TITLE: Converse of Sperner's lemma
QUESTION [2 upvotes]: The famous Sperner's lemma states that, if a labeling of a triangulation of a simplex satisfies certain conditions on the boundary, then there must exist a sub-simplex in which all labels are different. Here is an example, where the simplex and sub-simplices are right-angled triangles:
1
1   1
2   ?   3
2   2   2   3

I showed only the labels on the boundary. They satisfy Sperner's conditions, and indeed, whatever label we put instead of the ? in the interior (1, 2 or 3), we will have a fully-labeled sub-simplex.
Sperner's boundary conditions are sufficient but not necessary. For example, here:
1
1   2
1   ?   3
1   1   1   1

the labeling does not satisfy Sperner's boundary conditions, but still, whatever label we put instead of the ? in the interior (1, 2 or 3), we will have a fully-labeled sub-simplex.
My question is: are there conditions such that, if the labeling satisfies the conditions on the boundary, then not every completion of the labeling in the interior admits a fully-labeled simplex?

REPLY [3 votes]: Let $n$ be the dimension of the main simplex, then we are looking for a triangulation without simplices that have $n+1$ labels.
Triangular case ($n=2$)
Let $\ell(u)$ denote the label of vertex $u$, let $\partial$ the boundary of the main simplex, and let $k_\partial(a,b)$ be the number of (clockwise) edges $(u,v)$ along $\partial$ for which $\ell(u)=a$ and $\ell(v)=b$.
Condition 1. For any $a\neq b$, we must have $k_\partial(a,b)=k_\partial(b,a)$.
Necessity
To see that condition 1 is necessary for there not to be a fully-labeled simplex, consider a labeled triangulation without any fully-labeled simplex.
The edges with labels $a$ and $b$ dualize to a set of paths, each of which is either a cycle in the interior, or paths that start and end on an edge of the boundary. In particular, the latter type of path always has label $a$ on one side, and label $b$ on the other side, which means that it must use both a clockwise and a counter-clockwise boundary edge with labels $a$ and $b$.
Sufficiency
We also show that if condition 1 applies, then there exists a triangulation without a fully-labeled simplex.
We proceed by induction on $K(\partial):=k_\partial(1,2)+k_\partial(2,3)+k_\partial(1,3)$.
Let $\partial$ be a boundary satisfying condition 1, and assume that for any boundary $\partial'$ that satisfies condition 1 and $K(\partial')<K(\partial)$, there exists a labeled triangulation of $\partial'$ without a fully-labeled simplex.
If $K(\partial)=0$, then the boundary has a single label, and we can triangulate and label the interior arbitrarily using only that label without creating any fully-labeled simplex.
Next, suppose that for two labels $a\neq b$ and an integer $k\geq 1$, $\partial$ contains a path of labels $ab^ka$ (i.e. two vertices labeled $a$ connected by a path of vertices labeled $b$).
Introduce an interior vertex $v$ (labeled $a$) creating a triangle with each edge along the path $ab^ka$.
These triangles use only labels $a$ and $b$.
Let $\partial'$ be the cycle obtained from $\partial$ by replacing the path of $b$'s by the vertex $v$.
Then $\partial'$ satisfies condition 1 and $K(\partial')=K(\partial)-1$, so by induction we can triangulate and label $\partial'$, and hence the remainder of $\partial$, without creating fully-labeled simplices.
Finally, suppose that $K(\partial)>0$, but there are no two labels $a\neq b$ for which $\partial$ contains a path $ab^ka$.
Then the entire boundary is labeled according to the pattern $(a^+b^+c^+)^+$ with $a\neq b\neq c\neq a$. But then $k_\partial(b,a)=0<k_\partial(a,b)$, so $\partial$ violates condition 1 and we are done.
Prescribed triangulations
Note that the above necessary and sufficient condition does not imply that all triangulations satisfying condition 1 can be labeled without creating a fully-labelled simplex. In fact, the converse is true, as illustrated by the figure below: it satisfies condition 1, but no matter what label the interior vertex is given, a fully-labeled simplex will be created.

As a consequence, the labeling of the boundary is insufficient to determine whether any particular triangulation admits a labeling without fully-labeled simplices.
General case ($n>2$)
Let $\partial$ be a labeled triangulated $(n-1)$-sphere, and fix a consistent orientation of its $(n-1)$-simplices.
Say that an $(n-1)$-simplex $(v_1,v_2,\dots,v_n)$ has labeling $L=(l_1,l_2,\dots,l_n)$ if there is an even permutation $\pi$ of $L$ such that $\ell(v_i)=\pi(l_i)$ for all $i$.
Let $k_\partial(L)$ be the number of $(n-1)$-simplices of $\partial$ with labeling $L$.
Call $L$ critical if it has $n$ distinct labels, and call an $(n-1)$-simplex critical if it has a critical labeling.
Condition 2.
For any critical $L$ and any odd permutation $\bar{L}$ of $L$, we have $k_\partial(L)=k_\partial(\bar{L})$.
Necessity
Consider a labeled triangulated $n$-ball without any fully-labeled simplex.
Let $\sigma$ be one of its $n$-simplices.
If $\sigma$ has a face $F$ with a critical labeling $L$, then the vertex $v'$ of $\sigma$ that does not lie on $F$ shares a color with a vertex $v$ of $F$, so there is a second face of $\sigma$ whose labeling (oriented with respect to $\sigma$) is an odd permutation of $L$, and all the other faces of $\sigma$ contain both $v$ and $v'$ and are hence not critical.
Therefore, the $(n-1)$-simplices with critical labelings dualize to a set of disjoint paths, each of which is either cycle interior to the ball, or a path between two critical $(n-1)$-simplices on the boundary.
Moreover, the labelings of these two simplices are odd permutations of one another since the orientations with respect to consecutive primal $n$-simplices along the path cancel out.
Sufficiency
Pick an arbitrary matching between the simplices of the boundary with critical labels, such that for any pair in the matching, their labelings are odd permutations of one another.
Connect any pair of matched simplices by a path of $n$-simplices inside the $n$-ball, creating a prism-like tunnel whose base and top are the matched critical simplices, and whose interior $(n-1)$-simplices all have the same critical labeling as the matched pair.
Note that the boundary of this tunnel (except for the matched simplices) consists of only non-critical simplices.
We can dig the tunnels in such a way that their faces are disjoint.
Now the complement of the tunnels (with respect to the $n$-ball) has only non-critical simplices on its boundary, so we can triangulate the interior arbitrarily, and assign the label $1$ to all interior vertices to obtain a triangulated $n$-ball without any fully-labeled simplex.


TITLE: Graph Theory: What is the definition of the "Sorted Edge" algorithm?
QUESTION [0 upvotes]: I've been googling for a while and can't find a clear definition of the "sorted edge" algorithm--can anyone provide it please?
A description would be helpful, but a simple statement of the algorithm may be sufficient.
Thank you.

REPLY [1 votes]: The algorithm sorts the edges in ascending order by cost. You choose edges in greedy order to create a path. So no three edges are incident to the same vertex, and you don't close the path to a circuit unless it is a Hamiltonian path.
This looks like a heuristic to solve the traveling salesman problem based on Kruskal's algorithm.


TITLE: Understanding the notion of differentiability in euclidean plane
QUESTION [0 upvotes]: Suppose $f: R^2 \to R $ is differentiable. In my notes, it says that then one have
$$ f(x+h_1, y+h_2) - f(x,y) = \dfrac{ \partial f }{\partial x} \cdot h_1 + \dfrac{ \partial f }{\partial y } \cdot h_2  + |h| o(h) $$
where $o(h) \to 0$ as $|h| \to 0$. Is this a definition or a theorem? I am trying to make sense of it. Can someone give me some intuition?

REPLY [2 votes]: Definition: $f$ is differentiable in $(x,y) \iff$ there is $a=(a_1,a_2) \in \mathbb R^2$ such that
$$\frac{f(x+h_1,y+h_2)-f(x,y)-a_1h_1-a_2h_2}{|h|} \to 0$$
as $h \to 0.$
End definition.
In this case one can show that $a_1= \dfrac{ \partial f }{\partial x }(x,y)$ and  $a_2= \dfrac{ \partial f }{\partial y }(x,y), $ hence
$$f(x+h_1, y+h_2) - f(x,y) = \dfrac{ \partial f }{\partial x}(x,y) \cdot h_1 + \dfrac{ \partial f }{\partial y }(x,y) \cdot h_2  + |h| o(h).$$


TITLE: Poincaré inequality for a subspace of $H^1(\Omega)$
QUESTION [16 upvotes]: The following is the well known Poincaré inequality for $H_0^1(\Omega)$:

Suppose that $\Omega$ is an open set in $\mathbb{R}^n$ that is bounded. Then there is a constant $C$ such that
  $$
\int_\Omega u^2\ dx\leq C\int_\Omega|Du|^2\ dx\quad \textrm{ for all   }\ \ \color{red}{u\in H_0^1(\Omega)}.
$$

Here is my questions:
Can the condition $u\in H_0^1(\Omega)$ be relaxed so that the inequality is still true? More precisely, suppose $H$ is a subspace of $H^1(\Omega)$ such that for any $u\in H$, $u|_\Gamma=0$, where $\Gamma$ is a ([Added:]nonempty) closed proper subset of the boundary $\partial\Omega$ and the idenity is in the sense of trace. In other words, $u$ is only zero on a part of the boundary. Do we still have the Poincaré ineqaulity in $H$?

[Added:] This question is motivated by the following exercise

REPLY [18 votes]: I'm going to augment your statement with three assumptions in order to make the proof work.  First, I'm going to assume that $\partial \Omega$ is Lipschitz.  We're going to need that to invoke the usual trace theory in $H^1(\Omega)$.  Second, I'm going to assume that $\Gamma \subseteq \partial \Omega$ is such that $\mathcal{H}^{n-1}(\Gamma) >0$, i.e. $\Gamma$ has nonzero Hausdorff measure on the boundary.  Third, I'm going to assume that $\Omega$ is connected.
Now we modify one of the standard proofs of the Poincaré inequality, namely the proof by contradiction.  Suppose, by way of contradiction, that the inequality fails.  Then for each $n \ge 1$ we can find $u_n \in H^1$ such that $u_n =0$ on $\Gamma$ and
$$
\Vert u_n \Vert_{L^2} \ge n \Vert D u_n \Vert_{L^2}.
$$
Exploiting the homogeneity of the norm, we can assume WLOG that $\Vert u_n \Vert_{L^2} = 1$ for all $n$. Then
$$
\Vert D u_n \Vert_{L^2} \le \frac{1}{n} \text{ and } \Vert u_n \Vert_{L^2} =1,
$$
from which we deduce that $\{u_n\}_{n=1}^\infty$ is uniformly bounded in $H^1(\Omega)$.  Using weak compactness and Rellich's theorem we can extract a subsequence $u_{n_k} \rightharpoonup u$ weakly in $H^1$ such that $u_{n_k} \to u$ strongly in $L^2$.
It's clear that the limit satisfies
$$
\Vert D u \Vert_{L^2} =0, \Vert u \Vert_{L^2} =1, \text{ and } u = 0 \text{ on } \Gamma.
$$
The last equality can be verified using convexity and its relation to weak convergence, for instance.  The first and third conditions, together with the fact that $\Gamma$ has positive measure and $\Omega$ is connected, tell us that $u=0$, which contradicts the fact that $\Vert u \Vert_{L^2}=1$.
Thus there exists $C >0$ such that
$$
\Vert u \Vert_{L^2} \le C \Vert D u \Vert_{L^2} \text{ for all } u \in H^1 \text{ s.t. } u\vert_{\Gamma}=0.
$$
Note that we really need connectedness here, as otherwise we can take two disconnected open sets and look at $\Gamma$ only in the boundary of one of them.  The inequality clearly fails in this case because it fails in the set that $\Gamma$ doesn't touch.
EDIT: Requested extra details.
Trace theory shows that there exists $C>0$ such that
$$
\Vert u\Vert_{L^2(\partial \Omega)} \le C \Vert u \Vert_{H^1(\Omega)}
$$
for all $u \in H^1(\Omega)$.  Note that this uses the fact that the boundary is Lipschitz.  In particular we can further bound
$$
\Vert u\Vert_{L^2(\Gamma)} \le C \Vert u \Vert_{H^1(\Omega)}.
$$
This bound shows that the set $\{u \in H^1(\Omega) : u=0 \text{ on }\Gamma\}$ is strongly closed in $H^1(\Omega)$.  However, it's also clearly a convex set.  We know from functional analysis (ultimately due to Hahn-Banach) that strongly closed convex sets are also weakly closed, so the fact that $\{u_{n_k}\}_k$ converges weakly to $u$ shows that $u$ belongs to this same set, i.e. $u=0$ on $\Gamma$.  Note, though, that this is not the only way to prove this result.


TITLE: Why do the solutions of this recurrence have a limit independent of the initial value?
QUESTION [4 upvotes]: For $0<a_0<1$ the limit $\lim\limits_{n\rightarrow \infty}\sqrt{1-a_n}= \frac{\sqrt{5}-1}{2}$ for the sequence $a_{n+1}=\sqrt{1-a_n}$.
But the problem is that the choice of $a_0$ does not matter.
The question is why?

REPLY [2 votes]: We are considering a question equivalent to $f(x) = \sqrt{1 - x}$ for $x \in (0,1)$. It is not so hard to show that this function has a fixed point, and exactly one fixed point in the interval $(0,1)$. (In fact, this fixed point is our limit).
In your question, you seem to know that the limit exists and are able to calculate it (take the even and odd terms separately, for example). This means that you know that no matter what starting value, the iterative function $f(x)$ will converge. But there is only one fixed point in the interval of its range, and thus it must converge to that point.
This is the same reasoning that goes to show that $\cos(x) = x$, or rather $x_{n+1} = \cos(x_n)$, will always have the same limit regardless of starting point, more or less. One shows a limit exists, and then the limit must be the fixed point (for cosine, this number has a name, the Dottie Number).


TITLE: Helix around Helix around Circle
QUESTION [1 upvotes]: I'm trying to find the parametric equations for a helix around a helix around a circle (helix on helix on circle)
That is: I would like to start with a circle, add a helix around it and a helix around the helix.(See video)
I'm ok even if the second helix is not perfectly orthogonal to the first helix as long as we can have a simpler parametrization.
I'm ok also if the curve represents a helix around a helix around a helix.
I know the helix around a helix around an axis is quite easy but I was not able to find a solutions for this case.
I'm interested in this parametric curve as a way to represent time and I would like to write a program to show data attached to that curve.
Edit: I already know the parametric equations of a helix around a torus:
$$x(t) = (R+ r\cos(nt))  \cos(t)$$
$$y(t) = (R+ r\cos(nt))  \sin(t)$$
$$z(t) = s t + r  \sin(nt)$$
where $R$ is the radius of the torus
$r$ is the radius of the helix
$n$ is the winding number
$s$ vertical velocity ($0$ if we want a closed curve).
What I'm looking for is the next level helix on top of that.

