}+...+y^{n-1}}$ you need to bound the term $x^{n-1}+...+y^{n-1}$ but from below, not from above
because its in the denominator
@EE18 I don't like your solution so I didn't look at it in detail
it probably follows what I suggested but I just don't like the writing
its probably just your style that I don't like
Hope you don't mind me being honest
4:41 PM
oh, sure
thinking in terms of bounds is not necessary in the first place
cause we have a fractional expression with nominator converging to $0$ and denominator converging to something non-zero, so the fraction converges to $0$
that's how I would present it
5:13 PM
Can anyone take a look at what I wrote above pls, I just have that doubt
5:32 PM
@Thorgott I guess the work (at least for the uninitiated like me) relies on showing that the denominator does not converge to 0 right? Hence why it was necessary to show some fixed constant bounding the denominator from below?
@Jakobian No offense taken. My writing is not exactly the most parsimonious...it needs to improve lol
5:54 PM
@EE18 sorry, I confused the variables, it's not clear a priori the denominator converges
Still struggling with this guy here. I can use the info to show that $|p(z)| > 0$ for $|z|>R$ but I can't seem to show $|p(z) > R$. I'll post what I have so far I guess but i doubt it's useful
No worries at all :) there was a lot of notation going around
it is not hard to see that it is bounded below though: each summand is positive, so if the denominator were not bounded below, each summand would have a subsequence converging to $0$, forcing a subsequence of $(y_n)_n$ to converge to $0$, which would force a subsequence of $(x_n)_n$ to converge to $0$, contrary to assumption
Q: For $p\in\Bbb C[X]$ with $p_n=1$ show that $|p(z)|>R$ whenever $|z|>R$

Masacroso For $p\in\Bbb C[X]$ with $p_n=1$ show that $|p(z)|>R$ whenever $|z|>R$. Where $$R:=1+\sum_{k=0}^{n-1}|p_k|$$ (This is the exercise $11$ of Analysis I of Amann and Escher, on page $109$.) We have that $p$ is a polynomial of the kind $$p=X^n+p_{n-1}X^{n-1}+\ldots+p_1X+p_0$$ where the coefficient...

Got the right book and everything...can't believe I didn't find it when I searched
Maybe MSE search is better than google search
thanks very much Soumik
6:00 PM
I searched on google
This is what I got when i copy pasted the q lol: math.stackexchange.com/questions/1927806/…
maybe there were some extra characters messing the search up
if you're searching on google then use words, like I searched 'abs value of a polynomial is larger than sum of its coefficients'
If you're searching on approach zero then search by characters
I had never heard of approach zero
thank you again
@Thorgott yeah. That's why we bound it
I don't really like the argument by contradiction via subsequences here tbh. Its easy enough to just claim $x_n$ needs to be close enough to $x$ for large enough $n$, and the rest of the terms is just a finite amount
to say $x_n$ is bounded from below that is
that's why I was trying to suggest EE18 to prove there is $N$ such that $x_n > x/2$ for $n > N$
which is obvious to people that know any topology, since $(x/2, \infty)$ is an open neighbourhood of $x$
going from epsilon definition of convergence, taking $\varepsilon = x/2$
6:23 PM
Q: $A \in M_2(\mathbb{R}^2)$,$\langle x,y \rangle = x^TAy$ is an inner product iff $\alpha > 0, det(A) > 0$

eager2learnShow that given $A=\left( \begin{array}{cc}\alpha &\beta\\\beta&\delta\\\end{array}\right) \in M_2(\mathbb{R}^2)$, $\langle x,y \rangle = x^TAy, (x,y \in \mathbb{R}^2)$, defines an inner product on $\mathbb{R}^2$ if and only if $\alpha >0$ and det$(A)>0$. I tried to prove the $\Leftarrow$ direct...

@SoumikMukherjee You sent me this.
@EE18 helpful exercise for you. Suppose $x_n$ converges to $x$. Prove that this is the same to say, that for any open interval $U$ with $x\in U$, there exists $N$ such that $x_n\in U$ for $n \geq N$
However, I think the post has a serious typo. Shouldn't it be $M_{2} (\Bbb R)$ instead of $M_{ 2}(\Bbb R^2)$ in the theorem statement in the original post?
@SoumikMukherjee Thanks for confirming. I am fixing it, right away.
Oh, did you take a look at this: math.stackexchange.com/questions/4922735/…
@ThomasFinley To minimize $$ (Ax-b)^T(Ax-b)=x^T\left(A^TA\right)x-2x^TA^Tb+b^Tb $$ We need for all variations $\delta x$, $$ 2\delta x^T\left(A^TAx-A^Tb\right)=0 $$ which means $x=\left(A^TA\right)^{-1}A^Tb$.
