TITLE: If $\Vert f(x)-f(y) \Vert\leq k\Vert x-y\Vert,\;\;\forall\;x,y\in\Bbb{R}^n,$ then $x(t,x_0)=x_0,\forall \;t\geq 0,$
QUESTION [3 upvotes]: We consider the following O.D.E
\begin{align}(1)\;\;\;\begin{cases}x'(t)=f(x(t)) & t\geq 0,\\x(0)=x_0\in \Bbb{R}^n&\end{cases}\end{align}
where \begin{align}f: \Bbb{R}^{n}\to \Bbb{R}^{n}\end{align}
Assuming that \begin{align}\Vert f(x)-f(y) \Vert\leq  k\Vert x-y\Vert,\;\;\forall\;x,y\in\Bbb{R}^n.\end{align}
Assuming that $f(x_0)=0.$ Please, do I prove that $x(t,x_0)=x_0,\forall \;t\geq 0,$ where $x(\cdot\,,x_0)$ is the solution of $(1)$. If there are references, I also would appreciate!

REPLY [4 votes]: $f$ being Lipschitz means, any solution to an initial value problem is unique (see e.g. https://en.wikipedia.org/wiki/Initial_value_problem#Existence_and_uniqueness_of_solutions)
Let us define a function $y(t):=x_0$, then $y'(t)=0$, since it is constant. We also have $y(0)=x_0$.
Furthermore $f(y(t))=f(x_0)=0$, hence $y'(t)=0=f(y(t))$. That means $y$ satisfies the differential equation and the initial conditions. Since solutions are unique, we are done, because then $x(t,x_0)=x_0$.


TITLE: Riemann sum of infinite series
QUESTION [2 upvotes]: Let $f$ be a non-negative, bounded and continuous function such that $\int_\mathbb{R} f(x)\, \mathrm{d}x < \infty$.
Does it hold that
$$
\lim_{n \rightarrow \infty} \sum_{j \in \mathbb{Z}} \frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg)
=
\int_\mathbb{R} f(x)\, \mathrm{d}x
$$
Current approach
Due to $f \geq 0$ we can use Fubini's theorem to show that
$$
 \sum_{j \in \mathbb{Z}} \frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg)
=
 \frac{1}{n} \sum_{k = 1}^n \sum_{j \in \mathbb{Z}} f\bigg(j + \frac{k}{n} \bigg).
$$
Thus, the question boils down to whether the Riemann sum converges when the function that is being integrated is an infinite series, i.e. does it hold that
$$
 \frac{1}{n} \sum_{k = 1}^n \sum_{j \in \mathbb{Z}} f\bigg(j + \frac{k}{n} \bigg)
\rightarrow
\int_0^1  \sum_{j \in \mathbb{Z}} f(j + x) \, \mathrm{d}x
$$
as $n \rightarrow \infty$?
My current approach to show this looks as follows:
$$
\bigg\vert
    \frac{1}{n} \sum_{k = 1}^n \sum_{j \in \mathbb{Z}} f\bigg(j + \frac{k}{n} \bigg)
    -
    \int_0^1  \sum_{j \in \mathbb{Z}} f(j + x) \, \mathrm{d}x
\bigg\vert
\leq
 \sum_{j \in \mathbb{Z}} \bigg\vert
    \frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg)
    -
    \int_0^1 f(j + x) \, \mathrm{d}x
\bigg\vert
$$
Here, I can't figure out how to bound the absolute value such that the limit of the series converges to zero.

REPLY [3 votes]: Define $f$ (continuous nonnegative integrable) like this.  Let $f(x) = 0$ on $(-\infty,1]$.
For each $m \in \mathbb N$, on the interval $[m,m+1]$ the function has
$f\left(m+\frac{k}{m}\right) = 1$ for $k=1,2,\dots,m-1$,
$f(m)= f(m+1) = 0$,
$0\le f(x) \le 1$,
$\int_m^{m+1} f = \frac{1}{m^2}$.
For example, on $[4,5]$ it could look like this:
where the three triangles have height $1$ and width $1/24$.
Let
$$
S_n = \sum_{j \in \mathbb{Z}} \frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg) .
$$
We claim $S_n$ does not converge to $\int_{-\infty}^\infty f$.
Choose $j_1 \in \mathbb N$ so that
$$
\sum_{j=j_1}^\infty \frac{1}{j^2} < \frac14
\quad\text{and thus}\quad
\int_{j_1}^\infty f < \frac14 .
$$
Then choose $n_1 \ge 4$ so that, for each $j$ with $1 \le j \le j_1$,
and for each $n \ge n_1$,
$$
\left|\frac{1}{n}\sum_{k=1}^{n} f\left(j+\frac{k}{n}\right) - \int_{j}^{j+1} f\right| < \frac{1}{4j_1}
\quad\text{and thus} \quad
\sum_{j=1}^{j_1} \left|\frac{1}{n}\sum_{k=1}^{n} f\left(j+\frac{k}{n}\right) - \int_{j}^{j+1} f\right| < \frac{1}{4}.
$$
We claim that, for all $n > n_1$, we have
$$
\left|S_n - \int_{-\infty}^\infty f\right| > \frac{1}{4}
$$
[and thus $S_n$ does not converge to $\int_{-\infty}^\infty f$].
To prove this:  Fix $n > n_1$.  Then
$$
S_n = \sum_{j \in \mathbb{Z}} \frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg) = A_n+B_n,\quad\text{where}\\
A_n=\sum_{j=1}^{j_1}\frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg)\\
B_n=\sum_{j=j_1+1}^{\infty}\frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg) .
$$
Compute:
\begin{align}
A_n - \int_1^{j_1+1} f
&=
\sum_{j=1}^{j_1}\left[\frac{1}{n} \sum_{k = 1}^n f\bigg(j + \frac{k}{n} \bigg) - \int_j^{j+1} f\right]
\\ &
\ge -\sum_{j=1}^{j_1} \left|\frac{1}{n}\sum_{k=1}^{n} f\left(j+\frac{k}{n}\right) - \int_{j}^{j+1} f\right|
> -\frac{1}{4} ,
\\
B_n& \ge
\frac{1}{n} \sum_{k = 1}^n f\bigg(n + \frac{k}{n} \bigg)
=\frac{n-1}{n} \ge \frac34 ,
\\
- \int_{j_1+1}^\infty f & \ge - \frac{1}{4}.
\end{align}
Add to get
$$
S_n - \int_1^\infty f > \frac14 .
$$