REPLY [2 votes]: HINT: You are looking at a curve spiraling at a constant rate about a torus. Start by finding parametric equations of a torus. Now make the two angles linear functions of $t$. It appears you want one to go much faster than the other, so that suggests how you should relate those linear functions.
EDIT: Based on your comment, you want a curve that spirals around a given space curve. The way to do that is to take an orthonormal basis for the normal plane to the curve at each point and go around a circle as you move along the curve. In particular, take the Frenet frame $T,N,B$ for the curve, parametrized say by $\alpha(s)$. Now consider $$\alpha(s)+\cos\theta(s)N(s)+\sin\theta(s)B(s)$$ where $\theta$ is a linear function of $s$. (If you don't know about Frenet frames, see this or my differential geometry text, linked in my profile. It's most convenient to work with arclength-parametrized curves $\alpha(s)$, but the chain rule will do the heavy lifting for you if they're not.)


TITLE: Proving $\mathbb{R}$ with the discrete metric is complete
QUESTION [2 upvotes]: I'm trying to show that every Cauchy space in $\mathbb{R}$ under the discrete metric (I believe this is the correct terminology) is convergent. Here is my attempt:

Let $\{p_n\}$ be such a Cauchy sequence, so we can  find $N \in \mathbb{N}$ such that for all $m,n \geq N$, we have $d(p_m, p_n) < 1$, i.e., $d(p_m, p_n) = 0$, in which case $p_m = p_n$ for all $m,n \geq N$, so $\{p_n\}$ is eventually constant. Let $p_n := \alpha$ for all $n \geq N$. Given $\epsilon > 0$, we choose $N$ as above, and notice that for $i,j \geq N$, we have $d(p_i, p_j) < 1$, in which case $p_i = p_j$. Therefore, we in fact have $d(p_i, p_j) = d(\alpha, \alpha) = 0 < \epsilon$. Therefore, $p_n$ converges to $\alpha$, as required. Therefore, $(\mathbb{R}, d)$ is complete.

How does this look?

REPLY [1 votes]: Your idea is correct and so is the proof. Well, almost. Towards the end, you want to show that for each $\epsilon>0$ there exists $N\in\mathbb N$ such that $d(p_i,\alpha)<\epsilon$ for all $n\geq N$.
This is what you intended, but you instead write $d(p_i,p_j)<\epsilon$ for all $n\geq N$ which is the hypothesis (Cauchy). Instead, the last sentence should be framed somewhat like.

Let $\epsilon>0$ be given. Then there exists $N\in\mathbb N$ such that $$d(p_i,\alpha)=d(\alpha,\alpha)=0<\epsilon$$ for all $n\geq N$. Since $\epsilon>0$ was arbitrary, so $(p_n)\to\alpha$.


TITLE: Where should the 2$\pi$ go in the Fourier Transform?
QUESTION [6 upvotes]: In some lecture notes on Harmonic Analysis from Terence Tao here, he defines the fourier transform by $$\hat{f}(\tau)=\int_{\mathbb{R}}e^{-2\pi i t\tau}f(t)dt$$
and then says

This is really the best place to put the $2\pi$; listen to the harmonic analysts and
representation theorists on this one, and ignore the PDE people, physicists, and
engineers!

But why is this the best spot for it?  It seems cleaner than having a factor of $\frac{1}{2\pi}$ or $\frac{1}{\sqrt{2\pi}}$ in front of the integral.  But it isn't so clean that you  forget there is a $2\pi$ at all, like if you hide it in the measure $dt$.
Are there other reasons?  Better Reasons? Reasons that only reveal themselves after working in harmonic analysis for a while?  Any answers would be appreciated.

REPLY [7 votes]: There are conventions that are standard in different fields of application.  For example, mathematicians like the $1/\sqrt{2 \pi}$ factor because it makes the FT unitary.  Physicists like the $1/(2 \pi)$ factor because it helps them remember conservation laws that are deduced from Parseval's Theorem.  (I fit into this camp.)  And electrical engineers like the $2 \pi$ in the exponential because they like to operate in terms of cycles per second rather than with angular frequencies.
I may be over-generalizing, but as I have experience in all three camps, I think my opinion is grounded in some fact.


TITLE: How do I find a primitive element in $Z_7$?
QUESTION [0 upvotes]: I understand the definition of a primitive element. I also know that 3 is primitive element of $Z_7$. I was shown that 2 is not primitive element of $Z_7$ because $2^{3}$ = 8=  1. I do not understand why 2 is not a primitive element because of that reason. Can someone explain please?

REPLY [2 votes]: It's because a primitive root, k, modulo n is a number such that you can write every number in your modulus as a power of k.
So, here, you can write every element of Z7 as a power of 3 (various powers of 3 in modulo 7, in order, are 3, 2, 6, 4, 5, 1, at which point it loops) where you cannot write everything as a power of 2 because it loops at 2, 4, 1, 2, 4, 1.
In particular, the rigorous defition of a primitive root, k, modulo n is that, for every integer, j, coprime to k (ie they share no common factors but 1) there exists a power of k such that said power is congruent to j.
Hope that helps!


TITLE: What would be the Lyapunov exponent of this map
QUESTION [2 upvotes]: $p_i$ is the probability of the occurence of unique symbol $i$. In the case of Tent map, the Lyapunov exponent (LE) is log of the derivative of the tent map which is almost always 2. So, $\lambda = log(2)$.
For the probabilistic Bernoulli map, I tried to find the derivative like this:
$f'(x) = 1/p_1$ for $0<x<p_1$, or $f'(x) = 1-p_1/p_2$ for
$p_1<x\le p_2$ and so on. Please correct me if the derivative and limits are wrong.
In K Feltekh, D Fournier-Prunaret, S Belghith, Analytical expressions for power spectral density issued from one-dimensional continuous piecewise linear maps with three slopes. Signal Process. 94:, 149–157 (2014).
the anlytical expression for LE for any piecewise linear map in general is $\lambda = (1-p)\ln(2/(1-p)) + p\ln(1/p)$ where $p$ is a constant and $p \in (0,1)$.
Then, How can I apply the above result and calculate the LE for the map in Eq(6)? Since, Tent map is conjugate to Bernoulli map, so would the probabilistic Bernoulli map have the same LE as Tent Map which is log(2)?

REPLY [1 votes]: The Lyapunov exponent is
$$
\sum_{i=1}^m p_i \log (f'|I_i)=\sum_{i=1}^m p_i \log (1/p_i)=\text{entropy}.
$$
More precisely, this is the averaged Lyapunov exponent. Let me explain. You want to compute
$$
\int_0^1 \lambda(x)\,d\mu(x),
$$
where $\mu$ is the Bernoulli measure and
$$
\lambda(x)=\lim_{n\to\infty}\frac1n\log |(f^n)'(x)|
$$
when the limit exists. But since $\mu$ is invariant you get
$$
\begin{split}
\int_0^1 \lambda(x)\,d\mu(x)
&=\lim_{n\to\infty}\frac1n\int_0^1\log |(f^n)'(x)|\,d\mu(x)\\
&=\lim_{n\to\infty}\frac1n\sum_{j=0}^{n-1}\int_0^1\log |f'(f^j(x))|\,d\mu(x)\\
&=\int_0^1\log |f'(x)|\,d\mu(x)\\
&=\sum_{i=1}^m p_i \log (f'|I_i).
\end{split}
$$
Indeed, since $\mu$ is $f$-invariant we have
$$
\int_0^1\log |f'(f^j(x))|\,d\mu(x)
=\int_0^1\log |f'(x)|\,d\mu(x)
$$
for any integer $j\ge0$.


TITLE: Deriving a scattering amplitude of a loop diagram using path integral formulation
QUESTION [3 upvotes]: I am following Zee's QFT in a nutshell and his excerise I.7.2 states to derive
$$\frac{1}{2}(-i\lambda)^2 \int \frac{d^4k}{(2 \pi)^4}\frac{i}{k^2 -m^2 +i\epsilon}\frac{i}{(k_1+k_2-k)^2 -m^2 +i\epsilon}\tag{I.7.23}$$
from the first principles, that is from
$$Z(J) = \int D\phi e^{i \int d^4x \frac{1}{2}[(\partial \phi)^2 - m^2\phi^2] -(\lambda/4!)\phi^4 + J\phi}.\tag{I.7.11}$$
The amplitude written is just a basic loop diagram of order $\lambda^2$.
My attempt is to follow the Zee's approach in which i expand $Z(J)$ in series of $\lambda$ and $J$. So right before the expansion in the series I have
$$Z(J) = Z(0,0) e^{-(i/4!)\lambda\int d^4w \left (\frac{\delta}{i\delta J(w)}\right)^4}e^{-(i/2)\iint d^4x d^4y J(x) D(x-y)J(y)}$$
As i want 2 vertices and 6 lines, in the series expansion I choose $\lambda^2$ and $J^6$ and I have
$$Z(J) = Z(0,0) \frac{1}{2!}\left[\frac{-i}{3!}\lambda \int d^4w \left( \frac{\delta}{i\delta J(w)}\right)\right]^2\frac{1}{6!}\left[\frac{-i}{2}\iint d^4xd^4y J(x)D(x-y)J(y)\right]^6.$$
After some algebra i get
$$Z(0,0)\frac{i^8\lambda^2}{2^73!3!6!}\iint d^4w d^4z D(w-z)D(w-z)D(x_1-w)D(x_2-w)D(x_3-z)D(x_4-z)$$
And this, to me at least, does look like a loop diagram. We have 2 inner lines $D(w-z)$ and 4 external $D(x_i - w(z))$. But here is where my understanding stops. I dont know how to get from this to the amplitude itself. This part is not really covered in the Zee's book that well. I would appreciate if a bit more detailed explanation is given.

REPLY [0 votes]: I was writing this as a comment, but it got too long so I'm including it as an answer instead.
You might want to go through the calculation again, the term of interest in Z[J] before taking any derivatives to extract the 4-pt function should have a factor of:
$$
\frac{1}{16}\quad \textrm{instead of what you quote},\quad \frac{1}{2^73!3!6!}
$$
See in particular the $\lambda^2$ coefficient in $W[J]$ in eqn 2.16 in arxiv.org/abs/1512.02604. The quantity $W[J]$ is as usual the exponentiated version of $Z[J]$, so that it generates all connected and renormalised Green functions. Once you have the correct combinatorial factor you need the 4th $J$ derivative and then you need to go to momentum space by Fourier transforming all position space Green functions.
Slightly more precisely, the 4-pt 1-loop position-space Green function is (schematically):
$$
G\sim \delta_J^4 Z[J]\,\big|_{J=0},
$$
and the relevant term in $Z[J]$ is:
$$
Z[J] = \dots +\frac{1}{2}\Big(\frac{\lambda}{4!} \int \delta_J^4\Big)^2 Z_0[J]+\dots,
$$
where $Z_0[J]$ is the $\lambda=0$ version of $Z[J]$, so that $Z_0[J]$ is Gaussian in $J$. All in all, the derivative you need to calculate the 4-pt one-loop amplitude is:
$$
G\sim \delta_J^4\frac{1}{2}\Big(\frac{\lambda}{4!} \int \delta_J^4\Big)^2 Z_0[J]\,\big|_{J=0}.
$$
In fact, there's also a tadpole (or 'cephalopod') diagram at the same order (see the diagram immediately after the one of interest in eqn (2.16) in  arxiv.org/abs/1512.02604) so in your derivation you will find the associated extra terms in addition to the diagram of interest.


TITLE: Solving an equation involving the sum of square roots of a quadratic
QUESTION [1 upvotes]: So I am building a computer program.
In the program I need to build a function that takes this arguments {a, b, c, d, e, f, s, u}, and returns back the value of x in this equation:
$$
\sqrt{\left(x-a\right)^2+\left(\left(u\left(x-e\right)+f\right)-b\right)^2}+\sqrt{\left(x-c\right)^2+\left(\left(u\left(x-e\right)+f\right)-d\right)^2}=s
$$
So basically I just need to simplify and solve the equation to get this form: $$ x=\mathrm{everythingElse}$$
And then I can code it.
The problem is that I cant solve this equation, first of all I trying to do it by hand and there is to many arguments I just couldn't solve it, than I tried to use wolfram alpha but it failed as well due to "execution time exceeded".

REPLY [0 votes]: Just to show that this equation is linear precisely when $s$ is at most the distance between $(a,b)$ and $(c,d)$, and hence we are talking about a degenerate ellipsis (the line segment between $(a,b)$ and $(c,d)$ or an "empty" ellipsis, with no points.
Write $(x,y)=(e,f)+t(1,u)$. Then you want $(x,y)$ to be on the ellipse with foci $(a,b), (c,d)$ and major axis $s$.
Writing $\mathbf u=(a,b)-(e,f), \mathbf  v=(c,d)-(e,f)$, and $\mathbf x = (x,y)-(e,f)=t\mathbf m$, where $\mathbf m=(1,u)$ we are seeing a solution to:
$$|\mathbf u-t\mathbf m|+|\mathbf v-t\mathbf m|=s$$
Squaring, re-arranging, and squaring again gives us:
$$\begin{align}
0&=\left(s^2-|\mathbf u-t\mathbf m|^2-|\mathbf v-t\mathbf m|^2\right)^2- 4|\mathbf u-t\mathbf m|^2|\mathbf v-t\mathbf m|^2\\
&=s^4-2s^2\left(|\mathbf u-t\mathbf m|^2+|\mathbf v-t\mathbf m|^2\right)+\left(|\mathbf u-t\mathbf m|^2-|\mathbf v-t\mathbf m|^2\right)^2\\
&=s^4-2s^2\left(|\mathbf u|^2+|\mathbf v|^2 -2t(\mathbf u + \mathbf v)\cdot \mathbf  m+2t^2|\mathbf  m|^2\right)+\left(|\mathbf u|^2-|\mathbf v|^2+2t(\mathbf u-\mathbf v)\cdot\mathbf  m)\right)^2
\end{align}$$
So this equation is linear in $t$ precisely when $$\left|(\mathbf u-\mathbf v)\cdot \mathbf m\right|=s|\mathbf m|$$
But:
$$1\geq |\cos \theta |= \frac{|(\mathbf u-\mathbf v)\cdot \mathbf m|}{|\mathbf u-\mathbf v||\mathbf m|}=\frac{s}{|\mathbf u-\mathbf v|}$$
If $s> |\mathbf u-\mathbf v|$, we reach a contradiction.
(You can deal with the case $\mathbf u=\mathbf v$ separately.)


TITLE: Examples of incorrect arguments being fertilizer for good mathematics?
QUESTION [24 upvotes]: Sometimes (perhaps often?) vague or even outright incorrect arguments can sometimes be fruitful and eventually lead to important new ideas and correct arguments.

I'm looking for explicit examples of this phenomenon in mathematics.

