TITLE: Simulate a camera PSF?
QUESTION [1 upvotes]: Say I have a $3D$ object stored as a $3D$ $\texttt{NumPy}$ array in python. I want to view this $3D$ object as a camera would give: a particular aperture, focal length, etc. My understanding is I can blur each layer of this object by a certain amount by convoluting it with the camera's PSF. After blurring, all the slices can be summed together, resulting in a final image. The issue I am having is how to create this PSF, so it's $3D$ $(x,y,z)$. I see a lot of literature for deriving PSF for microscopes, but to my understanding, the PSF on a camera isn't symmetric across the focal plane; objects closer to the camera blur out faster.  I wondered if anyone has come across a $3D$ PSF for a conventional camera or how about I could derive it myself?

REPLY [2 votes]: If you wanted to do it in purely ray optics, it's easiest to proceed with ray tracing. The rays cast from camera sensor would go through the objective lenses and finally find their intersections with the nearest points of the object (unless absorbed inside the objective). Don't forget that the ray may partially reflect from the lenses, and the lenses have coatings that change reflectivity depending on wavelength and incidence angle from the simple Fresnel model. This process will let you reproduce depth of field effect, bokeh and most aberrations of the lens, as well as lens flare.
But this is not sufficient if your aperture is small enough. In this case you also need to take diffraction into account. In case of a point light source at infinity, focused at the sensor, you can calculate the pattern as Fraunhofer diffraction by the aperture. For a polygonal aperture the integral can be calculated analytically, see e.g. [1] for triangle, and use superposition for arbitrary polygon.
Now, if the source point isn't at infinity, it might be a good approximation to treat the rays from the object as if originating at infinity in the same direction. This will neglect their sphericity, but might not affect the result too much. This I cannot guarantee because I didn't check how much of an error this approximation introduces.
Of course, this isn't exact solution of wave propagation through the whole system. But it should get most of the details right.
References
1: R.M. Sillitto & Winifred Sillitto (1975) "A Simple Fourier Approach to Fraunhofer Diffraction by Triangular Apertures", Optica Acta: International Journal of Optics, 22:12, 999-1010,
DOI: 10.1080/713819012


TITLE: Prove that $\{x_n\}_n$ is convergent.
QUESTION [2 upvotes]: Here is an exercise:

Let $x_n=1+2+\dots+\frac1n-\ln n$. Prove that $\{x_n\}_n$ is convergent.

(I believe that this can be found in the site, however I cannot find immediately, so I post it here.)
The hints are much appreciated. I don't want complete proof.
Thanks for your help.

REPLY [4 votes]: Let's present a different proof than any I've seen on this site, via a picture and a hint.
Look carefully at the following and remember that $\displaystyle \int_1^x \frac 1t dt = \ln x$:

Justify that the area in blue is the limit you're looking for, and for that matter the first $n$ blue bits corresponds to the $n$th term of the sequence. Show the limit exists and is finite (which can be done entirely with the picture too).


TITLE: $\phi^4$-theory, S-matrix Feynman diagram to first order from Peskin and Schroeder
QUESTION [2 upvotes]: This relates to page 111 in Peskin and Schroeder.
We have the $\phi^4$ S-matrix for a 2-particle to 2-particle scattering reaction:
$$-i\frac{\lambda}{4!}\int d^4x \langle p_1p_2|\mathcal T\left(\phi(x)\phi(x)\phi(x)\phi(x)\right)|p_Ap_B\rangle \tag{4.90}$$
For which, using Wick's theorem we can rewrite the time-ordering as normal ordered contracted operators. This is all fine. To first order we obtain:
$$\langle p_1p_2|\mathcal N\left(-i\frac{\lambda}{4!}\int d^4x\text{ }\phi(x)\phi(x)\phi(x)\phi(x)+\text{contractions}\right)|p_Ap_B\rangle \tag{4.92}$$
Since we are evaluating S-matrix elements and not correlation functions we can no longer neglect terms which aren't fully contracted. We then note that if two fields are contracted with themselves and two fields remain, the normal-ordered product of the remaining two fields will be of the form $\mathcal N\left((a+a^\dagger)(a+a^\dagger)\right)=(a^\dagger a^\dagger+2a^\dagger a+aa)$, and as we commute these opeators past the $a$'s and $a^\dagger$'s of the initial and final states, we find that only a term with an equal number of $a$'s and $a^\dagger$'s can survive.
It is this last sentence that confuses me, I have written out in full what I expect this to look like, but I cannot see why an equal number of creation and annihilation operators is necessary for the term to not vanish.

REPLY [2 votes]: Suppose we have a general term of the form
$$\langle 0 | a_{p_1} a_{p_2} \cdots a_{p_{n-1}}a_{p_n} a_{p_1}^\dagger a_{p_2}^\dagger \cdots a^\dagger_{p_{m-1}}a_{p_m}^\dagger | 0 \rangle$$
We can eliminate $a_{p_1}^\dagger$ by commuting it past all the $a_{p_i}$ factors until it anhilitates the vacuum state. Along the way, we will end up with a bunch of terms of the form
$$\langle 0 | a_{p_1} \cdots [a_{p_i},a_{p_1}^\dagger] \cdots a_{p_n} a_{p_2}^\dagger \cdots a_{p_m}^\dagger | 0 \rangle.$$
In each of these terms, we can use the commutation relation
$$[a_p,a_{p'}^\dagger]=(2\pi)^3\delta^{(3)}(p-p')$$
to eliminate $a_{p_i}$ and $a_{p_1}^\dagger$ from the term. The net result is the sum of a bunch of terms, each with one less $a$ and one less $a^\dagger$. We can continue this process until we run out of either $a^\dagger$ factors or $a$ factors.
At this point, there are three possibilities. If there were more $a$ factors than $a^\dagger$ factors, we will have the sum of a bunch of terms, each the product of delta functions with a factor of the form $\langle 0|a_{p'_1}\cdots a_{p'_{n-m}}|0\rangle$. These terms are all zero because an annihilation operator appears next to $|0\rangle$. Similarly, if there were more factors of $a^\dagger$, we would end up with an $a^\dagger$ next to $\langle 0|$ in each term. The only way to avoid a trivial result is if there are an equal number of factors of $a$ and $a^\dagger$. Then we will end up with a sum of products of delta functions.


TITLE: Partial derivatives question?
QUESTION [1 upvotes]: I have to find $dw$ if $w=f(u,v,z)$ where $u=x^2+y^2,v=x^2-y^2,z=2xy$.
Now,I know that  $dw=  ( ∂w/ ∂z)*dz  + (∂w/ ∂u)*du +  (∂w/ ∂v)*dv$
The problem is that for example,if I want to find  $∂w/ ∂z$ I don't know how to relate $w$ to $z$?

REPLY [0 votes]: (I withdraw my earlier approach.)
We can "solve" the $u$ and $v$ equations for  $x$ and $y$ as
$$2x^2 \ = \ u + v \  \ \ , \ 2y^2  \ = \ u  -  v  \ ,$$
which will allow us to write $ \ z \ $ as
$$ \ z \ = \ 2xy \ = \ (4x^2y^2)^{1/2} \ =  \ [ \ (u+v)(u-v) \ ]^{1/2} \ = \ [ \ (u^2 -v^2) \ ]^{1/2}, $$
from which we can now extract
$$dz \ = \ \frac{\partial z}{\partial u} \cdot du \  + \  \frac{\partial z}{\partial v} \cdot dv \ , $$
to replace in your chain-rule equation for $dw$ .


TITLE: Prerequisites for Algebraic Geometry (Algebra)
QUESTION [2 upvotes]: I am a physics major who's studying mathematics, and I am interested in Topology and Geometry. I am mainly interested in Michael Atiyah's work in Algebraic Geometry and Chern-Simons Theories in physics.
What are the prerequisites in Abstract Algebra and Topology for Algebraic Geometry? I have already taken Real Analysis at Rudin Level, Group, Ring, and Module Theory at Dummit and Foote level. Are Field and Galois theories required to study Algebraic Geometry? Also, would Multivariate Analysis be helpful?

REPLY [1 votes]: Question: "Are Field and Galois theories required to study Algebraic Geometry? Also, would Multivariate Analysis be helpful?"
Answer: You should find a good book on field theory and Galois theory,  and also read some books on commutative algebra: The classical books Matsumura - "Commutative ring theory" and Atiyah-Macdonald - "Commutative algebra" are self contained (with some errors) and a good start. You will find lists of errata online.


TITLE: Proving that only one function solves Kepler's equation.
QUESTION [0 upvotes]: Prove that there exists only one continuous function $f:\mathbb{R}\to \mathbb{R}$ that satisfies Kepler's equation
$$f\left(x\right)-\epsilon\sin\left(f\left(x\right)\right)=x$$
where $\epsilon$ is between $0$ and $1$.
 Any hint will help a lot.

REPLY [2 votes]: Suppose that the equation possess two solutions say $y_1$ and $y_2$ and let $z=y_1-y_2$. Then we have $$z=\varepsilon \big (\sin y_1-\sin y_2\big).$$ By the mean value theorem $$|z|=\varepsilon|\sin y_1-\sin y_2|\leq \varepsilon |z|.$$ This implies that $\varepsilon \geq 1$ which is a conttradiction since  $0<\varepsilon<1$ given. Thus $y_1=y_2$ .


TITLE: Does uniformly boundedness implies boundedness of the union of images?
QUESTION [0 upvotes]: Suppose:

$X$ is a compact space
$(Y, d)$ is a metric space
$ \mathcal{F} \subseteq C(X, Y) $ is a collection of continuous maps
$\mathcal{F}$ is uniformly bounded, i.e. there exists a number $M$ such that $ d(f(x), g(x)) \leq M $ for any $x \in X$ and $ f, g \in \mathcal{F} $. This is equivalent to saying that $\mathcal{F}$ is a bounded subset in $C(X, Y)$ under the sup metric.

Question:
Can we say that the union of images
$$ \mathcal{F}(X) = \bigcup_{f \in \mathcal{F}} f(X) = \{ f(x) : x \in X, f \in \mathcal{F} \} $$
is a bounded subset in $(Y, d)$?
To prove the boundedness of $ \mathcal{F}(X) $, we need to find a number $M'$ such that $ d(f(x), g(y)) \leq M' $ for any $x, y \in X$ and $ f, g \in \mathcal{F} $. This involves one more variable $y \in X$ than the uniformly boundedness of $\mathcal{F}$.

REPLY [1 votes]: Fix $h \in \mathcal F$. Since $h$ is a continuous function on  a compact space it is bounded. Now $d(f(x),g(y))\leq d(f(x),h(x))+d(h(x),h(y))+d(h(y),g(y))$.


TITLE: Triangle Inequality?
QUESTION [0 upvotes]: I'm having trouble proving the following claim:
$\forall a, b, c \in \mathbb{R}_+: T(a, b, c) \Rightarrow [|a − b| < c$ and $|b − c| < a$ and $|a − c| < b]$
Where $T(a, b, c)$ is a predicate for a triangle with sides $a, b, c$.
Don't even know where to start. Can someonle please help me?

REPLY [0 votes]: Note: I assume that you do not allow a triangle to have zero area.
From the symmetry of the statement, it suffices to show that $T(a, b, c)$ implies $|a - b| < c$. Start from the triangle inequality:
\begin{align}
a & < b + c & & ; \text{ from } T(a, b, c) \\
a - b & < c.
\end{align}
Or we can start with another triangle inequality
\begin{align}
b & < a + c & & ; \text{ from } T(a, b, c) \\
b - a & < c
\end{align}
These two inequalities imply $|a - b| < c$.


TITLE: Proper notation for equation solving.
QUESTION [1 upvotes]: What is the proper notation when solving one equation by inserting another known relationship?
$$y=x^2, y+2x^2=1$$$$y+2x^2=1\Leftrightarrow x^2+2x^2=1 \Leftrightarrow x=\pm\sqrt{\frac{1}{3}}$$
I assume the first $\Leftrightarrow$ is wrong since we are inserting a relationship, and the $y+2x^2=1$ equation is not logically equal to $x^2+2x^2=1$ by it self.
Edit: I guess what I'm asking is: Is there a way of notating $f(x),g(x)\Rightarrow h(x)$ cleanly?

REPLY [0 votes]: \begin{align}&&y=x^2 \;&\text{ and }\; y+2x^2=1\\&\iff &y=x^2 \;&\text{ and }\; x^2+2x^2=1 \\&\iff &y=x^2 \;&\text{ and }\; x=\pm\sqrt{\frac{1}{3}}\\&\iff &y=\frac13 \;&\text{ and }\; x=\pm\sqrt{\frac{1}{3}}.\end{align}

REPLY [0 votes]: Why do you need any notation?

Since $y=x^2$ and $y+2x^2=1$, we have $3x^2=1$, which is equivalent to $x=\pm\sqrt{1/3}$.


TITLE: Does "in-between" multiplication preserve equality?
QUESTION [0 upvotes]: In a magma $(S;*)$, multiplication on the left and the right preserves equality. That is, if $a=b$, then $c*a=c*b$ and $a*c=b*c$. But what about "in-between" multiplication? That is, if $a*c=b*c$, does it follow that $(a*d)*c=(b*d)*c$? Or, as I think is more likely, is there a magma where that property does not hold?

REPLY [2 votes]: Let $a, b, c: \mathbb{R}^{2}\rightarrow \mathbb{R}^{2}$ be linear maps. Suppose that $c(x, y) = (x, 0)$.  Suppose that $a(x, y) = (0, y), b(x, y) = (0, -y)$.  Then $a\circ c = b\circ c = 0$.  Let $d(x, y) = (y, x)$.  Then $(a\circ d \circ c)(x,y) = (0, x)$ and $(b \circ d \circ c)(x, y) = (0, -x)$.


text:
\begin{document}

\date{}

\maketitle

\begin{abstract}
\noindent
We prove explicit formulae for $\alpha$-points of $L$-functions from the Selberg class.
Next we extend a theorem of Littlewood on the vertical distribution of zeros of the Riemann zeta-function $\zeta(s)$ to the case of $\alpha$-points of the aforementioned $L$-functions.
This result implies the uniform distribution of subsequences of $\alpha$-points and from this a discrete universality theorem in the spirit of Voronin is derived.
\end{abstract}

\section{Introduction and Statement of the Main Results}

The Riemann zeta-function $\zeta(s)$ plays a central role in number theory and, in particular, the distribution of its zeros is relevant for the error term in the prime number theorem and further asymptotic formulae in analytic number theory. In his invited talk at the fifth International Mathematical Congress at Cambridge in 1912, Landau \cite{astart} suggested that ``[g]iven an analytic function, the points for which this function is $0$ are very important; however, of equal interest are those points where the function assumes a given value''.

Given a complex number $\alpha$, the roots of the equation
$$
\zeta(s)=\alpha
$$
are called $\alpha$-points and will be denoted by $\rho_{\alpha}=\beta_{\alpha}+i\gamma_{\alpha}$.
Moreover, for the non-trivial zeros of $\zeta(s)$ we just use the notation $\rho=\beta+i\gamma$.
Our investigations have their origin in the work of Landau \cite{lan} who proved that if $x>1$, then
$$\sum\limits_{0<\gamma\leq T}x^\rho=-\dfrac{T}{2\pi}\Lambda(x)+O_x(\log T)$$
for any $T>1$, where $\Lambda(x)$ is the von Mangoldt function extended to the whole real line by setting it to be equal to zero if $x$ is not a positive integral power of a prime $p$.
The dependence of the implicit constant on $x$ was made explicit only much later by Gonek \cite{Gon}.

Such explicit formulae have been proven very useful to the study of the vertical distribution of the non-trivial zeros of $\zeta(s)$.
For instance, Radamacher \cite{rad} employed Landau's formula and proved that the sequence $(a\gamma)_{\gamma>0}$ is uniformly distributed mod 1 for any real number $a\neq0$, provided that the Riemann Hypothesis is true.
Later Elliott \cite{Ell} remarked that the latter condition can be removed, and (independently) Hlawka \cite{Hl} obtained the aforementioned result unconditionally.
On the other hand,
Ford and Zaharescu \cite{F.Z} and Ford, Soundararajan and Zaharescu \cite{F.S.Z} made use of Gonek's explicit version of Landau's formula to prove quantitative results regarding the distribution of $\left(a\gamma\right)_{\gamma>0}$ with respect to its discrepancy.
For related results we also refer to Fujii \cite{F1,F2,F3,F4,F5,F6,F7}.

