TITLE: Why is the sample mean a random variable?
QUESTION [6 upvotes]: I know that a random variable is a measurable function from some measurable space $(\Omega, \mathscr{F})$ to some borel space $(R, \mathscr{B})$.
Suppose I am given some random sample $X_1, ..., X_n$. From the definition of a random sample, each $X_i:\Omega \rightarrow R$ are a measurable function with common domain the sample space $\Omega$.
I know that the sample mean is a random variable. Hence it is a measurable function.
So what type of function is the sample mean exactly? From which measurable space to where? And what exactly does the notation $\bar{X} = n^-1 \sum _i^n X_i$ mean given that it does not mean that $\bar{X} (s)= n^{-1} \sum _i^n X_i(s)$?
I ask because usually defining a function $f$ by $f = v+g$ where $g:A \rightarrow R, v:A \rightarrow R$ implies that $f(x) = v(x) + g(x) \forall x \in A$ holds.

REPLY [9 votes]: With this setup the sample mean is another measurable function $\Omega \to \mathbb{R}$ and it is just given by $\bar{X}(s) = \frac{1}{n} \sum_{i=1}^n X_i(s)$. The entire subtlety of this question, which you've glossed over, is how one actually defines the sample space $\Omega$ and the functions $X_i$ in it in general!
For example, suppose I want the $X_i$ to be $n$ independent samples from a normal distribution $N(\mu, \sigma)$. What is the sample space? It is not the sample space $\mathbb{R}$ of a single sample from a normal distribution. In fact it is $\mathbb{R}^n$, the product of $n$ copies of the sample space of a single sample, equipped with the product measure, and the $X_i$ are the $n$ coordinate projections $\mathbb{R}^n \to \mathbb{R}$. This construction is how we guarantee independence. So the sample mean is again another function $\mathbb{R}^n \to \mathbb{R}$ given by the mean of the $n$ coordinates.
Generally - and this is a surprisingly subtle point I've only seen explained well by Terence Tao, here and here - thinking of random variables as measurable functions on a fixed sample space is something of a distraction, because in probability theory we always retain the freedom to enlarge the sample space as necessary to accommodate new sources of randomness (e.g. in this case adding another sample $X_{n+1}$). As Tao says:

At a purely formal level, one could call probability theory the study of measure spaces with total measure one, but that would be like calling number theory the study of strings of digits which terminate. At a practical level, the opposite is true: just as number theorists study concepts (e.g. primality) that have the same meaning in every numeral system that models the natural numbers, we shall see that probability theorists study concepts (e.g. independence) that have the same meaning in every measure space that models a family of events or random variables [emphasis mine]. And indeed, just as the natural numbers can be defined abstractly without reference to any numeral system (e.g. by the Peano axioms), core concepts of probability theory, such as random variables, can also be defined abstractly, without explicit mention of a measure space [emphasis mine]; we will return to this point when we discuss free probability later in this course.

And:

In order to have the freedom to perform extensions every time we need to introduce a new source of randomness, we will try to adhere to the following important dogma: probability theory is only “allowed” to study concepts and perform operations which are preserved with respect to extension of the underlying sample space. (This is analogous to how differential geometry is only “allowed” to study concepts and perform operations that are preserved with respect to coordinate change, or how graph theory is only “allowed” to study concepts and perform operations that are preserved with respect to relabeling of the vertices, etc..)

As long as one is adhering strictly to this dogma, one can insert as many new sources of randomness (or reorganise existing sources of randomness) as one pleases; but if one deviates from this dogma and uses specific properties of a single sample space, then one has left the category of probability theory and must now take care when doing any subsequent operation that could alter that sample space. This dogma is an important aspect of the probabilistic way of thinking, much as the insistence on studying concepts and performing operations that are invariant with respect to coordinate changes or other symmetries is an important aspect of the modern geometric way of thinking. With this probabilistic viewpoint, we shall soon see the sample space essentially disappear from view altogether [emphasis mine], after a few foundational issues are dispensed with.

The sample space and measurable function are analogous to a representation of a group; it's a "representation" of a random variable, not the invariant content of the random variable itself (given by its CDF or its moments, for example). This can be made precise in various ways using algebras of random variables; see e.g. this old blog post of mine on noncommutative probability.


TITLE: On Elliot's classification of AF-algebras
QUESTION [2 upvotes]: I have two questions with a relation to Elliot's classification of Approximately Finite $C*$-algebras (from now on referred to as AF-algebras).
Elliot's classical result yields that whenever $A$ and $B$ are AF-algebras and $\sigma \colon K_0(A) \to K_0(B)$ is a an isomorphism of scaled ordered $K_0$-groups, then this isomorphism induces a $*$-ismorphism $\varphi \colon A \to B$. Furthermore, it holds that $K_0(\varphi) = \sigma$. A proof of this is to be found in Blackadar's "$K_0$-theory for operator algebras".
I have been told that the above result is in fact slightly stronger. Namely that if $\tau \colon A \to B$ is another $*$-isomorphism then $K_0(\varphi) = K_0(\tau) = \sigma$, meaning that the lift is in fact uniquely determined. How does one obtain this fact?
My second question is of a slightly different kind. I have found a result saying that whenever one have $*$-homomorphisms $\iota_1, \iota_2 \colon C \to D$ where $C$ and $D$ are finite-dimensional $C*$-algebras, then $\iota_1$ are $\iota_2$ are unitarily equivalent. Again, I have been told that this may be generalised in the following sense: If we instead suppose for $C$ and $D$ to be AF-algebras, then $\iota_1$ and $\iota_2$ are approximately unitarily equivalent. It is my impression that Wikipedia mentions this result here, but I cannot find any reference for it. Does anyone have a reference or an idea as to how one obtains the result?

REPLY [1 votes]: In your second question, I guess you mean $\iota_1,\iota_2 \colon C \to D$ are such that $K_0(\iota_1) = K_0(\iota_2)$.
Note that the following is true (Lemma 7.3.2. in "An Introducion to K-theory for C*-algebras" by Rordam, Larsen, Lausten)
Lemma. Assume $A$ is finite dimensional and $B$ is a unital C*-algebra with the cancellation property (for example an AF-algebra). Then the following holds: If $\varphi, \psi \colon A \to B$ are $*$-hom., then $K_0(\varphi) = K_0(\psi)$ if and only if there exists a unitary $u \in B$ such that $\varphi = u^* \psi(-) u$.
This lemma  immediately answers your second question.
It also answers your first question: Whenever $A,B$ are AF algebras and $\varphi,\psi \colon A \to B$ induce the same map on $K$-theory, $\varphi$ and $\psi$ are approximately unitarily equivalent.