Of course, most proof ideas start out vague and eventually crystallize. So I think the more incorrect/vague the original argument or idea, and the more important the final fruit, the better, as long as there is still a pretty direct connection from the vague idea to the final fruit.
Note: Many "paradoxes" sort-of are like this, but I think aren't what I'm looking for. (William Byers's book "How Mathematicians Think" has several examples and lots of discussion of the important role of paradox in mathematical research.) For example, the relationship between Russell's paradox, Godel's Incompleteness Theorem, and the undecidability of the halting problem (Church; Turing). But I think, unless the paradox has some other aspects of the vague-idea-as-fertilizer phenomenon, that I'm not looking for examples of paradoxes, though I am willing to be convinced otherwise.
Edit: It's been suggested that this a duplicate of this other question, but I really think it is not. I am more interested in examples of outright  incorrect (or nearly so) original statements that nonetheless lead to fruitful mathematics, whereas the other question seems to essentially be asking about ideas that start out intuitive, non-rigorous, or ill-defined and then are turned into rigorous arguments but along the same intuitive lines. (And, as I said above, I think I agree with one of the answers there that that is simply much of mathematics.) By comparing the answers to the other question to the three great answers already on this question (knot theory rising because Kelvin thought atoms were knotted strings; Lame's erroneous proof of FLT leading Kummer to develop algebraic integers; Lebesgue's incorrect proof that projections of Borel sets are Borel leading to Suslin's development of analytic sets), one can get a sense of the difference.

REPLY [5 votes]: Obviously König's theorem should appear on this page. König suggested a proof by which the real numbers cannot be well-ordered. Unfortunately, he misunderstood some of the work he relied on, and thence we have this wonderful theorem known as König's theorem or Zermelo–König's theorem:

If $I$ is any set, and for each $i\in I$, $|A_i|<|B_i|$, then $\left|\bigcup_{i\in I}A_i\right|<\left|\prod_{i\in I}B_i\right|$.


TITLE: Integrals of forms are equal implies they differ by $d\mu$
QUESTION [2 upvotes]: The problem is about the proof of the following result.
If $\omega_1,\omega_2 \in \Omega^n_c(X)$ (where $X$ is smooth manifold) are such that $\int_X\omega_1=\int_X \omega_2$ then there is $\mu\in\Omega^{n-1}_c(X)$ such that $\omega_1-\omega_2=d\mu$.
I am using text at link. Part where I am stuck is on numerical page 202 which is page 30 in the file. It says that using partition of unity argument in Step 4 we can assume $\omega_1,\omega_2$ are supported in some parametrizable open sets.
Can someone explain to me why is this true?

REPLY [2 votes]: Note that $\omega\in \Omega_c^n(X)$ means $\omega$ is a $\mathcal{C}^\infty$ $n$-form, compactly supported on the manifold $X$.
The notation is introduced on page 106 of the document.
Partitions of unity are discussed starting on page 188.
Regarding the beginning of step 4:
By assumption, $\int_X \omega_i = \int_{U_i}\omega_i = 0$.
Step 1 tells us that $\omega_i = d\mu_i$ for $\mu_i \in \Omega_c^{n-1}(U_i)$.
This implies $\omega_1\sim \omega_2$.
Another way to look at this proof is to ignore the argument regarding the special case $c=0$.
Assume $\int_X\omega_1 = \int_X\omega_2 = c$ where $c$ can be zero.
Don't divide by $c$.
Go through step 5 but instead choose $\int_X \alpha_i = c$.
The rest of step 5 goes through as written, giving $(a)\Rightarrow (b)$.


TITLE: Characterize the sum of subspaces
QUESTION [1 upvotes]: Let $U,V$ be the following subspaces of $\mathbb R^3$
$U= \{(a,b,c) \in \mathbb R^3,a+b+c=0 \}$
$V=\{(a,b,c) \in \mathbb R^3,a=c  \} $
$W=\{(0,0,c)\in \mathbb R^3\}$
Carachterize :
a)$U+V$
I think of using new variables such as : $c=-a-b$ and then
\begin{align}
& = \{(a,b,c) \in \mathbb R^3 , c=-a-b\} \\U
& =\{(a,b,-a-b)\} \\
\end{align}
and V:
\begin{align}
V& = \{(a,b,c) \in \mathbb R^3,a=c  \}                     \\
V& =\{c,b,c\}                      \\
\end{align}
And then $U+V$
$U+V=(-a-b,b,-a-b) $
What other option should i have consider?
b)$U+W$
I basically think the same as this example, to characterize means to me that i must consider the second vector constraint, to find the sum in this case.
$U+W=\{(0,0,-a-b)\}$ , c is still $-a-b$
I would appreciate guidance

REPLY [0 votes]: Remark that
$x \in U+V \implies x=u+v=(a,b,-a-b)+(a,b,a)$ since $u\in U$ and $v\in V$


TITLE: A question about pullback of a smooth form.
QUESTION [0 upvotes]: In Lee book, "Introduction to Smooth Manifold, 2nd edition" there is a proof about the existence and Uniqueness of the Exterior differentiation (page 365). He define the exterior differentiation of a smooth $k$-form $\omega\in \Omega^k(M)$ locally on each smooth chart $(U, \phi)$ by using pullbacks:
$$d\omega = \phi^*d(\phi^{-1*}\omega).$$
My question is how do we pullback $\omega$ by $\phi^{-1}$? why I ask because by definition of pullback we could pull back by $\phi^{-1}$ only for forms on $U$, but here $\omega$ is a smooth form on $M$ not $U$. Something is not clear here for me please help.

REPLY [1 votes]: Properly speaking, Lee is not pulling back $\omega$ but its restriction to $U$. We can define the restriction to $U$ of $\omega$ via $\omega|_U = i^*\omega$ where $i: U \longrightarrow M$ is the inclusion map. So really, this defines $(d\omega)|_U$. You can show that these glue together to properly define $d\omega$.


text:
\begin{document}
	\maketitle
	\begin{abstract}
		We prove that the minimum of the modulus of a random trigonometric polynomial with Gaussian coefficients, properly normalized, has limiting exponential distribution.
	\end{abstract}

	\section{Introduction}
	Let $n\ge 1$ and consider the random trigonometric polynomial given as
	\[
	P_n(x):= \frac{1}{\sqrt{2n+1}} \sum_{j=-n}^{n} \zeta_j e^{ijx}
	\]
	where $i :=\sqrt{-1}$ and $\{\zeta_j\}$ are standard independent complex Gaussian coefficients; that is, the density of the random varible $\zeta_j$ with respect to the Lebesgue measure in the complex plane is $\frac{1}{\pi} e^{-|z|^2}$. We note that with this choice of coefficients,
	the polynomial $P=P_n$ is a mean-zero (complex-valued) stationary Gaussian process on $\bT = \bR/2\pi \bZ$ with covariance kernel given by
	\begin{equation}
	\label{eq:covariance_kernel_of_polynomial}
	r_n(x) := \bE\left[P(0)\overline{P(x)}\right] = \frac{1}{2n+1} \sum_{j=-n}^{n} e^{-ijx} = \frac{\sin\left(\left(n+\frac{1}{2}\right)x\right)}{(2n+1) \sin\left(x/2\right)}.
	\end{equation}
	In this paper we study the random variable
	\begin{equation}
	\label{eq:definition_of_minimum}
	m_n:=\min_{x\in \bT}|P_n(x)|
	\end{equation}
	and its limiting distribution as $n\to\infty$. The main result is the following.
	\begin{theorem}
		\label{thm:limiting_distribution_of_minimum}
		For all $\tau>0$ we have that
		\[
		\lim_{n\to\infty} \bP\left(m_n \ge \frac{\tau}{n}  \right) =  e^{- \lambda \tau}
		\]
		where $\lambda = 2\sqrt{\pi/3}$.
	\end{theorem}
	\subsection{Background}
	The study of random polynomials (and in particular, their zeros) has a long history. Consider the \emph{Kac polynomial}
	\[
	F(z):= \sum_{j=0}^{n} \eta_j z^j,\quad z\in \bC,
	\]
	where $\{\eta_j\}$ is an i.i.d. sequence of complex random variables. It is well known that if $\bE\log\left(1+ |\eta_0|\right)<\infty$ then
	  the zeros of $F$ concentrate uniformly around the unit
	  circle as the degree $n$ tends to infinity
	  \cite{erdos_turan,sparo_sur} (see \cite{hughes_nikeghbali}
	  for a more modern perspective).  For finer
	results  and additional references, see
	\cite{shepp_vanderbei} for the Gaussian coefficients case (i.e. when $\eta_j = \zeta_j$) or	\cite{ibragimov_zeitouni} for the general case.

	In view of these results, it is natural to expect that
	the random variable
	$$m_n(F) := \min_{ |z| = 1} |F(z)| $$
	tends to zero as $n\to\infty$, and to study the order of magnitude at which this random variable decay. A particular case of this problem, when the coefficients $\eta_j$ are Rademacher random variables (that is, $\eta_j$ takes the values $\{\pm 1\}$ with equal probability), was posed already by Littlewood in \cite{littlewood}. In \cite{konyagin}, Konyagin proved that in the Rademacher case, for all $\eps>0$
	\[
	\bP\left( m_n(F) \ge \frac{1}{n^{1/2-\eps}} \right) \xrightarrow{n\to\infty } 0.
	\]
	In a later paper, Konyagin and Schlag \cite{konyagin_schlag} proved
	that for either  the Rademacher or  Gaussian cases, there exists some absolute constant $C>0$ such that
	\[
	\limsup_{n\to\infty} \bP\left(m_n(F) < \frac{\eps}{\sqrt{n}}\right) \leq C\eps
	\]
	for all $\eps>0$.
	Note that for the case of complex Gaussian coefficients, $m_n(F)$ is exactly $m_n$ (up to a normalization by $1/\sqrt{n}$) as defined in (\ref{eq:definition_of_minimum}), so Theorem \ref{thm:limiting_distribution_of_minimum} resolves this question for the Gaussian case. The same method of proof works
(after some minor modifications) in the case of real Gaussian coefficients, see Section \ref{sec:real_gaussian} for more details.

After the completion of this paper, we learnt \cite{cook_nguyen}
that N. Cook and H. Nguyen proved a universality result
for the minimum modulus of random polynomials  with i.i.d. coefficients,
by a comparison method. Their starting point is our Theorem
\ref{thm:limiting_distribution_of_minimum} for the Gaussian case.
In particular, their work settles  the
problem for the Rademacher coefficients case.
	\subsection{Structure of the proof}
		The proof of Theorem
		  \ref{thm:limiting_distribution_of_minimum} is based on the
		 observation  that locally (within intervals of length
		much smaller than $1/n$), the polynomial $P_n$ is well
		approximated by its linear interpolation. This observation
		is a consequence of a-priori bounds on the second
		derivative, see Lemma \ref{lemma:second_derivative_is_small}.
		In particular, by
		the ``high-school" exercise in Section \ref{subsec-highschool},
		the value and the location of local minima of $P_n$ can be well
		predicted by linear interpolation from an appropriate
		net of points. Crucially,
		this observation also implies that points which
		are candidates for being global minima are well separated,
		see
		Lemma \ref{lemma:points_in_the_process_are_separated}.

		Introduce a net of points $x_\alpha\in \bT$, and set
		$X_\alpha$ to be
		a signed version
		of $n|P_n(x_\alpha^*)|$,

		where
		$x_\alpha^*$ is the location of the minimum of
		$P_n(\cdot)$ based
		on linear interpolation from $(P_n(x_\alpha), P_n'(x_\alpha))$.
		Introduce a
	 ``good'' event	$\cA_{\al}$  that is typical for
		global minima, see (\ref{eq-defA}) for the precise definition.
		The global minimum  $nP_n(\cdot)$ is then well approximated
		by the point closest to $0$ of the
		point process
		$\mathcal{M}_n := \sum_{\al} \delta_{X_\al}
		\mathbbm{1}_{\cA_\al} $.
		Theorem  \ref{thm:limiting_distribution_of_minimum}
		is then a consequence of the fact that $\mathcal{M}_n$ converges to a Poisson point process of intensity $\sqrt{\pi/3}$
		(the intensity is computed in Corollary \ref{cor:limit_intensity_of_point_process}).
		The Poisson convergence, in turn, is based on
		a characterization of Poisson processes due to Liggett
	      \cite{liggett}, and uses
		a technique introduced by Biskup and
		Louidor in \cite{biskup_louidor}:
		one
		exploits the fact that $P_n$ is a Gaussian process
		and that minima are well separated to deduce
		an invariance property of $\mathcal{M}_\infty$ with respect
		to additive i.i.d. perturbations of the points $X_\alpha$.
		The details of this argument appear in Section
		\ref{sec-liggett}.

		We remark that most of the above argument does
		not use the Gaussian nature of the coefficients
		in any essential way.
		The only place in the argument where the coefficients of
		$P_n$ are required to be Gaussian
		(or rather, have a `not too small' Gaussian component) is in
		Section \ref{sec-liggett}, where we extract a Poisson limit
		from Liggett's characterization.
		In Section \ref{sec-small_gaussian_component} we give a sketch
		of how our result can be generalized to
		a more general choice of random coefficients,
		having a small Gaussian component. Of course, this extension
		is covered by the Cook-Nguyen
		theorem mentioned above, and the sketch just serves to
		illustrate the flexibility, together with the limitations, of our approach.

	\subsection{A high-school exercise}
\label{subsec-highschool}
	Suppose we are given two (non-zero) planar vectors $A= (a_1,a_2)$ and $B=(b_1,b_2)$. We want to find the distance between the origin and the straight line $\left\{A+tB\mid t\in \bR \right\}$. Set $F(t):= A + tB$ and let $t_{min}$ be defined via the relation
	\[
	|F(t_{min})| = \min_{t\in \bR} |F(t)|.
	\]
	We denote by $\gamma$ the angle between $A$ and $B$. It is evident (see Figure \ref{fig:minimum_of_linear_function}) that
	\[
	\frac{\avg{A,B}}{|B|}  = |A| \cos(\gamma)  = |F(t_{min}) - A| = -t_{min}|B|
	\]
	and so $t_{min} = -\avg{A,B}/|B|^2$. Now, simple algebra yields that $$F(t_{min}) = \frac{a_1b_2 - a_2 b_1}{\sqrt{b_1^2 + b_2^2}} = \frac{\avg{A,B^\perp}}{|B|} $$
	where $B^\perp$ is an anti-clockwise rotation of the vector $B$ by $90^\circ$ (see again Figure \ref{fig:minimum_of_linear_function}). In complex notation, by considering $A = a_1 + i a_2$ and $B=b_1 + i b_2$, we have
	\[
	t_{\min} = -\frac{\re{(A\overline{B})}}{|B|^2}, \quad F(t_{\min}) = \frac{\im{(A\overline{B})}}{|B|}.
	\]
\begin{figure}[H]
	\begin{center}
		\scalebox{0.25}{\includegraphics{minimum_of_linear_function_1.eps}}
		\caption{The dashed line is $\left\{A+tB \mid t\in \bR\right\}$. We see that $\cos(\gamma)|A| = |A-F(t_{min})|$ and that $F(t_{min})$ is the projection of the vector $A$ onto the straight line perpendicular to $B$.}
		\label{fig:minimum_of_linear_function}
	\end{center}
\end{figure}

	\subsection*{Notation}
	We write $f\ll g$ or $f= \cO(g)$ if there exist a constant $C>0$ that does not depend on $n$ such that $f\leq C g$. We will also write $f = o(g)$ if $f/g\to 0$ as $n\to\infty$. We denote by $dm(\cdot)$ the Lebesgue measure on $\bC$, and by
	$C_c(\bR)$  the space of continuous, compactly supported functions
      on $\bR$. We write
    $\mathcal{N}_{\bR}(a,b)$ for the Gaussian law with mean $a$ and variance
    $b$. For random variables $X$ and $Y$, we
    write $X\stackrel{\mbox{\scriptsize{\rm law}}}{=} Y$ if they are identically distributed. For a sequence of random variables $X_n$, we write $X_n\xrightarrow{\ d \ } X$ if
  $X_n$ converges in distribution to $X$ as $n\to\infty$. Finally, for
  $N\in \bN$ even we write $[N] := \left\{-N/2,-N/2+1,\ldots,N/2-1\right\}$.