Motivated by such considerations, the third author \cite{st} studied the distribution of the $\alpha$-points of $\zeta(s)$.
He first proved the analogue of Landau's formula, that is, if $\alpha\neq1$ is complex number, $x\neq1$ is a positive number and $\epsilon>0$, then
\begin{align}\label{Reh}
\mathop{\sum\limits_{0<\gamma_\alpha\leq T}}_{\beta_\alpha>0}x^{\rho_\alpha}
=\dfrac{T}{2\pi}\left(a(x)-x\Lambda\left(\dfrac{1}{x}\right)\right)+O_{\alpha,x,\epsilon}\left(T^{1/2+\epsilon}\right)
\end{align}
for any $T>1$, where $a(x)$ is defined on the positive integers by the coefficients of the ordinary Dirichlet series
$$\dfrac{\zeta(s)}{\zeta(s)-\alpha}=\sum\limits_{n=1}^{\infty}\dfrac{a(n)}{n^s},$$
and is equal to zero for any other real number $x$.\footnote{The special case of $\alpha=1$ will be discussed in the remark following Theorem 3.}
He then employed a classic result due to Levinson \cite{levia}, regarding the clustering of the $\alpha$-points of $\zeta(s)$ around the critical line, and proved that the sequence $(a\gamma_\alpha)_{\gamma_\alpha>0}$ is uniformly distributed mod 1 for any real number $a\neq0$.
Recently,
Rehberg \cite{Reh} and Gonek and Baluyot \cite{bg} proved (independently) relation \eqref{Reh} and made the dependence on $x$ explicit; in accordance they proved discrepancy estimates for the sequence $(a\gamma_\alpha)_{\gamma_\alpha>0}$.

We improve upon these results in Theorem \ref{MAIN3}.
As a matter of fact, we start by proving analogous results for functions $\mathcal{L}(s)$ in the Selberg class $\mathcal{S}$ of a given degree $d_\mathcal{L}$.
The definition and properties of $\alpha$-points of such functions are similar to the one of $\zeta(s)$, which belongs to $\mathcal{S}$ with $d_\zeta=1$.
For more details we refer to Section 2.

To state our theorems, we only need to define here for a given complex number $\alpha\neq1$ and $\mathcal{L}\in\mathcal{S}$, the function $\Lambda(x;\mathcal{L},\alpha)$ which for positive integers $x=n$ is given by the coefficients of the  Dirichlet series
\begin{align}\label{coeff}
\dfrac{\mathcal{L}'(s)}{\mathcal{L}(s)-\alpha}=-\sum\limits_{n=1}^{\infty}\dfrac{\Lambda(n;\mathcal{L},\alpha)}{n^s},
\end{align}
and is equal to zero for any other real number $x$.
The error terms in the subsequent theorems depends on the real quantity $\sigma(\mathcal{L},\alpha)$, which is related to the convergence of the latter Dirichlet series and is explicitly given by (\ref{aabscissa}).
Laslty, the {\it Landau symbols} $o(\cdot)$ and $O(\cdot)$  have their usual meaning, and if the implied constant depends on some parameter $\epsilon$ (say), then we write $O_\epsilon$.
The implicit constants may depend also on $\mathcal{L}$ and $\alpha$ but we drop such subscripts.
The same comments apply to the {\it Vinogradov symbols} $\ll$ and $\gg$.

\begin{theorem}\label{MAIN1}
Let $\mathcal{L}\in \mathcal{S}\setminus\lbrace1\rbrace$, $\alpha\neq1$ be a complex number and $\epsilon>0$. Then
\begin{align*}
\mathop{\sum\limits_{0<\gamma_\alpha\leq T}}_{\beta_\alpha\geq0}x^{\rho_\alpha}
=&-\dfrac{T}{2\pi}\Lambda\left(x;\mathcal{L},\alpha\right)+O_\epsilon\left(x^{\sigma(\mathcal{L},\alpha)+\epsilon}\left(1+\mathbbm{1}_{\mathbb{R}\setminus\mathbb{Z}}(x)\min\left\{\frac{T}{x},\dfrac{1}{\|x\|}\right\}\right)\right)+\\
&+O_\epsilon\left(x^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon}\log T\left(1+\dfrac{1}{\log x}\right)\right)+\\
&+O\left(\frac{1}{x^2}\left(\log T\left(1+\dfrac{1}{\log x}\right)+\min\left\{T,\dfrac{1}{\log x}\right\}\right)\right)
\end{align*}
for any $x,T>1$, where $\mathbbm{1}_\mathcal{A}$ is the characteristic function of a set $\mathcal{A}$ and $\|x\|$ denotes the distance from $x$ to the nearest integer.
\end{theorem}

\begin{theorem}\label{MAIN2}
Let $\mathcal{L}\in \mathcal{S}$ be such that $d_{\mathcal{L}}\geq2$, $\alpha\neq1$ be a complex number and $\epsilon\in(0,1/2)$.  Then
\begin{align*}
\mathop{\sum\limits_{0<\gamma_\alpha\leq T}}_{\beta_\alpha\geq0}x^{-\rho_\alpha}
=&\,-\dfrac{T}{2\pi x}\overline{\Lambda\left(x;\mathcal{L},0\right)}+O_\epsilon\left(x^\epsilon\left(1+\mathbbm{1}_{\mathbb{R}\setminus\mathbb{Z}}(x)\min\left\{\dfrac{T}{x},\dfrac{1}{\| x\|}\right\}\right)\right)+\\
&+O\left(x^{\epsilon}+\dfrac{1}{x^{\sigma\left(\mathcal{L},\alpha\right)+1}\log x}+x^{\epsilon}\log T\left(\dfrac{1}{\log x}+\log\dfrac{1}{\epsilon}\right)\right)
\end{align*}
for any $x,T>1$.
\end{theorem}

\begin{theorem}\label{MAIN3}
Let  $\mathcal{L}\in S$ be such that $d_\mathcal{L}=1$ and $\alpha\neq1$ be a complex number.
Let also $m=1$ if $\mathcal{L}\equiv\zeta$ and $m=7$, otherwise.
Then
\begin{align*}
\mathop{\sum\limits_{0<\gamma_\alpha\leq T}}_{\beta_\alpha\geq0}x^{-\rho_\alpha}
=&-\dfrac{T}{2\pi x}\overline{\Lambda\left(x;\mathcal{L},0\right)}+O\left(\log x\min\left\{\dfrac{T}{x},\dfrac{1}{\langle x\rangle}\right\}\right)+O\left(\left(\log T\right)^{2m}\log(3x)\right)+\\
&+O\left((1+\log x)\log(2x)\log\log(3x)+\dfrac{1}{x^{\sigma\left(\mathcal{L},\alpha\right)+1}\log x}\right)+\\
&+O\left(\log T\left(\dfrac{1}{\log x}+\log x+\log\log(3x)\right)\right)
\end{align*}
for any $x,T>1$.
\end{theorem}

We do not treat the case of $\alpha=1$ since it seems to be rather complicated.
The major difficulty arises from the fact that the function on the right-hand side of \eqref{coeff} can not be represented as an ordinary Dirichlet series if $\alpha=1$ (this will become apparent in Lemma 1 of Section 3).
There have been attempts in \cite{JMS}, \cite{Reh},  and \cite{st} but in each case the aforementioned lack of an ordinary Dirichlet series represantation has been overseen.
It was first Baluyot and Gonek \cite{bg} who noticed these inaccuracies and gave a partial answer also for the case of $\alpha=1$.

Explicit formulae in the case of $L$-functions in the Selberg class have been proven also by Murty and Perelli \cite{MP} with respect to their non-trivial zeros, as well as by Jakhlouti, Mazhouda and Steuding \cite{JMS} with respect to $\alpha$-points.
In particular those formulae in \cite{JMS} hold under some hypotheses which are not yet proven to be true.
Moreover, their formulae are not uniform in $x$ and there is a $T^{1/2+\epsilon}$ in the error term.
Our results are unconditional, uniform in $x$ and have smaller error terms.

As we mentioned earlier, such formulae are useful in the study of the vertical distribution of $\alpha$-points of a given function.
However, our poor knowledge of the horizontal distribution of $\alpha$-points of functions $\mathcal{L}$ in the Selberg class which have degree $d_\mathcal{L}\geq2$, prevents us from deriving results similar to the ones that have been proved for $\zeta(s)$.
We would have to assume certain hypotheses such as the Lindel\"of hypothesis for $L$-functions in the Selberg class.
For more details we refer to \cite{JMS} and its bibliography.
We are optimistic that our explicit formulae and these additional hypotheses can produce discrepancy estimates for sequences $(a\gamma_\alpha)_{\gamma_\alpha>0}$, but we will not persue this direction further.
Instead we prove an unconditional result regarding the vertical-distribution of such $\alpha$-points.

\begin{theorem}\label{Main1}
Let $\mathcal{L}\in\mathcal{S}\setminus\lbrace1\rbrace$ and $\alpha$ be a complex number.
Then there exists a positive constant $A=A(\alpha)$ such that for every $T\geq \exp(3)$ one can find an $\alpha$-point $\beta_\alpha+i\gamma_\alpha$ of $\mathcal{L}(s)$ satisfying
\begin{align*}
|\gamma_\alpha-T|<\dfrac{A}{\log\log\log T}.
\end{align*}
\end{theorem}
This result is a generalization of a theorem due to Littlewood \cite{ little } regarding the non-trivial zeros of $\zeta(s)$, and it yields the following corollary.

\begin{corollary}\label{Main2}
Let $\mathcal{L}\in\mathcal{S}\setminus\lbrace1\rbrace$, $b>0$ be a real number and $\alpha$ a complex number.
Then there exists a subsequence of $\alpha$-points $\left(\rho_{\alpha,n_k}\right)_{k\in\mathbb{N}}$, of $\mathcal{L}(s)$, such that $\gamma_{\alpha,n_k}=bk+o(1)$, and the sequence $\left(a \gamma_{\alpha, n_{k^m}}\right)_{k\in\mathbb{N}}$, is uniformly distributed mod 1 for every real number $a\notin b^{-1}\mathbb{Q}$ and every positive integer $m$.
\end{corollary}

Lastly, with the aid of the latter corollary, we prove a theorem which combines the vertical-distibution of $\alpha$-points with the universality of $\zeta(s)$.
\begin{theorem}\label{universality}
Let $\mathcal{L}\in\mathcal{S}\setminus\lbrace{1}\rbrace$ and $\alpha\in\mathbb{C}$.
Then there exists a subsequence of $\alpha$-points $\left(\rho_{\alpha,n_k}\right)_{k\in\mathbb{N}}$, of $\mathcal{L}(s)$ such that for any compact set with connected complement $K\subseteq\mathcal{D}:=\left\{s\in\mathbb{C}:1/2<\sigma<1\right\}$, any non-vanishing function $f$ which is continuous on $K$ and analytic in its interior, and any $\varepsilon>0$
$$\liminf\limits_{N\to\infty}\dfrac{1}{N}\sharp\left\{1\leq k\leq N:\max\limits_{s\in K}|\zeta\left(s+i\gamma_{\alpha, n_k}\right)-f(s)|<\varepsilon)\right\}>0,$$
where $\sharp A$ denotes the cardinality of a set $A\subseteq\mathbb{N}$.
\end{theorem}

In zeta-function theory, `universality' stands for an approximation property as in the previous theorem; this name has been coined with respect to the fact that a wide class of functions can be uniformly approximated by certain shifts of a single function. The first universality theorem is due to Voronin \cite{voro}; in his result there was no condition on the shifts. The first discrete universality theorem is due to Reich \cite{reich}; in his result the shifts are taken from an arithmetic progression.

The statement of Theorem \ref{universality} is not new in the case of $\alpha=0$; however, different from previous versions, it is unconditional. First, Garunk\v stis, Laurin\v cikas and Macaitien\.e \cite{glm} proved the above universality theorem under assumption of what they call a {\it weak Montgomery conjecture} on the spacing of the imaginary parts of the nontrivial zeros (as it would follow from the pair correlation conjecture). Secondly, Garunk\v stis and Laurin\v cikas \cite{gl} obtained the same universality result assuming the Riemann hypothesis. It appears that the statement holds unconditionally and even in the wider context of $\alpha$-points of $L$-functions from the Selberg class.

\section{The Selberg Class: Definition and First Properties}

For a survey on the Selberg class and the value-distribution of its functions we refer to Kaczorowski and Perelli \cite{KP}, Murty and Murty \cite{MM}, and Perelli \cite{Per}.
Here we give a few definitions and poperties of $L$-functions from the Selberg class which is usually denoted by $\mathcal{S}$ and was introduced by Selberg \cite{sel}.
It consists of Dirichlet series
\begin{align*}
\mathcal{L}(s):=\sum\limits_{n=1}^{\infty}\dfrac{f(n)}{n^s},
\end{align*}
satisfying the following hypotheses:
\begin{enumerate}
\item{\it Ramanujan hypothesis.} $f(n)\ll_\epsilon n^\epsilon$.
\item{\it Analytic continuation.} There exists a non-negative integer $k$ such that $(s-1)^k\mathcal{L}(s)$ is an entire function of finite order.
\item{\it Functional equation.} $\mathcal{L}(s)$ satisfies a functional equation of type
\begin{align*}
\mathfrak{L}(s)=\omega\overline{\mathfrak{L}(1-\overline{s})},
\end{align*}
where
\begin{align*}
\mathfrak{L}(s):=\mathcal{L}(s)Q^s\prod\limits_{j=1}^{J}\Gamma\left(\lambda_js+\mu_j\right)
\end{align*}
with positive real numbers $Q$, $\lambda_j$, and complex numbers $\mu_j$, $\omega$ with $\Re\mu_j\geq0$ and $|\omega|=1$.
\item{\it Euler product.} $\mathcal{L}(s)$ has a product represantation
\begin{align*}
\mathcal{L}(s)=\prod\limits_{p}\mathcal{L}_p(s),
\end{align*}
where
\begin{align*}
\mathcal{L}_p(s)=\exp\left(\sum\limits_{k=1}^{\infty}\dfrac{b(p^k)}{p^{ks}}\right)
\end{align*}
with suitable coefficients $b(p^k)$ satisfying $b(p^k)\ll p^{k\theta}$ for some $\theta<1/2$.
\end{enumerate}
Axioms (i) and (ii) imply that a function $\mathcal{L}(s)$ from $\mathcal{S}$ is a Dirichlet series which is absolutely convergent for $\sigma>1$ and it has an analytic continuation to the whole complex plane except for a possible pole at $s=1$.
From axiom (iii) one obtains the quantity
$$d_{\mathcal{L}}:=2\sum\limits_{j=1}^J\lambda_j,$$
which is called the degree of $\mathcal{L}(s)$ and, although the data from the functional equation is not unique, $d_{\mathcal{L}}$ is well-defined (by its appearance in a zero-counting formula).
It has been conjectured that the degree is always a non-negative integer.
This deep conjecture has been verified so far only for $d_\mathcal{L}\leq 2$ by work of Kaczorowski and Perelli \cite{KP2}.
The only element of degree zero is the function constant $1$;
 the degree one elements are the Riemann zeta-function $\zeta(s)$ in addition with real shifts of Dirichlet $L$-functions to primitive residue class characters.
Typical examples of degree two are Dedekind zeta-functions to quadratic number fields and $L$-functions to normalized newforms.
All these examples posses an Euler product and, indeed, axiom (iv) implies that if $\mathcal{L}\in\mathcal{S}$, then it is zero-free in the half-plane $\sigma>1$ and $f(n)$ is a multiplicative function.