TITLE: How to find the asymptotic distribution of $Y_{n} X_{n}$
QUESTION [1 upvotes]: Given that $X_{n}$ converges in probability to $2$.
Also if $Y_{n}$ converges in distribution to $N(0,5)$.
Then what is the asymptotic distribution of $Y_{n} X_{n}$?
If $X_{n} \overset{p}{\rightarrow} 2$ then this implies that $X_{n} \overset{d}{\rightarrow} 2$
And then if $Y_{n} \overset{d}{\rightarrow} N(0,5)$, then does this mean that $$Y_{n} X_{n} \overset{d}{\rightarrow} 2N(0,5)=N(0,10)?$$

REPLY [1 votes]: $g(x) := f(2,x)$ is a continuous bounded funtion for every continuous bounded function $f:\Bbb R^2 \to \Bbb R $. Thus due to $Y_n \to  Z \sim \mathcal N (0,5)$ in distribution
$$\Bbb E[f(2,Y_n)] = \Bbb E [g(Y_n)] \to \Bbb E[g(Z)] = \Bbb E[ f(2,Z)]$$
Thus $(2,Y_n) \to (2,Z)$ in distribution.
Furthermore for any $\varepsilon > 0 $ we have
$$\Bbb P ( \Vert (X_n , Y_n) - (2, Y_n) \Vert > \varepsilon) = \Bbb P (\vert X_n - 2\vert > \varepsilon) \to 0$$
For a continuous $f$ with compact support (thus uniformly continuous) and $\delta , \varepsilon > 0$ such that $\Vert (x,y) - (u,v)\Vert \le \varepsilon \Rightarrow \vert f(x,y) -f(u,v)\vert < \delta$
$$\vert\Bbb E [f(X_n , Y_n)] - \Bbb E[ f(2,Z)] \vert \\\le \vert \Bbb E [f(2, Y_n)] - \Bbb E[f(2,Z)] \vert + \vert \Bbb E[f(X_n , Y_n )] - \Bbb E[f(2, Y_n)] \vert\\\le \vert \Bbb E [f(2, Y_n)] - \Bbb E[f(2,Z)] \vert + \Bbb E[\vert f(X_n , Y_n )-f(2, Y_n) \vert ; \Vert (X_n , Y_n) - (2, Y_n) \Vert > \varepsilon]+ \Bbb E[\vert f(X_n , Y_n )-f(2, Y_n) \vert ; \Vert (X_n , Y_n) - (2, Y_n) \Vert \le \varepsilon]\\\le \vert \Bbb E [f(2, Y_n)] - \Bbb E[f(2,Z)] \vert + 2\Vert f\Vert_{\infty}\Bbb P( \Vert (X_n , Y_n) - (2, Y_n) \Vert > \varepsilon)+ \delta
$$
All in all
$$\limsup_{n\to\infty} \vert\Bbb E [f(X_n , Y_n)] - \Bbb E[ f(2,Z)] \vert \le \delta \to 0 $$,
as $\delta\to 0$. Thus $(X_n , Y_n) \to (2, \mathcal N(0,5))$ in distribution by the portmanteau theorem.
Now for a continuous bounded function $f:\Bbb R \to \Bbb R$, the function $g(x,y) := f(xy)$ is continuous and bounded. Therefore
$$\Bbb E[f(X_n Y_n)] \to \Bbb E[f(2Z)]$$
and thus $X_nY_n\to 2Z \sim \mathcal N (0, 20)$ in distribution
By the way: In my notation $Z\sim \mathcal N(0,5)$ means $\Bbb E[Z^2] = 5$


TITLE: Proof of $\Phi^\alpha =\{x\in\mathbb R\mid f(x)\leq \alpha \}$ bounded imply $f$ lower semi continuous.
QUESTION [1 upvotes]: Let $f:\mathbb R\to \mathbb R$. We suppose that for all $\alpha \in\mathbb R$, $\Phi^\alpha =\{x\in\mathbb R\mid f(x)\leq \alpha \}$ closed. Prove that $f$ is Lower semi continuous. I think that my proof is very complicated, and I was wondering if there is easier or not ?

Proof Let $(x_n)$ a sequence that converge to $x_0$. Suppose by contradiction that $\liminf_{n\to \infty }f(x_n)<f(x_0)$. Set $y_n=\inf_{k\geq n}x_k$ and $\ell=\liminf_{n\to \infty }f(x_n)$. Let $\alpha \in\mathbb R$ s.t. $$\ell<\alpha <f(x_0).$$
By definition of $y_n$, for all $n\in\mathbb N^*$ there is $k_n\geq n$ s.t. $$f(x_{k_n})\leq y_n+\frac{1}{n}\leq \ell+\frac{1}{n},$$
where the last inequality comes from the fact that $(y_n)$ is increasing. So, let $N$ s.t. $\frac{1}{n}\leq \alpha -\ell$ for all $n\geq N$. In particular, $$f(x_{k_n})\leq \alpha <f(x_0),$$
for all $n\geq N$, and thus $(x_{k_n})_{n\geq N}$ is a sequence of $\Phi^\alpha $ that doesn't converges in $\Phi^\alpha $. Contradiction.

I think my proof is quite complicate. Is there easier ?

REPLY [1 votes]: A very little simplification could be to remark that $$y_n+\frac{1}{n}\underset{n\to \infty }{\longrightarrow } \ell<\alpha,$$
and thus, there is $N$ s.t. $$f(x_{k_n})\leq y_n+\frac{1}{n}<\alpha $$ for all $n\geq N$. So no need to use the fact that $(y_n)$ is increasing.   But except this point, I'm not sure there is an easier proof. Well, I don't think that it's a very complicate proof neither...


TITLE: How can we show this inequality?
QUESTION [2 upvotes]: Let $E$ be a nonempty set and $d,\rho,\rho_r$ be metrics on $E$ with $$d\le1+\beta\rho\tag1$$ for some $\beta\in(0,1]$. Assume there are $k,\delta>0$ such that $$d(x,y)=1+\beta\rho(x,y)\;\;\;\text{for all }x,y\in E\text{ with }\rho_r(x,y)\ge\delta\tag2$$ and $$d(x,y)\ge1+\beta k\tag3$$ for all $x,y\in E$ with $\rho(x,y)\ge k$ and $\rho_r(x,y)\ge\delta$.

How can we show that $$1-\alpha+\alpha d(x,y)\le\underbrace{\left(\frac{1+\alpha\beta k}{1+\beta k}\right)}_{=:\:\alpha_1}d(x,y)\tag4$$ for all $x,y\in E$ with $\rho(x,y)\ge k$ and $\rho_r(x,y)\ge\delta$?