6:27 PM
@ThomasFinley Sorry I have not, but you got an answer
@SoumikMukherjee :D yup, seems what I did is correct.
We are facing a cyclone rn
oh. hope you survive
@SoumikMukherjee really?! Is it creatin a havoc?
@Jakobian The open interval is a neighborhood so its immediate from the definition
6:29 PM
whats your definition?
Stay safe, @SoumikMukherjee
I thought you were using $|x-x_n|<\varepsilon$ one
The weather down here is pleasant
@robjohn Thanks for your insight! As an answerer seems to suggest my solution is valid, I want another clarification from some authoritative voice like you, whether is it really the case, I mean, Is my solution correct? What do you think about it?
Currently having heavy rain but i think i am in a safe distance from the eye, so won't get directly affected
@SoumikMukherjee Ah, great!
6:34 PM
That's the one Jakobian
oh okay so you already have that
well then my whole talk about taking $\varepsilon = x/2$ is useless
just take $U = (x/2, \infty)$ as a neighbourhood instead
Sure, but I guess there's still work in showing that indeed almost all terms are in U
Which I believe gets us back to epsilon
no this is literally the definition
$U$ is a neighbourhood of $x$
then $r = \min(x_1, x_2, ..., x_N, x/2)$ works as a positive lower bound for the sequence $x_n$
For each U in U(a), there is some some some N? what's that about?
@EE18 once you reach topology, there is no going back to epsilon deltas
6:50 PM
@ThomasFinley what allows you to multiply by $A^\ast$ to get the minimum?
Otherwise, it looks good
@leslietownes $\mathcal U(a)$ is the set of all neighborhoods of $a$
not sure if that was the question?
@robjohn I was using this above theorem. Sorry, I should have mentioned it. What do you think now?
hmm cant quite figure out the font, not mathcal or mathscr
EE18: joke, punning on interpreting the footnote 3 as a cubing of the word "some"
Your jokes might be too high-brow for lowly old me leslie
i did the old "scan the line" and didn't even see you wrote some three times
6:57 PM
Totally new to this, but let $S$ be a surface. Is the definition of $\int_S \omega$, where say $\omega=f(\alpha,\beta)d\alpha\wedge d\beta$ is a two-form, simply $\int_S f(\alpha,\beta)\,d\alpha d\beta$? If there is a parametrization $(u,v)$ of $S$ ranging over $D$, then I know that we get an integral over $D$ involving the Jacobian, but I'm wondering if it's correct to write $\int_S \omega:=\int_S f(\alpha,\beta) \ d\alpha d\beta$?
@robjohn Does it make sense to call my solution justified, now? If you need more context of this theorem, I can provide a proof of this in a screenshot but I don't think our original question in OP requires us to prove this, imho.
@SoumikMukherjee So the argument was that for some $N$ we get all $x_n > x/2$ (because $x_n \in U$ and then from there we just use that qth roots is an increasing function of its argument we get $(x/2)^{1/q}$ a lower boud on the $y_n$
I see. Thanks for clarifying Jakobian
@SoumikMukherjee Still kinda wilded out that the textbook would leave out n>1 as a requirement
seems moderately important...
What book is that?
Amann Escher Vol I
Was discussing yesterday w Leslie, I suspect this will get used in their proof of FTA
So n = 1 and n = 0 can be ignored via other arguments
but still...
@EE18 Yes, this is used in proof of FTA
7:30 PM
@psie I guess some authors prefer to write $\int_S \omega :=\int_S f|d\alpha d\beta|$.
I'm trying to make the connection to the Lebesgue integral, i.e. how the integral of a two-form becomes a Lebesgue integral.
I am asked to "Describe the function $C^\times → C^\times$ , z → 1/z geometrically. My sense is sort of vague...since $1/z = \overline z / |z|^2$ I can kind of say "flip across the real axis and scale down by magnitude", but that seems vague. Is there a better answer to this sort of thing?
8:15 PM
@ThomasFinley yes. Most of my answer was proving that theorem.
Is the following correct: One can show that for two $r,s \in K(X)$, $\underline{rs} = \underline{r}\cdot\underline{s}$ and $\underline{r+s} = \underline{r}+\underline{s}$ on the subset on which both $\underline{r}$ and $\underline{s}$ are defined (namely the intersection $U_r \cap U_s$).