The functional equation of a function from the Selberg class will play a crucial role in the proofs presented here.
We prefer to rewrite it as
\begin{align}\label{functequa}
\mathcal{L}(s)=H_\mathcal{L}(s)\overline{\mathcal{L}(1-\overline{s})},
\end{align}
valid for all $s\in\mathbb{C}$, where
\begin{align*}
H_\mathcal{L}(s)=\omega Q^{1-2s}\prod\limits_{j=1}^J\dfrac{\Gamma(\lambda_j(1-s)+\overline{\mu_j})}{\Gamma(\lambda_js+\mu_j)}.
\end{align*}
By Stirling's formula, it follows that
\begin{align}\label{Stril}
H_\mathcal{L}(\sigma+it)=\left(\lambda Q^2 t^{d_\mathcal{L}}\right)^{1/2-\sigma}\exp\left(-it\log\left(\lambda Q^2\left(\dfrac{t}{e}\right)^{d_\mathcal{L}}\right)\right)e^{i\pi(\mu-d_\mathcal{L})/4}\left(\omega+O_{\sigma_1,\sigma_2}\left(\dfrac{1}{t}\right)\right)
\end{align}
and
\begin{align}\label{fS}
-\dfrac{H'_\mathcal{L}}{H_\mathcal{L}}(\sigma+it)=\log\left(\lambda Q^2\left(\dfrac{t}{e}\right)^{d_\mathcal{L}}\right)+O_{\sigma_1,\sigma_2}\left(\dfrac{1}{t}\right),
\end{align}
uniformly in vertical strips $\sigma_1\leq\sigma\leq\sigma_2$, where
$$\mu:=2\sum\limits_{j=1}^J(1-2\mu_j)\,\,\,\text{ and }\,\,\,\lambda:=\prod\limits_{j=1}^J\lambda_j^{2\lambda_j}.$$
In addition to axiom (iii), we have
\begin{align}\label{order}
|\mathcal{L}(\sigma+it)|\asymp_{\sigma_1,\sigma_2}|t|^{(1/2-\sigma)d_\mathcal{L}}|\mathcal{L}(1-\sigma+it)|,\,\,\,|t|\geq t_0>0,
\end{align}
uniformly in $\sigma_1\leq\sigma\leq\sigma_2$.

Concerning the distribution of the $\alpha$-points of a non-constant element $\mathcal{L}\in\mathcal{S}\setminus\lbrace1\rbrace$ of the Selberg class, it is worth to mention that their location in general is pretty similar to the case of the Riemann zeta-function.
There may be an $\alpha$-point in the neighbourhood of a pole of a Gamma-factor in the functional equation (by Rouch\'e's theorem), however, those so-called trivial $\alpha$-points lie not too distant from the negative real axis and with finitely many further exceptions there are no other in the left half-plane $\sigma\leq0$.
All other $\alpha$-points are said to be non-trivial and they lie in the right half-plane $\sigma\geq0$.
Their number $\mathcal{N}(\alpha,T)$ is asymptotically given by a Riemann-von Mangoldt--type formula: if $\alpha\neq1$ is a complex number, then
\begin{align}\label{RMf}
\mathcal{N}(\alpha,T):=\left\{s\in\mathbb{C}:\mathcal{L}(s)=\alpha\text{ and }T<t\leq2 T\right\}=\dfrac{d_\mathcal{L}}{2\pi}T\log\dfrac{4T}{e}+\dfrac{T}{2\pi}\log(\lambda Q^2)+O(\log T).
\end{align}
The proofs of \eqref{Stril}-\eqref{RMf} can be found in \cite[Lemma 6.7, Theorem 6.8, Corollary 7.4]{S}).

\section{Auxiliary Lemmas}

Before stating our first lemma we remind the Dirichlet convolution of two arithmetical functions.
If $f,g:\mathbb{N}\to\mathbb{C}$, then their Dirichlet convolution (or multiplication) is defined by
$$(f*g)(n):=\sum\limits_{d\mid n}f(d)g\left(\dfrac{n}{d}\right)$$
for all $n\in\mathbb{N}$.
This operation is associative and we denote by $f^{* k}$ the Dirichlet convolution of $f$ with itself $k$ times for some $k\in\mathbb{N}$.
Lastly, we denote by $f_0$ the arithmetical function which is identical to $f$ except at $n=1$, where we define $f_0(1):=0$.

\begin{lemma}\label{Init}
Let $\mathcal{L}\in \mathcal{S}$ and $\alpha\neq1$ be a complex number.
 If
 \begin{align}\label{aabscissa}
 \sigma(\mathcal{L},\alpha)
 :=
 \left\{ \begin{array}{ll}
1+\sup_{\rho_\alpha}\beta_\alpha,&\alpha\neq0,\\
 1,&\alpha=0.
 \end{array}
 \right.
 \end{align}
 then $1\leq\sigma(\mathcal{L},\alpha)<\infty$ and the function $\mathcal{L}'(s)/\left(\mathcal{L}(s)-\alpha\right)$ can be represented as an ordinary Dirichlet series which is absolutely convergent in the half-plane $\sigma>\sigma\left(\mathcal{L},\alpha\right)$. In particular,
\begin{align*}
\dfrac{\mathcal{L}'(s)}{\mathcal{L}(s)-\alpha}=-\sum\limits_{n=1}^{\infty}\dfrac{\Lambda(n;\mathcal{L},\alpha)}{n^s}
\end{align*}
for any $\sigma>\sigma\left(\mathcal{L},\alpha\right)$, where
\begin{align}\label{vMal}
\Lambda\left(n;\mathcal{L},\alpha\right):=\left(f\cdot\log\right)*\left(\sum\limits_{k=0}^{\infty}\dfrac{(-1)^kf_0^{*k}}{(1-\alpha)^{k+1}}\right)(n),
\end{align}
for any positive integer $n$.
Moreover, $\Lambda\left(n;\mathcal{L},\alpha\right)\ll_\epsilon n^{\sigma(\mathcal{L},\alpha)-1+\epsilon}$.
\end{lemma}

\begin{proof}
When $\alpha=0$ there is nothing to prove because for $\mathcal{L}\in S$ we know that
$$\dfrac{\mathcal{L}'(s)}{\mathcal{L}(s)}=-\sum\limits_{n=1}^{\infty}\dfrac{\Lambda(n;\mathcal{L},0)}{n^s},\,\,\,\sigma>1,$$
where $\Lambda(n;\mathcal{L},0)=f(n)\Lambda(n)$ and $\Lambda(n)$ is the von Mangoldt function.
Since $\Lambda(n)\ll\log n$, the Ramanujan hypothesis for the coefficients $f(n)$ imply that $\Lambda(n;\mathcal{L},0)\ll_\epsilon n^\epsilon$.

Now let $\alpha\neq0$.
Then
\begin{align*}
\mathcal{L}(s)-\alpha=(1-\alpha)\left(1+\sum\limits_{n=2}^\infty\dfrac{f(n)}{(1-\alpha)n^s}\right)
\end{align*}
is an ordinary Dirichlet series and the uniqueness theorem for Dirichlet series (see \cite[\S 9.6]{Titch1}) implies that there is an $A>0$ such that $\mathcal{L}(s)-\alpha\neq0$ for any $\sigma>A$.
Therefore, $$0\leq\sigma(\mathcal{L},\alpha)-1\leq A.$$
Moreover, it follows from \cite[Appendix, pg. 89-90]{Lan} that there is $B>0$ such that
\begin{align}\label{inv}
\dfrac{1}{\mathcal{L}(s)-\alpha}=\dfrac{1}{1-\alpha}\sum\limits_{k=0}^\infty(-1)^k\left(\sum\limits_{n=2}^\infty\dfrac{f(n)}{1-\alpha}\dfrac{1}{n^s}\right)^k=\sum\limits_{n=1}^\infty\left(\sum\limits_{k=0}^{\infty}\dfrac{(-1)^kf_0^{*k}(n)}{(1-\alpha)^{k+1}}\right)\dfrac{1}{n^s}
\end{align}
for any $\sigma>B$, while from \cite[Satz 12]{Lan} we know that the Dirichlet series on the right hand side of the latter relation converges for every $\sigma>\sigma(\mathcal{L},\alpha)-1$.
Hence, the aforementioned series is a Dirichlet series representation of $1/(\mathcal{L}(s)-\alpha)$ in the  half-plane $\sigma>\sigma(\mathcal{L},\alpha)-1$.
This implies that it is absolutely convergent in the half-plane $\sigma>\sigma(\mathcal{L},\alpha)$ and that
\begin{align}\label{g}
g(n):=\sum\limits_{k=0}^{\infty}\dfrac{(-1)^kf_0^{*k}(n)}{(1-\alpha)^{k+1}}\ll_\epsilon n^{\sigma\left(\mathcal{L},\alpha\right)-1+\epsilon}.
\end{align}
On the other hand, the series
\begin{align}\label{der}
\mathcal{L}'(s)=-\sum\limits_{n=1}^{\infty}\dfrac{f(n)\log n}{n^s}
\end{align}
is absolutely convergent for any $\sigma>1$.
The first assertion of the lemma follows now by mutliplying the Dirichlet series \eqref{inv} and \eqref{der}.
In addition, relation \eqref{g} and the Ramanujan hypothesis yield that the coefficients of the resulting series satisfy
\begin{align*}
\Lambda\left(n;\mathcal{L},\alpha\right)=-\sum\limits_{d\mid n}g\left(\dfrac{n}{d}\right)f(d)\log d \ll_\epsilon \sum\limits_{d\mid n}\left(\dfrac{n}{d}\right)^{\sigma\left(\mathcal{L},\alpha\right)-1+\epsilon/2}d^{\epsilon/2}\ll_\epsilon n^{\sigma\left(\mathcal{L},\alpha\right)-1+\epsilon}.
\end{align*}
\end{proof}

\begin{lemma}\label{hori}
Let $\mathcal{L}\in\mathcal{S}$ and $\alpha$ be a complex number.
Then
\begin{align*}
\dfrac{\mathcal{L}'(s)}{\mathcal{L}(s)-\alpha}=\sum\limits_{|\gamma_{\alpha}-t|\leq1}\dfrac{1}{s-\rho_\alpha}+O(\log|t|),\,\,\,|t|\geq1,
\end{align*}
uniformly in $-1\leq\sigma\leq\sigma(\mathcal{L},\alpha)$.
\end{lemma}
\begin{proof}
The proof follows in exactly the same way, except of some minor changes, as the one of \cite[Lemma 8]{GS} in the case of $\zeta(s)$.
\end{proof}

\begin{lemma}\label{lb}
Let $\mathcal{L}\in\mathcal{S}$ and $\sigma_1\leq\sigma_2<0$.
Then
\begin{align*}
|\mathcal{L}(\sigma+it)|\gg_{\sigma_1,\sigma_2} |t|^{(1/2-\sigma)d_\mathcal{L}},\,\,\,|t|\geq t_0>0,
\end{align*}
uniformly in $\sigma_1\leq\sigma\leq\sigma_2$.
\end{lemma}
\begin{proof}
We refer to \cite[Chapter III, \S 2]{bes} for an introduction to the theory of analytic almost periodic functions and their elementary properties.
Since the Dirichlet series $\mathcal{L}(s)$ is uniformly convergent in the strip $1<1-\sigma_2\leq\sigma\leq1-\sigma_1$,
it is a uniformly almost periodic function in that strip.
In addition, $\mathcal{L}(s)$ is non-vanishing in $\sigma>1$ by assumption.
Hence,
$\inf\left\{\left|\mathcal{L}(s)\right|:1-\sigma_2\leq\sigma\leq1-\sigma_1\right\}>0$ by \cite[\S 2, Corollary 1]{bes}.
 Taking also into account relation \eqref{order} the lemma follows.
\end{proof}

The following lemma originates from the work of Levinson \cite[Lemma 3.3, Lemma 3.4]{Lev} and it has been improved by Gonek in \cite[Lemma 1, Lemma 2]{gon}.
Although the proof there is given for a fixed real number $a$, one can easily see that it holds uniformly in any finite interval $[a_1,a_2]$.

\begin{lemma}\label{Gonek}
Let $a_1, a_2$ be fixed real numbers such that $a_1\leq a_2$.
Then, for all  sufficiently large numbers $A,B\gg_{a_1,a_2}1$ with $A<B\leq2A$, we have that
\begin{align}\label{GoN}
\mathcal{I}(r,a,A,B):=\int\limits_{A}^B\exp\left(it\log\dfrac{t}{re}\right)\left(\dfrac{t}{2\pi}\right)^{a-1/2}\mathrm{d}t=M(r,A,B)+E(r,A,B),
\end{align}
uniformly in $a_1\leq a\leq a_2$,
where
\begin{align*}
M(r,a,A,B)=
\left\{
\begin{array}{ll}
(2\pi)^{1-a}r^ae^{-ir+i\pi/4},&A<r\leq B,\\
0,&\text{otherwise},
\end{array}
\right.
\end{align*}
and
\begin{align*}
E(r,a,A,B)\ll_{a_1, a_2} A^{a-1/2}+\dfrac{A^{a+1/2}}{|A-r|+A^{1/2}}+\dfrac{B^{a+1/2}}{|B-r|+B^{1/2}}.
\end{align*}
\end{lemma}

\begin{lemma}[Gallagher's lemma]\label{gall}
Let $T_0$ and $T\geq\delta>0$ be real numbers and $A$ be a finite subset of $\left[T_0+\delta/2, T+T_0-\delta/2\right]$.
Define $N_\delta(x)=\sum_{t\in A,|t-x|<\delta}1$ and assume that $f(x)$ is a complex-valued continuous function on $\left[T_0,T+T_0\right]$ continuously differentiable on $\left(T_0,T+T_0\right)$. Then
\begin{align*}
\sum\limits_{t\in A}N_{\delta}^{-1}(t)|f(t)|^2\leq\dfrac{1}{\delta}\int\limits_{T_0}^{T+T_0}|f(x)|^2\mathrm{d}x+\left(\int\limits_{T_0}^{T+T_0}|f(x)|^2\mathrm{d}x\int\limits_{T_0}^{T+T_0}|f'(x)|^2\mathrm{d}x\right)^{1/2}.
\end{align*}
\end{lemma}

\begin{proof}
For a proof see \cite[Lemma 1.4]{M}.
\end{proof}

\section{Explicit Formulae}

\begin{proof}[Proof of Theorem \ref{MAIN1}]
Let $T_0:=\min\left\{1,\gamma_\alpha:\gamma_\alpha>0\right\}/2$.
Such number exists as follows from the Riemann-von Mangoldt formula \eqref{RMf}.
Recall the definition of $\sigma\left(\mathcal{L},\alpha\right)$ in \eqref{aabscissa} and
assume that $T$ is not an ordinate of an $\alpha$-point of $\mathcal{L}(s)$.
Since there are only finitely many $\alpha$-points in the plane $\left\{s\in\mathbb{C}:\sigma\leq0,\,t\geq T_0\right\}$, by the calculus of the residues, we have that
\begin{align}\label{init}
\begin{split}
\mathop{\sum\limits_{0<\gamma_\alpha\leq T}}_{\beta_\alpha\geq0}x^{\rho_\alpha}
&=\dfrac{1}{2\pi i}\left\{\,\int\limits_{\sigma\left(\mathcal{L},\alpha\right)+\epsilon+iT_0}^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon+iT}+\int\limits_{\sigma\left(\mathcal{L},\alpha\right)+\epsilon+iT}^{-2+iT}+\int\limits_{-2+iT}^{-2+iT_0}+\int\limits_{-2+iT_0}^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon+iT_0}\right\}\hspace{-1pt}x^s\dfrac{\mathcal{L}'(s)}{\mathcal{L}(s)-\alpha}\mathrm{d}s\hspace*{-1pt}+\hspace{-1pt}O(1)\\
&=:\mathcal{I}_1+\mathcal{I}_2+\mathcal{I}_3+\mathcal{I}_4+O(1).
\end{split}
\end{align}