It guess we need to use $(3)$ to obtain $(4)$. But the only thing I'm able to obtain $$1-\alpha+\alpha d(x,y)\ge\alpha_1(1+\beta k)\tag5$$ and $$\alpha_1d(x,y)\ge1+\alpha\beta k\tag6$$ for all $x,y\in E$ with $\rho(x,y)\ge k$ and $\rho_r(x,y)\ge\delta$, which is not useful.

REPLY [1 votes]: Multiplying both sides of the inequality you want to obtain by $1+\beta k$ and then removing $\alpha\beta kd(x,y)$, we see that it is equivalent to prove
$$
(1-\alpha) (1+\beta k) + \alpha d(x,y) \leq d(x,y)
$$
And this is exactly inequality (3) multiplied by $(1-\alpha)$.


TITLE: Modulos race, which formula reach 100 first?
QUESTION [5 upvotes]: I would like to know if there was a way to determine according to this formula:
\begin{equation*}
\sum_{i=0} \frac{A*i+B \pmod{100}}{100}
\end{equation*}
and the same with different values of A and B
\begin{equation*}
\sum_{i=0} \frac{A'*i+B'\pmod{100}}{100}.
\end{equation*}
If there is a way to determine which one reach 100 the fastest?
Assuming $A, B \in \Bbb N$ and $ A \neq 0$

REPLY [2 votes]: Not a full answer, but it will sometimes solve the problem.  We can recast the question to be for what $n$ does $\sum_{i=0}^n (Ai+B) \pmod {100}$ reach $10000$  If $A$ is coprime to $100$, when we have summed $100$ terms we will have one each of $0$ through $99$, for a sum of $4950$.  After two cycles, so at $n=199$ the sum will be $9900$  We then have to ask after how many more terms we have accumulated the missing $100$.  I don't see a simple way to get there.
If $A$ is not coprime to $100$, let $d=\gcd(A,100)$.  After $\frac {100}d$ terms we have gone through a cycle.  The $Ai$ terms will contribute $d\frac 12\cdot \frac {100}d(\frac {100}d-1)=50(\frac {100}d-1)$ to the sum, while the $B$'s will contribute $(B \pmod d)\frac {100}d$  For example, let $A=35$, so $d=5$.  Each cycle of $20$ will get $0,5,10\dots 95$ from the $Ai$, summing to $950$.  We also get $B \pmod 5$ from the $B$'s, so if $B=8$ this will contribute another $60$ and we get a total of $1010$ from each cycle of $20$  That means that $10$ cycles of $20$ terms $(n=199)$ will sum to $10100$ and we need to stop a few terms earlier.  Again I don't see a simple formula to determine how many terms earlier.  The one simple observation is that you want $B \pmod d$ to be large to get there quickly.


TITLE: A graph that is tough but not Hamiltonian.
QUESTION [2 upvotes]: A graph G is tough if the number of $c(G-s) \le |S|$ for all $S \in V(G)$.
And I read that every tough graph is Hamiltonian, but the other way is not true. I was wondering if there is an example of a graph that is tough but is not Hamiltonian.
I think Petersen's graph is one example, but are there any other such graphs?
Any help would be appreciated.

REPLY [2 votes]: Chvátal's original paper defining toughness gives an example of a $7$-vertex graph which is tough but not Hamiltonian:

This is easy to verify, since a Hamiltonian cycle would have to use both edges out of each of the vertices with degree $2$, which includes three edges out of the center vertex.


TITLE: How can I simplify $\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$?
QUESTION [1 upvotes]: This problem got me thinking hard:
$$\frac{3^{n+1} + 3^{n-1}}{3^{n+2} + 3^n}$$
Considering the fact that the common base is '$3$', I decided to do this:
$$\log_3\left(\frac{(n+1)(n-1)}{(n+2)(n)}\right)$$
$$\log_3\left(\frac{n^2-n+n+1}{n^2+2n}\right)$$
$$\log_3\left(\frac{n^2 + 1}{n^2 + 2n}\right)$$
$$\log_3\left(\frac{1}{2n}\right)$$
I am not 100% sure if have simplified the problem. I need some clarification.

REPLY [2 votes]: $$\frac{3^{n+1}+3^{n-1}}{3^{n+2}+3^n}=\frac{3^{n-1}(3^2+1)}{3^{n-1}(3^3+3)}=\frac{10}{30}=\frac 13$$
This shows your simplification is incorrect for a number of reasons:

$\log(3^{n+1}+3^{n-1})\neq \log(3^{n+1})\log(3^{n-1})$. The $\log$ rule works the other way: $\log(a)+\log(b)=\log(ab)$
You cannot simplify $\frac{n^2+1}{n^2+2n}$ to $\frac{1}{2n}$: big no-no. You can simplify multiplied expressions, not $summed$ expressions. For instance $\frac{n(n+1)}{(n+3)(n+1)}=\frac{n}{n+3}$ is fine.


TITLE: Determine the coordinates of the vector $(4,3,2,1)$ in the basis $\{(1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1)\}$
QUESTION [0 upvotes]: I did this
$$(4)(1,0,0,0)+(3)(1,1,0,0)+(2)(1,1,1,0)+(1)(1,1,1,1) = (10,6,3,1)$$
The right answer is $(1,1,1,1)$. I don't understand. Where did it go wrong?

REPLY [1 votes]: The question is perhaps ambiguously worded. There are two interpretations:

A vector has coordinates $(4,3,2,1)$ in the basis $((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1))$. What are its Cartesian coordinates? (This is the version that you answered.)
A vector has Cartesian coordinates $(4,3,2,1)$. What are its coordinates in the basis $((1,0,0,0),(1,1,0,0),(1,1,1,0),(1,1,1,1))$? (This is the version with answer $(1,1,1,1)$.)

I automatically assumed version $2$ when I read the question; you automatically assumed version $1$. I think the question is at fault here.

REPLY [0 votes]: The coordinates $(a,b,c,d)$ of $(4,3,2,1)$ in that basis are the real numbers such that $a(1,0,0,0)+b(1,1,0,0)+c(1,1,1,0)+d(1,1,1,1)=(4,3,2,1)$. Then we arrive at the following system of equations:
$$\begin{cases}
a+b+c+d=4\\
b+c+d=3\\
c+d=2\\
d=1 \end{cases}$$
Solving the system $Ax=(4,3,2,1)^T$ where $A=\begin{pmatrix} 1&1&1& 1\\
0&1&1&1\\0&0&1&1\\0&0&0&1 \end{pmatrix}$ gives the answer $(1,1,1,1)$.