The underlines denote the function in $K^{U_r}$ induced by $r \in K(X)$ by taking any element $\langle p ,q \rangle$ of the equivalence class in $r$ with $q(k) \neq 0$ for $k \in \Bbb K$ and setting $r(k) := p(k)/q(k)$
Sorry, I should mention that I defined the subset of $K$, $U_r$, for which there exists an element $\langle p,q \rangle$ with $q(k) \neq 0$
8:51 PM
EE18: 'describe geometrically' is such a vague instruction, i would need to read the surrounding sections of book to know what they even expect. it's pretty common for complex analysis books to discuss geometric properties/interpretations of moebius transformations (of which 1/z is one) in a way that would give you language to answer that. but without that context, it wouldn't make much sense as an instruction.
Totally fair
Ya there hasn't been enough discussion of complex analysis for me to answer this i feel like
EE18: if you aren't familiar with inversion in a circle as a general geometric thing, you can make sense of it without coordinates. en.wikipedia.org/wiki/Inversive_geometry
Here's I guess one thought: to your point about inversion, I can interpret $f: z \mapsto z/|z^2|$ as reflection across the unit circle right?
Then $z \mapsto 1/z = c \circ f$ where $c$ is the complex conjugation map
$c$ is a field automorphism which I've shown already
but i guess that's all useless elaboration of the same thing I said initially
9:18 PM
@EE18 yes. The nice thing about inversion is that it sends lines and circles to lines and circles.
Thank you robjohn :)
i dont understand this answer math.stackexchange.com/a/1337361/234396
It seems to be talking abt totality which if so is already defined for functions anyway so what distinguishes operations from functions
In this context, an operation is a function which has multiple inputs, and a single output.
An operation acts on a Cartesian product. That's what distinguishes it from other kinds of functions.
Think "addition", which acts on $X^2$.
A partial operation might be something like subtraction on the natural numbers, or division on the integers.
9:35 PM
How does it "act" on cartesian products?
And how come functions dont also do that?
"acts on" means "has as it's domain", as I have used it just now.
An operation IS a function.
It is a kind of function.
Every operation is a function. But not all functions are operations.
What is an example of when a function isn't an operation?
$x\mapsto x^2$.
In the spirit on the definition, operations take more than one variable.
They can take just one, like multiplicative inverses
The thing is, this is a highly context dependent definition. Why do you care about it right now?
9:47 PM
Which part is context dependent? Functions are typically pretty consistently defined thru mathematics I thought
@Obliv that really isn't the spirit of the definition. But if you didn't like that, the Identity function on $X^2$, where $X$ is your favorite space.
Function is well defined. Operation is not.
The term "operation" is trying to capture things like multiplication and addition.
Assume that A is finite and $f: A \to A$. Show that f is one-to-one iff ran f = A
Can't believe this is difficult to prove, but i can't think of anything other than an induction argument. All of the MSE answers seem to take things for granted like "f not injective implies |f(A)| < |A|"
they're super circular
See e.g. Asaf comments here: math.stackexchange.com/questions/4501772/…
But I'm miffed because i feel like Enderton wants me to be able to do this quickly given what he's developed to this point.
What is your definition of $|A|<|B|$?
@XanderHenderson I feel like I'd be able to see the difference better if I had the set theoretic definitions of the two idk
I don't even have one yet. Maybe I'll just keep thinking about it because I don't want to waste your time by posting all of the facts I have at the moment
If you do have access to Enderton these facts are in Chapter 6.2
(on Finite Sets)
9:53 PM
@Obliv You are being too rigid in your thinking.
An operation is a function which acts on a Cartesian product. In your head, you you should be thinking about addition and multiplication. The word is meant to distinguish these kinds of functions from more general functions.
And if you aren't working in that context, why do you care?
Why do you need to know what an "operation" is?
@Obliv Xander is just saying that, as a matter of convention, we use "operation" iff we have a function with the domain a Cartesian product of some sets (or, perhaps, some subset of a Cartesian product) while function can be defined set -theoretically as a relation obeying the single-valued proeprty
As used, operations are functions but the converse need not hold. But some people might have a sufficiently general definition of "operation" that, also, every function is an operation
@EE18 Without knowing what theory you have access to, it is hard to know how you prove the thing.
i feel like i'm forgetting something. suppose i have a set of vectors $v_1=(1,0,0), v_2=(x_1,x_2), v_3=(y_1,y_2,y_3)$. Is there a name for such a basis? The best I can think of is being "upper-triangular" in the sense that, if i take these as column vectors of a matrix, then said matrix is upper-triangular.
@Semiclassical I've never heard a name for such a basis.