	\section{Point process of near-minima values}
	\label{sec-point_process_of_near_minima}
	Fix some $\e>0$ small (that will not depend on $n$; $\e=1/100$ is good enough) and set $N:=2\lfloor n^{2-\e} /2\rfloor$ so that
	$N$ is even. We consider $N$ equidistributed points on the unit circle given by
	\[
	x_\al  = \frac{2\pi \al}{N}, \quad \al = -N/2,\ldots,N/2-1.
	\]
	Denote the interval of length $2\pi/N$ centered at the point $x_\al$ by $I_\al$, namely,
	\[
	\bT = \bigcup_{\al = -N/2}^{N/2-1} I_\al, \quad I_\al = \left[x_\al - \frac{\pi}{N}, x_\al + \frac{\pi}{N}\right].
	\]
	The linear approximation for the polynomial at the point $x_\al$ is
	\begin{equation}
	  \label{eq-Falpha}
      F_\al(x) := P(x_\al) + (x-x_\al) P^\prime(x_\al).
    \end{equation}
    Following the high-school exercise from Section \ref{subsec-highschool}, we set
	\begin{equation*}
		Y_\al := -\frac{\re\left(P(x_\al)\overline{P^\prime(x_\al)}\right)}{|P^\prime(x_\al)|^2}, \qquad Z_\al := n\cdot \frac{\im\left(P(x_\al)\overline{P^\prime(x_\al)}\right)}{|P^\prime(x_\al)|}.
	\end{equation*}
	And so, $Z_\al$ is the minimal modulus (kept with a sign and scaled by $n$) of the linear approximation $F_\al$ and $Y_\al$ is the unique point such that $|F_\al(Y_\al)| = |Z_\al|/n$. The event that the interval $I_\al$ produce a candidate for the minimal value is given by
	\begin{align}
	  \label{eq-defA}
		\cA_\al = &\cA_\al^\prime \cap \cA_\al^\dprime ,\qquad
	\mbox{\rm where} \nonumber \\
		&\cA_\al^\prime = \left\{Y_\al \in I_\al, \ |Z_\al| \leq \log n \right\}  \\ &\cA_\al^\dprime =\left\{|P(x_\al)| \leq n^{-1/2}, \ |P^\prime(x_\al)| \in \left[n^{1-\e/2}, C_0 n\sqrt{\log n} \right]  \right\}.\nonumber
	      \end{align}
	      $C_0$ in the definition above is a large absolute constant which we specify in Lemma \ref{lemma:probability_of_single_point_to_be_near_minima}; $C_0=10$ is good enough. The event $\cA_\al^\prime$ tells us that the interval $I_\al$ gives a candidate for the minimum and the event $\cA_\al^\dprime$ is just the typical values of $\left(P(x_\al),P^\prime(x_\al)\right)$ so that the interval $I_\al$ gives a candidate. We can now define the point process on $\bR$ of near-minima values as
	\begin{equation}
		\label{eq:definition_of_extremal_process}
		\mathcal{M}_n := \sum_{\al=-N/2}^{N/2-1} \delta_{X_\al}, \quad \text{where } \  X_\al = Z_\al \cdot \mathbbm{1}_{\cA_\al} + \infty \cdot {\mathbbm{1}}_{\cA_\al^c}.
	\end{equation}
	 Here and throughout, we consider $\mathcal{M}_n$ as an element
	 of the space of locally finite, integer valued positive
	 measures on $\bR$, equipped with the local weak$^*$
       topology generated by bounded, compactly supported functions. Thus, we never consider the points at infinity that are contributed
     by the events $\cA_\al^c$.

     Essentially, the linear approximations captures the global minimum of the polynomial since the second derivative is small. In what follows we make this idea precise. For $\beta>0$, define the event
	 \[
	   \mathcal{G}_{\beta} = \left\{ \sup_{x\in \bT} |P^\dprime_n(x)| \leq n^{2+\beta}\right\}.
	 \]
	 \begin{lemma}
	 	\label{lemma:second_derivative_is_small}
		For any $\beta>0$ we have $\bP(\mathcal{G}_{\beta}^c) \ll \exp(-n^\beta)$.
	 \end{lemma}
	 \begin{proof}
	 	Denote by $g(x) := n^{-2} \re{P^\dprime(x)}$ just for the proof. Then $g$ is a trigonometric polynomial such that for all $x\in \bT$,
		  $g(x)$ is distributed $\mathcal{N}_{\bR} (0,\sigma^2)$ where $\sigma^2 = \frac{1}{10} + o(1)$. Hence, for all $\theta\in \bR$ we have that
	 	\begin{equation}
	 	\label{eq:moment_generating_function_of_second_derivative}
	 	\bE\left[e^{\theta g(x)}\right] = e^{\theta^2/(5+o(1))}.
	 	\end{equation}
	 	Let $\ti{x}$ be the point such that $|g(\ti{x})| = \norm{g}_\infty$, and recall that $\norm{g^\prime}_\infty \leq (2n+1) \norm{g}_\infty$ by Bernstein's inequality. Provided
		that $|x-\ti{x}|\leq 1/4n$, we have that
	 	\begin{align*}
	 	|g(x)| &\ge |g(\ti{x})| - |g(\ti{x}) - g(x)| \ge \norm{g}_\infty - |\ti{x}-x| \norm{g^\prime}_\infty \geq \frac{1}{2} \norm{g}_\infty.
	 	\end{align*}
		Combining this with (\ref{eq:moment_generating_function_of_second_derivative}) and Fubini, we get the bound
	 	\begin{align*}
		  \bE\left[e^{\theta \norm{g}_\infty/2}\right] &\leq 2n\bE\left[\int_{|x-\ti{x}| \le 1/4n} \left( e^{\theta g(x)} + e^{-\theta g(x)} \right) dx\right] \\ &\leq 2n \bE\left[\int_{\bT} \big(e^{\theta g(x)} + e^{-\theta g(x)}\big) dx \right] \leq 8\pi n e^{\theta^2/5}.
	 	\end{align*}
	 	Now, we can use the
		Markov inequality with $\theta = 2\sqrt{2}$ and see that
	 	\begin{align*}
	 	\bP\left(\mathcal{G}_{\beta}^c\right) \leq 2\bP\left(\norm{g}_\infty \ge \frac{n^\beta}{\sqrt{2}}\right) \leq \bE\left[e^{\theta\norm{g}_\infty} \right]e^{-\theta n^{\beta}/\sqrt{2} } \ll e^{-n^\beta}.
	 	\end{align*}
	 \end{proof}
 	We now turn to compute the probability that the interval $I_\al$ contributed a point to $\cM_n$.
 	\begin{lemma}
 		\label{lemma:probability_of_single_point_to_be_near_minima}
		For any interval $[a,b]\subset \bR$ and for all $\al \in [N]$ we have
 		\begin{equation*}
 			\bP\left(X_\al \in [a,b] \right) = \sqrt{\frac{\pi}{3}}\cdot \frac{b-a}{N} + o\left(\frac{1}{N}\right),
 		\end{equation*}
 		as $n\to\infty$.
 	\end{lemma}
 	\begin{proof}
 		The proof is a simple Gaussian computation. By stationarity, we may assume that $x_\al = 0$. Set $\sigma_n:=\sqrt{\frac{n(n+1)}{3}}$, so that
		$\left(P(0),P^\prime(0)/\sigma_n\right)$ are independent standard complex Gaussian random variables. Indeed, we note that
		\begin{align*}
			\bE\left[P(0)\overline{P^\prime(0)}\right] &= \frac{1}{2n+1}\sum_{j=-n}^{n} -ij = 0, \quad \text{and that } \\
			\bE\left[|P^\prime(0)|^2\right]& = \frac{1}{2n+1} \sum_{j=-n}^{n} j^2 = \frac{n(n+1)}{3}.
		\end{align*}
		Moreover, a straight forward computation shows that $\bE\left[P(0)^2\right] = \bE\left[P(0)P^\prime(0)\right] = \bE\left[P^\prime(0)^2\right] = 0$. By the bounded convergence theorem
 		\begin{align*}
		  \bP\big(\cA_\al^\prime &\cap \left\{Z_\al \in [a,b] \right\}\big) \\ &= \bP\left(\left|Y_\al\right| \leq \frac{\pi}{N}, Z_\al \in [a,b] \right) \\ &= \int_{\bC} e^{-|w|^2} \bP\left(\frac{\left|\re\left(P(0)\wbar{w}\right)\right|}{|w|^2} \leq \frac{\pi \sigma_n}{N} , \ n\cdot \frac{\im \left(P(0)\wbar{w}\right)}{|w|} \in \left[a,b\right] \right)\frac{dm(w)}{\pi} \\ &= \int_{\bC} e^{-|w|^2} \bP\left(\left|\re\left(P(0)\right)\right| \leq \frac{\pi \sigma_n}{N}|w|\right)\cdot \bP\Big(\im\left(P(0)\right) \in \left[\frac{a}{n},\frac{b}{n}\right]\Big) \frac{dm(w)}{\pi} \\ &= \frac{2\pi \sigma_n}{\pi^2 N} \cdot \frac{b-a}{n} \int_{\bC} e^{-|w|^2}|w| dm(w) + o(N^{-1}) \\ &= \sqrt{\frac{\pi}{3}} \cdot \frac{b-a}{N} + o(N^{-1}),
 		\end{align*}
 		where in the third equality we used the rotational symmetry of the Gaussian distribution. To conclude the lemma, we show that $\bP\left(\cA_{\al}^\prime \cap (\cA_\al^\dprime)^c  \cap \left\{Z_\al \in [a,b]\right\} \right)$ is negligible. Since $|Z_\al|/n = |P(0) + Y_\al P^\prime(0)|$,
		the triangle inequality yields that on the event $\cA_\al^\dprime \cap \{ Z_\al \in [a,b]\},$
 		\begin{equation}
 			\label{eq:bound_on_value_of_poly_assuming_crossing}
 			|P(0)| \leq |Z_\al|/n  + |Y_\al||P^\prime(0)| \ll \frac{1}{n} + C_0\frac{n\sqrt{\log n}}{n^{2-\e}}  \ll \frac{1}{n^{1-2\e}}.
 		\end{equation}
 		From (\ref{eq:bound_on_value_of_poly_assuming_crossing}) we conclude that
 		\[
 		\left\{|P(0)| \ge n^{-1/2},\ Z_\al \in [a,b]\right\} \subset \left\{ |P^\prime(0)|\ge C_0 n\sqrt{\log n} ,\ Z_\al \in [a,b]\right\}
 		\]
 		for all fixed $C_0>0$ and for $n$ large enough.
		We thus get the upper bound
 		\begin{align*}
 			\bP&\left(\cA_{\al}^\prime \cap (\cA_\al^\dprime)^c \cap \left\{Z_\al \in [a,b]\right\} \right) \\ & \leq \bP\left(\cA_{\al}^\prime \cap \left\{Z_\al \in [a,b]\right\} \cap \left\{|P^\prime(0)| \leq n^{1-\e/2} \right\} \right) + \bP\left(|P^\prime(0)| \ge C_0 n\sqrt{\log n}\right) \\
			&= \cO\left(\frac{1}{n^2}\right).
 		\end{align*}
 		The first probability is $\cO(n^{-2})$ from the
		same Gaussian computation as done above and
		the second probability is $o(n^{-2})$ for a large absolute constant $C_0$ because $P'(0)$ is a complex Gaussian variable
		of variance bounded by $n^2$; one sees that $C_0=10$ will do.
		Altogether,
 		\begin{align*}
 			\bP\left(X_\al \in [a,b]\right) &=
			\bP\left(\left\{Z_\al \in [a,b] \right\} \cap \cA_\al\right)\\
			&=
			\bP\left(\left\{Z_\al \in [a,b] \right\} \cap \cA_\al^\prime \right)
		      - \bP\left(\left\{Z_\al \in [a,b] \right\} \cap \cA_\al^\prime \cap (\cA_\al^\dprime)^c\right) \\ &= \sqrt{\frac{\pi}{3}} \cdot \frac{b-a}{N} + o\left(\frac{1}{N}\right).
 		\end{align*}
 		\end{proof}
 	\begin{corollary}
 		\label{cor:limit_intensity_of_point_process}
		For any interval $[a,b]\subset \bR$ we have, as $n\to\infty$,
 		\[
 		\bE\left[\cM_n([a,b]) \right] = \sqrt{\frac{\pi}{3}} (b-a) + o(1).
 		\]
 	\end{corollary}
 	\begin{corollary}
 		\label{cor:tightness_of_point_process}
		The sequence of point processes $\{ \cM_n \}$ is tight. That is, for any interval $I\subset \bR$ and for all $n\ge n_0$,
 		\[
 		\lim_{K\to\infty} \bP\left(\cM_n(I) \ge K \right) = 0.
 		\]
 	\end{corollary}
 	Both Corollary \ref{cor:limit_intensity_of_point_process} and Corollary \ref{cor:tightness_of_point_process} are immediate consequences of Lemma \ref{lemma:probability_of_single_point_to_be_near_minima}. We turn to prove that the extremal process $\cM_n$ captures the minimum modulus of our polynomial $P_n$.
 	\begin{lemma}
 		\label{lemma:extremal_process_captures_minimum}
 		For all $\tau>0$ we have that
 		\[
 		\lim_{n\to\infty} \left|\bP\left(m_n \ge \frac{\tau}{n}  \right) - \bP\left(\cM_n((-\tau,\tau)) = 0 \right)\right| =0.
 		\]
 	\end{lemma}
 	\begin{proof}
 		Clearly,
 		\begin{equation*}
 		\bP\left(m_n \ge \frac{\tau}{n}\right) = \bP\left(\bigcap_{\al} \left\{\min_{x\in I_\al}|P_n(x)| \ge \frac{\tau}{n} \right\}\right).
 		\end{equation*}
		Recall the definition of the linear approximation (\ref{eq-Falpha}). For each $\al$ we use Taylor expansion and see that on the event $\cG_\e$,
 		\[
 		\left|P_n(x) - F_\al(x) \right| \ll \frac{\sup_{x\in \bT}|P^\dprime(x)|}{N^2} \ll \frac{1}{n^{2-3\e}}.
 		\]
		Hence, for large enough $n$, we have the upper bound
 		\begin{align*}
		  &\left|\bP\left(m_n \ge \frac{\tau}{n}\right) - \bP(\cM_n((-\tau,\tau)) = 0)\right|  \\ & \leq \sum_{\al=-N/2}^{N/2-1} \bP\left(\left\{\min_{x\in I_\al}|P_n(x)| \ge \frac{\tau}{n},\ |X_\al| < \tau \right\}\cap \cG_\e
		  \right) \\ & \qquad + \sum_{\al=-N/2}^{N/2-1} \bP\left(\left\{\min_{x\in I_\al}|P_n(x)| < \frac{\tau}{n},\ |X_\al| \geq \tau \right\}\cap
		  \cG_\e \right) + 2\bP\left(\mathcal{G}^c_\e\right) \\ &\leq \sum_{\al=-N/2}^{N/2-1} \bP\left( \tau - \frac{1}{2n^{1-3\e}}<|X_\al| \leq \tau\right) + \bP\left( \tau <|X_\al| \leq \tau + \frac{1}{2n^{1-3\e}}\right) + \cO\left(e^{-n^\e}\right).
 		\end{align*}
 		Following the same computation we did in the proof of Lemma \ref{lemma:probability_of_single_point_to_be_near_minima}, we see that
 		\[
 		\bP\left( \tau - \frac{1}{n^{1-3\e}}<|X_\al| \leq \tau\right) = 2\sqrt{\frac{\pi}{3}}\frac{\tau}{N n^{1-3\e}}(1+o(1))
 		\]
 		and a similar equality for the other term in the sum. Altogether, we see that
 		\[
 		\left|\bP\left(m_n \ge \frac{\tau}{n}\right) - \bP(\cM_n((-\tau,\tau)) = 0)\right| \ll \frac{1}{n^{1-3\e}}
 		\]
 		and we are done.
 	\end{proof}
 	By Lemma \ref{lemma:extremal_process_captures_minimum}, the limit distribution of $m_n$ (Theorem \ref{thm:limiting_distribution_of_minimum}) will follow if we can show that
	$\cM_n$ converge in distribution (as a point process) to a Poisson point process with the desired intensity. This is established in what follows. We first prove that the points in the extremal process come from well separated intervals.
 	\begin{lemma}
 		\label{lemma:points_in_the_process_are_separated}
 		For all $\e>0$ we have that
 		\begin{equation*}
 			\lim_{n\to\infty} \bP\bigg(\bigcup_{\stackrel{\al\not=\al^\prime}{|x_\al - x_\al^\prime|\leq n^{-\e}}} \cA_\al \cap \cA_{\al^\prime} \bigg) = 0.
 		\end{equation*}
 	\end{lemma}
 	To prove Lemma \ref{lemma:points_in_the_process_are_separated} we will need two claims. We first prove the lemma assuming both claims, and then turn to prove each claim separately. Denote by
 	\begin{equation}
 	\label{eq:two_sums_in_separation_lemma}
 	S_\RomanNumeralCaps{1} := N \sum_{\al=1}^{\lfloor n^{1-2\e}/2\pi \rfloor} \bP\left(\cA_0 \cap\cA_\al \right), \qquad S_\RomanNumeralCaps{2} := N \sum_{\al=\lceil n^{1-2\e}/2\pi \rceil}^{\lfloor n^{2-2\e}/2\pi \rfloor} \bP\left(\cA_0 \cap\cA_\al \right).
 	\end{equation}
 	\begin{claim}
 		\label{claim:first_sum_in_seperatin_lemma_is_small}
 		$S_\RomanNumeralCaps{1} = o(1)$ as $n\to\infty$.
 	\end{claim}