We start by estimating $\mathcal{I}_1$.
From Lemma 1 we know that the Dirichlet series expression of $\mathcal{L}'(s)/\left(\mathcal{L}(s)-\alpha\right)$ is absolutely convergent in the half-plane $\sigma\geq \sigma\left(\mathcal{L},\alpha\right)+\epsilon$ and $\Lambda\left(n;\mathcal{L},\alpha\right)\ll_\epsilon n^{\sigma(\mathcal{L},\alpha)-1+\epsilon/2}$.
 Therefore, interchanging integration and summation yields
 \begin{align}\label{I1}
 \begin{split}
 \mathcal{I}_1=-\dfrac{1}{2\pi}\sum\limits_{n=1}^{\infty}\Lambda(n;\mathcal{L},\alpha)\left(\dfrac{x}{n}\right)^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon}\int\limits_{T_0}^T\left(\dfrac{x}{n}\right)^{it}\mathrm{d}t=-\dfrac{T-T_0}{2\pi}\Lambda(x;\mathcal{L},\alpha)+E,
\end{split}
 \end{align}
 where
 \begin{align*}
 E\hspace*{-1.1pt}\ll\hspace*{-1pt}\mathop{\sum\limits_{n=1}^{\infty}}_{n\neq x}\hspace*{-1pt}|\Lambda(n;\mathcal{L},\alpha)|\hspace*{-1pt}\left(\dfrac{x}{n}\right)^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon}\min\hspace*{-1pt}\left\{\hspace*{-1pt}T,\left|\log\frac{x}{n}\right|^{-1}\right\}
 \hspace*{-1pt}\ll_\epsilon \mathop{\sum\limits_{n=1}^{\infty}}_{n\neq x}\dfrac{x^{\sigma(\mathcal{L},\alpha)+\epsilon}}{n^{1+\epsilon/2}}\min\hspace*{-1pt}\left\{\hspace*{-1pt}T,\dfrac{\max\left\{x,n\right\}}{|x-n|}\hspace*{-1pt}\right\}.
 \end{align*}
 Assume that $x$ is not an integer.
 Let also $\lfloor x\rfloor$ and $\lceil x\rceil$, denote the largest integer which is less than or equal to $x$ and the smallest integer which is greater than or equal to $x$, respectively.
 Splitting the latter sum into sums where the indices of their summands range over $n\leq x/2$, $n\geq 2x$, $x/2<n<\lfloor x\rfloor$,  $\lceil x\rceil<n<2x$, $n=\lfloor x\rfloor$ and $n=\lceil x\rceil$, we have that
 \begin{align}\label{E}
 \begin{split}
 E&\ll_\epsilon x^{\sigma(\mathcal{L},\alpha)+\epsilon}+ x^{\sigma(\mathcal{L},\alpha)+\epsilon/2}\mathop{\sum\limits_{\frac{x}{2}< n<2x}}_{n\neq\lfloor x\rfloor,\lceil x\rceil}\dfrac{1}{|x-n|}+x^{\sigma(\mathcal{L},\alpha)-1+\epsilon/2}\min\left\{T,\dfrac{x}{ x-\lfloor x\rfloor},\dfrac{\lceil x\rceil}{\lceil x\rceil-x}\right\}\\
 &\ll_\epsilon  x^{\sigma(\mathcal{L},\alpha)+\epsilon}+x^{\sigma(\mathcal{L},\alpha)+\epsilon/2}\log x+x^{\sigma(\mathcal{L},\alpha)-1+\epsilon/2}\min\left\{T,\dfrac{x}{\| x\|}\right\}\\
 &\ll_\epsilon x^{\sigma(\mathcal{L},\alpha)+\epsilon}+x^{\sigma(\mathcal{L},\alpha)+\epsilon/2}\min\left\{\frac{T}{x},\dfrac{1}{\|x\|}\right\}.
 \end{split}
 \end{align}
 If on the other hand $x$ is a positive integer, then we do not need to consider the cases $n=\lfloor x\rfloor$ and $n=\lceil x\rceil$ seperately, in which event we just have
 \begin{align}\label{error1}
 E\ll_\epsilon x^{\sigma(\mathcal{L},\alpha)+\epsilon}.
 \end{align}
 Since $T_0\Lambda(x;\mathcal{L},\alpha)\ll_\epsilon x^{\sigma(\mathcal{L},\alpha)-1+\epsilon/2}$, it follows from relations \eqref{I1}-\eqref{error1} that
 \begin{align}\label{I1f}
 \mathcal{I}_1=-\dfrac{T}{2\pi}\Lambda(x;\mathcal{L},\alpha)+O_\epsilon\left(x^{\sigma(\mathcal{L},\alpha)+\epsilon}\left(1+\mathbbm{1}_{\mathbb{R}\setminus\mathbb{Z}}(x)\min\left\{\frac{T}{x},\dfrac{1}{\| x\|}\right\}\right)\right).
 \end{align}

 We continue with estimating $\mathcal{I}_2$.
 It follows from Lemma \ref{hori} that
 \begin{align}\label{I2}
2\pi i \mathcal{I}_2-\sum\limits_{|\gamma_\alpha-T|\leq 1}\,\int\limits_{\sigma\left(\mathcal{L},\alpha\right)+\epsilon+iT}^{-2+iT}\dfrac{x^{s}}{s-\rho_\alpha}\mathrm{d}s
\ll\log T\int\limits_{-2}^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon}x^{\sigma}\mathrm{d}\sigma
\ll x^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon}\dfrac{\log T}{\log x}.
 \end{align}
 Moreover, by the calculus of the residues, we know that
 \begin{align*}
 \int\limits_{\sigma\left(\mathcal{L},\alpha\right)+\epsilon+iT}^{-2+iT}\dfrac{x^{s}}{s-\rho_\alpha}\mathrm{d}s=&\,-2\pi i R(\rho_\alpha)+\\
 &+\left\{
 \int\limits_{\sigma\left(\mathcal{L},\alpha\right)+\epsilon+iT}^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon+i(T+2)}+
 \int\limits_{\sigma\left(\mathcal{L},\alpha\right)+\epsilon+i(T+2)}^{-2+i(T+2)}+\int\limits_{-2+i(T+2)}^{-2+iT}\right\}\dfrac{x^{s}}{s-\rho_\alpha}\mathrm{d}s,
 \end{align*}
 where $R(\rho_\alpha)$ is 0 or 1, depending on whether $\rho_\alpha$ lies or not, respectively, in the rectangle with vertices $\sigma\left(\mathcal{L},\alpha\right)+\epsilon+iT$,  $\sigma\left(\mathcal{L},\alpha\right)+\epsilon+i(T+2)$, $-2+i(T+2)$ and $-2+iT$.
 Then
 \begin{align}\label{smal}
 \begin{split}
  \int\limits_{\sigma\left(\mathcal{L},\alpha\right)+\epsilon+iT}^{-2+iT}\dfrac{x^{s}}{s-\rho_\alpha}\mathrm{d}s
 & \ll 1+\int\limits_{T}^{T+2}\dfrac{x^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon}}{\sqrt{\epsilon^2+(t-\gamma_\alpha)^2}}\mathrm{dt}+\dfrac{x^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon}}{\log x}+\int\limits_{T}^{T+2}\dfrac{\mathrm{dt}}{x^2\sqrt{4+(t-\gamma_\alpha)^2}}\\
  &\ll_\epsilon x^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon}.
  \end{split}
 \end{align}
 Since $\sum_{|\gamma_\alpha-T|\leq 1}1\ll\log T$ by relation \eqref{RMf}, it follows from \eqref{I2} and \eqref{smal} that
 \begin{align}\label{I2f}
 \mathcal{I}_2\ll_\epsilon  x^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon}\log T\left(1+\dfrac{1}{\log x}\right).
 \end{align}

In order to estimate $\mathcal{I}_3$, we firstly observe that Lemma \ref{lb} implies that
$$|\mathcal{L}(-2+it)|>2 |\alpha|$$
for any $t\geq T_1$, where $T_1\gg 1$ is a sufficiently large positive number such that it is not an ordinate of an $\alpha$-point of $\mathcal{L}(s)$.
Therefore, for $T\geq T_1$ we have that
\begin{align}\label{I3}
\begin{split}
2\pi x^2\mathcal{I}_3+O(1)
&=-\int\limits_{T_1}^Tx^{it}\dfrac{\mathcal{L}'}{\mathcal{L}}(-2+it)\left(1-\frac{\alpha}{\mathcal{L}(-2+it)}\right)^{-1}\mathrm{d}t\\
&=-\int\limits_{T_1}^Tx^{it}\dfrac{\mathcal{L}'}{\mathcal{L}}(-2+it)\left(1+\sum\limits_{k=1}^{\infty}\left(\dfrac{\alpha}{\mathcal{L}(-2+it)}\right)^k\right)\mathrm{d}t\\
&=:-\left(\mathcal{I}_{31}+\mathcal{I}_{32}\right).
\end{split}
\end{align}
In view of the functional equation \eqref{functequa} and relation \eqref{fS}, it follows that
\begin{align}\label{forla}
\begin{split}
\mathcal{I}_{31}
&=\int\limits_{T_1}^Tx^{it}\left(-\dfrac{\overline{\mathcal{L}}'}{\overline{\mathcal{L}}}(3+it)+\dfrac{H_\mathcal{L}'}{H_\mathcal{L}}(-2+it)\right)\mathrm{d}t\\
&=\sum\limits_{n=1}^{\infty}\dfrac{\overline{\Lambda\left(n;\mathcal{L},0\right)}}{n^{3}}\int\limits_{T_1}^T(xn)^{it}\mathrm{d}t-\int\limits_{T_1}^Tx^{it}\log\left(\lambda Q^2\left(\dfrac{t}{e}\right)^{d_\mathcal{L}}\right)\mathrm{d}t+\int\limits_{T_1}^Tx^{it}O\left(\dfrac{1}{t}\right)\mathrm{d}t\\
&=:\mathcal{I}_{3111}+\mathcal{I}_{312}+\mathcal{I}_{313}.
\end{split}
\end{align}
Observe that
\begin{align}
\mathcal{I}_{3111}\ll\min\left\{T,\dfrac{1}{\log x}\right\}\,\,\,\text{ and }\,\,\,\mathcal{I}_{313} \ll\log T.
\end{align}
To estimate $\mathcal{I}_{312}$ we integrate by parts.
Then
\begin{align}\label{forla1}
\mathcal{I}_{312}=\left[-\dfrac{x^{it}}{\log x}\log\left(\lambda Q^2\left(\dfrac{t}{e}\right)^{d_\mathcal{L}}\right)\right]_{T_1}^T+\int\limits_{T_1}^T\dfrac{x^{it}ed_\mathcal{L}}{t\log x}\mathrm{d}t\ll\dfrac{\log T}{\log x}.
\end{align}
Lastly, Lemma \ref{lb} implies that
\begin{align}\label{I32}
\mathcal{I}_{32}\ll \int\limits_{T_1}^{T}\sum\limits_{k=1}^{\infty}\left(\frac{\alpha}{t^{3d_\mathcal{L}/2}}\right)^k\mathrm{d}t\ll\int\limits_{T_1}^{T}\dfrac{1}{t^{5/2}}\mathrm{d}t\ll1.
\end{align}
In view of relations \eqref{I3}-\eqref{I32}, we obtain that
\begin{align}\label{I3f}
\mathcal{I}_3\ll\dfrac{1}{x^2}\left(\min\left\{T,\dfrac{1}{\log x}\right\}+\log T\left(1+\dfrac{1}{\log x}\right)\right),
\end{align}
as long as $T\geq T_1$.
For the remaining $T$, that is for $T_0\leq T<T_1$, we can bound $\mathcal{I}_3$ from above trivially by $x^{-2}$ which is absorbed  by the error term appearing in \eqref{I1f} since $x>1$.

It remains to estimate $\mathcal{I}_4$ which is independent of $T$ and, therefore, can be bounded trivially from above by $x^{\sigma\left(\mathcal{L},\alpha\right)+\epsilon}$.

The theorem now follows from \eqref{init}, \eqref{I1f}, \eqref{I2f} and \eqref{I3f}.
Recall that we assumed in the beginning that $T$ is not an ordinate of an $\alpha$-point.
Otherwise, we choose a $T_2\in[T,T+1]$ such that it is not an ordinate of an $\alpha$-point and observe that
$$\mathop{\sum\limits_{0<\gamma_\alpha\leq T}}_{\beta_\alpha\geq0}x^{\rho_\alpha}=\mathop{\sum\limits_{0<\gamma_\alpha\leq T_2}}_{\beta_\alpha\geq0}x^{\rho_\alpha}+O\left(x^{\sigma\left(\mathcal{L},\alpha\right)}\sum\limits_{T\leq\gamma_\alpha\leq T+1}1\right)
=\mathop{\sum\limits_{0<\gamma_\alpha\leq T_2}}_{\beta_\alpha\geq0}x^{\rho_\alpha}+O\left(x^{\sigma\left(\mathcal{L},\alpha\right)}\log T\right).$$
Then we estimate the latter sum as described above, while the error term is absorbed by the error term appearing in \eqref{I2f}.
\end{proof}

\begin{proof}[Proof of Theorem \ref{MAIN2}]
Let $T_0$ be as in the proof of Theorem \ref{MAIN1} and $\epsilon\in(0,1/2)$.
If $T$ is not an ordinate of an $\alpha$-point,
then by the calculus of the residues it follows that
\begin{align}\label{init1}
\begin{split}
\mathop{\sum\limits_{0<\gamma_\alpha\leq T}}_{\beta_\alpha\geq0}x^{-\rho_\alpha}
=&\dfrac{1}{2\pi i}\left\{\,\int\limits_{\sigma\left(\mathcal{L},\alpha\right)+1+iT_0}^{\sigma\left(\mathcal{L},\alpha\right)+1+iT}+\int\limits_{\sigma\left(\mathcal{L},\alpha\right)+1+iT}^{-\epsilon+iT}+\int\limits_{-\epsilon+iT}^{-\epsilon+iT_0}+\int\limits_{-\epsilon+iT_0}^{\sigma\left(\mathcal{L},\alpha\right)+1+iT_0}\right\}\hspace{-1pt}x^{-s}\dfrac{\mathcal{L}'(s)}{\mathcal{L}(s)-\alpha}\mathrm{d}s+\\
&+O(x^\epsilon)\\
=:&\,\mathcal{I}_1+\mathcal{I}_2+\mathcal{I}_3+\mathcal{I}_4+O(x^\epsilon).
\end{split}
\end{align}
The error term occurs from the possible existence of finitely many $\alpha$-points in the vertical strip $-\epsilon\leq\sigma<0$.

The absolute convergence of $\mathcal{L}'(s)/(\mathcal{L}(s)-\alpha)$ in the half-plane $\sigma\geq\sigma(\mathcal{L},\alpha)+1$ and our assumption that $x>1$ imply that
\begin{align}\label{I1i}
2\pi x^{\sigma\left(\mathcal{L},\alpha\right)+1}\mathcal{I}_1=\sum\limits_{n=1}^{\infty}\dfrac{\Lambda\left(n;\mathcal{L},\alpha\right)}{n^{\sigma\left(\mathcal{L},\alpha\right)+1}}\int\limits_{T_0}^T(xn)^{-it}\mathrm{d}t\ll\sum\limits_{n=1}^{\infty}\dfrac{|\Lambda\left(n;\mathcal{L},\alpha\right)|}{n^{\sigma\left(\mathcal{L},\alpha\right)+1}\log(xn)}\ll\dfrac{1}{\log x}.
\end{align}

Once more, $\mathcal{I}_4$ is easily seen to be bounded by $x^\epsilon$ which is abosrbed in the error term appearing in \eqref{init1}.