REPLY [0 votes]: Well, you should start with
$(4,3,2,1) = 4(1,0,0,0) + 3(0,1,0,0) + 2(0,0,1,0) + 1(0,0,0,1) \ (*)$,
which I suspect is what you meant to write when you wrote
$(4)(1,0,0,0)+(3)(1,1,0,0)+(2)(1,1,1,0)+(1)(1,1,1,1)=(10,6,3,1)$.
$$$$
So you could write each of the bases $\ (1,0,0,0),\ (0,1,0,0),\ (0,0,1,0)\ $ and $\ (0,0,0,1)\ $
in terms of (i.e. as a linear combination of) the vectors you were given,
$(1,0,0,0), (1,1,0,0), (1,1,1,0)\ $ and $\ (1,1,1,1)$,
and then solve $(*)$.


TITLE: prove that using uniform bounded theorem
QUESTION [1 upvotes]: Let $y=(\eta_j),\eta_j\in \mathbb C$, be such that $\sum \xi_j\eta_j$ converges for every $x=(\xi_j)\in c_0$ where $c_0\subset l^\infty$ is the subspace of all complex sequences converging to zero. Show that $\sum |\eta_j| < \infty$

using theorem ( Let (Tn) be a sequence of bounded linear operators Tn:X->Y from a Banach space X into a normed space Y such that ‖T(n)x‖ is bounded for every  x∈X,say ,Absolutely (T(n)x) less than or equal cx, where cx is real number. Then the sequence of the norms ‖T(n)‖is bounded that is , there is a c such that Absolutely (T(n)) less than or equal c)
This question in book Introductory Functional Analysis with Applications page 255 question (10)

REPLY [3 votes]: First, we assume that $\eta_j$ are real numbers. Define for an integer $n$ and $a\in c_0$, $$T_n(a):=\sum_{j=1}^n\eta_ja_j.$$
Then $T_n$ is a bounded linear functional on $c_0$. Taking for $j\leqslant n$, $a_j=\operatorname{sgn}(a_j)$ and $a_j=0$ if $j\gt n$, we can see that $\lVert T_n\rVert=\sum_{j=1}^n|\eta_j|$. The wanted conclusion follows from the uniform boundedness principle.
In the general case, write $\eta_j=b_j+ib'_j$, where $b_j$ and $b'_j$ are real numbers and use the previous reasoning with $(b_j)_{j\geqslant 1}$ and $(b'_j)_{j\geqslant 1}$.


TITLE: Beginner Physics - Explaining longitudinal waves
QUESTION [0 upvotes]: I am having difficulty grasping the concept of a longitudinal wave. My textbook definition "In longitudinal waves, the vibration is backwards and forwards in the direction of motion of the wavefront"
If it vibrates backwards and then forwards, would it not be in the same position it originally was?
Do we assume it vibrates forward at a rate faster than that at which it vibrates backwards?
Further, what am I supposed to 'visualize' when I think of waves, in a physics sense?
What is the purpose of a longitudinal wave?
Sorry if I am asking a lot of questions,

REPLY [0 votes]: There are two different but connected motions one can speak of when discussing pulse and wave propagation. I suspect you may not have a clear separation of these two ideas.
One type of motion is the disturbance motion or wave motion. When you watch a wave move, this is typically what catches your eye. If you see a wave and you watch it travel, you're watching the disturbance travel. There's not actually a physical object that travels, but it's the disturbance itself that does.
Another type of motion is the particle motion. This is the motion that each individual bit of material is undergoing. Imagine you tie a small bit of string to a slinky. The particle motion is what the bit of string does.
The key idea is that for a longitudinal wave, the particle motion will be "forward and backward" while the motion of the wave will only be "forward".
Take a look at this video of a longitudinal wave. Try to identify the two different types of motion. You'll see the backward and forward motion of the "particles", as well as the forward-only motion of the disturbances.
If you're still unsure about this, that video link ends with a transverse wave. In those waves, the particle motion is actually perpendicular to the wave motion. If the wave is moving to the right, the particles can be moving up and down. This might actually be an easier example for separating the two types of motion.


TITLE: If $a$ is a group element and $a^{6}=e$, why can't $|a|=4?$
QUESTION [3 upvotes]: Suppose that $a$ is a group element and $a^6=e$. What are the possibilities for $|a|?$ Intuitively, I know that the possibilities would be $1$, $2$, $3$ and $6$. But I  still can't disprove the possibility that $|a|=4$, for example.

REPLY [4 votes]: If $a^4 =e$ then $a^2 = a^2e = a^2a^4 = a^6 =e$, so $|a| \le 2$, a contradiction.

REPLY [3 votes]: If $a^6 = e$ and $a^4=e$, then
$$e = a^6 = (a^2)(a^4) = (a^2)e = a^2$$
so the order of $a$ is not $4$.

REPLY [1 votes]: If so then we need $6$ is divisible by $4$ because $<a>$ is a subgroup, which is a contradiction.

REPLY [1 votes]: If $a=e$, then $|a| = 1$.
If $a\neq e$, $|a|=4$, then $a^4 = e$. But $e =a^6 = a^2e=a^2$, so $|a|=2$.


TITLE: Proof of derivative of convolution $(h * f)' = h' * f$
QUESTION [0 upvotes]: I am writing an article related to Computer Vision but to complete the math under the technique I should prove the following equality:
$$\frac{\mathrm d}{\mathrm dx}(h*f) = \left(\frac{\mathrm d}{\mathrm d x}h \right)*f$$
I am not a mathematician and I am not used to convolution analysis. So, how can I prove it?

REPLY [4 votes]: Recall that the convolution is defined as
$$(h\ast f)(x) = \int_{-∞}^∞ h(x - τ)f(τ) \mathrm d τ \quad \text{for $x ∈ ℝ$}.$$
Now use the special case of Leibniz’s integral rule.
To apply this, you only need $h$ to be continuously differentiable.

REPLY [2 votes]: \begin{align}
\frac{d}{d x}(h*f) &= \frac{d}{d x}\int f(t)h(x-t)dt\\
&= \int \frac{\partial}{\partial x}\left(f(t)h(x-t)\right)dt \\
&= \int f(t)\frac{\partial h(x-t)}{\partial x}dt \\
&= \left(\frac{d}{d x}h\right)*f
\end{align}


text:
\section{Triangulation criteria}
\label{sec:triangulation}

We are interested in the following scenario. Suppose we have a finite set of points $\pts \subset \man$ in a compact Riemannian manifold, and an (abstract) simplicial complex $\acplx$ whose vertex set is $\pts$, and such that every simplex in $\acplx$ defines a non-degenerate Riemannian simplex. When can we be sure that $\acplx$ triangulates $\man$? Consider a convex ball $\bmapball$ centred at $p \in \pts$. We require that, when lifted to $T_p\man$, the simplices near $p$ triangulate a neighbourhood of the origin. If we require that the simplices be small relative to $\bmaprad$, and triangulate a region extending to near the boundary of the lifted ball, then Riemannian simplices outside of $\bmapball$ cannot have points in common with the simplices near the centre of the ball, and it is relatively easy to establish a triangulation.