But it isn't really my field.
looking around, the answers here link it to the concept of flags which is sorta neat: math.stackexchange.com/questions/3285513/…
but "sorta neat" is pretty thin gruel
10:00 PM
It sucks that I still can't see a difference without a formal definition. The identity map "acts" on a cartesian product of sets as well, but it's not an operation for some reason.
The codomain of an operation is generally a bare set, not a Cartesian product. That's the difference
Lots of these definitions come from books which are not as careful as a set theory text would be. I think Xander's advice is good: don't worry about words like "operation" outside of the given context you're working because it's so overloaded
@Obliv In what context do you care about this definition?!
@XanderHenderson Fair play. I will refrain from posting all the relevant theorems (8 or so this section) here
I've asked you that question three times, and you haven't answered it, yet...
@EE18 Where did you get that?
@XanderHenderson Set theory
10:07 PM
@Obliv You aren't being specific.
What, specifically are you studying? In the context of that thing you are studying, how has this definition come up?
One possible definition which would be used in something like Enderton: given a set $X$, an $n$-ary operation $n > 1$ (but you could choose $n \geq 1$ if you wanted for your definition) on $X$ is a function $X^n \to X$.
@EE18 That is definitely the spirit of the given definition.
@XanderHenderson I saw a relational definition of functions given in munkres so I wanted to extend that idea to have a relational definition on operations in set theory
But that definition fails to capture a lot of important things which I would want to call operations.
Such as subtraction on $\mathbb{N}$, or division on $\mathbb{Z}$.
@Obliv An operation is a function.
@EE18 ur codomain isn't explicitly a cartesian product as you claimed before
10:11 PM
It is a special kind of function, where the domain is a Cartesian product, and the codomain is "just" a set.
I understand that but that's also what a function is
But every function can regard its domain as a one-fold Cartesian product with itself, and its domain as a set---underlying products can be "forgotten".
The point is that "operation" is not a set-theoretic definition.
@EE18 $n=0,1$ should definitely be allowed
@Obliv I said precisely the opposite. The codomain is NOT a Cartesian product. That's what makes it an "operation"
10:12 PM
It is a definition in a more restricted context, where we care about they underlying structure of the sets. There is more structure than "just" sets.
Oh my bad
@Thorgott Granted, but that would be a matter of taste/what you want to capture with your definition no?
With the word "operation" that is
@EE18 if you haven't defined cardinality, how did you define finite
Bijection with some natural
@EE18 well, in any context (I've come across) where you want "operations" to look like $X^n\rightarrow X$, you want to allow $n=0,1$
@EE18 well, you have to know some other facts. like what kinds of injections or surjections can exist between finite sets of various sizes
10:19 PM
this seems interesting
can't think about a counterexample right now
@SineoftheTime It is a standard example. Does $L^2$ imply $L^4$?
The answer is "no".
Consider a function with a "spike".
maybe it's enough to take the standard countexample and 'smooth' it
E.g. $x \mapsto x^{1/2-\varepsilon}$ is square integrable but not quartic integrable for any small $\varepsilon$.
Ik I even wrote it in a comment but OP wants a smooth function
Well, that is a little different, depending on what is meant by "smooth".
$1/x$ is smooth on its (real) domain, in that it has continuous derivatives of all orders wherever it is defined.
10:24 PM
Even so, take my example, and put it on the interval $(0,1)$.
It is smooth there.
But if the function must be defined on all of $\mathbb{R}$, then the answer is "yes", because smoothness implies bounded, which prevents there from being spikes.
Choose some $\varepsilon > 0$. Since $f$ is square integrable, there exists some $r > 0$ such that $|x| > r$ implies that $|f(x)|<\varepsilon$. On $[-r,r]$, $f$ is continuous, hence bounded.
Then, assuming that $\varepsilon$ (above) is less than $1$, $$\int_{\mathbb{R}\setminus [-r,r]} f^2 > \int_{\mathbb{R}\setminus[-r,r]} f^4. $$
The guy on the left is finite (since $f$ is square integrable), so the guy on the right is finite.
And $$\int_{[-r,r]} f^4 \le \int_{[-r,r]} M^4 < \infty, $$ where $M$ is whatever (constant) bound of $f$ you like.
$f$ is bounded because it's smooth?
@SineoftheTime It is bounded because it is smooth on all of $\mathbb{R}$ and square integrable.
So smoothness (and square integrable) implies bounded. In precisely the way I outlined above.
I'm confused :)
@SineoftheTime Square integrable means that the "tails" go to zero. So for any $\varepsilon > 0$, there exists some $r$ such that $|x|>r$ implies that $|f(x)| < \varepsilon$. This is a consequence of (square) integrability.