To estimate $\mathcal{I}_2$ we work as in \eqref{I2} and \eqref{smal}, where instead of $x^s$ we have $x^{-s}$.
Then
\begin{align}\label{I2i}
\begin{split}
\mathcal{I}_2
&\ll\sum\limits_{|\gamma_\alpha-T|\leq 1}\left(1+\int\limits_{T}^{T+2}\dfrac{x^{-\left(\sigma\left(\mathcal{L},\alpha\right)+1\right)}}{\sqrt{1+(t-\gamma_\alpha)^2}}\mathrm{dt}+\dfrac{x^\epsilon}{\log x}+\int\limits_{T}^{T+2}\dfrac{x^\epsilon}{\sqrt{\epsilon^2+(t-\gamma_\alpha)^2}}\mathrm{d}t\right)+x^\epsilon\dfrac{\log T}{\log x}\\
&\ll\sum\limits_{|\gamma_\alpha-T|\leq 1}\left(1+\dfrac{x^\epsilon}{\log x}+x^\epsilon\int\limits_{T}^{T+2}\min\left\{\frac{1}{\epsilon},\dfrac{1}{|t-\gamma_\alpha|}\right\}\mathrm{d}t\right)+x^\epsilon\dfrac{\log T}{\log x}\\
&\ll x^\epsilon\log T\left(\dfrac{1}{\log x}+\log\dfrac{1}{\epsilon}\right).
\end{split}
\end{align}
It remains to estimate $\mathcal{I}_3$.
Lemma \ref{lb} implies that
\begin{align*}
|\mathcal{L}(-\epsilon+it)|>2|\alpha|
\end{align*}
for any $t\geq T_1$, where $T_1\gg_\epsilon1$ is a sufficienlty large positive number such that it is not an ordinate of an $\alpha$-point.
Therefore, for $T\geq T_1$ we have that
\begin{align}\label{I3i}
\begin{split}
\dfrac{2\pi}{x^\epsilon}\mathcal{I}_3+O(1)
&=-\int\limits_{T_1}^Tx^{-it}\dfrac{\mathcal{L}'}{\mathcal{L}}(-\epsilon+it)\left(1-\frac{\alpha}{\mathcal{L}(-\epsilon+it)}\right)^{-1}\mathrm{d}t\\
&=-\int\limits_{T_1}^Tx^{-it}\dfrac{\mathcal{L}'}{\mathcal{L}}(-\epsilon+it)\left(1+\sum\limits_{k=1}^{\infty}\left(\dfrac{\alpha}{\mathcal{L}(-\epsilon+it)}\right)^k\right)\mathrm{d}t\\
&=:-\left(\mathcal{I}_{31}+\mathcal{I}_{32}\right).
\end{split}
\end{align}
Working as in \eqref{forla}-\eqref{forla1}, it follows that
\begin{align*}
\mathcal{I}_{31}=\sum\limits_{n=1}^{\infty}\dfrac{\overline{\Lambda\left(n;\mathcal{L},0\right)}}{n^{1+\epsilon}}\int\limits_{T_1}^T\left(\dfrac{n}{x}\right)^{it}\mathrm{d}t+O\left(\log T\left(1+\dfrac{1}{\log x}\right)\right).
\end{align*}
By estimating the latter sum as in \eqref{I1}-\eqref{error1}, we obtain that
\begin{align}\label{I31i}
\mathcal{I}_{31}=\dfrac{T}{x^{1+\epsilon}}\overline{\Lambda\left(x;\mathcal{L},0\right)}+O_\epsilon\left(1+\dfrac{\mathbbm{1}_{\mathbb{R}\setminus\mathbb{Z}}(x)}{x^{\epsilon/2}}\min\left\{\dfrac{T}{x},\dfrac{1}{\| x\|}\right\}\right)+O\left(\log T\left(1+\dfrac{1}{\log x}\right)\right)
\end{align}

To estimate $\mathcal{I}_{32}$ we simply note that $d_{\mathcal{L}}\geq2$.
Thus, in view of Lemma \ref{lb}, we have that
\begin{align}\label{I32i}
\mathcal{I}_{32}
\ll_\epsilon\int\limits_{T_1}^T\sum\limits_{k=1}^{\infty}\left(\dfrac{\alpha}{t^{(1/2+\epsilon)d_\mathcal{L}}}\right)^k\mathrm{d}t
\ll_\epsilon\int\limits_{T_1}^T\dfrac{1}{t^{1+2\epsilon}}\ll_\epsilon 1.
\end{align}
Combining relations \eqref{I3i}-\eqref{I32i} yields that
\begin{align}\label{I3if}
\mathcal{I}_3
=-\dfrac{T\overline{\Lambda\left(x;\mathcal{L},0\right)}}{2\pi x}+O_\epsilon\hspace*{-2pt}\left(x^\epsilon\left(1+\mathbbm{1}_{\mathbb{R}\setminus\mathbb{Z}}(x)\min\hspace*{-0.7pt}\left\{\dfrac{T}{x},\dfrac{1}{\| x\|}\right\}\right)\hspace*{-1pt}\right)+O\hspace*{-2pt}\left(x^{\epsilon}\log T\left(1+\dfrac{1}{\log x}\right)\hspace*{-1pt}\right),
\end{align}
as long as $T\geq T_1$.
For the remaining $T$, that is for $T_0\leq T<T_1$, we have that
\begin{align}\label{remain}
\mathcal{I}_3=\pm\dfrac{T\overline{\Lambda\left(x;\mathcal{L},0\right)}}{2\pi x}+\int\limits_{-\epsilon+iT}^{-\epsilon+iT_0}x^{-s}\dfrac{\mathcal{L}'(s)}{\mathcal{L}(s)-\alpha}\mathrm{d}s=-\dfrac{T\overline{\Lambda\left(x;\mathcal{L},0\right)}}{2\pi x}+O_\epsilon\left(x^\epsilon\right).
\end{align}

The theorem now follows from relations \eqref{init1}-\eqref{I2i}, \eqref{I3if} and \eqref{remain} and by arguing similarly as in the end of the proof of the Theorem \ref{MAIN1} for the case of $T$ being an ordinate of an $\alpha$-point.
\end{proof}

\begin{proof}[Proof of Theorem \ref{MAIN3}]
We set $\epsilon=1/\log 3x$ and we work exactly as in the proof of Theorem \ref{MAIN2} to estimate $\mathcal{I}_1$, $\mathcal{I}_2$ and $\mathcal{I}_4$.
In particular, we have that $1\leq x^{\epsilon}\leq e$ and
\begin{align}\label{firste}
\mathcal{I}_1\ll\dfrac{1}{x^{\sigma\left(\mathcal{L},\alpha\right)+1}\log x},\,\,\,\mathcal{I}_2\ll\log T\left(\dfrac{1}{\log x}+\log\log (3x)\right)\,\,\,\text{ and }\,\,\,\mathcal{I}_4\ll1.
\end{align}

The estimation of $\mathcal{I}_3$ is more laborious.
Since $d_\mathcal{L}=1$, $\mathcal{L}(s)$ is either the Riemann zeta-function $\zeta(s)$ or a shift $L\left(s+i\theta;\chi\right)$ of a Dirichlet $L$-function for some $\theta\in\mathbb{R}$ and some primitive Dirichlet character $\chi$.
Then, it follows from \cite[Theorem 3.22, Theorem 8.20, Theorem 8.22]{Ten} that
\begin{align}
|\mathcal{L}(\sigma+it)|\gg\left(\log |t|\right)^{-m},\,\,\,|t|\geq t_0>1,
\end{align}
uniformly in $1\leq\sigma\leq2$, where $m=1$ if $\mathcal{L}(s)=\zeta(s)$ and $m=7$, otherwise.
In view of relation \eqref{order} we deduce that
\begin{align}\label{gorder}
|\mathcal{L}(\sigma+it)|\gg\dfrac{|t|^{1/2-\sigma}}{\left(\log |t|\right)^{m}},\,\,\,|t|\geq t_0>1,
\end{align}
uniformly in $-1\leq\sigma\leq0$.
Hence, there is a sufficiently large positive number $T_1$, only this time independent of $\epsilon$, such that it is not an ordinate of an $\alpha$-point and
$$|\mathcal{L}(\sigma+it)|>2|\alpha|$$
for any $t\geq T_1$ and  $-1\leq\sigma\leq0$.
Therefore, for $T\geq T_1$ we have that
\begin{align}\label{I3ii}
\begin{split}
\frac{2\pi}{ x^{\epsilon}}\mathcal{I}_3+O(1)
&=-\int\limits_{T_1}^Tx^{-it}\dfrac{\mathcal{L}'}{\mathcal{L}}(-\epsilon+it)\left(1-\frac{\alpha}{\mathcal{L}(-\epsilon+it)}\right)^{-1}\mathrm{d}t\\
&=-\int\limits_{T_1}^Tx^{-it}\dfrac{\mathcal{L}'}{\mathcal{L}}(-\epsilon+it)\left(1+\dfrac{\alpha}{\mathcal{L}(-\epsilon+it)}+\sum\limits_{k=2}^{\infty}\left(\dfrac{\alpha}{\mathcal{L}(-\epsilon+it)}\right)^k\right)\mathrm{d}t\\
&=:-\left(\mathcal{I}_{31}+\mathcal{I}_{32}+\mathcal{I}_{33}\right).
\end{split}
\end{align}

Once more, we have that
\begin{align}\label{I31ii}
\mathcal{I}_{31}=\sum\limits_{n=1}^{\infty}\dfrac{\overline{\Lambda\left(n;\mathcal{L},0\right)}}{n^{1+\epsilon}}\int\limits_{T_1}^T\left(\dfrac{n}{x}\right)^{it}\mathrm{d}t+O\left(\log T\left(1+\dfrac{1}{\log x}\right)\right).
\end{align}
In this case, however, we know that $|\Lambda\left(n;\mathcal{L},0\right)|=\Lambda(n)$.
Thus,
\begin{align}\label{I31part}
\begin{split}
\sum\limits_{n=1}^{\infty}\dfrac{\overline{\Lambda\left(n;\mathcal{L},0\right)}}{n^{1+\epsilon}}\int\limits_{T_1}^T\left(\dfrac{n}{x}\right)^{it}\mathrm{d}t\
&=\frac{T-T_1}{x^{1+\epsilon}}\overline{\Lambda\left(x;\mathcal{L},0\right)}+\mathop{\sum\limits_{n=1}^{\infty}}_{n\neq x}\dfrac{\Lambda(n)}{n^{1+\epsilon}}\min\left\{T,\left|\log\dfrac{x}{n}\right|^{-1}\right\}\\
&=\hspace*{-0.55pt}\dfrac{T\overline{\Lambda\left(x;\mathcal{L},0\right)}}{x^{1+\epsilon}}
\hspace*{-0.55pt}+\hspace*{-0.55pt}O\left(\log(2 x)\log\log (3x)+\log x\min\hspace*{-0.5pt}\left\{\dfrac{T}{x},\dfrac{1}{\langle x\rangle}\right\}\right),
\end{split}
\end{align}
as follows by \cite[Lemma 2]{Gon}.

It is simple to estimate $\mathcal{I}_{33}$ since by relation \eqref{gorder} it follows that
\begin{align}\label{I33i}
\mathcal{I}_{33}\ll\int\limits_{T_1}^{T}\sum
\limits_{k=2}^{\infty}\left(\dfrac{\alpha(\log t)^m}{t^{1/2+\epsilon}}\right)^k\mathrm{d}t\ll(\log T)^{2m}\int\limits_{T_1}^T\dfrac{1}{t^{1+2\epsilon}}\mathrm{d}t\ll(\log T)^{2m}\log(3x).
\end{align}

To estimate $\mathcal{I}_{32}$ we start with integration by parts, that is,
\begin{align}\label{I32e}
\dfrac{x^{\epsilon}}{\alpha i}\mathcal{I}_{32}=-\int\limits_{-\epsilon+iT_1}^{-\epsilon+iT}x^{-s}\dfrac{\mathcal{L}'}{\mathcal{L}^2}(s)\mathrm{d}s
=\left[\dfrac{x^{-s}}{\mathcal{L}(s)}\right]_{-\epsilon+iT_1}^{-\epsilon+iT}+\log x\int\limits_{-\epsilon+iT_1}^{-\epsilon+iT}\dfrac{x^{-s}}{\mathcal{L}(s)}\mathrm{d}s.
\end{align}
The first term on the right-hand side is bounded above by $T_1^{-1/2}(\log T_1)^m\ll1$ by relation \eqref{gorder}.
For the second term we split the path of integration in a dyadic manner:
\begin{align}\label{split}
\int\limits_{-\epsilon+iT_1}^{-\epsilon+iT}\dfrac{x^{-s}}{\mathcal{L}(s)}\mathrm{d}s=O(1)+\sum\limits_{k=1}^{\left\lfloor\log (T/T_0)/\log 2\right\rfloor}\int\limits_{-\epsilon+iT/2^{k}}^{-\epsilon+iT/2^{k-1}}\dfrac{x^{-s}}{\mathcal{L}(s)}\mathrm{d}s=:O(1)+\sum\limits_{k=1}^{\left\lfloor\log (T/T_1)/\log 2\right\rfloor}I_k.
\end{align}
 Observe that $2^k\leq T$ for all $k=1,\dots,\left\lfloor\log (T/T_1)/\log 2\right\rfloor$.
 To estimate $I_{k}$ we use the functional equation  \eqref{functequa} of $\mathcal{L}(s)$:
 \begin{align*}
I_{k}
=\int\limits_{-\epsilon+iT/2^k}^{-\epsilon+iT/2^{k-1}}\dfrac{x^{-s}}{H_\mathcal{L}(s)\overline{\mathcal{L}(1-\overline{s})}}\mathrm{d}s
 =ix^{\epsilon} \int\limits_{T/2^k}^{T/2^{k-1}}\dfrac{x^{-it}}{H_{\mathcal{L}}(-\epsilon+it)\overline{\mathcal{L}(1+\epsilon+it)}}\mathrm{d}t.
 \end{align*}
 Since $\mathcal{L}(s)$ is absolutely convergent in the half-plane $\sigma>1$ and $f(n)$ is a completely multiplicative arithmetical function, we have that
 \begin{align*}
 \dfrac{1}{\mathcal{L}(s)}=\sum\limits_{n=1}^{\infty}\dfrac{\mu(n)f(n)}{n^s},\,\,\,\sigma>1,
 \end{align*}
 where $\mu(n)$ is the M\"obius function and the series converges absolutely in the half-plane $\sigma>1$ (see for example \cite[Section 11.4, Example 3]{apo}.
 In addition to relation \eqref{Stril}, it follows that
 \begin{align*}
I_{k}
=&\,\dfrac{ix^{\epsilon}e^{i\pi(1-\mu)/4}}{\left(2\pi\lambda Q^2\right)^{1/2+\epsilon}}\sum\limits_{n=1}^{\infty}\dfrac{\mu(n)\overline{f(n)}}{n^{1+\epsilon}}\times\\
&\times\int\limits_{T/2^k}^{T/2^{k-1}}\exp\left(id_\mathcal{L}t\log\left(\left(\dfrac{n\lambda Q^2}{x}\right)^{1/d_\mathcal{L}}\dfrac{t}{e}\right)\right)\left(\dfrac{t}{2\pi}\right)^{-\epsilon-1/2}\dfrac{1}{\omega+O\left(\frac{1}{t}\right)}\mathrm{d}t.
\end{align*}
Observe that the implicit constant in the latter relation is independent of $\epsilon$ since $0<\epsilon<1/\log3$ and we applied relation \eqref{Stril} in the strip $-1/\log 3\leq\sigma\leq0$.
If we take now $T_1\gg1$ to be a sufficiently large positive number then
$$\left(\omega+O\left(\dfrac{1}{t}\right)\right)^{-1}=\dfrac{1}{\omega}+O\left(\dfrac{1}{t}\right)$$
for any $t\geq T_1$.
Then integrating by substitution will yield
\begin{align}\label{Ik2}
 I_{k}
 \ll &\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{1+\epsilon}}\left|\,\int\limits_{d_\mathcal{L}T/2^k}^{d_\mathcal{L}T/2^{k-1}}\exp\left(it\log\left(\left(\dfrac{n\lambda Q^2}{x}\right)^{1/d_\mathcal{L}}\dfrac{t}{d_\mathcal{L}e}\right)\right)\left(\dfrac{t}{2\pi}\right)^{-\epsilon-1/2}\mathrm{d}t\right|+\\
 &+\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{1+\epsilon}}\int\limits_{T/2^k}^{T/2^{k-1}}\dfrac{1}{t^{\epsilon+3/2}}\mathrm{d}t,
 \end{align}
 since $\left|\mu(n){f(n)}\right|=1$ for all $n\in\mathbb{N}$.
 From the last term of the relation above we get
 \begin{align}\label{Ik2p}
 \sum\limits_{n=1}^{\infty}\dfrac{1}{n^{1+\epsilon}}\int\limits_{T/2^k}^{T/2^{k-1}}\dfrac{1}{t^{\epsilon+3/2}}\mathrm{d}t
\ll \left(\dfrac{2^k}{T}\right)^{1/2}\log\log(3x).
 \end{align}
 For the first term of \eqref{Ik2}, which we denote by $J_k$, we employ Lemma \ref{Gonek}.
 To this end, we consider first $T_1\gg1$ sufficiently large, with the implicit constant being independent of $\epsilon$ as it ranges over the interval $(0,1/\log3)$.
 Then, following our notation from Lemma \ref{Gonek}, we obtain that
 \begin{align*}
 J_k=&\mathop{\sum\limits_{n=1}^{\infty}}_{\frac{d_\mathcal{L}T}{2^{k}}<d_\mathcal{L}\left(\frac{x}{\lambda Q^2n}\right)^{1/d_\mathcal{L}}\leq \frac{d_\mathcal{L}T}{2^{k-1}}}\dfrac{1}{n^{1+\epsilon}}M\left(d_\mathcal{L}\left(\dfrac{x}{\lambda Q^2n}\right)^{1/d_\mathcal{L}},-\epsilon,\dfrac{d_\mathcal{L}T}{2^{k}},\dfrac{d_\mathcal{L}T}{2^{k-1}}\right)+\\
 &+\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{1+\epsilon}}E\left(d_\mathcal{L}\left(\dfrac{x}{\lambda Q^2n}\right)^{1/d_\mathcal{L}},-\epsilon,\dfrac{d_\mathcal{L}T}{2^{k}},\dfrac{d_\mathcal{L}T}{2^{k-1}}\right)\\
 &=:M_k+E_k.
 \end{align*}
 We estimate firstly $M_k$:
 \begin{align*}
M_k\ll\mathop{\sum\limits_{n=1}^{\infty}}_{\frac{d_\mathcal{L}T}{2^{k}}<d_\mathcal{L}\left(\frac{x}{\lambda Q^2n}\right)^{1/d_\mathcal{L}}\leq \frac{d_\mathcal{L}T}{2^{k-1}}}\dfrac{1}{n^{1+\epsilon}}\left(\dfrac{x}{\lambda Q^2n}\right)^{-\epsilon/d_\mathcal{L}}
\ll\sum\limits_{\left(\frac{2^{k-1}}{T}\right)^{d_\mathcal{L}}\frac{x}{\lambda Q^2}\leq n<\left(\frac{2^{k}}{T}\right)^{d_\mathcal{L}}\frac{x}{\lambda Q^2}}\dfrac{1}{n}
\ll 1.
 \end{align*}
We estimate now $E_k$:
 \begin{align*}
 E_k&\ll\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{1+\epsilon}}\left[\left(\dfrac{d_\mathcal{L}T}{2^k}\right)^{-\epsilon-1/2}+\dfrac{\left(\frac{d_\mathcal{L}T}{2^k}\right)^{-\epsilon+1/2}}{\left|\frac{d_\mathcal{L}T}{2^k}-d_\mathcal{L}\left(\frac{x}{\lambda Q^2n}\right)^{1/d_\mathcal{L}}\right|+\left(\frac{d_\mathcal{L}T}{2^k}\right)^{1/2}}+\right.\\&\hspace*{7cm}\left.+\dfrac{\left(\frac{d_\mathcal{L}T}{2^{k-1}}\right)^{-\epsilon+1/2}}{\left|\frac{d_\mathcal{L}T}{2^{k-1}}-d_\mathcal{L}\left(\frac{x}{\lambda Q^2n}\right)^{1/d_\mathcal{L}}\right|+\left(\frac{d_\mathcal{L}T}{2^{k-1}}\right)^{1/2}}\right]\\
 &\ll\left(\dfrac{2^k}{T}\right)^\epsilon\sum\limits_{n=1}^{\infty}\dfrac{1}{n^{1+\epsilon}}\left[\left(\dfrac{2^k}{T}\right)^{1/2}+1+2^{-\epsilon}\right]\\
 &\ll\left(\dfrac{2^k}{T}\right)^{\epsilon}\log\log(3x).
 \end{align*}
 Summing up we will have that
 \begin{align*}
 \sum\limits_{k=1}^{\left\lfloor\log (T/T_1)/\log 2\right\rfloor}J_k
 &\ll\sum\limits_{k=1}^{\left\lfloor\log (T/T_1)/\log 2\right\rfloor}\left(1+\left(\dfrac{2^k}{T}\right)^{\epsilon}\log\log(3x)\right)\\
 &\ll\log T+\dfrac{2^{\epsilon\log (T/T_1)/\log 2}}{(2^\epsilon-1)T^\epsilon}\log\log(3x)\\
 &\ll\log T+\log(3x)\log\log(3x).
 \end{align*}
From this and relations \eqref{Ik2} and \eqref{Ik2p} it follows that
\begin{align*}
 \sum\limits_{k=1}^{\left\lfloor\log (T/T_1)/\log 2\right\rfloor}I_{k}
 \ll\,& \sum\limits_{k=1}^{\left\lfloor\log (T/T_1)/\log 2\right\rfloor}\left(\dfrac{2^{k}}{T}\right)^{1/2}\log\log(3x)+\log T+\log(3x)\log\log(3x)\\
 \ll\,&\log T+\log(2x)\log\log(3x).
\end{align*}
Then relations \eqref{I32e} and \eqref{split} imply that
\begin{align*}
\mathcal{I}_{32}\ll1+\log x\left(\log T+\log(2x)\log\log(3x)\right)
\end{align*}
and, thus, from \eqref{I31ii}-\eqref{I33i} we have that
\begin{align}\label{laste}
\begin{split}
\mathcal{I}_3=&\,-\dfrac{T\overline{\Lambda\left(x;\mathcal{L},0\right)}}{2\pi x}
+O\left(\log(2 x)\log\log (3x)+\log x\min\left\{\dfrac{T}{x},\dfrac{1}{\langle x\rangle}\right\}\right)+\\
&+O\left((\log T)^{2m}\log(3x)+\dfrac{\log T}{\log x}+\log x\left(\log T+\log(2x)\log\log(3x)\right)\right),
\end{split}
\end{align}
as long as $T\geq T_1$.