 	\begin{claim}
 		\label{claim:second_sum_in_seperatin_lemma_is_small}
 		$S_\RomanNumeralCaps{2} = o(1)$ as $n\to\infty$.
 	\end{claim}
 	\begin{proof}[Proof of Lemma \ref{lemma:points_in_the_process_are_separated}]
 		Applying the union bound and stationarity we see that
 		\begin{align}
 		\bP\bigg(\bigcup_{\stackrel{\al\not=\al^\prime}{|x_\al - x_\al^\prime|\leq n^{-\e}}} \cA_\al \cap \cA_{\al^\prime} \bigg) & \leq \sum_{\stackrel{\al\not=\al^\prime}{|x_\al - x_\al^\prime|\leq n^{-\e}}} \bP\left(\cA_\al \cap \cA_{\al^\prime} \right) \nonumber \\ \nonumber &\ll N\sum_{\al=1}^{\lfloor n^{2-2\e}/2\pi \rfloor} \bP\left(\cA_0 \cap\cA_\al \right) = S_\RomanNumeralCaps{1} + S_\RomanNumeralCaps{2} = o(1),
 		\end{align}
 		where
		the last equality is due to Claims \ref{claim:first_sum_in_seperatin_lemma_is_small} and \ref{claim:second_sum_in_seperatin_lemma_is_small}.
 	\end{proof}
 	\begin{proof}[Proof of Claim \ref{claim:first_sum_in_seperatin_lemma_is_small}]
 		Fix some $\beta<\epsilon/2$, and consider first the term $\al = 1$ in the sum $S_\RomanNumeralCaps{1}$. Observe that for all $x\in I_1$,
 		\begin{equation*}
 		\left|F_0(x) - F_1(x)\right| \leq \left|F_0(x) - P(x)\right| + \left|F_1(x) - P(x)\right| \ll \frac{\sup_{x\in \bT}|P^\dprime(x)|}{N^2},
 		\end{equation*}
 		which yields that on the event $\cG_\beta$, for all $x\in I_1$ we have that
 		\begin{equation}
 			\label{eq:difference_between_two_adjacent_linear_interpolations_is_small}
 			|F_0(x) - F_1(x)| \ll n^{-2+3\e}.
 		\end{equation}
 		Following the same computation as done in the proof of Lemma \ref{lemma:probability_of_single_point_to_be_near_minima}, we see that
 		\[
 		\bP\left(\cA_0 , \ |Y_0| \in \left[\frac{\pi}{N} - \frac{\pi}{Nn^{\e/4}},\frac{\pi}{N}\right]\right) \ll \frac{1}{Nn^{\e/4}},
 		\]
 		whence, on the event $\cA_0 \cap \left\{|Y_0| \leq \frac{\pi}{N}-\frac{\pi}{Nn^{\e/4}}\right\}$ we have that
 		\begin{equation}
 			\label{eq:lower_bound_for_linear_interpolation_end_of_interval}
 			\left|F_0\left(\frac{\pi}{N}\right)\right| = \left|\frac{Z_0}{n} + \left(\frac{\pi}{N}-Y_0\right)P^\prime(0)\right| \gg \frac{n^{1-\e/2}}{Nn^{\e/4}} - \frac{\log n}{n} \gg \frac{1}{n^{1-\e/4}}.
 		\end{equation}
 		Therefore, we combine (\ref{eq:difference_between_two_adjacent_linear_interpolations_is_small}) and (\ref{eq:lower_bound_for_linear_interpolation_end_of_interval}) to conclude that on the event $\cG_\beta \cap \cA_0 \cap \left\{|Y_0| \leq \frac{\pi}{N}-\frac{\pi}{Nn^{\e/4}}\right\}$ we have that for all $x\in I_1$,
 		\[
 		|F_1(x)| \geq |F_0(x)| - |F_1(x) - F_0(x)| \geq \left|F_0\left(\frac{\pi}{N}\right)\right| - |F_1(x) - F_0(x)| \gg \frac{1}{n^{1-\e/4}},
 		\]
 		which implies that the event $\cA_1$ does not hold.
		Altogether, we apply Lemma \ref{lemma:second_derivative_is_small} and obtain that
 		\begin{align*}
 		N\bP\left(\cA_0 \cap \cA_1 \right) &\leq N\bP\left(\cA_0 \cap \left\{ |Y_0| \in \left[\frac{\pi}{N} - \frac{\pi}{Nn^{\e/4}},\frac{\pi}{N}\right] \right\}\right) \\ & \qquad  + N\bP\left( \cA_0 \cap \cA_1 \cap \left\{|Y_0| \leq \frac{\pi}{N}-\frac{\pi}{Nn^{\e/4}} \right\}\right) \\
		&\ll n^{-\e/4} + N\bP\left(\cG^c_\beta\right) \ll n^{-\e/4}.
 		\end{align*}
 		The treatment of the rest of the sum $S_\RomanNumeralCaps{1}$ is similar, only that we do not have to impose the extra separation within the interval. We have for all $x\in I_\al$ that
 		\begin{equation}
 		\label{eq:difference_between_linear_approximations_separation_lemma}
 		\left|F_0(x) - F_\al(x)\right| \le \left|F_0(x) - P(x)\right| + \left|F_\al(x)  - P(x)\right| \ll |x_\al|^2 \sup_{x\in \bT}|P^\dprime(x)|.
 		\end{equation}
 		Furthermore, on the event $\cA_0$, we have the lower bound
 		\begin{equation}
 		\label{eq:lower_bound_for_linear_approximation_separation_lemma}
 		|F_0(x)| \ge \left|F_0\left(x_\al - \frac{\pi}{N}\right)\right| \geq \left|x_{\al-1}\right||P^\prime(0)| - \frac{\log n}{n} \gg |x_{\al-1}| n^{1-\e/2}
 		\end{equation}
 		for all $x\in I_\al$. Recalling that
 		$\beta< \e/2$, we combine (\ref{eq:difference_between_linear_approximations_separation_lemma}) and (\ref{eq:lower_bound_for_linear_approximation_separation_lemma}) to see that on the event $\cG_\beta$,
 		\[
 		\forall x\in I_\al, \qquad |F_\al(x)| \gg \left|x_{\al-1}\right|n^{1-\e/2} + |x_\al|^2 n^{2+\beta} \gg \frac{1}{n^{1-\e/2}}
 		\]
 		(here we use that $n^{-1-\e} \gg |x_{\al-1}| \ge 2\pi/N$) which implies that $\cA_\al$ does not hold. We thus have,
 		\[
 		S_{\RomanNumeralCaps{1}} \ll n^{-\e/4} + N
 		\cdot \lfloor n^{1-2\e}/2\pi \rfloor\cdot \bP\left(\cG^c_\beta\right) \ll n^{-\e/4} +  n^{3-3\e} e^{-n^{\beta}} = o(1).
 		\]
 	\end{proof}
 	\begin{proof}[Proof of Claim \ref{claim:second_sum_in_seperatin_lemma_is_small}]
 		Denote by $\Id$ the identity matrix. The random vector
 		$
 		V_\al:=\left(P(0),\frac{P^\prime(0)}{\sigma_n}, P(x_\al),\frac{P^\prime(x_\al)}{\sigma_n} \right)
 		$ is a mean zero complex Gaussian with independent real
		and imaginary component; the covariance matrix of
		both the real and imaginary parts is
		$\frac{1}{2}\begin{pmatrix}
 		\Id & \Sigma_\al \\ \Sigma_\al & \Id
 		\end{pmatrix},$ where
 		\[
 		\Sigma_\al := \begin{pmatrix}
 		r_n(x_\al) & -r_n^\prime(x_\al)/\sigma_n \\  r_n^\prime(x_\al)/\sigma_n & -r_n^\dprime(x_\al)/\sigma_n^2
 		\end{pmatrix},
 		\]

 		and $r_n$ is given by
		(\ref{eq:covariance_kernel_of_polynomial}).

		The density of the vector $V_\al$ in $\bC^4$ (or $\bR^8$) is bounded from above by a constant multiple of
 		\begin{align}
 		\label{eq:determinant_computation_for_density_V}
 		\left[\det\begin{pmatrix}
 		\Id & \Sigma_\al \\ \Sigma_\al & \Id
 		\end{pmatrix}\right] ^{-4} &= \left[\det\left(\Id - \Sigma_\al^2\right) \right]^{-4} \\ &= \left[\det\left(\Id - \Sigma_\al\right) \det\left(\Id + \Sigma_\al\right) \right]^{-4}. \nonumber
 		\end{align}
 		Differentiating (\ref{eq:covariance_kernel_of_polynomial}) twice and applying
		some trigonometric identities yield that
 		\begin{align*}
 		&r_n(x) = \frac{\sin\left(\left(n+\frac{1}{2}\right)x\right)}{(2n+1)\sin(x/2)}, \\ &r_n^\prime (x) = \frac{1}{2\sin(x/2)} \left[\cos\left(\left(n+1/2\right)x\right) - \frac{\cos(x/2) \sin\left(\left(n+\frac{1}{2}\right)x\right)}{(2n+1)\sin(x/2)}\right],  \\ & r_n^\dprime (x) = \frac{n^2 \sin \left(\left(n+\frac{3}{2}\right) x\right) + (n+1)^2 \sin \left(\left(n-\frac{1}{2}\right) x\right) - (2 n (n+1)-1) \sin \left(\left(n+\frac{1}{2}\right) x\right)}{\sin^3(x/2)( 8 n+4)}.
 		\end{align*}
 		Using Taylor's approximation and some algebra it is evident that for $n^{-1-2\e} \leq |x_\al| \leq \e/n$ we have
 		\begin{align}
 		\label{eq:taylor_expansion_for_correlations}
 		r_n(x_\al) &= 1- \frac{(n x_\al) ^2}{6} + \frac{(nx_\al)^4}{120} + \cO\left(|nx_\al|^6\right), \nonumber \\ \frac{r_n^\prime(x_\al)}{\sigma_n} &= - \frac{nx_\al}{\sqrt{3}} + \frac{(nx_\al)^3}{10\sqrt{3}} + \cO\left(|nx_\al|^5\right), \\ -\frac{r_n^\dprime(x_\al)}{\sigma_n^2} &= 1 - \frac{(nx_\al)^2}{10} + \cO\left(|nx_\al|^4\right). \nonumber
 		\end{align}
 		Using (\ref{eq:taylor_expansion_for_correlations}), we see that
 		\begin{align*}
 		\det\left(\Id - \Sigma_\al \right) &= \big(1- r_n(x_\al)\big)\left(1 + r_n^\dprime(x_\al)/\sigma_n^2\right) + \big(r_n^\prime(x_\al)/\sigma_n \big)^2 = \frac{(n x_\al)^2}{3} + \cO(|n x_\al|^4), \\ \det\left(\Id + \Sigma_\al \right) &= \big(1 + r_n(x_\al)\big)\left(1 - r_n^\dprime(x_\al)/\sigma_n^2\right) + \big(r_n^\prime(x_\al)/\sigma_n \big)^2 = 4 + \cO(|n x_\al|^2),
 		\end{align*}
 		which together with (\ref{eq:determinant_computation_for_density_V}) implies that
 		\begin{equation}
 		\label{eq:bound_on_det_density_of_V}
 		\left[\det\begin{pmatrix}
 		\Id & \Sigma_\al \\ \Sigma_\al & \Id
 		\end{pmatrix}\right] ^{-4} \ll  \frac{1}{(nx_\al)^8} \qquad \text{for } \  n^{-1-2\e} \leq |x_\al| \leq \e/n.
 		\end{equation}
 		Furthermore, for $|x_\al| \ge \e/n$, the density of $V_\al$ is uniformly bounded from above by a constant $C=C_\e$. Combining this observation with (\ref{eq:bound_on_det_density_of_V}), we can bound the sum $S_{\RomanNumeralCaps{2}}$ (recall (\ref{eq:two_sums_in_separation_lemma})) as
 		\begin{align*}
 		S_{\RomanNumeralCaps{2}} &\leq N \sum_{\al=\lceil n^{1-2\e}/2\pi \rceil}^{\lfloor \e N/2\pi n \rfloor} \bP\left(\cA_0 \cap\cA_\al \right) + N \sum_{\al=\lceil \e N/2\pi n \rceil}^{\lfloor n^{2-2\e}/2\pi \rfloor} \bP\left(\cA_0 \cap\cA_\al \right) \\ & \ll \frac{(\log n)^2}{N} \sum_{\al=\lceil n^{1-2\e}/2\pi \rceil}^{\lfloor \e N/2\pi n \rfloor} n^{8\e} + \frac{(\log n)^2}{N}  \sum_{\al=\lceil \e N/2\pi n \rceil}^{\lfloor n^{2-2\e}/2\pi \rfloor} C_\e \\ & \ll \frac{(\log n)^2 n^{8\e}}{n} + \frac{(\log n)^2}{n^\e} = o(1).
 		\end{align*}
 	\end{proof}
 	\section{Liggett's invariance and proof of Theorem
	\ref{thm:limiting_distribution_of_minimum}}
	\label{sec-liggett}
	In this section, we prove the convergence in distribution
	of $\mathcal{M}_n$ to a Poisson process of constant intensity,
      following a characterization of the latter due to Liggett. In this, we follow a method developed by Biskup and Louidor in their study of the two dimensional
      Gaussian free field \cite{biskup_louidor}. For more background, see
      the lecture notes \cite{biskup}.

