TITLE: The alternating sum of primes defines an injection
QUESTION [1 upvotes]: Define $\displaystyle\alpha(n)=\sum^n_{k=1}(-1)^{n-k}p_k$, where $p_k$ is the $k$:th prime. Show that $\alpha$ is an injection
$\mathbb Z_+\to\mathbb Z_+$.
It's easy to see while considering sums as $(3-2)+(7-5)+(13-11)+\cdots$, but I would like a short and nice proof.

REPLY [1 votes]: First, show that $\alpha(2k+2) > \alpha(2k)$ and $\alpha(2k+3) > \alpha(2k+1)$, so that $\alpha$ is injective when restricted to either even or odd $n$.
Second, show that $\alpha(2k) \neq \alpha(2l+1)$ for any $k,l$ using a parity argument.


TITLE: Are functions in Lp space always bounded?
QUESTION [9 upvotes]: I know that functions in $L^2$ space have finite norms by definition, but are they also bounded "almost everywhere" ?
So say for instance the following functions norm is finite but it is not bounded.
$$ f(x) = \frac{1}{(x-\frac{1}{2})^2} ~~ ; ~~~ f(x) \in L^2[(0,1),\mu] $$
$$ {\|f(x)\|}_2 = \left(\int_0^1{\frac{1}{(x-\frac{1}{2})^2}}\right)^{1/2}$$
$$ {\|f(x)\|}_2 = \left(\left. {\frac{1}{(x-\frac{1}{2})}}\right]_0^1\right)^{1/2}$$
$$ {\|f(x)\|}_2 = 2$$

REPLY [14 votes]: Let's fix the definitions first.

A function $f:(0,1)\to\mathbb R$ is called bounded if there is a number $M$ such that $|f(x)|\le M$ for all $x\in (0,1)$.

Since the elements of $L^p$ are not functions, but equivalence classes of functions, the concept of boundedness must be adjusted.

A function $f:[0,1]\to\mathbb R$ is called essentially bounded if there is a number $M$ such that $|f(x)|\le M$ for almost all $x\in (0,1)$. (That is, the inequality holds on some set $E$ such that $(0,1)\setminus E$ has zero measure.)

If two functions $f,g$ belong to the same equivalence class and $f$ is essentially bounded, then $g$ is also essentially bounded. Thus, the property of being essentially bounded makes sense for elements of $L^p$.
Does it hold for all elements of $L^p$? Yes if $p=\infty$ (by definition), no if $p<\infty$. For example, the function $f(x)=x^{-1/(p+1)}$ defines an element of $L^p((0,1))$ which is not essentially bounded. For every $M$, the set $\{x:|f(x)|>M\}$ contains an interval, and therefore is not a null set.

Remark. One often  allows functions to take values in the extended real line $[-\infty, \infty]$. In this case, one can say that every element of $L^p$ is finite almost everywhere.


TITLE: Determine the orbits of isometries of the plane on unordered pairs of points
QUESTION [0 upvotes]: "Let the group M2 of isometries of the Euclidean plane act on the set S consisting of pairs of unordered points of the plane. Determine the orbits of this action, and for each orbit, the stabilizer group."
I would appreciate some help in understanding what the question is asking!  Does the set S contain all points of the plane?  If so, why is the question different from M2 acting on the plane?  If S doesn't contain all points of the plane, then what happens if we apply a translation from M2 which takes a point in S to a point outside S?
Part of the problem is that I'm unfamiliar with the term "unordered".  How would you explicitly write the representation of the set out?

REPLY [0 votes]: The set $S$ consists of all unordered pairs of points of the plane. That's different from the set of all points on the plane. In symbols: $S = \{\{x,y\} \mid x,y\text{ points}\}$.
The word "unordered" tells us that the pair $(x,y)$ should not be distinguished from the pair $(y,x)$. For that reason I wrote $\{x,y\}$, since $\{x,y\} = \{y,x\}$.
An element of the set $S$ is for example $\{(0,0),(1,0)\}$, which is the pair consisting of the point with the coordinates $(0,0)$ and the point with the coordinates $(1,0)$.