The theorem now follows by combining \eqref{firste} and \eqref{laste} and by arguing similarly as in the end of the proof of the Theorem \ref{MAIN1} for the case of $T$ being an ordinate of an $\alpha$-point.
\end{proof}

\section{Littlewood's theorem and Applications to Universality}

\begin{proof}[Proof of Theorem \ref{Main1}]
Let $n_0>1$ be the first positive integer $n$ such that $f(n)\neq0$.
We know that such a number exists, because $\mathcal{L}(s)$ differs from the only constant function in $\mathcal{S}$.
This also implies that $d_{\mathcal{L}}\geq1$.
 In the sequel we make use of the notation $a_n:=f(n)$, for all $n\in\mathbb{N}$.

Now we define the function
$$g(s):=\left\{
\renewcommand{\arraystretch}{1.5}
\begin{array}{lllll}
\dfrac{|a_{n_0}|n_0^s}{a_{n_0}}(\mathcal{L}(s)-1),&\text{ if }\alpha=1,\\
\mathcal{L}(s)-\alpha,&\text{ if }\Re \alpha<1,\\
\alpha-\mathcal{L}(s),&\text{ if }\Re \alpha>1,\\
-i\left(\mathcal{L}(s)-\alpha\right),&\text{ if }\Re \alpha=1\text{ and }\Im \alpha>0,\\
i\left(\mathcal{L}(s)-\alpha\right),&\text{ if }\Re \alpha=1\text{ and }\Im \alpha<0,
\end{array}
\right.$$
for every $s\in\mathbb{C}\setminus\lbrace1\rbrace$ .
Then there exists $\sigma_0=\sigma_0(\alpha)>1$ such that
\begin{align}\label{gasy}
g(s)\asymp\Re(g(s))\asymp1,
\end{align}
uniformly in the half-plane $\sigma\geq\sigma_0-1$.
Indeed, the absolute convergence of $\mathcal{L}(s)$ in the half-plane $\sigma>1$ implies that
$$\Re(g(s))\leq|g(s)|\ll1,$$
uniformly in any half-plane $\sigma\geq\sigma_1>1$.
On the other hand, for $\sigma>1$,
$$\Re(g(s))=\left\{
\renewcommand{\arraystretch}{1.5}
\begin{array}{lllll}
|a_{n_0}|+\sum\limits_{n>n_0}\left(n_0/n\right)^{\sigma}E(n,t),&\alpha=1,\\
1-\Re \alpha+\sum\limits_{n\geq n_0}n^{-\sigma}F(n,t),&\Re \alpha<1,\\
\Re \alpha-1-\sum\limits_{n\geq n_0}n^{-\sigma}F(n,t),&\Re \alpha>1,\\
\Im \alpha+\sum\limits_{n\geq n_0}n^{-\sigma}G(n,t),&\Re \alpha=1\text{ and }\Im \alpha>0,\\
-\Im \alpha-\sum\limits_{n\geq n_0}n^{-\sigma}G(n,t),&\Re \alpha=1\text{ and }\Im \alpha<0,
\end{array}
\right.$$
where
\begin{align*}
\renewcommand{\arraystretch}{1.5}
\begin{array}{lll}
E(n,t):=|a_{n_0}|^{-1}\left[\Re \left(\overline{a_{n_0}}a_n\right)\cos\left(t\log{{\dfrac{n_0}{n}}}\right)-\Im \left(\overline{a_{n_0}}a_n\right)\sin\left(t\log\dfrac{n_0}{n}\right)\right],\\
F(n,t):=\Re a_n\cos\left(t\log n\right)+\Im a_n\sin\left(t\log n\right),\\
G(n,t):=\Im a_n\cos\left(t\log n\right)-\Re a_n\sin\left(t\log n\right).
\end{array}
\end{align*}
Since $E(n,t),F(n,t),G(n,t)\ll_\epsilon n^{\epsilon}$ for all $n\in\mathbb{N}$ and $t\in\mathbb{R}$, there is a sufficiently large $\sigma_0>0$ such that
$$|g(s)|\geq\Re(g(z))\gg1,$$
uniformly in $\sigma\geq\sigma_0-1$.
Moreover, by arguing similarly, we can take $\sigma_0>0$ large enough such that also
\begin{align}\label{order1}
|\mathcal{L}(s)|\gg1,
\end{align}
uniformly in $\sigma\geq\sigma_0-1$.

Lastly, we  define for every $s\in\mathbb{C}\setminus\lbrace1\rbrace$ the function $h(s):=\log g(s),$ where the logarithm takes its principal value for $\sigma\geq\sigma_0-1$, and for other points $s$ we define $h(s)$ to be the value obtained from $h(\sigma_0)$ by continuous variation along the line segments $[\sigma_0,\sigma_0+it]$ and $[\sigma_0+it,\sigma+it]$, provided that the path does not cross a zero or a pole of $g(s)$; if it does, then we take $h(s)=h(s+0)$.
From the latter definition and \eqref{gasy} it follows that
\begin{align}\label{or}
h(s)\ll1,
\end{align}
uniformly in $\sigma\geq\sigma_0-1$.

For the second part of the proof we employ a technique used by Titchmarsh (see \cite[Theorem 9.12]{Titch2}).

It suffices to show that for every sufficiently large positive number $T$, if there is no $\alpha$-point of $\mathcal{L}(s)$ such that $|\gamma_\alpha-T|\leq\delta<1/2$, then $\delta\ll\left(\log\log\log T\right)^{-1}$.
To this end, let $T$ be a large positive number and assume that $\mathcal{L}(s)$ has no $\alpha$-point such that $|\gamma_\alpha-T|\leq\delta<1/2$.
Then, the function $h(s)$, as it was defined above is regular in the rectangle $$\mathcal{R}:=\left\{s\in\mathbb{C}:-(\sigma_0+1)\leq\sigma\leq\sigma_0+1,\,\,\,\,\,|t-T|\leq\delta\right\}.$$
We set
\begin{align}\label{N}
N:=\left\lfloor\dfrac{8\sigma_0}{\delta}\right\rfloor+1,
\end{align}
and we consider the circles
$$\mathcal{K}_{m,n}:=\left\{s\in\mathbb{C}:\left|s-\left(\sigma_0-\frac{n\delta}{4}+iT\right)\right|=\dfrac{m\delta}{4}\right\};$$
we define the maximum value of $h$ on them by
$$\mathbf{M}_{m,n}:=\max\limits_{s\in\mathcal{K}_{m,n}}|h(s)|$$
for every $n=0,\dots,N$ and $m=1,2,3,4$.
Observe that all circles $\mathcal{K}_{m,n}$ lie inside the strip $-(\sigma_0+1)\leq\sigma\leq\sigma_0+1$ and, in particular, in $\mathcal{R}$.
Now the proof follows by arguing similarly as in \cite[Theorem 9.12]{Titch2}.

\end{proof}

\begin{proof}[Proof of Corollary \ref{Main2}]
Let $(\rho_{\alpha,n})_{n\in\mathbb{N}}$ be the sequence of $\alpha$-points of $\mathcal{L}(s)$.
Then, for every $k\geq k':=\lfloor\exp(3)/b\rfloor$, Theorem \ref{Main1} yields the existence of a positive integer $n_k$ such that $$\gamma_{\alpha,n_k}=bk+O\left(\dfrac{1}{\log\log\log (bk)}\right)=bk+o(1).$$
We also set $\gamma_{\alpha,n_k}=\gamma_{\alpha,n_{k'}}$ for every positive integer $k<k'$.
Now, if $a\notin b^{-1}\mathbb{Q}$ is a real number and $m$ a positive integer, then we know that the sequence $\left(a b k^m\right)_{k\in\mathbb{N}}$, is uniformly distributed mod 1  (see for example \cite[Chapter 1, Theorem 3.2]{K-N}) and $$\lim\limits_{k\to\infty}\left(a bk^m-a\gamma_{\alpha,n_{k^m}}\right)=0.$$
Hence, it follows from \cite[Chapter 1, Theorem 1.2]{K-N} that the sequence $\left(a\gamma_{\alpha,n_{k^m}}\right)_{k\in\mathbb{N}}$ is uniformly distributed mod 1.
\end{proof}

\begin{proof}[Proof of Theorem \ref{universality}]
Let $\alpha\in\mathbb{C}$ and $m=1$.
 According to Corollary \ref{Main2}, we can find a subsequence of $\alpha$-points of $\mathcal{L}(s)$ such that
\begin{align}\label{In}
|\gamma_{\alpha, n_k}-2\pi k|<\dfrac{\pi}{2},
\end{align}
for every $k\geq k_1$, where $k_1\gg1$ is a sufficiently large and fixed positive integer.
In addition, the sequence $\left(a\gamma_{\alpha, n_k}\right)_{k\in\mathbb{N}}$, is uniformly distributed modulo one or, equivalently from Weyl's criterion \cite[Chapter 1, Theorem 2.1]{K-N},
\begin{align}\label{W1}
\lim\limits_{N\to\infty}\dfrac{1}{N}\sum\limits_{k=1}^N\e\left(a\gamma_{\alpha, n_k}\right)=0,
\end{align}
for every $a\notin(2\pi)^{-1}\mathbb{Q}$.

Let $\mathbb{P}$ denote the set of prime numbers. Since the numbers
$$1\,\,\,\text{ and }\,\,\,\dfrac{\log p}{2\pi},\,\,\,p\in\mathbb{P},$$
are linearly independent over $\mathbb{Q}$, it follows from\cite[Chapter 1, Theorem 6.2]{K-N} and relation \eqref{W1}  that the sequence
\begin{align}\label{W2}
\left(\gamma_{\alpha, n_k}\dfrac{\log p}{2\pi}\right)_{p\in M},\,\,\,k\in\mathbb{N},
\end{align}
is uniformly distributed modulo one in $\mathbb{R}^{\sharp M}$, for any finite set $M\subseteq\mathbb{P}$.
If we define now the truncated Euler products of $\zeta(s)$,
$$\zeta_M\left(s,\left(\theta_p\right)_{p\in M}\right):=\mathlarger\prod\limits_{p\in M}\left(1-\dfrac{\e\left(\theta_p\right)}{p^s}\right)^{-1}$$
for every $\left(s,\left(\theta_p\right)_{p\in M}\right)\in\left\{s\in\mathbb{C}:{\sigma>0}\right\}\times\mathbb{R}^{\sharp M}$ and finite $M\subseteq\mathbb{P}$, then \eqref{W2} and \cite[Lemma 4]{P} yield the existence of positive numbers $v$ and $d$ such that
\begin{align}\label{App1}
\liminf\limits_{N\to\infty}\dfrac{1}{N}\sharp\left\{1\leq k\leq N:\max\limits_{s\in K}|\zeta_{\left\{p:p\leq y\right\}}(s+i\gamma_{\alpha, n_k},\underline{0})-f(s)|<\varepsilon\right\}>d,
\end{align}
for every $y\geq v$.

Let  $\delta:=\pi$, $T_0:=\gamma_{\alpha, n_{k_1}}-\pi/2$, $T:=\gamma_{\alpha, n_N}-\gamma_{\alpha, n_{k_1}}+\pi$, $A:=\lbrace \gamma_{\alpha, n_{k_1}},\dots,\gamma_{\alpha, n_N}\rbrace$
  and
 \begin{align*}
 f(t):=\zeta(s+it)-\zeta_{\lbrace p:p\leq y\rbrace}(s+it,\underline{0}).
 \end{align*}
Then inequality \eqref{In} and Lemma \ref{gall} yield
 \begin{align}\label{G}
 \begin{split}
\sum\limits_{k=k_1}^N\hspace{-1pt}|\zeta(s+i\gamma_{\alpha, n_k})\hspace{-1pt}-\hspace{-1pt}\zeta_{\lbrace p:p\leq y\rbrace}(s+i\gamma_{\alpha, n_k},\underline{0})|^2\leq&\,2\int\limits_{0}^{2\pi(N+1)}|f(x)|^2\mathrm{d}x+\\
&+\hspace{-1pt}\left(\int\limits_{0}^{2\pi(N+1)}|f(x)|^2\mathrm{d}x\hspace{-1pt}\int\limits_{0}^{2\pi(N+1)}\left|f'(x)\right|^2\mathrm{d}x\right)^{1/2}.
\end{split}
\end{align}
Bohr \cite[Hilfssatz 2]{Bohr} proved that
\begin{align}\label{Bo}
\lim\limits_{y\to\infty}\limsup\limits_{T\to\infty}\dfrac{1}{T}\int\limits_{0}^{T}|\zeta(s+i\tau)-\zeta_{\left\{p:p\leq y\right\}}(s+i\tau,\underline{0})|^2\mathrm{d}\tau=0,
\end{align}
uniformly on compact subsets of $\mathcal{D}$. It is clear that an application of Cauchy's integral formula would yield \eqref{Bo} also in the case of $\zeta'-\zeta'_{\left\{p:p\leq y\right\}}$.
Therefore, in view of  \eqref{G} and \eqref{Bo} it follows that
\begin{align}\label{A}
\lim\limits_{y\to\infty}\limsup\limits_{N\to\infty}\dfrac{1}{N}\sum\limits_{k=1}^{N}|\zeta(s+i\gamma_{\alpha, n_k})-\zeta_{\lbrace p:p\leq y\rbrace}(s+i\gamma_{\alpha, n_k},\underline{0})|^2=0,
\end{align}
uniformly on compact subsets of $\mathcal{D}$.