      The first step is to rewrite $P_n$ as a sum of two independent polynomials.
 	Denote by $Q=Q_n$ an independent copy of the random polynomial $P=P_n$, and consider the random polynomial (of degree $n$) given by
 	\begin{equation}
 	\label{eq:def_of_polynomial_with_perturbation}
 	\widehat{P}_n(x) := \sqrt{1-\frac{1}{n^2}}P_n(x) + \frac{1}{n} Q_n(x).
 	\end{equation}
 	Clearly, $\what{P}_n \stackrel{\mbox{\scriptsize{\rm law}}}{=} P_n$, and hence
 	\begin{equation}
 		\label{eq:point_processes_have_the_same_law}
		\what{\cM}_n \stackrel{\mbox{\scriptsize{\rm law}}}{=} \cM_n,
 	\end{equation}
 	where $\cM_n$ is the extremal process (\ref{eq:definition_of_extremal_process}) and $\what{\cM}_n$ is the extremal process that corresponds to the polynomial $\widehat{P} = \widehat{P}_n$. The goal of this section is to study the relation between these two point processes. Let $\widehat{X}_\al$ and $\what{Y}_\al$ be the analogous variables to $X_\al$ and $Y_\al$, which correspond to the polynomial $\what{P}$ instead on $P$, see (\ref{eq-defA}).

 	By Corollary \ref{cor:tightness_of_point_process} (tightness of the sequence $\{\cM_n\}$), we can find a subsequence $\{n_k\}\subset \bN$ so that
 	\begin{equation}
 		\label{eq:subsequential_convergence_of_both_processes}
 		\cM_{n_k}\xrightarrow{\ d \ } \cM_\infty, \quad \what{\cM}_{n_k} \xrightarrow{\ d \ } \ \what{\cM}_\infty,
 	\end{equation}
 	as $k\to\infty$. We denote the law of the limiting point process $\cM_\infty$ by $\eta$, then by (\ref{eq:point_processes_have_the_same_law}) it is evident that
 	\[
	  \eta \stackrel{\mbox{\scriptsize{\rm law}}}{=} \what{\cM}_\infty.
 	\]
	The following lemma is the key element in
	extracting the Poisson limit, which is a consequence of $\eta$ having an invariance property. As usual, for $f\in C_c(\bR)$ we denote the linear statistics of a point process $W$ by
	\[
	\avg{W,f} := \sum_{w\in W} f(w).
	\]
 	\begin{lemma}
 		\label{lemma:limiting_pp_satisfy_invariance}
 		Let $\eta$ be given as above and let $f\in C_c(\bR)$ be a non-negative function. Then
 		\begin{equation}
 			\label{eq:liggett_property_for_pp}
 			\bE\left[e^{-\avg{\eta,f}}\right] = \bE\big[ e^{-\langle\eta,\ti{f}\rangle}\big],
 		\end{equation}
 		where
 		\[
		  \ti{f}(x):= -\log \bE_{G} \left[e^{-f(x + G)} \right]
 		\]
		and $G\sim \mathcal{N}_{\bR} (0, 1/2)$. Here $\bE_{G}$ denotes the expectation with respect to the Gaussian variable $G$.
 	\end{lemma}
 	Before proving Lemma \ref{lemma:limiting_pp_satisfy_invariance}, we will need two simple results. Lemma \ref{lemma:almost_independent_gaussians_are_like_gaussians} gives a quantitative approximation for ``almost-independent" normal variables as truly independent normal variables. Lemma \ref{lemma:points_in_both_processes_are_similar} tells us that with high probability, the perturbation (\ref{eq:def_of_polynomial_with_perturbation}) did not introduce any new points into the extermal process, nor did it delete the points that where present before the perturbation.

	\begin{lemma}[{\cite[Lemma 4.9]{biskup_louidor}}]
 		\label{lemma:almost_independent_gaussians_are_like_gaussians}
 		Fix $\sigma^2>0$ and $m\ge 1$. Suppose that $G_1,\ldots, G_m$ are i.i.d. $\mathcal{N}_{\bR}(0,\sigma^2)$. For each $f\in C_c(\bR)$ and for each $\e>0$ there exist $\delta>0$ such that if $\left(W_1,\ldots,W_m\right)$ are multivariate normal with $\bE[W_i] = 0$ and
 		\[
 		\max_{1\leq i,j\leq m} \left|\bE\left[W_i W_j\right] - \sigma^2\delta_{i,j}\right|\leq \delta
 		\]
 		then
 		\[
 		\left|\log \frac{\bE \exp \left\{\sum_{i=1}^{m} f(X_i)\right\}}{\prod_{i=1}^{m}\bE (e^{f(Y_i)})}\right| < \e.
 		\]
 	\end{lemma}

 	\begin{lemma}
 		\label{lemma:points_in_both_processes_are_similar}
 		We have that
 		\[
 		\lim_{K\to\infty} \limsup_{n\to\infty} \bP\left(\bigcup_{\al = -N/2}^{N/2-1} \left\{ |X_\al| \leq K, \ |\widehat{X}_\al| > 2K \right\} \right) = 0.
 		\]
 	\end{lemma}
 	\begin{proof}
 		We turn to bound $\bP\left( |X_\al| \leq K, \ |\widehat{X}_\al| > 2K \right)$ for each $\al$, and assume by stationarity that $x_\al = 0$. We have,
 		\begin{align*}
 			\bP&\left( |X_\al| \leq K, \ |\widehat{X}_\al| > 2K \right) \\ &\leq \bP\left( |X_\al| \leq K, \ \what{Y}_\al \not \in I_\al\right) + \bP\left(|X_\al| \leq K, \ |\widehat{X}_\al| > 2K, \ \what{Y}_\al \in I_\al \right)  =: \bP(E_1) + \bP(E_2).
 		\end{align*}
 		To bound the probabilities $\bP(E_i)$, $i=1,2$ we exploit the relation between the polynomial and its small perturbation (\ref{eq:def_of_polynomial_with_perturbation}). Denote by
 		\[
 		{\begin{array}{cc}
 		F(x) = A+ Bx  \\ \widehat{F}(x) = \widehat{A}+ \widehat{B}x
 		\end{array}}
 		 \quad \text{where, } \quad
 		 { \begin{array}{cc}
 		 	A = P_n(0), & B = P_n^\prime(0), \\ \widehat{A} = \widehat{P}_n(0),  & \widehat{B} = \widehat{P}_n^\prime(0).
 		 	\end{array}}
 		\]
 		We see that
 		\begin{align*}
 		\widehat{A} = \sqrt{1-\frac{1}{n^2}} A + \frac{1}{n} Q(0), \quad \widehat{B} = \sqrt{1-\frac{1}{n^2}} B + \frac{1}{n} Q^\prime(0).
 		\end{align*}
 		By Taylor expanding the square root and using the fact that $|A|\leq n^{-1/2}$ and $|B|\in \left[n^{1-\e/2}, C_0 n\sqrt{\log n} \right] $ on the event $\cA_\al$,
		we see that,
 		\begin{equation}
 	\label{eq:difference_in_linear_approximation_after_perturbation}
 			\left|A - \what{A} + \frac{Q(0)}{n} \right| \ll \frac{1}{n^{3/2}}, \qquad \left|B - \what{B}\right| \ll \frac{|Q^\prime(0)|}{n} + \frac{\sqrt{\log n}}{n}.
 		\end{equation}
 		Therefore, on the event $\cA_\al \cap \left\{|Q(0)|\leq n^\e , \ |Q^\prime(0)| \le n^{1+\e} \right\}$, we have that
		$|B| = |\what{B}|(1+o(1))$ and thus
 		\begin{align*}
 			\left|Y_\al - \what{Y}_\al\right| &= \left| \frac{\re(A\wbar{B})}{|B|^2} - \frac{\re(\what{A}\wbar{\what{B}})}{|\what{B}|^2} \right| \\ & \ll \frac{\left|\re\left(A \wbar{Q^\prime}(0) + Q(0)\wbar{B} \right)\right|}{n|B|^2} \ll \frac{1}{n^{3/2-2\e}} + \frac{|Q(0)|}{n^{3-2\e}} \ll \frac{1}{n^{3/2 - 2\e}}.
 		\end{align*}
 		Therefore,
 		\begin{align}
 			\bP\left(E_1\right)
			&\leq \bP\left(|X_\al|  \leq K , \ |Y_\al| \in \left[\frac{\pi}{N} - \frac{1}{n^{3/2 - 2\e}}, \frac{\pi}{N}\right]  \right)  \nonumber\\
			&\qquad + \bP\left(\left\{|Q(0)|\ge n^\e \right\} \cup \left\{|Q^\prime(0)| \ge n^{1+\e}  \right\} \right) \nonumber \\ \label{eq:bound_on_prob_of_E_1} &\ll \frac{1}{N n^{3/2 - 2\e}} + e^{ - n^\e}.
 		\end{align}
 		To bound $\bP(E_2)$, we use (\ref{eq:difference_in_linear_approximation_after_perturbation}) once more and see that on the event $E_2$,
 		\begin{align*}
 			\left|X_\al - \what{X}_\al\right| &= \left|\frac{\im(A \wbar{B})}{|B|} - \frac{\im(\what{A} \wbar{\what{B}})}{|\what{B}|} \right| \ll \frac{\left|\im\left( Q(0)\wbar{B}  \right)\right|}{|B|} \ll |Q(0)|,
 		\end{align*}
 		which implies that
 		\begin{align}
 			\label{eq:bound_on_prob_of_E_2}
 			\bP\left(E_2\right)& \leq \bP\left(|X_\al| \leq K , \ |Q(0)| \geq \frac{K}{2} \right)\\ \nonumber &= \bP\left(|X_\al|\leq K\right) \cdot \bP\left(|Q(0)| \geq K/2\right) \ll \frac{Ke^{-K^2/4}}{N}.
 		\end{align}
 		Combining the bounds (\ref{eq:bound_on_prob_of_E_1}) and (\ref{eq:bound_on_prob_of_E_2}) together with the union bound, we see that
 		\begin{align*}
 			\bP\left(\bigcup_{\al = 1}^{N} \left\{ |X_\al| \leq K, \ |\widehat{X}_\al| > 2K \right\} \right) &\leq N\bP\left( |X_\al| \leq K, \ |\widehat{X}_\al| > 2K \right) \\ &\ll \frac{1}{n^{3/2-2\e}} + Ke^{-K^2/4}.
 		\end{align*}
 	\end{proof}
 	\begin{proof}[Proof of Lemma \ref{lemma:limiting_pp_satisfy_invariance}]
 		Fix $f\in C_c(\bR)$, and assume that the support of $f$ is strictly contained in $(-K,K)$ for some $K>0$ (large enough). Let $\mathcal{F}_P$ denote the $\sigma$-algebra generated by the coefficients $\left\{\zeta_j \right\}_{j=-n}^{n}$ of the polynomial $P$.