Instead, we aim for finer local control of the geometry. We establish geometric conditions (\Lemref{lem:embed.star}) that ensure that the complex consisting of simplices incident to $p$, (i.e., the star of $p$) is embedded by a given map into the manifold. In order to achieve this result we require a strong constraint on the differential of the map in question. Since we work locally, in a coordinate chart, we consider maps $F: \R^n \supseteq U \to \R^n$. We demand that for some linear isometry $T:\R^n \to \R^n$ we have
\begin{equation}
  \label{eq:strong.diff.bnd}
  \onorm{dF_u - T} \leq \eta,
\end{equation}
for some $0 \leq \eta \leq 1$, and all $u \in U$. This is stronger than the kind of bounds found, for example, in the Rauch theorem (\Lemref{lem:poly.rauch}), which have the form
\begin{equation}
  \label{eq:bound.sing.vals}
  (1-\eta)\norm{w} \leq \norm{dF_uw} \leq (1 + \eta)\norm{w}.
\end{equation}
Whereas \eqref{eq:bound.sing.vals} implies that $dF_u$ is close to a linear isometry at every $u\in U$, \Eqnref{eq:strong.diff.bnd} means that $dF_u$ is close to \emph{the same} linear isometry for all $u \in U$.

Using a local constraint of the form~\eqref{eq:strong.diff.bnd} to establish the embedding of vertex stars, we demonstrate, in \Secref{sec:generic.tri}, generic criteria which ensure that a map from a simplicial complex to a Riemannian manifold is a homeomorphism. We then turn our attention to the specific case where the map in question is the barycentric coordinate map on each simplex.

Using a refinement of the Rauch theorem established by Buser and Karcher~\cite{buser1981} we show, in \Secref{sec:diff.exp.trans}, that transition functions arising from the exponential map (\Eqnref{eq:exp.transition}) are subject to bounds of the form~\eqref{eq:strong.diff.bnd}.  Then in \Secref{sec:triang.riem.splxs} we observe that the barycentric coordinate map is also subjected to such bounds, and thus yield \Thmref{thm:triangulation} as a particular case of the generic triangulation criteria.

\subsection{Generic triangulation criteria}
\label{sec:generic.tri}

\label{sec:smooth}

We say that a map $F: \rem \to \ren$ is \defn{smooth} if it is of class $\cinfty$.  If $A \subset \rem$, then $F: A \to \ren$ is \defn{smooth} on $A$ if $F$ can be extended to a function that is smooth in an open neighbourhood of $A$. (I.e., there exists an open neighbourhood $U \in \rem$ and a smooth map $\tilde{F}: U \to \ren$ such that $A \subseteq U$, and $\tilde{F}|_A = F$.)  This definition is independent of the ambient space $\rem$ containing $A$. In particular, if $A \subseteq \R^k \subseteq \rem$, then the smoothness of $F$ does not depend on whether we consider $A$ to be a subset of $\R^k$ or of $\rem$. In the case that $A$ is the closure of a non-empty open set, continuity of the partial derivatives implies that they are well defined on all of $A$ and independent of the chosen extension. See Munkres~\cite[\S 1]{munkres1968} for more details.

For our purposes, we are interested in smooth maps from non-degenerate closed Euclidean simplices of dimension $n$ into an $n$-dimensional manifold $\man$. We will work within coordinate charts, so our primary focus will be on maps of the form
\begin{equation*}
  F: \splxsE^n \to \R^n,
\end{equation*}
such that \Eqnref{eq:strong.diff.bnd} is satisfied for all $u \in \splxsE^n$. As an example of how we can exploit this bound, we observe that a map satisfying \Eqnref{eq:strong.diff.bnd} is necessarily an embedding with bounded metric distortion if its domain is convex:
\begin{lem}
  \label{lem:metric.distort.convex}
  Suppose $A \subset \ren$ is convex, and
  $F: A \to \ren$ is a smooth map such that, for some non-negative $\eta <1$,
  \begin{equation*}
    \onorm{dF_u - T} \leq \eta,
  \end{equation*}
  for all $u \in A$, and some linear isometry $T:\ren \to \ren$.
  Then
  \begin{equation*}
    \abs{\,  \norm{F(u) - F(v)} - \norm{u - v} \,} \leq \eta \norm{u - v}
    \quad \text{for all } u,v \in A.
  \end{equation*}
\end{lem}
\begin{proof}
  We observe that it is sufficient to consider the case $T = \Id$, because if $\tilde{F} = T^{-1} \circ F$, then $\onorm{d\tilde{F} - \Id} = \onorm{dF - T}$, and $\norm{\tilde{F}(u) - \tilde{F}(v)} = \norm{F(u) - F(v)}$.

  Assume $u\neq v$. For the lower bound we consider the unit vector $\hat{u} = \frac{u-v}{\norm{u-v}}$, and observe that
  \begin{equation*}
    \dotprod{dF(u-v)}{\hat{u}} \geq (1-\eta)\norm{u-v} > 0,
  \end{equation*}
  so, by integrating along the segment $\seg{u}{v}$, we find
  \begin{equation*}
    \norm{F(u)- F(v)} \geq \dotprod{\big(F(u) - F(v) \big)}{\hat{u}} \geq (1-\eta) \norm{u-v}.
  \end{equation*}
  For the upper bound we employ the unit vector $\hat{w} = \frac{F(u) - F(v)}{\norm{F(u) - F(v)}}$:
  \begin{equation*}
    \dotprod{ \big( F(u)-F(v) \big)}{\hat{w}} = \norm{F(u) - F(v)} \leq (1 + \eta) \dotprod{(u-v)}{\hat{w}} \leq (1 + \eta) \norm{u-v}.
  \end{equation*}
\end{proof}

For our purposes we will be free to choose a coordinate system so that $F$ keeps a vertex fixed. We will have use for the following observation, which can be demonstrated with an argument similar to the proof of \Lemref{lem:metric.distort.convex}:
\begin{lem}
  \label{lem:bound.disp}
  If $A \subseteq \ren$ is a convex set and $F: A \to \ren$ is a smooth map with a fixed point $p \in A$ and
  \begin{equation*}
    \onorm{dF_u - \Id} \leq \eta, \qquad \text{for all } u \in A,
  \end{equation*}
  then
  \begin{equation*}
    \norm{F(u) - u } \leq \eta{\norm{u-p}} \qquad \text{for all } u \in A.
  \end{equation*}

\end{lem}

\paragraph{Embedding complexes}
\label{sec:embed.cplx}

In preparation for considering triangulations we first consider the problem of mappings of complexes into $\ren$.