TITLE: If $A \subseteq X$ and $B \subseteq Y$ then $A \coprod B \subseteq X \coprod Y$.
QUESTION [4 upvotes]: I am beginning to study algebraic topology. Suppose $X$ and $Y$ are topological spaces. In the lecture notes, the topology of $X \coprod Y$ was defined in the following way:
"the open subsets are precisely those of the form $A \coprod B$ with $A \in \tau_X$ and $B \in \tau_Y$".
This tacitly implies that if $A \subseteq X$ and $B \subseteq Y$ then $A \coprod B \subseteq X \coprod Y$. I drew the following diagram for this situation:
$\require{AMScd}$
\begin{CD}
A @>{i_A}>> A \coprod B @<{i_B}<< B\\
@V{\iota_A}VV @V{\phi}VV @V{\iota_B}VV\\
X @>{i_X}>> X \coprod Y @<{i_Y}<< Y
\end{CD}
Here $i_A, \iota_A$ etc. are all the normal inclusion maps, and $\phi$ is defined in the natural way, associating $i_B(b)$ with $i_Y(\iota_B(b))$ for all $b \in B$ and $i_A(a)$ with $i_X(\iota_A(a))$ for all $a \in A$. Clearly $\phi$ is an injection, so I can see how $A \coprod B \cong \phi(A \coprod B) \subseteq X \coprod Y$, but I don't see how $A \coprod B \subseteq X \coprod Y$.
In other places on the internet (e.g. Definition of Disjoint Union Spaces) the topology is defined in a slightly different way that seems not to have this issue.
Does the definition in my lecture notes actually make sense? Am I just being too pedantic about pointless details?

REPLY [5 votes]: The core of your question has not much to do with topology - it is a question about sets. Before you can define a topology on $X \coprod Y$, you have to understand what this object is on the level of sets. Already on that level you can ask (without any reference to topology)

If $A \subset X$ and $B \subset Y$, is it true that $A \coprod B \subset X \coprod Y$?

There are two ways to introduce a new mathematical object:

Direct constructive approach: Simply give an explicit construction of the object.

Conceptual approach: Specify a characteristic property determining the desired object up to unique isomorphism and then prove existence via giving an explicit construction.

In my opinion the conceptual approach should be preferred because it explains the purpose of what is introduced. This approach is of course more abstract - and it also requires to construct the new object.
I think you know that the concept of disjoint union of sets can be introduced as follows:

Given two sets $X,Y$, we say that a triple $(X \coprod Y, i_X, i_Y)$ consisting of a set $X \coprod Y$ and functions $i_X : X \to X \coprod Y, i_Y : Y \to X \coprod Y$ has the disjoint union property if for each pair of functions $f_X : X \to Z, f_Y :Y \to Z$ having the same codomain $Z$ there exists a unique function $f : X \coprod Y \to Z$ such that $f \circ i_X = f_X, f \circ i_Y = f_Y$.