		We have that
 		\begin{align}
 			\nonumber
 			\bE\left[e^{-\avg{\eta,f}}\right] &= \lim_{k\to\infty} \bE\left[\exp\left(-\avg{\what{\cM}_{n_k},f} \right)\right] \\ &= \label{eq:conditional_expectation_for_linear_statistics} \lim_{k\to\infty} \bE\bigg[\bE\Big[\exp\Big(-\sum_{\al=1}^{N_k} f(\what{X}_\al)\Big) \mid
			\mathcal{F}_P\Big] \bigg],
 		\end{align}
 		where here $N_k = 2\lfloor n_k^{2-\e}/2\rfloor$. From the estimates
		(\ref{eq:difference_in_linear_approximation_after_perturbation}), we see that outside of an event of $o(1)$ probability, we have that
		for all $\alpha \in [N]$,
 		\[
 		\what{X}_\al = X_\al + \frac{\re(Q(x_\al)\wbar{P^\prime(x_\al)})}{|P^\prime(x_\al)|} + \cO\left(\frac{1}{\sqrt{n}}\right),
 		\]
		and the error term is uniform in $\al$. Denote by $G_\al := \re(Q(x_\al)\wbar{P^\prime(x_\al)})/|P^\prime(x_\al)|$. Then, conditioned on $\mathcal{F}_P$,
		the random variables $\{G_\al\}$ are jointly normal,
		and each $G_\alpha$ has law $\mathcal{N}_{\bR}(0,1/2)$. Using Lemmas \ref{lemma:points_in_the_process_are_separated} and \ref{lemma:points_in_both_processes_are_similar},
		we obtain that outside of an event of probability $o(1)$, we have for all $\al,\al^\prime\in\left\{ \al\in [N] :  |X_\al|\leq 2K \right\}$ that
		\begin{equation}
		  \label{eq-cov}
		  |\bE\left[G_\al G_{\al^\prime} \mid \mathcal{F}_P\right]| \ll \left|\bE\left[Q(x_\al) \wbar{Q(x_{\al^\prime})}\right]\right| = \left|r_n(x_\al-x_{\al^\prime})\right|
		\ll \frac{1}{n^{1-\e}},
	      \end{equation}
	      where $r_n$ is again as in (\ref{eq:covariance_kernel_of_polynomial}). Putting everything together, we use Lemma \ref{lemma:almost_independent_gaussians_are_like_gaussians} and the uniform continuity of $f$ to see that
 		\begin{align*}
		  \bE\bigg[\exp\Big(-\sum_{\al=1}^{N_k} f(\what{X}_\al)\Big) \mid \mathcal{F}_P\bigg] &= e^{o(1)} \bE\bigg[\exp\Big(-\sum_{\al=1}^{N_k} f(X_\al + G_\al)\Big) \mid \mathcal{F}_P\bigg] \\ &= e^{o(1)} \exp\Big(-\sum_{\al=1}^{N_k} \ti{f}(X_\al)\Big)
 		\end{align*}
 		where the $o(1)$ term may be random (measurable on $\mathcal{F}_P$, but still, it is of order $o(1)$ with probability approaching $1$ as $n\to\infty$.)
	      Plugging into (\ref{eq:conditional_expectation_for_linear_statistics}) and using (\ref{eq:subsequential_convergence_of_both_processes}) we get that
 		\begin{equation*}
 			\bE\left[e^{-\avg{\eta,f}}\right] = \lim_{k\to\infty} \bE\Big[\exp\Big(-\sum_{\al=1}^{N} \ti{f}(X_\al)\Big)\Big] = \bE\big[ e^{-\langle\eta,\ti{f}\rangle}\big]
 		\end{equation*}
 		as desired.
 	\end{proof}
 	Finally, we extract the Poisson limit from relation (\ref{eq:liggett_property_for_pp}) and with that the proof of Theorem \ref{thm:limiting_distribution_of_minimum}.
 	\begin{proposition}
 		\label{prop:invariance_property_implies_poisson}
 		Suppose that $\eta$ is a point process on $\bR$ such that
		(\ref{eq:liggett_property_for_pp}) holds for
		all non-negative $f\in C_b(\bR)$. Then $\eta$ is a Poisson point process whose intensity measure $\mu$ is a constant multiple of the Lebesgue measure.
	\end{proposition}
	\begin{proof}

	  From (\ref{eq:liggett_property_for_pp}) we know that the law of $\eta$ is an invariant measure for the transformation that
	  consists of adding to each point in the support of
	  $\eta$ an independent mean-zero Gaussian variable of variance $1/2$. Therefore, by \cite[Theorem 4.11]{liggett}, the law of $\eta$ is a mixture of Poisson processes
		of intensities $\mu$ that  satisfy the relation
		\begin{equation}
		  \label{eq-mu}
		  \mu \star \mathcal{N}_{\bR}(0,1/2) = \mu
		\end{equation}
		where $\star$ denotes the convolution of two measures. By the result of Deny \cite[Theorem~3']{deny} (based on Choquet-Deny \cite{choquet_deny}),

		we know that any solution of (\ref{eq-mu}) is of the form
		\[
		\mu = \left(\int_{\bR} e^{-\rho x} d\nu(\rho) \right) dx
		\]
		where $\nu$ is a measure supported on those exponential functions $e_\rho(x):= e^{-\rho x}$ which satisfy $e_\rho\star \mathcal{N}(0,1/2) = e_{\rho}$. A straight forward computation shows that,
		\[
		 1 = e^{\rho x} \int_{\bR} e^{-\rho (x-y)} e^{-y^2} \frac{dy}{\sqrt{\pi}} = e^{\rho^2/4}
		\]
		which in turn implies that $\rho =0$. Thus, we conclude that the measure $\nu$ is a constant multiple of a delta point mass at $\rho =0$. That is, $\mu$ is some multiple of the Lebesgue measure on $\bR$. Since convex combinations of Poisson processes with constant intensity yield a Poisson process of some (constant) intensity, we conclude that $\eta$ is a Poisson process with a constant intensity, which proves the proposition.
  	\end{proof}
 	\begin{proof}[Proof of Theorem \ref{thm:limiting_distribution_of_minimum}]
 		By Proposition \ref{prop:invariance_property_implies_poisson}, we know that $\{\cM_n\}$ converges on a subsequence
		to a Poisson process with intensity that is a multiple of the Lebesgue measure. By Corollary \ref{cor:limit_intensity_of_point_process}, we know that the limit of the intensity of $\cM_n$ is $\sqrt{\pi/3}$ times the Lebesgue measure. Since the limiting process does not depend on the subsequence, we use the tightness once more and conclude that $\cM_n$ converge to a Poisson point process with this given intensity. It remains to apply Lemma \ref{lemma:extremal_process_captures_minimum} and see that
 		\[
 		\lim_{n\to\infty} \bP\left(m_n \ge \frac{\tau}{n}  \right) = \lim_{n\to\infty} \bP\left(\cM_n((-\tau,\tau))\right) = \exp\left(-2\sqrt{\frac{\pi}{3}}\tau \right).
 		\]
 	\end{proof}

 	\section{Real Gaussian coefficients}
 	\label{sec:real_gaussian}
 	In this section we briefly comment on the analogous result to Theorem \ref{thm:limiting_distribution_of_minimum} in the case of real Gaussian coefficients. Let $\{X_j\}$ be an i.i.d. sequence of $\mathcal{N}_{\bR} (0,1)$ random variables and consider the random trigonometric polynomial given by
	\begin{align}
		\label{eq:definition_of_real_polynomial}
 		T_{n}(x) &:= \frac{1}{\sqrt{2n+1}} \sum_{j=-n}^{n} X_j e^{ijx} \\ &= \frac{1}{\sqrt{2n+1}} \left(\sum_{j=-n}^{n} X_j \cos(jx) + i\sum_{j=-n}^{n} X_j \sin(jx) \right) =: R_n(x) + i I_n(x). \nonumber
 	\end{align}
 	As before, we denote by $m_n(T) = \min_{x\in \bT} |T_n(x)|$.
 	\begin{theorem}
 		\label{thm:real_gaussian_case}
 		For any $\tau>0$ we have that
 		\[
 		\lim_{n\to\infty} \bP\left(m_n(T) \ge \frac{\tau}{n}\right) = e^{-\lambda \tau}
 		\]
 		where $\lambda= 2\sqrt{\pi/3}$.
 	\end{theorem}
 	The proof the Theorem \ref{thm:real_gaussian_case} is almost identical to that of Theorem \ref{thm:limiting_distribution_of_minimum}, only the computations are more cumbersome. The reason for this complication is that the polynomial $T_n$ is no longer stationary (as opposed to $P_n$). Still, $T_n$ is a complex-valued Gaussian process on $\bT$ so computations are possible.

 	The correlations of the real and imaginary part of $T=T_n$ are given by
 	\begin{align}
 		\label{eq:correlations_in_the_real_case}
 		&\bE\left[R(x) R(y)\right] = \frac{1}{2n+1} \sum_{j=-n}^{n} \cos(jx)\cos(jy) = \frac{r_n(x-y) + r_n(x+y)}{2} \nonumber \\ & \bE\left[I(x) I(y)\right] = \frac{1}{2n+1} \sum_{j=-n}^{n} \sin(jx)\sin(jy) = \frac{r_n(x-y) - r_n(x+y)}{2} \\ & \bE\left[R(x) I(y)\right] = \frac{1}{2n+1} \sum_{j=-n}^{n} \cos(jx)\sin(jy) = 0, \nonumber
 	\end{align}
 	where $r_n$ is given by (\ref{eq:covariance_kernel_of_polynomial}) and $x,y\in \bT$. Recall the definition of the event that the interval $I_\al$ produced a candidate for a minimal value (\ref{eq-defA}). It follows from (\ref{eq:correlations_in_the_real_case}) that as long as $|x_\al|\in [n^{-1+\e}, \pi - n^{-1+\e}] $ then
 	\[
 	\bE\left[R(x_\al)^2\right] = \frac{1}{2} + o(1) \quad \text{and } \quad \bE\left[I(x_\al)^2\right] = \frac{1}{2} + o(1).
 	\]
 	That is, the random variable $T(x_\al)$ scale as a standard complex Gaussian and similar computations as in Lemma \ref{lemma:probability_of_single_point_to_be_near_minima} can be carried out with no problems. Still, we need to show that with probability tending to 1 the minimum of $T$ does not occur inside $\left\{|x|\leq n^{-1+\e}\right\} \cup \left\{|x-\pi|\leq n^{-1+\e}\right\} $, this we do in Lemma \ref{lemma:minimum_does_not_occur_inside_small_interval_real_coefficients} below.

 	In proving that the points of the extremal process are obtained from well separated intervals (i.e. the analogous result to Lemma \ref{lemma:points_in_the_process_are_separated}), we remark that proving Claim \ref{claim:first_sum_in_seperatin_lemma_is_small} for the real coefficients case is straight forward. To prove Claim \ref{claim:second_sum_in_seperatin_lemma_is_small} for the real case, one can use (\ref{eq:correlations_in_the_real_case}) while noticing that
 	\begin{equation*}
 		r_n(x+y) \ll r_n(2x) + |x-y|r_n^\prime (2x) \ll \frac{1}{n|x|} + |x-y| = o\left( (n|x-y|)^2\right)
 	\end{equation*}
 	provided that $|x|\ge n^{-1+\eps}$ and that $n^{-1-2\eps}\leq |x-y|\le n^{-\eps}$. The proof of Liggett's invariance (Section \ref{sec-liggett}) also translates to the case of real coefficients with no problems.

 	It remains to prove that with high probability the intervals $$\left\{|x|\leq n^{-1+\e}\right\} \cup \left\{|x-\pi|\leq n^{-1+\e}\right\}$$ will not contribute a point to the extermal process that corresponds to $T$. Since $T(x)$ and $T(x+\pi)$ have the same distribution, it suffices to consider the interval centered around $0$.
 	\begin{lemma}
 		\label{lemma:minimum_does_not_occur_inside_small_interval_real_coefficients}
 		Let $T = T_n$ be given as (\ref{eq:definition_of_real_polynomial}), then
 		\[
 		\lim_{n\to\infty }\bP\left(\min_{|x|\leq n^{-1+\e}}|T(x)| \leq \frac{\log n}{n}\right) = 0.
 		\]
 	\end{lemma}
 	\begin{proof}
 		By repeating the exact same argument as in Lemma \ref{lemma:second_derivative_is_small}, we can show that for any $\e>0$
 		\[
 		\bP\left(\norm{T^\prime}_\infty \ge n^{1+\e} \right) \ll \exp\left(-n^{\e}\right).
 		\]
 		We cover the interval $\left\{n^{-1-2\e} \leq |x|\leq n^{-1+\e} \right\}$ by non-overlapping intervals $\left\{J_\ell\right\}_{\ell = 1}^{L}$ so that $|J_\ell| \leq n^{-5/4}$, and let $y_\ell\in J_\ell$ be some point. Notice that $L\leq n^{1/4+\e}$. On the event $\left\{\norm{T^\prime}_\infty \le n^{1+\e}\right\}$, we know by Lagrange's mean value theorem that
 		\[
 		\left|T(x) - T(y_\ell)\right| \leq n^{-5/4} \norm{T^\prime}_{\infty} \leq n^{-1/4+\e}, \quad \text{for all } x\in J_\ell.
 		\]
 		Thus, on the event $\left\{\min_{x\in J_\ell} |T(x)| \leq \frac{\log n}{n} \right\} \cap \left\{ \norm{T^\prime}_\infty \le n^{1+\e} \right\}$, we know that
 		\begin{equation}
 			\label{eq:real_case_bound_on_point_if_minimum_occur}
 			|T(y_\ell)| \leq \frac{\log n}{n} + n^{-5/4} n^{1+\e} \le 2n^{-1/4+\e}
 		\end{equation}
 		for $n$ large enough. Recall from (\ref{eq:correlations_in_the_real_case}) that $T(y_\ell) = (R(y_\ell) ,I(y_\ell))$ is a bivariate mean-zero Gaussian with correlations given by
 		\[
 		\begin{bmatrix}
 		\frac{1}{2n+1} \sum_{j=-n}^{n} \cos^2(jy_\ell) & 0 \\ 0 & \frac{1}{2n+1} \sum_{j=-n}^{n} \sin^2(jy_\ell)
 		\end{bmatrix} =: \begin{bmatrix}
 		1-\Lambda_\ell & 0 \\ 0 & \Lambda_\ell
 		\end{bmatrix}.
 		\]
 		For $|n y_\ell| \leq \pi/2$ we have that
 		\begin{align}
		  \label{eq-Lam}
 			\Lambda_\ell \geq \frac{4y_\ell^2}{\pi^2(2n+1)} \sum_{j=-n}^{n} j^2 \gg n^{-2-4\e} n^2 \gg n^{-4\e}.
 		\end{align}
		Moreover, for $|ny_\ell| \leq \pi/2$, the density of $T(y_\ell)$ is uniformly bounded from above. Using (\ref{eq:real_case_bound_on_point_if_minimum_occur}) and (\ref{eq-Lam}), we see that
 		\begin{align*}
 			\bP\left(\min_{|x|\leq n^{-1+\e}}|T(x)| \leq \frac{\log n}{n}\right) & \leq \bP\left(\min_{|x|\leq n^{-1-2\e}} |T(x)| \leq \frac{\log n}{n},\  \norm{T^\prime}_\infty \le n^{1+\e}\ \right) \\ &\quad + \sum_{\ell =1}^{L} \bP\left(|T(y_\ell)| \leq 2 n^{-1/4+\e}\right) + \bP\left(\norm{T^\prime}_\infty \ge n^{1+\e} \right) \\ & \ll \bP\left(|T(0)|\leq 2n^{-\e}\right) + n^{-1/4+7\e} +e^{-n^\e} = o(1).
 		\end{align*}
 	\end{proof}

	\section{Coefficients with small Gaussian component}
	\label{sec-small_gaussian_component}
	In this section we briefly sketch an extension of Theorem \ref{thm:real_gaussian_case} to a more general choice of coefficients. Recall that a
	real-valued
	random variable $\xi$ satisfies \emph{Cram\'{e}r's condition} if
	\begin{equation}
		\label{eq:cramer_condition}
		\limsup_{t\to\infty} \left|\bE\left[e^{it\xi}\right]\right| < 1.
	\end{equation}
	In particular, (\ref{eq:cramer_condition}) is satisfied when the
	distribution of $\xi$ possesses
	an absolutely continuous density,
	but there are also many singular distributions
	which satisfy (\ref{eq:cramer_condition}). Now,
	let $\delta_n\in (0,1)$ be any sequence such that
	$\delta_n \gg (\log n) /n$ and consider the random polynomial
	(analogous to (\ref{eq:definition_of_real_polynomial}))
	\begin{equation}
	  \label{eq:perturbed}
		T_n(x) = \frac{1}{\sqrt{2n+1}} \sum_{j=-n}^{n} \left(\xi_j + \delta_n X_j\right) e^{ijx}
	\end{equation}
	where $\xi_j$ are i.i.d. random variables whose

	common distribution satisfies (\ref{eq:cramer_condition})
	and where $X_j$ are i.i.d. $\mathcal{N}_{\bR} (0,1)$
	and independent of the $\xi_j$'s. We will
	further assume the normalization conditions
	$\bE[\xi] = 0$ and $\bE[\xi^2] = 1-\delta_n^2$.