If $G\subseteq\mathbb{C}$ is a bounded domain and $g$ an analytic and square integrable  function on $G$, then
$$|g(z)|\leq\left(\sqrt{\pi}d(z,\partial G)\right)^{-1}\left(\mathlarger\iint_G|g(\sigma+it)|^2\mathrm{d}\sigma\mathrm{d}t\right)^{1/2},$$
for every $z\in G$ (see for example \cite[Chapter 1, Theorem 1]{D-S}). Thus, by considering a bounded domain $G$ such that $K\subseteq G\subseteq\overline{G}\subseteq\mathcal{D}$, relation \eqref{A} and Chebyshev's inequality yield the existence of a positive number $u$ such that
\begin{align}\label{App2}
\liminf\limits_{N\to\infty}\dfrac{1}{N}\sharp\left\{1\leq k\leq N:\max\limits_{s\in K}\left|\zeta\left(s+i\gamma_{\alpha, n_k}\right)-\zeta_{\lbrace p: p\leq y\rbrace}\left(s+i\gamma_{\alpha, n_k},\underline{0}\right)\right|<\tilde{\varepsilon}
\right\}>1-\tilde{\varepsilon},
\end{align}
for every $y\geq u$, where $\tilde{\varepsilon}=\min\lbrace\varepsilon,d/2\rbrace$.

Now the theorem follows from \eqref{App1} and \eqref{App2} by taking $y=\max\lbrace u,v\rbrace$.
\end{proof}

config: arxiv


TITLE: Counterexample: For real functions existence of all higher order derivatives doesn't imply analycity.
QUESTION [3 upvotes]: In the lecture we had an example for a function $f: \mathbb R \to \mathbb R$, which is not analytic. We defined, that a function is said to be analytic at some point $x_0$ if a Taylor series expansion is valid at that point.
The following example is supposed to demonstrate that for real functions the existence of all higher order derivatives does not imply analycity.
Let $f: \mathbb R \to \mathbb R$ be a real function defined by $f(x) = e^{-1/x^2}$. Then $f$ is infinitely many times differentiable and $f^{(n)}(0) = 0$, $\forall n \in \mathbb N \cup \{0\}$.
But $f \ne 0$ in some neighbourhood of $x = 0$ and therefore $f$ is not analytic.
I understand the logic behind that counterexample, but I don't understand why we are allowed to take the point $x = 0$, since $f$ is actually not defined at that point?!
Thanks for help.

REPLY [4 votes]: Actually you define $f(x):=e^{-\frac{1}{x^2}}$ for all $x\neq 0$ and $f(0):=0$. Then you check continuity at $x=0$ and go on with the proof.


TITLE: What is called the property of function that it does not change value when you transform arguments.
QUESTION [0 upvotes]: My question is probably rather simple, but I cannot find appropriate name or am too stupid to find a definition of such property.
I'd like to give an example:
$f({\bf r}_1, ..., {\bf r}_N, ) = \sum^{N}_{i=1}{\left| {\bf r}_{i} \right|}$
where ${\bf r}_i$ - is a position of $i$-th point in a $N$ dimensional space
It basically is a total distance of all points from the origin. So this function has such property that I can transform arguments, and the function does not change it's value. E.g: I can rotate whole system around origin.
My question is what is called such a property? I have two candidates but I don't think they are right:

function is symmetrical (it has rotational symmetry round the origin)
function is degenerated in respect to transformation (rotation round the origin)

There is also another transformation, that does not change the value - swapping two positions.
What is the proper mathematical term to describe it.

REPLY [0 votes]: Another term used in physics is to say that the function is isotropic.


TITLE: Show that $x^3 + 3x+2$ is irreducible in $\mathbb{Z}[5]$
QUESTION [3 upvotes]: I am reading Fraleigh in which Fraleigh proves that $f(x) = x^3 + 3x+ 2$ is irreducible over $\mathbb{Z}[5]$. He explains that the if $f(x) = x^3 + 3x+ 2$ were reducible over $\mathbb{Z}[5]$, then there would exist ar least one linear factor of $f(x)$ of the form $x-a$ for some $a \in \mathbb{Z}[5]$. Then, $f(a)$ would be $0$. However, $f(0) = 2, f(1) = 1, f(2) = 1, f(-1) = -2,f(-2) = -2$.
I don't understand why Fraleigh didn't analyze $f(4)$ and $f(3)$ instead of $f(-1)$ and $f(-2)$ respectively. Can someone please explain? Thanks!

REPLY [0 votes]: When multiplying modulo $n$ for relatively small $n$ if you choose to use negative numbers instead of higher positive numbers you keep the numbers small at the expense of having to keep track of the sign. This is often easier when using brute force to check for irreducibly of a polynomial and other such computations. For example it can make proofs of quadratic reciprocity easier to follow.


TITLE: Colour of Radiation emitted by a object when heated up?
QUESTION [0 upvotes]: I read this in my textbook and want to confirm if this is right what I am writing.
Here , we have a metal which when heated up (Temp keeps on rising ) on a furnace turns red.
After that , it emits white and blue radiations as temp keeps on increasing.
Q1 Why does the radiation emit blue and white colour when the body is becoming more and more red due to intense heat of furnace ?
Why does it emit blue and white radiation if body colour is becoming red?

REPLY [1 votes]: The color of light depends on the frequency or wavelength of the light. Red is the color of the lowest frequency a human eye can see,  blue light  has more energy. white light is a mixture of all colors or frequencies a human eye can detect.  So the first color is red, when you heat something, the frequency before it is hot enough you can not see but feel it as warmth. the higher the temperature, the more energy, so you see more and more colors, until it seems white, you will never see blue alone, but the light may seem more blueish when the temperature and so the energy goes up.


TITLE: Real world applications of the study of Egyptian Fractions
QUESTION [4 upvotes]: I recently read about the study of Egyptian Fractions on the Good Math, Bad Math blog. The references to this article show that many years of research have gone into trying to find efficient ways to calculate the minimum length forms.
Are there any real-world applications that computing minimum-length Egyptian forms can be applied to?

REPLY [3 votes]: Electronic circuit design is one area where Egyptian fractions have practical use. When electrical resistors are added to a parallel circuit, the reciprocals are added to find the total resistance. An example is a parallel circuit with 1,2,3, and 6 Ohm resistors.
$$\frac{2}{1\Omega}=\frac{1}{1\Omega}+\frac{1}{2\Omega}+\frac{1}{3\Omega}+\frac{1}{6\Omega}$$
Inductance (L) in Henrys (H) in parallel circuits is calculated in the same way.
$$\frac{1}{2 H}=\frac{1}{4 H}+\frac{1}{5 H}+\frac{1}{36 H}+\frac{1}{45 H}$$


TITLE: Position as a function of time with given Potential
QUESTION [0 upvotes]: Ok, so I have a given potential with
$$V(x) = \frac12\left(hx^2 + \frac{l}{x^2}\right)$$
where $h$ and $l$ are positive constants. Now I need to calculate the position $x(t)$ as a function of time.
I also know that $$\frac12m\dot x^2 + V(x) = E$$ where $E$ is just the total energy of the system.
This leaves me with the equation:
$$\frac{1}{2}m\dot{x}^2 = E - \frac12\left(hx^2 + \frac{l}{x^2}\right)$$
I have no clue how I can solve a differential equation like this.

REPLY [0 votes]: Mathematica gives
$$\int \frac{1}{\sqrt{a-bx^2 - c/x^2}} =
-\frac{\sqrt{-c + a x^2 - b x^4}
  }{(
 2 \sqrt{b} x \sqrt{a - (c + b x^4)/x^2})}  \arctan (a - 2 b x^2)/(2 \sqrt{b} \sqrt{-c + a x^2 - b x^4)}$$
Can you use that?


TITLE: A bound on the $\mathrm{L}^p$ norm in terms of the $\mathrm{L}^2$ norm in periodic Sobolev spaces
QUESTION [1 upvotes]: (The question was originally posted on Math StackExchange.)
Preliminaries: Given ${s \geq 0}$, let ${\mathrm{H}_P^s}$ denote the ${\mathrm{L}^2}$-based fractional-order Sobolev space of ${P}$-periodic functions on the line with norm
\begin{equation*}
\| u \|_{\mathrm{H}_P^s} = \left( \sum_{n \in \mathbb{Z}} \bigg(1 + \frac{4 \pi^2 n^2}{P^2}\bigg)^{s} |\widehat{u}_n|^2 \right)^\frac{1}{2},
\end{equation*}
where ${\widehat{u}}$ is the ${n}$th Fourier coefficient. We set ${\mathrm{L}_P^2 \equiv \mathrm{H}_P^0}$ and recall Parseval's theorem giving the equivalent $\mathrm{H}^s(-\frac{P}{2},\frac{P}{2})$ norm when $s$ is a nonnegative integer. Moreover, let ${\| \cdot \|_{\infty}}$ express the supremum norm on an interval of length ${P}$.

Question: Fix ${s > \frac{1}{2}}$ and ${p > 2}$ and consider functions ${u}$ in an open ball of ${\mathrm{H}_P^s}$, that is, ${\| u
\|_{\mathrm{H}_P^s} < R}$. Is it possible to estimate
  \begin{equation*} \| u \|_{\mathrm{L}_P^p}^{p} := \int_{-\frac{P}{2}}^{\frac{P}{2}} u^{p} \,
 \mathrm{d}x \lesssim \| u \|_{\mathrm{L}_P^2}^{2 + \epsilon}
\end{equation*} for some ${\epsilon > 0}$ (likely depending on ${p}$)?

Ideas: We have
\begin{equation*}
\| u \|_{\mathrm{L}_P^p}^{p} \leq \| u \|_{\infty}^{p - 2} \| u \|_{\mathrm{L}_P^2}^{2},
\end{equation*}
so it is perhaps easiest to establish ${\| u \|_{\infty}^{p  - 2} \lesssim \| u \|_{\mathrm{L}_P^2}^{\epsilon}}$.

That is, do we have
  $${\| u \|_{\infty} \lesssim \| u \|_{\mathrm{L}_P^2}^{\tilde{\epsilon}}}$$
  for some $\tilde{\epsilon} > 0$ given that ${\| u
\|_{\mathrm{H}_P^s} < R}$?

When ${s = 1}$ this is true based on the estimate
\begin{equation}
\| u \|_{\infty} \lesssim \| u' \|_{\mathrm{L}_P^2}^{\frac{1}{2}} \| u \|_{\mathrm{L}_P^2}^{\frac{1}{2}} \lesssim \| u \|_{\mathrm{L}_P^2}^{\frac{1}{2}}
\end{equation}
(since ${\| u' \|_{\mathrm{L}_P^2} \leq \| u \|_{\mathrm{H}_P^1} < R}$). This estimate is found in these [lecture notes, page 21] when the domain is the real line ${\mathbb{R}}$, but it would be nice to see an argument/reference to the periodic case. And does it extend to the general setting with $s > \frac{1}{2}$?

REPLY [0 votes]: Let me expand my comment into an answer.
Let $s>\frac12$ and pick any $r\in(\frac12,s)$.
Interpolating between Sobolev spaces gives $\|u\|_{H^r_P}\lesssim\|u\|_{H^0_P}^\delta\|u\|_{H^s_P}^{1-\delta}$ for some $\delta>0$.
As was shown to you in the MSE answer to the question, $\|u\|_{L^\infty}\lesssim\|u\|_{H^r_P}$.
Putting these estimates together gives $\|u\|_{L^\infty}\lesssim\|u\|_{L^2_P}^\delta\|u\|_{H^s_P}^{1-\delta}$.
Assuming $\|u\|_{H^s_P}\lesssim1$, this homogeneous estimate gives you the inhomogeneous one you need: $\|u\|_{L^\infty}\lesssim\|u\|_{L^2_P}^\delta$.


text:
\section{Supermaximal models and cohomology}
\subsection{The Model  \(Y_{\Bc(n-1)}\) is a Supermaximal Model}
\label{subsec:supermaximalsvelato}

We can     now answer  to the question,  raised in  Section \ref{sec:supermaximal}, about how to construct a model  that is `bigger' than the maximal model, admits the extended \(S_{n+1}\) action and is minimal with these properties.

Let us state this in a more formal way.

Let us consider  the poset  \(\Bc(n-1)\) that    indicizes  the  strata of the minimal model \(Y_{minA_{n-1}}\), and let us denote  by \({\mathcal L}\) the family of the
models  obtained by blowing up all the  building subsets of these strata. We observe that \({\mathcal L}\) has a natural poset structure given by the relation
\(Y_{\G_1}\leq Y_{\G_2}\) if and only if   \(\G_1\subseteq \G_2\) (by Li's definition, this also means that there is a birational projection of \(Y_{\G_2}\) onto \(Y_{\G_1}\)).

 Let us denote by \(\Tc\) the set the elements of \(\Bc(n-1)\) with  depth 1.
From Theorem \ref{teo:listabuilding} and Remark \ref{rem:ordinescoppiamenti} it follows that if  we blowup in \(Y_{minA_{n-1}}\)  the strata  that correspond to the elements of \(\Tc\) (in a suitable order, i.e. first the strata with smaller dimension) we obtain the model \(Y_{maxA_{n-1}}\).

\begin{definition}
The supermaximal model $Y_{supermaxA_{n-1}}$ associated with the root arrangement \(A_{n-1}\) is the minimal model \(Y_\Kc\) in the poset \({\mathcal L}\) that admits the \(S_{n+1}\) action and such that \(\Kc\supseteq \Tc\).
\end{definition}

We notice that this last property means that the supermaximal $Y_{supermaxA_{n-1}}$ model admits a birational projection onto \(Y_{maxA_{n-1}}\).

As a consequence of Theorem \ref{teo:chiusoazionebuilding} we have proven the following result:
\begin{theorem}
The model \(Y_{\Bc(n-1)}\) is the supermaximal model associated with the root arrangement \(A_{n-1}\).

\end{theorem}

\begin{remark}
There is a family of \(S_n\)-invariant building sets that are intermediate between $\Fc_{A_{n-1}}$ and
$\Cc_{A_{n-1}}$

(these building sets have been classified in \cite{GaiffiServenti2}).
Let $\Ec$
be such a building set and let \(Y_\Ec\) be the corresponding  model.
We will denote by \(Y_{super\Ec}\) the minimal model in  \({\mathcal L}\) among the models \(Y_{\Kc}\) that  admit the \(S_{n+1}\)-action and such that \(\Kc\supseteq \Ec\).
Depending on the choice of $\Ec$, it may be $Y_{super\Ec} \lneqq Y_{\Bc(n-1)}$. This happens for
instance when $\Ec$ is the  building set that contains $\Fc_{n-1}$
and all the triples $\{ V, A_1, A_2\} \in \Bc(n-1)$ such that the sum of $A_1$ and $A_2$ is direct and
has dimension $n-2$ (this building set is denoted by \(\G_2(A_{n-1})\) in  \cite{GaiffiServenti2}).

 \end{remark}

\subsection{The Cohomology of a Complex Supermaximal Model}
The discussion in the preceding sections  points out the interest of the supermaximal models and of the corresponding symmetric group actions.

 Let \(\Fc\) be the  building set of irreducible subspaces associated with a subspace arrangement  in a complex vector space \(V\) of dimension \(n\).
 In Section \ref{sec:buildingcombinatorial} we defined the building set
$\Bc(n-1) = \Nc'(\Fc_{A_{n-1}})$, the building set for the supermaximal model for $A_{n-1}$. By  analogy with this notation we write $\Bc(\Fc)$ for the building set $\Nc'(\Fc)$, that
generalizes in a way the idea of supermaximal model.