TITLE: Prove that this limit equals to 2.
QUESTION [0 upvotes]: Prove $$\lim_{(x,y)\to(0,0)}\frac{\sin(x^2+y^2)}{1-\cos(\sqrt {x^2+y^2})}=2$$

REPLY [2 votes]: You can use polar coordinates, that is,
\begin{equation*}
 \begin{cases}
x = r \cos(\theta) \\
y = r \sin(\theta)
\end{cases}
\end{equation*}
so the statement $(x,y) \rightarrow (0,0)$ is equivalent to $r \rightarrow 0$ (whatever path you choose, it is determined also by the angle, which you don't fix). Then this limit becomes
\begin{equation*}
\lim_{r \rightarrow 0} \frac{\sin(r^2)}{1-\cos(r)}
\end{equation*}
and now you can use, for example, l'Hôpital, or the Taylor expressions
\begin{equation*}
\begin{cases}
\sin(x) = x + O(x^3) \\
\cos(x) = 1 - \frac{x^2}{2!} + O(x^4)
\end{cases}
\end{equation*}
(with this orders you will get enough).


TITLE: Stochastic integral of squared process
QUESTION [0 upvotes]: It may be that this question has been asked before, but the search function did not revel anything in this area.
In reading some old lecture notes on stochastic calculus I have encountered the following reasoning:
$$
\int H^2d\langle M\rangle=\int H^2d\langle M,M\rangle=\int H d\langle H \boldsymbol{\cdot} M,M\rangle=\langle H \boldsymbol{\cdot} M, H \boldsymbol{\cdot} M\rangle=\langle H \boldsymbol{\cdot} M\rangle
$$
I am not quite sure I understand where the second and third equality signs come from.
In this particular material the stochastic integral of $H\in L^2(M)$ with respect to $M\in\mathcal{M}^c_{\text{0,loc}}$ is defined as the unique process $H\boldsymbol{\cdot}M\in\mathcal{H}^{2,c}_0$ satisfying the relation:
$$
\langle H\boldsymbol{\cdot}M, N\rangle = \int H d \langle M,N\rangle, (\forall)
 N\in\mathcal{M}^c_{\text{0,loc}},
$$
where $\mathcal{M}^c_{\text{0,loc}}$ is the set of continuous local martingales null at zero and for each $M\in \mathcal{M}^c_{\text{0,loc}}$ I define $L^2(M)$ to be the set of all equivalence classes of predictable processes $H=(H_t)_{t\geq t}$ with the property that:
$$
\newcommand{\norm}[1]{\left\lVert#1\right\rVert}
\mathbb{E}_M[H^2]=\norm {H}^2_{L^2(M)}=\mathbb{E}\Big[ \int_0^\infty H^2_sd\langle M\rangle_s \Big] < +\infty,
$$
where two processes $H$ and $H'$ are considered to be the members of the same equivalence class if $\norm{H-H'}_{L^2(M)}=0$. Finally, $\mathcal{H}^{2,c}_0$ is the set of all continuous RCLL $(P,\mathbb{F})$-martingales, null at $t=0$ with the property that
$$
\sup_{t \geq 0} \mathbb{E}[M_t^2]<\infty.
$$
I would be very grateful if you could point me to a reference that can help in proving the above.
The integral in the RHS of the definition of $M\boldsymbol{\cdot}N$ (which is actually a theorem), is taken to be pathwise ($\omega$-by-$\omega$) Lebesgue-Stieltjes integral with respect to the finite variance process which is in this case the quadratic covariation defined as
$$
\langle M, N \rangle := \frac{1}{4}(\langle M+N \rangle - \langle M-N \rangle).
$$

REPLY [0 votes]: Your problems deal with the stochastic integral with respect to locally square-integrable martingale, please refer to S. N. Cohen and R. J. Elliott, Stochastic Calculus and Applications, 2nd ed.(2015), Th.12.2.1, p.264.
If you concern to stochastic integral with respect to locally continuous martingale only, please refer to D. Revuz, M. Yor, Continuous Martingales and Brownian Motion, 3rd ed. Springer, 2005, Th.2.2, p.137.
For more general semimartingale case, please refer to J. Jacod and  A. N. Shiryayev, Limit Theory for Stochastic Processes, 2ed. Springer, 2003, Th.I.4.40, p.68.