10:39 PM
but if $f\in L^2$ we can't say that $\lim_{|x|\to \infty} f(x)=0$ right?
"Smooth" means that it is continuous. Continuous implies "bounded" on any compact set. The interval $[-r,r]$ is compact, so $f$ is bounded on that interval, i.e. there exists some $M > 0$ such that $|f(x)| < M$ for any $x \in [-r,r]$.
You can assume without loss of generality that $M > \varepsilon$, so $M$ is a global bound on $f$.
@SineoftheTime Sure you can.
Again, because the function is smooth, therefore continuous.
If it doesn't go to zero, then it isn't integrable.
Q: Square integrable function that doesn't go to zero?

MikhailI'm reading through some elementary quantum mechanics textbooks and a few authors mention that "there exist pathological functions that are square-integrable but do not go to zero at infinity." (Griffiths) I am having trouble coming up with one, does anybody have an example? The only one I can ...

Oh, there's the counterexample, then.
nate eldridge's answer is the most conceptually clear, imvho, although i guess for some students the devil would be in the details of "with a little more work..."
I think that I am in "complex analysis" mode, where "smooth" and "analytic" are the same thing.
Though I think that you could come up with something analytic with similar issues...
10:51 PM
it's also kind of conceptually weird to me to think about 'going to zero' in the sense of a pointwise limit on L^2, like, why would you do that.
i can imagine some old grandmother saying "Well of course if you do that, that'll happen, what do you expect" in the same tone she'd respond to you shooting your eye out with a bb gun
@leslietownes I dare you to like that frozen pole!
@leslietownes the bump function comes to the rescue
OK so I came up with one direction given what I have but the other direction is evading me
Suppose $f$ injective but not surjective. Then $f(A) \subset A$ (proper) so that by Lemma 6F we have $f(A) \approx m$ for some $m \in n$. It follows that $n$ injects into $m$: take the maps $\varphi: n \to A$ and $\psi: A \to m$ which obtain because of the equinumerosity between the given sets and consider $g:= \psi \circ f \circ \varphi$. $g$ is an injection because the composition of injections is an injection.
Further, it's easy to see that $g$ is a surjection: for any $l \in m$ there is some $a \in f(A)$ with $\psi(a) = a$. By definition of $f(A)$ we have some $b \in A$ with $a = f(b)$. And since $\varphi$ a surjection there is some $t \in n$ with $\varphi(t) = b$. Thus $g(t) = l$. But then $g$ is a bijection and so we've shown that $n$ is equinumerous to aproper subset of itself, contradicting the Pigeonhole Principle.
i do kind of wonder if there is some physical significance or consideration behind the existence of "lim_{x to infty}" on L^2 or if it's just a thing of "doctor it hurts when i do this"
The other direction I'll keep thinking, but these are the sorts of facts I have Thorgott
10:57 PM
i would ask the outside member on my thesis committee about stuff like that and there was always a really good answer
How would one suggest deriving a contradiction from there being a surjection of $n$ onto $n$ which is not an injection
you can easily circle around this drain of stuff for the rest of time, but, this is going to be some kind of proof by induction. if that formulation doesn't lend itself to induction, maybe logically rearrange it in some equivalent form where the inductive hypothesis is more useful
may i recommend the book concrete mathematics by graham knuth and patashnik, it has a lot of complicated problems in it, including many involving induction, but differs from enderton and amann-escher and whoever else in that if you work through them all you will actually know something
I will look into it it
I figured induction was around the corner :( hoped to avoid. Probably could w sufficiently clever use of what enderton has given me
literally anything about the naturals in enderton is going to be, even has to be, proved by induction
that's the cheat code to those miserable sections of that book
11:10 PM
For sure, I guess what I’m saying is that some of the lemmas I’ve been given use induction in the proof already
So I was hoping to massage them assuming that they encode already what I need
(This is Ch 6 on cardinality)
like if you could just start working through the first google match to 'concrete mathematics pdf' instead of this
while remaining in compliance with all laws of your jurisdiction of course
what a wonderful world it would be
11:46 PM
I'm not sure what formal statement of the pigeonhole principle you have in mind
it's all very highly specific to a sequencing of lemmas in enderton
a scarlet letter E should be attached to these questions
Perhaps a cursive E indeed
not a mathcal E but something similar enough to provoke confusion with it
My statement of PHP is "no natural is equinumerous to a proper subset of itself"
$\mathscr E$
hahah my five year old kid just used the f-word in reference to something that happened in a video game she was playing, and i can't even be mad about it, because it was contextually appropriate
her phrasing was "what the ---- was that"

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