A simplicial complex $\acplx$ is a set of abstract simplices such that if $\splxs \in \acplx$, then $\splxt \in \acplx$ for every face $\splxt \subset \splxs$. We will only consider finite simplicial complexes. A subcomplex of $\acplx$ is a subset that is also a simplicial complex. The \defn{star} of a simplex $\splxs \in \acplx$ is the smallest subcomplex of $\acplx$ consisting of all simplices that have $\splxs$ as a face, and is denoted $\str{\splxs}$.

In particular, if $p$ is a vertex of $\acplx$, then $\str{p}$ is the set of simplices that contain $p$, together with the faces of these simplices.

The \defn{carrier} (``geometric realisation'') of $\acplx$ is denoted $\carrier{\acplx}$.  We are interested in complexes endowed with a \defn{piecewise flat metric}. This is a metric on $\carrier{\acplx}$ that can be realised by assigning lengths to the edges in $\acplx$ such that each simplex $\splxs \in \acplx$ is associated with a Euclidean simplex $\splxsE \subset \carrier{\acplx}$ that has the prescribed edge lengths. Certain constraints on the edge lengths must be met in order to define a valid piecewise flat metric, but for our current purposes we will have a metric inherited from an embedding in Euclidean space.

We say that $\acplx$ is \defn{embedded} in $\ren$ if the vertices lie in $\ren$ and the convex hulls of the simplices in $\acplx$ define a geometric simplicial complex. In other words, to each $\splxs,\splxt \in \acplx$ we associate $\splxsE = \convhull{\splxs}$, $\splxtE = \convhull{\splxt}$, and we have $\splxsE \cap \splxtE =  \convhull{\splxs \cap \splxt}$. The topological boundary of a set $B \subset \ren$ is denoted by $\bdry{B}$, and its topological interior by $\intr{B}$. If $\acplx$ is embedded in $\R^n$, and $p$ is a vertex of $\acplx$, we say that $\str{p}$ is a \defn{full star} if $p \in \intr{\carrier{\str{p}}}$.

The \defn{scale} of $\acplx$ is an upper bound on the length of the longest edge in $\acplx$, and is denoted by $\scale$.

We say that $\acplx$ is $\thickbnd$-thick if each simplex in $\acplx$ has thickness greater than $\thickbnd$.  The \defn{dimension} of $\acplx$ is the largest dimension of the simplices in $\acplx$. We call a complex of dimension $n$ an \defn{$n$-complex}. If every simplex in $\acplx$ is the face of an $n$-simplex, then $\acplx$ is a \defn{pure $n$-complex}.

A map $F: \carrier{\acplx} \to \ren$ is \defn{smooth on $\acplx$} if for each $\splxs \in \acplx$ the restriction $F \big|_{\splxsE}$ is smooth. This means that $d(F\big|_{\splxsE})$ is well defined, and even though $dF$ is not well defined, we will use this symbol when the particular restriction employed is either evident or unimportant. When the underlying complex on which $F$ is smooth is unimportant, we simply say that $F$ is \defn{piecewise smooth}.

$F$ is \defn{piecewise linear} if its restriction to each simplex is an affine map. The \defn{secant map} of $F$ is the piecewise linear map defined by the restriction of $F$ to the vertices of $\acplx$.

We are interested in conditions that ensure that $F: \carrier{\acplx} \to \ren$ is a topological embedding. Our primary concern is with the behaviour of the boundary. The reason for this is captured by the following variation of a lemma by Whitney~\cite[Lem AII.15a]{whitney1957}:
\begin{lem}[Whitney]
  \label{lem:whitney.cplx.embed}
  Let $\acplx$ be a (finite) simplicial complex embedded in $\ren$ such that $\intr{\carrier{\acplx}}$ is non-empty and connected, and $\bdry{\carrier{\acplx}}$ is a compact $(n-1)$-manifold. Suppose $F: \carrier{\acplx} \to \ren$ is smooth on $\acplx$ and such that $\det d(F\big|_{\splxsE}) > 0$ for each $n$-simplex $\splxs \in \acplx$. If the restriction of $F$ to $\bdry{\carrier{\acplx}}$ is an embedding, then $F$ is a topological embedding.
\end{lem}
\begin{proof}
  The assumptions on $\intr{\carrier{\acplx}}$ and $\bdry{\carrier{\acplx}}$ imply that $\acplx$ is a pure $n$-complex, and that each $(n-1)$ simplex is either a boundary simplex, or the face of exactly two $n$-simplices.  Whitney showed \cite[Lem AII.15a]{whitney1957} that any $x \in \intr{\carrier{\acplx}}$ admits an open neighbourhood $U \subset \intr{\carrier{\acplx}}$ such that the restriction of $F$ to $U$ is a homeomorphism. In particular, $F(\intr{\carrier{\acplx}})$ is open.

  By the Jordan-Brouwer separation theorem~\cite[\S IV.7]{outerelo2009degree}, $\R^n \setminus F(\bdry{\carrier{\acplx}})$ consists of two open components, one of which is bounded. Since $F(\carrier{\acplx})$ is compact, $F(\intr{\carrier{\acplx}})$ must coincide with the bounded component, and in particular $F(\intr{\carrier{\acplx}}) \cap F(\bdry{\carrier{\acplx}}) = \emptyset$, so $F(\intr{\carrier{\acplx}})$ is a single connected component.

  We need to show that $F$ is injective. First we observe that the set of points in $F(\intr{\carrier{\acplx}})$ that have exaclty one point in the preimage is non-empty. It suffices to look in a neigbhourhood of a point $y \in F(\bdry{\carrier{\acplx}})$. Choose $y = F(x)$, where $x$ is in the relative interior of $\splxsE^{n-1} \subset \bdry{\carrier{\acplx}}$. Then there is a neighbourhood $V$ of $y$ such that $V$ does not intersect the image of any other simplex of dimension less than or equal to $n-1$. Let $\splxs^n$ be the unique $n$-simplex that has $\splxs^{n-1}$ as a face. Then $F^{-1}(V \cap F(\carrier{\acplx}) \subset \splxsE^n$, and it follows that every point in $V \cap \intr{\carrier{\acplx}}$ has a unique point in its image.