To show that Theorem \ref{thm:real_gaussian_case} remains true with
(\ref{eq:perturbed}) replacing (\ref{eq:definition_of_real_polynomial}),
we follow the same steps as in the proof of
Theorem \ref{thm:real_gaussian_case}. First,

	we need to prove results analogous to Section
	\ref{sec-point_process_of_near_minima}
	for the process of near-minima values. This amounts to proving
	a local limit theorem for the vector
	$\left(T_n(x_\alpha),T_n^\prime(x_\alpha)\right)\in \bR^4$
	for $|x_\alpha|\in [n^{-1+\e}, \pi - n^{-1+\e}]$
	(and, for Claim \ref{claim:second_sum_in_seperatin_lemma_is_small},
	the vector $\left(T_n(x_{\alpha}),T_n^\prime(x_{\alpha}),T_n(x_{\beta}),T_n^\prime(x_{\beta})\right)\in \bR^8$ for
	$\alpha\not=\beta$).
	As the correlations of $T_n$ scale exactly like
	(\ref{eq:correlations_in_the_real_case}), this local limit result
	can be extracted from an Edgeworth expansion of order three,
	see for example \cite[Corollary~20.4]{bha_rao}.
	The condition (\ref{eq:cramer_condition}) allows us to apply
	the corresponding Edgeworth expansion.
	To deal with the intervals $\left\{|x|\leq n^{-1+\e}\right\}$ and
	$\left\{|x-\pi|\leq n^{-1+\e}\right\}$
	one obtains the analogue of Lemma
	\ref{lemma:minimum_does_not_occur_inside_small_interval_real_coefficients} by using the basic Berry-Esseen inequality, see
	\cite[Lemma~3.3]{konyagin_schlag}.

	It remains to explain how one can prove
	the results from Section \ref{sec-liggett} for $T_n$ of
	(\ref{eq:perturbed}).
	Write $$X_j = X_j^{\prime} + \frac{1}{\delta_n n}X_j^{\prime\prime}$$
	where $X_j^\prime,X_j^{\prime\prime}$ are independent mean-zero Gaussians and $\text{Var}(X_j^{\prime\prime}) = 1$. Now, the polynomial $T_n$ splits into three polynomials
	\begin{equation*}
		T_n(x) = \ti{T}_n(x) + R_n(x) + \frac{1}{n} G_n(x),
	\end{equation*}
	where $T_n \stackrel{\mbox{\scriptsize{\rm law}}}{=} \ti{T}_n$, $G_n$ is a Gaussian polynomial given as in (\ref{eq:definition_of_real_polynomial}) independent of $\ti{T}_n$ and
	\begin{equation*}
		R_n(x) = \frac{\delta_n\left(\sqrt{1-1/(\delta_n n)^2}-1\right)}{\sqrt{2n+1}}\sum_{j=-n}^{n} X_j^{\prime} e^{ijx}.
	\end{equation*}
	(Note that $R_n$ is \textit{not} independent of $\ti{T}_n$.)
	By virtue of the
	Salem-Zygmund inequality \cite[Chapter~6, Theorem~1]{kahane},
	we get that
	\begin{align*}
		n \cdot \sup_{x\in \bT} |R(x)| &\ll n \sqrt{\log n} \cdot \delta_n\left(\sqrt{1-1/(\delta_n n)^2}-1\right)

		\ll \frac{\sqrt{\log n}}{n \delta_n} = o(1),
	\end{align*}
	with probability tending to 1 as $n\to\infty$.
	Hence, the polynomial $R_n(x)$ does not affect the
	limiting process of near minima values. Clearly, the independent ``kicks" given by the polynomial $n^{-1}G_n(x)$ correspond to the i.i.d. perturbations of the limiting process and we arrive at the setting of Liggett's theorem.

	\subsection*{Acknowledgments} O. Y.  thanks Alon Nishry for introducing him to this problem, encouraging him to work on it and for many fruitful discussions.
	O. Z. thanks Hoi Nguyen for suggesting this problem
	at an AIM meeting in August 2019, and
	Pavel Bleher, Nick Cook and Hoi Nguyen for stimulating
	discussions at that meeting concerning this problem.
	In particular, the realization that a linear approximation suffices
	when working on a net with spacing  $o(1/n)$ came out of those discussions.
	\vspace{1cm}
	\noindent

config: arxiv


TITLE: A problem of regular distribution
QUESTION [2 upvotes]: This is a exercise of Shiryaev's Probability on page 233:

Suppose that the random elements $(X, Y)$ are such that there is a regular distribution $P_x(B)=P(Y\in B\mid X=x)$. Show that if $E|g(X,Y)|<\infty$ then
  $$E[g(X,Y)\mid X=x]=\int g(x,y)P_x(dy) \text{  ($P_x$-a.s.)}$$

PS: I think there is a typo. Should we replace "$P_x$-a.s." by "$P_X$-a.s."?
I firstly tried to prove it for indicator function. For $g(x,y)=I_B(x,y)$, $B\in \mathcal{B}(R^2)$, I should prove for any $A\in \mathcal{B}(R)$,
$$\int_A E[I_B(X,Y)\mid X=x]P_X(dx)=\int_A\int I_B(x,y)P_x(dy)P_X(dx)$$
Then,
$$LHS=\int_{\{X\in A\}}I_B(X,Y)dP$$
For the RHS, let $B_x=\{y\mid (x,y)\in B\}$, I deduce that
$$\begin{align}
RHS &=\int_A\int I_{B_x}(y)Px(dy)P_X(dx)\\
    &=\int_A P_x(B_x)P_X(dx)\\
    &=\int_A P(Y\in B_x\mid X=x)P_X(dx)\\
    &=\int_A E(I_{B_x}(Y)\mid X=x)P_X(dx)\\
    &=\int_A E(I_{B}(x,Y)\mid X=x)P_X(dx)
\end{align}$$
This problem is the origin of my another problem: "An equality about conditional expectation"
In that thread, did has pointed out that $E(I_{B}(x,Y)\mid X=x)$ is meaningless, I agree with him. But what's wrong with my deduction above?
By the way, in the thread mentioned above, StefanHansen have recommend me the book Probability With a View Towards Statistics, Volume II by J. Hoffmann-Jørgensen to study the concept of regular conditional probabilities/distributions, but I can't find a copy of it around me. Can anyone introduce me other materials which has an excellent treatment of regular conditional probabilities/distributions?
Thank you very much!

REPLY [0 votes]: I want to answer my question by myself:
$\forall B\in \mathcal{B}(\mathbb{R}^2)$ and $\forall A\in \mathcal{B}(\mathbb{R})$, let
$$\mathcal{C}=\{B\in \mathcal{B}(\mathbb{R}^2)\mid x\mapsto\int_\mathbb{R}I_B(x,y)P_x(dy)\text{ is $\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable function and }\int_A E[I_B(X,Y)\mid X=x]P_X(dx)=\int_A\int_\mathbb{R}I_B(x,y)P_x(dy)P_X(dx)\}$$
$\forall A_1, A_2\in \mathcal{B}(\mathbb{R})$, then
$$\int_\mathbb{R}I_{A_1\times A_2}(x,y)P_x(dy)=I_{A_1}(x)\int_\mathbb{R}I_{A_2}(y)P_x(dy)=I_{A_1}P_x(A_2)$$
is $\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable function. Then
$$\int_A E[I_B(X,Y)\mid X=x]P_X(dx)=\int_{\{X\in A\}}I_{A_1\times A_2}(X,Y)P(d\omega)=P(\{Y\in A_2\}\cap\{X\in A\cap A_1\})$$
$$\int_A\int_\mathbb{R}I_{A_1\times A_2}(x,y)P_x(dy)P_X(x)=\int_A I_{A_1}(x)P(Y\in A_2\mid X=x)P_X(dx)=P(\{Y\in A_2\}\cap\{X\in A\cap A_1\})$$
So $A_1\times A_2\in\mathcal{C}. $
Let
$$\mathcal{L}=\{A_1\times A_2\mid A_1,A_2\in\mathcal{B}(\mathbb{R})\}$$
then $\mathcal{L}$ is a $\pi$-system and $\mathcal{L}\subset\mathcal{C}$.
$\forall B_1,B_2\in\mathcal{C}$ and $B_1\subset B_2$, then
$$\int_\mathcal{R}I_{B_2\backslash B_1}(x,y)P_x(dy)=\int_\mathcal{R}I_{B_2}(x,y)P_x(dy)-\int_\mathcal{R}I_{B_1}(x,y)P_x(dy)$$
is $\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable function, and
$$\int_A E[I_{B_2\backslash B_1}(X,Y)\mid X=x]P_X(dx)=\int_A E[I_{B_2}(X,Y)\mid X=x]P_X(dx)-\int_A E[I_{B_1}(X,Y)\mid X=x]P_X(dx)=\int_A\int_\mathcal{R}I_{B_2}(x,y)P_x(dy)P_X(dx)-\int_A\int_\mathcal{R}I_{B_1}(x,y)P_x(dy)P_X(dx)=\int_A\int_\mathcal{R}I_{B_2\backslash B_1}(x,y)P_x(dy)P_X(dx)$$
so $B_2\backslash B_1\in\mathcal{C}$.
Let $B_1,B_2,...\in\mathcal{B}(\mathbb{R^2})$ and $B_n\uparrow B$, then by MCT
$$\int_\mathbb{R}I_B(x,y)P_x(dy)=\int_\mathbb{R}\lim_{n\to\infty} I_{B_n}(x,y)P_x(dy)=\lim_{n\to\infty}\int_\mathbb{R}I_{B_n}(x,y)P_x(dy)$$
is $\mathcal{B}(\mathbb{R})/\mathcal{B}(\mathbb{R})$-measurable function and
$$\int_A E[I_B(X,Y)\mid X=x]P_x(dx)=\lim_{n\to\infty}\int_A E[I_{B_n}(X,Y)\mid X=x]P_X(dx)=\lim_{n\to\infty}\int_A\int_\mathbb{R}I_{B_n}(x,y)P_x(dy)P_X(dx)=\int_A\int_\mathbb{R}I_{B}(x,y)P_x(dy)P_X(dx)$$
so $B\in\mathcal{C}$.
Then $\mathcal{C}$ is a $\lambda$-system. So $\mathcal{C}=\mathcal{B}(\mathbb{R}^2)$.
We have prove the problem for all indicator functions, then by the routin method, we can extense to all simple functions, then to all absolutly integrable funciton.


TITLE: Recurrence relations Discrete Math
QUESTION [0 upvotes]: Instructions say: Find the solution to the recurrence relation and prove the correctness of your solution using mathematical induction.
$$
a_{n} = a_{n - 1} + 3\,,\quad n \geq 1\,;\qquad a_{0} = 1
$$
Please help I am completely confused, I've read the textbook section on mathematical induction and I'm so confused please be as descriptive in your explanations as possible.

REPLY [1 votes]: As $\displaystyle a_n-a_{n-1}=3$ which is independent of $n$
it is an Arithmetic Series with common difference $=3$
So, $\displaystyle a_n=a_0+n\cdot3$
For the inductive proof, let $\displaystyle a_n=a_0+n\cdot3$ holds true for $n=m$
i.e., $\displaystyle a_m=a_0+m\cdot3$
$\displaystyle\implies a_{m+1}=a_m+3=\cdots$


TITLE: Shift model - satisfy - Inference
QUESTION [0 upvotes]: Positive random variables $X$ and $Y$ satisfy a scale model with parameters $\delta > 0$, if $\mathbb P(Y\le t) = \mathbb P(\delta X\le t), \forall t > 0$, or equivalently, $G(t) = F\left(\dfrac t\delta\right), \delta > 0, t > 0$.
To Prove:
b) Show that if $X$ and $Y$ satisfy a shift model with the parameter $\Delta$, then $e^X$ and $e^Y$ satisfy a scale model with parameter $e^\Delta$.
c) Suppose a scale model holds for $X$, $Y$. Let $c > 0$ be a constant. Does $X'= X^c, Y' = Y^c$ satisfy a scale model? Does $\log X', \log Y'$satisfy a shift model?

My work:
b) $\mathbb P(e^Y \leq t) = \mathbb P(e^X + e^\Delta \leq t)$
$\mathbb P(Y \leq\log(t)) = \mathbb P(X  \leq \log(t-e^\Delta))$
Is $\Delta$ the same as $\delta$ or is it totally different?
c)
No idea about how to start.

REPLY [2 votes]: b. $$P(e^Y \le t) = P(e^{X+\Delta} \le t) = P(e^\Delta e^X \le t)$$
c. $$P(Y^c \le t) = P(Y \le t^{1/c}) = P(\delta X \le t^{1/c}) = P(\delta^c X^c \le t)$$
$$P(\log(Y^c) \le t) = P(Y \le e^{t/c}) = P(\delta X \le e^{t/c}) = P(\log((\delta X)^c) \le t) = P(\log(\delta^c) + \log(X^c) \le t)$$


TITLE: Setting up triple integral with order $\int_x\int_z\int_y dy dz dx$ and $y=4-x^2-4z^2$ and $ y = 0$
QUESTION [0 upvotes]: I was able to find the x and y bounds, $\int^2_{-2}\int_z\int_0^{4-x^2-4z^2} dy dz dx$, but I'm having trouble with z. Manipulating the y equation with y set to zero I got $4-x^2=4z^2->\sqrt{1-\frac{x^2}{4}}= z^2$ but they have (+/-)$\sqrt{4-x^2}/2$ and I'm not sure how they worked it out.

REPLY [1 votes]: First take the plus and minus square root, then  divide by $2$.$$4z^2=4-x^2\implies 2z = \pm \sqrt{4-x^2}\implies z=\pm \frac{1}{2}\sqrt{4-x^2}$$