Let \(Y_{\Bc(\Fc)}\) be the model obtained by blowing up all the strata of the minimal model \(Y_{\Fc}\).
We recall from Example \ref{es:boundary} that the

strata of \(Y_{\Bc(\Fc)}\) are in bijection with the nested sets in $\Bc(\Fc)$, according to the constructions given in \cite{procesimacpherson, li}.

Let us denote by \(\upi^{\Bc(\Fc)}_\Fc\) the projection from \(Y_{\Bc(\Fc)}\) onto \(Y_\Fc\).

\begin{theorem}\label{thm:cohomology}
A basis of the  integer cohomology of the complex model \(Y_{\Bc(\Fc)}\) is given by the following monomials:
\[  \eta\ c_{\Sc_1}^{\delta_1}c_{\Sc_2}^{\delta_2}\cdots c_{\Sc_k}^{\delta_k}  \]
where
\begin{enumerate}
\item  \(\Sc_1\subsetneq \Sc_2 \subsetneq\cdots \subsetneq \Sc_k\) is a chain of  \(\Fc\)-nested sets (possibly empty, i.e. \(k=0\)),  with \(\{V\}\subsetneq \Sc_1\);
\item  the exponents \(\delta_i\), for \(i=1,\ldots, k\),  satisfy the following inequalities: \(1\leq \delta_i \leq |\Sc_i|-|\Sc_{i-1}|-1\), where we put \(\Sc_0=\{V\}\);
\item \(\eta \) belongs to \((\upi^{\Bc(\Fc)}_\Fc)^*H^*(\Dc_{\Sc_1})\) (if \(k\geq 1\)) or to \((\upi^{\Bc(\Fc)}_\Fc)^*H^*(Y_\Fc)\) (if \(k=0\))
and is the image, via $(\upi^{\Bc(\Fc)}_\Fc)^*$, of a monomial in the  Yuzvinsky basis (see Section \ref{constru});
\item the element $c_{\Sc_i}$ is the Chern class of the normal bundle of $L_{\Sc_i}$ (the proper transform of  $\Dc_{\Sc_i}$) in \(Y_{\Bc(\Fc)}\).

\end{enumerate}
\end{theorem}

\begin{proof}
Let us fix some notation.
We construct \(Y_{\Bc(\Fc)}\) starting from \(Y_\Fc\) by choosing a sequence of blowups (we start by blowing up the 0-dimensional strata, then the 1-dimensional strata, and so on, see Remark \ref{rem:ordinescoppiamenti}). At a certain step of this blowup process we have blown up a  set of strata indicized by a subset \(\A\) of \(\Bc(\Fc)\). We then denote by \(Y_\A\) the variety we have obtained.
  This means that \(\A\) has  the following property: there exists an integer \(k\), with  \(2\leq k\leq n+1\) such that \(\A\)  contains all the nested sets with cardinality \(\geq k+1\) and doesn't contain any nested set with cardinality \(<k\).
\begin{remark}
According to this notation, the variety \(Y_\Fc\) can also be denoted by \(Y_{\emptyset}\), i.e., it is the variety we have when no stratum has been blown up.
\end{remark}

As a further notation, if \(Y_{\A_1}\) and \(Y_{\A_2}\) are  two varieties  obtained during the blowup process with \(\A_1\subset \A_2\subset \Bc(\Fc)\),  we denote by \(\pi^{\A_2}_{\A_1}\: : \:  Y_{\A_2}\rightarrow Y_{\A_1} \) the blow up map. Let us recall the following lemma.

\begin{lemma}[Keel, \cite{keel}] \label{lem:keel}
Given \(\A\) as before, let us suppose that the next stratum that we have to blow up is indicized by \(\Sc\in \Bc(\Fc)\setminus  \A\)  with \(|\Sc|=k\). We call $\Dc_\Sc'$ the proper transform of $\Dc_\Sc$ in $Y_\A$ and
$\widetilde{\Dc_\Sc}$ the proper transform of $\Dc_\Sc$ in $Y_{\A \cup \{\Sc\}}.$ Then
\[H^*(Y_{\A\cup \{\Sc\}})\cong

H^*(Y_{\A})[\zeta] /(J \cdot \zeta, P_{Y_\A/\Dc_\Sc'}(-\zeta))
\]

where $\zeta$ is the class of the proper transform
\(\widetilde{\Dc_\Sc}\)

of \(\Dc_\Sc\) in $H^2(Y_{\A\cup \{\Sc\}}),$
$J$ is the kernel of the restriction map $H^*(Y_\A) \to H^*(\Dc_\Sc')$ and
$P_{Y_\A/\Dc_\Sc'}[x]$ is any polynomial in $H^*(Y_\A)[x]$ whose restriction to $H^*(\Dc_\Sc')[x]$ is the Chern polynomial of the

normal bundle of

$\Dc_\Sc'$ in $Y_\A.$
\end{lemma}

Using the lemma we can prove the claim of the theorem by induction. We fix a sequence of blowups
$$X_0 \ot^\pi X_1 \ot^\pi \cdots \ot X_N$$ that constructs \(Y_{\Bc(\Fc)}=X_N\) starting from \(Y_\Fc = X_0\).
We assume by inductive hypothesis that a basis of the integer cohomology of $X_n$ is given by the monomials
\[  \eta\ c_{\Sc_1}^{\delta_1}c_{\Sc_2}^{\delta_2}\cdots c_{\Sc_k}^{\delta_k}  \]
where
\begin{enumerate}
  \item \(\Sc_1\subsetneq \Sc_2 \subsetneq\cdots \subsetneq \Sc_k\) is a possibly empty chain of  \(\Fc\)-nested sets already blown up;
  \item \(\eta \) belongs to \((\pi^{X_n}_{X_0})^*\iota^*H^*(\Dc_{\Sc_1})\) if \(k\geq 1\) ($\iota$ is the inclusion map of $\Dc_{\Sc_i}$ in $X_0$), and to \((\pi^{X_n}_{X_0})^*H^*(X_0)\) if \(k=0\) and is the image of a monomial in the  Yuzvinsky basis;
  \item the term $c_{\Sc_i}$ is the Chern class of the normal bundle of $\Dc'_{\Sc_i}$, that is the proper transform of $\Dc_\Sc$ in $X_n;$
  \item the exponents $\delta_i$ satisfy the inequalities \(1\leq \delta_i \leq |\Sc_i|-|\Sc_{i-1}|-1\), where we put \(\Sc_0=\{V\}.\)
\end{enumerate}

Let $\Dc_\Sc'$ be the stratum in $X_n$ that we have to blowup in order to get $X_{n+1}.$
We assume that  ${\Dc_\Sc'}$ is the proper transform in $X_n$ of the stratum $\Dc_\Sc$ in $X_0,$
corresponding to the $\Fc$-nested set $\Sc \in \Bc(\Fc).$ Let $d= | \Sc |.$
Since at the step $X_n$ we have already blown up all the strata $\Dc_\Tc$ with $| \Tc |> d,$ the submanifold
$\Dc_\Sc'$ is a complex supermaximal model of smaller dimension.
In fact, since we have that, as in Definition \ref{def:simple},
$$T(\Dc_\Sc \cap \Dc_\Tc) = T(\Dc_\Sc)_{\Dc_\Sc \cap \Dc_\Tc} \cap T(\Dc_\Tc)_{\Dc_\Sc \cap \Dc_\Tc},$$
the stratum
$\Dc_\Sc'$ can be obtained as a blowup of
$\Dc_\Sc$ along the strata $\Dc_\Tc \cap \Dc_\Sc$ with $| \Tc |> d.$
Hence we can assume by induction on the dimension of $\Dc_\Sc'$, that we know the cohomology $H^*(\Dc'_\Sc),$ according to the statement of the theorem.

Then, by Lemma \ref{lem:keel} we have that the cohomology of $X_{n+1}$ is given by

\[H^*(X_{n+1})\cong
H^*(X_n)[\zeta] /(J \cdot \zeta, P_{X_n/\Dc_\Sc'}(-\zeta))
\]
where $J$ is the kernel of the projection induced by the inclusion of $\Dc_\Sc'$ in $X_n$
and $\zeta = c_{\Sc}$ is the Chern class of the normal bundle of the proper transform $\widetilde{\Dc_\Sc}$ of $\Dc_\Sc$ in $X_{n+1}.$
The polynomial $P_{X_n/\Dc_\Sc'}(-\zeta)$ is the Chern polynomial of the normal bundle of $\Dc_\Sc'$ in $X_n,$ that has rank $d =|\Sc|$
and hence $\deg P_{X_n/\Dc_\Sc'}(-\zeta) = d - 1.$

It follows that a basis of $H^*(X_{n+1})$ is given by the union of two set of generators:
\begin{enumerate}
  \item the monomials
\[ \mu_1= \eta\ c_{\Sc_1}^{\delta_1}c_{\Sc_2}^{\delta_2}\cdots c_{\Sc_k}^{\delta_k}  \]
that are already in the base of $H^*(X_n)$ and that we can identify with the
corresponding generators of the cohomology of $X_{n+1}$ via the pull back
$\pi^*: H^*(X_n) \to H^*(X_{n+1});$
\item the monomials
\[ \mu_2 = \eta\ c_\Sc^{\delta} c_{\Sc_1}^{\delta_1}c_{\Sc_2}^{\delta_2}\cdots c_{\Sc_k}^{\delta_k} \]
where $0 < \delta < |\Sc|-1$ and $\eta\ c_{\Sc_1}^{\delta_1}c_{\Sc_2}^{\delta_2}\cdots c_{\Sc_k}^{\delta_k}$ is a monomial among the generators of the cohomology
$H^*(\Dc_\Sc');$ we identify $\mu_2$ with a generator of the cohomology of $H^*(X_{n+1})$ as follows:
given the monomial $\eta\ c_{\Sc_1}^{\delta_1}c_{\Sc_2}^{\delta_2}\cdots c_{\Sc_k}^{\delta_k} \in H^*(\Dc_\Sc'),$
we identify it with the coset $\eta\ c_{\Sc_1}^{\delta_1}c_{\Sc_2}^{\delta_2}\cdots c_{\Sc_k}^{\delta_k}+J \in H^*(X_n)$ by the projection induced by the inclusion of $\Dc_\Sc'$ in $X_n;$ hence when we multiply it by $c_\Sc,$ that is the Chern class of the normal bundle of the divisor $\widetilde{\Dc_\Sc} \in X_{n+1},$ this gives a well defined class \[\eta\ c_\Sc^{\delta} c_{\Sc_1}^{\delta_1}c_{\Sc_2}^{\delta_2}\cdots c_{\Sc_k}^{\delta_k} \in H^*(X_{n+1}).\]
In particular, since we are considering the class
$c_{\Sc_1}$ that is the Chern class of the normal bundle of the proper transform $\Dc_{\Sc_1}'$ of $\Dc_{\Sc_1}$ in $\Dc_\Sc'$, the exponent
$\delta_1$ will be at most $|\Sc_1| - |\Sc| -1,$ that is the dimension of the projectivized normal bundle
of $\Dc_{\Sc_1}'$ in $\Dc_\Sc'.$
\end{enumerate}

Let \(\Sc, \Tc\), with $\Sc \subset \Tc$ be two $\Fc$-nested sets.   Let $\Dc_\Sc'$ (resp. $\Dc_\Tc'$) be the proper transform of $\Dc_\Sc$ (resp. $\Dc_\Tc$) in $X_n.$

We assume
that in $X_n$ we have already performed the blowup of the stratum associated to $\Dc_\Tc.$
Since we have the inclusion $\Dc_\Tc \subset \Dc_\Sc,$ we can also consider the proper transform $\Dc_\Tc' \cap \Dc_\Sc'$
of $\Dc_\Tc$ in $\Dc_\Sc'$.

Let $\zeta_\Tc$ be the Chern class of the normal bundle of $\Dc_\Tc'$ in $X_n$ and let $\zeta_\Tc^\Sc$ be
the Chern class of the normal bundle of $\Dc_\Tc'\cap \Dc_\Sc'$ in $\Dc_\Sc'.$
We claim that the projection $i^*$ induced by the inclusion $i:\Dc_\Sc' \subset X_n$ maps
$i^*:\zeta_\Tc \mapsto \zeta_\Tc^\Sc.$
This follows since the Thom class $\tau_\Tc$ of the normal bundle of $\Dc_\Tc'$ in $X_n$ restricts to the Thom class $\tau_\Tc^S$ of the normal bundle of $\Dc_\Tc'\cap \Dc_\Sc'$ in $\Dc_\Sc$ and we have that the Chern class of a line bundle is the pullback of the zero-section of its Thom class.

Finally we notice that, since Chern classes are functorial, whenever we perform a blowup $\pi:X_{n+1} \to X_n$, the Chern class $c_\Tc$ of a divisor $\Dc_\Tc'$ in $X_n$ pulls back to the Chern class of its proper transform $\widetilde{\Dc_\Tc}$  and hence we can identify them in our notation.

\end{proof}


TITLE: Sum of probabilities or mean of probability
QUESTION [1 upvotes]: My question is about being confused about two way of approaching a problem, which in this case lead me to the same solution. One method is very verbose, the other one is fast and clean.
Let's consider this problem where there are three only components of a computer with probability of being defective 0.1, 0.05, 0.02. Now, I want the expected value E(X), where X is number of defective components.
Method 1
Sum of probabilities, a computer has a piece of each then:
$$E(X) = 0.1+0.05+0.02=0.17 $$
Method 2
Mean of probabilities, I calculate $ \Sigma\ x_i p_i$. Therefore:
$$p_1 = 0.1\times0.95\times0.98+0.9\times0.05\times0.98+ ... = 0.1543$$
$$p_2 = 0.9\times0.05\times0.02+0.1\times0.95\times0.02+... = 0.0077$$
$$p_3 = 0.1\times0.05\times0.02 = 0.0001$$
To then do $E(X) = 1\times0.1543+2\times0.0077+3\times0.0001 = 0.17$
In summary, both the methods give the same answer. The first one seems logical to me, but I don't have in my hands any proper formula, like I have it for the second method. Is there any flow that I am not seeing?

REPLY [4 votes]: Both methods are correct!
To see that Method 1 is correct, define the independent Bernoulli random variables
$$X_i=\begin{cases} 1, &\text{component $i$ is defective} \\ 0, &\text{component $i$ is not defective} \end{cases}$$ for $i=1,2,3$. Then $$X_i \sim \mathrm{Bern}(p_i)$$ where the $p_i$'s are the given probabilities. Now, obviously $$X=\sum_{i=1}^{3}X_i$$ so that due to linearity of expectation $$E[X]=E\left[\sum_{i=1}^{3}X_i\right]=\sum_{i=1}^{3}E[X_i]$$ But, since $E[X_i]=p_i$ we have that $$E[X]=\sum_{i=1}^{3}p_i$$ which yields the result that you found with Method 1, i.e. $E[X]=0.17$.

Just to verify that Method 2 yields also the correct result:
Since you have defined your random variable $X$ then - in order to calculate it's expected value - you should proceed as follows

Firstly, determine the sample space of $X$. There can be from $0$ up to $3$ defective components, so that $$X \in \{0,1,2,3\}$$
Secondly, determine the probability mass function of $X$, i.e. the probability of $X$ taking each of these values. You have that $$P(X=0)=(1-0.1)\cdot(1-0.05)\cdot(1-0.02)$$
and $$\begin{align*}P(X=1)=&0.1(1-0.05)(1-0.02)+(1-0.1)0.05(1-0.02)+(1-0.1)(1-0.05)0.02\end{align*}$$
and
$$\begin{align}P(X=2)=&(0.1)(0.05)(1-0.02)+(0.1)(1-0.05)(0.02)+(1-0.1)(0.05)(0.02)\end{align}$$
and
$$P(X=3)=0.1\cdot0.05\cdot0.02$$

This gives the following probabilities $$P(X=x)=\begin{cases}0.8379, & x=0 \\ 0.1543, & x=1 \\ 0.0077, & x=2 \\ 0.0001, & x=3 \\ \end{cases}$$ (note that these probabilities add up to $1$). Now the expected value of $X$ is given by $$E[X]=\sum_{x=0}^{3}xP(X=x)=0P(X=0)+P(X=1)+2P(X=2)+3P(X=3)=0.17$$