  Now the injectivity of $F$ follows from the fact that the number of points in the preimage is locally constant on $F(\intr{\carrier{\acplx}}) \setminus F(\bdry{\carrier{\acplx}})$, which in our case is connected. This is a standard argument in degree theory~\cite[Prop. IV.1.2]{outerelo2009degree}: A point $a \in F(\intr{\carrier{\acplx}})$ has $k$ points $\{x_1, \ldots, x_k\}$ in its preimage. There is a neighbourood $V$ of $a$ and disjoint neighbourhoods $U_i$ of $x_i$ such that $F|_{U_i}: U_i \to V$ is a homeomorphism for each $i \in \{1,\ldots,k\}$. It follows that the number of points in the preimage is $k$ for every point in the open neighbourhood of $a$ defined as
  \begin{equation*}
    W = V \setminus  F(\carrier{\acplx} \setminus \bigcup_i U_i).
  \end{equation*}
\end{proof}

\begin{lem}[Embedding a star]
  \label{lem:embed.star}
  Suppose $\acplx = \str{p}$ is a $\thickbnd$-thick, pure $n$-complex embedded in $\ren$ such that all of the $n$-simplices are incident to a single vertex, $p$, and $p \in \intr{\carrier{\acplx}}$ (i.e., $\str{p}$ is a full star). If $F: \carrier{\acplx} \to \ren$ is smooth on $\acplx$,
  and satisfies
  \begin{equation*}
    \onorm{dF - \Id}  < n\thickbnd
  \end{equation*}
  on each $n$-simplex of $\acplx$, then $F$ is an embedding.
\end{lem}
\begin{proof}
  If $\carrier{\acplx}$ is convex, then the claim follows immediately from \Lemref{lem:metric.distort.convex}.

  From the definition of thickness, we observe that $n\thickbnd \leq 1$, and therefore $\onorm{dF} > 0$.  By \Lemref{lem:whitney.cplx.embed}, it suffices to consider points $x,y \in \bdry{\carrier{\acplx}}$. Rather than integrating the differential of a direction, as we did implicitly in the proof of \Lemref{lem:metric.distort.convex}, we will integrate the differential of an angle.

  Let $Q$ be the $2$-dimensional plane defined by $p$, $x$, and $y$. We define the angle function $\phi: \ren \to \R$ as follows: For $z \in \ren$, let $\check{z}$ be the orthogonal projection of $z$ into $Q$. Then $\phi(z)$ is the angle that $\check{z}-p$ makes with $x-p$, where the orientation is chosen so that $\phi(y) < \pi$. (We can assume that $x$ and $y$ are not colinear with $p$, since in that case $[x,y]$ must be contained in $\carrier{\acplx}$, and the arguments of \Lemref{lem:metric.distort.convex} ensure that we would not have $F(x)=F(y)$.)

  Let $\alpha$ be the piecewise linear curve obtained by projecting the segment $[x,y] \subset Q$ onto $\bdry{\carrier{\acplx}}$ via the radial rays emanating from $p$. Parameterise $\alpha$ by the angle $\phi$, i.e., by the arc between $\frac{x-p}{\norm{x-p}}$ and $\frac{y-p}{\norm{y-p}}$ on the unit circle in $Q$. Then $d\phi(\alpha') = 1$, and we have
  \begin{align*}
    \phi(F(y)) - \phi(F(x)) &= \int_{F(\alpha)}d\phi = \int_0^{\phi(y)} d\phi(dF\alpha'(s)) \, ds.
  \end{align*}
  We will show that $d\phi(dF\alpha'(s)) > 0$; it follows that $\phi(F(y)) > \phi(F(x))$ and hence $F(y) \neq F(x)$.

  We need an observation about thick simplices: Suppose that $p$ is a vertex of a $\thickbnd$-thick $n$-simplex $\splxsE$, and $\splxtE$ is the facet opposite $p$. Then a line $r$ through $p$ and $\splxtE$ makes an angle $\theta$ with $\splxtE$ that is bounded by
  \begin{equation*}
    \sin \theta \geq n\thickbnd.
  \end{equation*}
  Indeed, the altitude of $p$ satisfies $\gsplxalt_{p} \geq n\thickbnd \glongedge$ by the definition of thickness, and the distance between $p$ and the point of intersection of $r$ with $\splxtE$ is less than $\glongedge$.

  If $\alpha(s)$ is the point of intersection of $r$ with $\splxtE$, we observe that $d\phi(\alpha'(s)) = \norm{\alpha'(s)}\sin \theta$, i.e., the magnitude of the component of $\alpha'(s)$ orthogonal to $r$. The angle $\vartheta$ between $dF\alpha'(s)$ and $\alpha'(s)$ satisfies $\sin \vartheta \leq \eta < n\thickbnd$. Therefore
  \begin{align*}
    d\phi(dF\alpha'(s))
    &\geq \norm{dF\alpha'(s)}(\sin \theta - \sin \vartheta)\\
    &\geq \norm{dF\alpha'(s)}(n\thickbnd - \eta)\\
    &> 0.
  \end{align*}
\end{proof}

\paragraph{Triangulations}

A simplicial complex $\acplx$ is a  \defn{manifold simplicial complex} if $\carrier{\acplx}$ is a topological manifold (without boundary).

A \defn{triangulation} of a manifold $\man$ is a homeomorphism $H: \carrier{\acplx} \to \man$, where $\acplx$ is a simplicial complex.

If $\man$ is a differentiable manifold, then $H$ is a \defn{smooth triangulation} if it is smooth on $\acplx$, i.e., the restriction of $H$ to any simplex in $\acplx$ is smooth.  We are concerned with smooth triangulations of compact Riemannian manifolds (without boundary).

Our homeomorphism argument is based on

the following observation:
\begin{lem}
  Let $\acplx$ be a manifold simplicial complex of dimension $n$ with finite vertex set $\pts$, and let $\man$ be a compact $n$-manifold.  Suppose $H: \carrier{\acplx} \to \man$ is such that for each $p \in \pts$, $H \big|_{\carrier{\str{p}}}: \carrier{\str{p}} \to \man$ is an embedding. If for each connected component $\man_i$ of $\man$ there is a point $y \in \man_i$ such that $h^{-1}(y)$ contains exactly one point in $\carrier{\acplx}$, then $H$ is a homeomorphism.
\end{lem}
\begin{proof}
  The requirement that the star of each vertex be embedded means that $H$ is locally a homeomorphism, so it suffices to observe that it is bijective. It is surjective by Brouwer's invariance of domain; thus $H$ is a covering map. The requirement that each component of $\man$ has a point with a single point in its pre-image implies that $H: \carrier{\acplx} \to \man$ is a single-sheeted covering, and therefore a homeomorphism.
\end{proof}

The following proposition generically models the situation we will work with when we describe a triangulation by Riemannian simplices:
\begin{prop}[Triangulation]
  \label{prop:generic.triangulation}
  Let $\acplx$ be a  manifold simplicial $n$-complex with finite vertex set $\pts$, and $\man$ a compact Riemannian manifold with an atlas $\{(W_p,\phi_p)\}_{p \in \pts}$ indexed by $\pts$. Suppose
  \begin{equation*}
    H:\carrier{\acplx} \to \man
  \end{equation*}
  satisfies:
  \begin{enumerate}
  \item For each $p \in \pts$ the secant map of $\phi_p \circ H$ restricted to $\carrier{\str{p}}$ is a piecewise linear embedding $\lmap_p:\carrier{\str{p}} \to \ren$ such that each simplex $\splxs \in \ccplx_p = \lmap_p(\str{p})$ is $\thickbnd$-thick, and $\carrier{\ccplx_p} \subset \ballEn{\lmap_p(p)}{h}$, with $\lmap_p(p) \in \intr{\carrier{\ccplx_p}}$.  The scale parameter $h$ must satisfy $h < \frac{\injradM}{4}$, where $\injradM$ is the injectivity radius of $\man$.
  \item For each $p \in \pts$, $\phi_p: W_p \stackrel{\cong}{\longrightarrow} U_p \subset \ren$ is such that $\close{B} = \cballEn{\lmap_p(p)}{\frac{3}{2}h} \subseteq U_p$, and $\onorm{(d\phi_p^{-1})_u} \leq \frac{4}{3}$, for every $u \in \close{B}$.
  \item
    The map
    \begin{equation*}
      F_p = \phi_p \circ H \circ \lmap_p^{-1}: \carrier{\ccplx_p} \to \ren
    \end{equation*}
    satisfies
    \begin{equation*}
      \onorm{(dF_p)_u-\Id} \leq \frac{n\thickbnd}{2}
    \end{equation*}
    on each $n$-simplex  $\splxs \in \ccplx_p$, and every $u \in \splxsE$.
  \end{enumerate}
  Then $H$ is a smooth triangulation of $\man$.
\end{prop}
\begin{proof}
  By \Lemref{lem:embed.star}, $F_p$ is a homeomorphism onto its image. It follows then that $H\big|_{\carrier{\str{p}}}$ is an embedding for every $p\in \pts$. Therefore, since $\carrier{\acplx}$ is compact, $H: \carrier{\acplx} \to \man$ is a covering map.

  Given $x \in \carrier{\acplx}$, with $x \in \splxsE$, and $p$ a vertex of $\splxsE$, let $\tilde{x} = \lmap_p(x) \in \carrier{\ccplx_p}$. Then the bound on $dF$ implies that $\norm{F_p(\tilde{x}) - \lmap_p(p)} \leq \left(1 + \frac{n\thickbnd}{2} \right)h \leq \frac{3}{2}h$, so $F_p(\tilde{x}) \in \close{B}$. Since $\phi_p^{-1} \circ F_p(\tilde{x}) = H(x)$, and
  \begin{align*}
    \norm{(d\phi_p^{-1})_{F(u)}(dF_p)_u} \leq \frac{4}{3} \left( 1 + \frac{n\thickbnd}{2} \right) \leq 2
  \end{align*}
  for any $u \in \splxsE \subset \carrier{\ccplx_p}$, we have that $\distM{H(p)}{H(x)} \leq 2h$.

  Suppose $y \in \carrier{\acplx}$ with $H(y) = H(x)$. Let $\splxt \in \acplx$ with $y \in \splxtE$, and $q \in \splxt$ a vertex. Then $\distM{H(p)}{H(q)} \leq 4h < \injradM$. Thus there is a path $\gamma$ from $H(x)$ to $H(p)$ to $H(q)$ to $H(y) = H(x)$ that is contained in the topological ball $\ballM{H(p)}{\injradM}$, and is therefore null-homotopic. Since $H$ is a covering map, this implies that $x = y$. Thus $H$ is injective, and therefore defines a smooth triangulation.
\end{proof}

config: arxiv


TITLE: Find and rigorously prove the supremum of the set:
QUESTION [0 upvotes]: S = $\{a \in \Bbb{Z} | a < 2+1/2$}
Given that a is in the set of all integers ($\mathbb{Z}$), would the supremum of this set be equal to $(2 + 1/2)$? If so, given an upper bound of $(2 + 1/2)$, how would I prove that for an $\epsilon < (2 + 1/2)$ this couldn't possibly be an upper bound and hence $\sup(S) = (2 + 1/2)$?

REPLY [0 votes]: Just because you can write something doesn't mean that is the must efficient way to write it.
To prove that for any $r < \sup S$ is not an upper bound you prove that for any $r < \sup S$ there must be an element of $s \in S$ so that $r < s \le \sup S$.
If of $\sup S = 2\frac 12$ then that means for any $r < 2\frac 12$ there must be an integer $n$ so that $r < n \le 2\frac 12$.
Is that true.  What if $r = 2.49$.  Is there an integer $n$ so that $2.49 < n \le 2\frac 12$?
Obviously not.  There aren't any integers between $2$ and $3$ so if $n$ is an integer and $n < 2\frac 12 < 3$ then $n \le 2$.  So there are no integers and therefore no elements of $S$ that are between $2$ and $2\frac 12$.  So we can't have any number between $2$ and $2\frac 12$ actually be the $\sup S$
Can you finish this?
Here's food for thought.  Let $S=\{$ all even primes less than $2000\}$.  So how does $\sup S$ compare to $2000$.  If we notice that $S = \{2\}$ what does that tell us.

  Claim 1:  $2$ is an upper bound of $S$.  Pf: If $s \in S$ then $s\in \mathbb Z$ and $s < 2\frac 12$.  So $s < 3$.  ANd $s\in \mathbb Z$ so it is not true that $2 < s <3$.  And as $s < 3$ and it is not true that $s < $ and $s > 2$ it must not be true that $s > 2$. So $s \le 2$.  So $2$ is an upper bound of $S$.

 Claim 2: If $r < 2$ then $r$ is not an upper bound. Pf: $2\in \mathbb Z$.  And $2< 2\frac 12$.  So $2 \in S$.  So $r < 2\in \mathbb S$.  So $r$ is not an upper bound.....


TITLE: Performing elementary row operations on matrices
QUESTION [0 upvotes]: Suppose you’re doing elementary row operations on matrices which have real entries and have 3 rows.
(a) Write down the elementary matrix which correspond to the elementary row operation r3 → r3 + π · r2.
(b) What elementary row operation is performed by left multiplying by the following elementary matrix?
$$
        \begin{matrix}
1 &0 &0 \\
−5 &1 &0 \\
0 &0 &1 \\
        \end{matrix}
$$
To find (a), would I just do the row operation r3 → r3 + π · r2 on the 3x3 identity matrix?
For (b), any guidance would be helpful.
Thanks!

REPLY [0 votes]: (a) Correct.
(b) r2=r2-5*r1
Try it out by left-multiplying the matrix in (b) to an arbitrary/anonymous 3x3 matrix. You will be able to verify that above is correct and know why it should be the case as you try the multiplication.