It is easy to verify that all triples having the disjoint union property are isomorphic via a unique "structure preserving" isomorphism. Moreover, the functions $i_X, i_Y$ must necessarily be injective and we have $i_X(X) \cup i_Y(Y) = X \coprod Y, i_X(X) \cap i_Y(Y) = \emptyset$. Therefore they embed $X, Y$ as disjoint subsets into $X \coprod Y$, and frequently authors regard $X,Y$ as genuine subsets of $X \coprod Y$ although this in general not literally true.
It remains to construct a triple with the disjoint union property. A standard approach is to define
$$X \coprod Y =  X \times \{1\} \cup Y \times \{2\},$$
$$i_X(x) = (x,1), i_Y(y) = (y,2).$$
With this definition the answer to your question is "yes".
Note that if $X \cap Y = \emptyset$ we can also take $X \coprod Y =  X \cup Y$; then $i_X, i_Y$ are the genuine inclusions. Then the answer is again "yes". However, we could use a more exotic definition like
$$X \coprod Y =  X \times \{(X,1)\} \cup Y \times \{(Y,2)\}$$
with the obvious $i_X, i_Y$. In that case the answer is "no".
You see that the answer to your question depends on the specific construction of disjoint unions. You can do it such that you get genuine inclusions $A \coprod B \subset X \coprod Y$, and this is a particlar nice feature, but can also do it in another way.
In my opinion the property $A \coprod B \subset X \coprod Y$ is not that important. It suffices to know that $A \coprod B$ embeds canonically into $X \coprod Y$, and that is all we need.
Remark:
There are many similar cases, for example the Cartesian product of sets. It also depends on the specific construction whether we have $A \times B \subset X \times Y$.
Update concerning the canonical embedding of $A \coprod B$ into $X \coprod Y$:
Given functions $u : X \to X', v : Y  \to Y'$, consider $f'_X = i_{X'} \circ u : X \to X' \coprod Y', f'_Y = i_{Y'} \circ v : Y \to X' \coprod Y'$. By the universal property we get a unique function $u \coprod v : X \coprod Y \to X' \coprod Y'$ such that $(u \coprod v) \circ i_X = f'_X =  i_{X'} \circ u, (u \coprod v) \circ i_Y = f'_Y =  i_{Y'} \circ v$.
Apply this to the inclusions $j_A : A \hookrightarrow X,  j_B : B \hookrightarrow Y$.


TITLE: Simple problem about Laplace Equation in a domain
QUESTION [0 upvotes]: Suppose that "$u$", is solution of the problem $$\triangle u=0, r<R $$ $$u_{r}(R,\phi)=f(\phi), 0<\phi\ < 2 \pi$$
Show that $$\int_{0}^{2 \pi}{f(\phi)d\phi}=0$$
I know what this question seem very simple, but I need a help, I recently learned this, thanks

REPLY [0 votes]: You are told to prove that $\int_0^{2\pi} f(\phi)\>d\phi=0$ is a necessary condition for the existence of a solution.
A hint:
Apply Gauss' Theorem (also called Divergence Theorem) to the vector field ${\bf v}:=\nabla u$ defined on the closed disk $B_R:=\{(x,y)\>|\>x^2+y^2\leq R^2\}$.


TITLE: Transition monoid membership for DFAs
QUESTION [9 upvotes]: Given a complete DFA $A=(Q, \Gamma, \delta, F)$, we can define a collection of functions $f_a$ for each $a\in \Gamma$and with $f_a:Q\rightarrow Q$, $f_a(q)=\delta(q, a)$. We can generalize this notion to a word $w=a_1, \cdots, a_m$ and $f_w=f_{a_1}\circ \cdots \circ f_{a_m}$ where $\circ$ denotes function composition. Furthermore we denote $G=\{f_w\mid w\in \Gamma^*\}$ and $G$ is monoid.
[$G$ is usually called transition monoid in the standard textbook, but here I reproduce the definition for clarity.]
The question is, given a function $f:Q\rightarrow Q$, can we decide $f\in G$ (ideally in polynomial time), and if this is the case (i.e., there exists a $w$ such that $f=f_w$), whether $w$ is only polynomially long, or can be exponentially long?
[I guess that indeed such a word could be exponentially long, but I am looking for a simple example.]

REPLY [9 votes]: Decidability
It's decidable.  There are only finitely many possible functions $f:Q \to Q$, so you can model this as a graph reachability problem, with one vertex per function and an edge $g \to h$ if there exists $a \in \Gamma$ such that $h = f_a \circ g$.  Then, testing whether a function $g$ is in $G$ reduces to testing whether $g$ is reachable in the graph from $f_\epsilon$.  You can find the shortest such word using breadth-first time.  The running time might be exponential in $Q$, though.
Length of the word
The shortest such word might be exponentially long.  Here is an example of such a DFA.  Let $p_1,\dots,p_k$ be the first $k$ primes.  Then a state will be of the form $(i,x)$ where $i \in \{1,\dots,k\}$ and $x_i \in \{0,1,\dots,p_i-1\}$.  Define a DFA with unary alphabet $\Gamma=\{0\}$ and the transition function $\delta((i,x),0 = (i,x+1 \bmod p_i)$.  The function $f_0 :Q \to Q$ is given by
$$f_0(i,x) = (i,x+1 \bmod p_i).$$
Now consider the function $g:Q \to Q$ given by
$$g(i,x) = (i,x-1 \bmod p_i).$$
It is possible to use the Chinese remainder theorem to show that $g = f_{0^n}$ where $n=p_1 \times p_2 \times \cdots \times p_k -1$, and that $0^n$ is the shortest such word.  Moreover, $|Q|= p_1 + \dots + p_k$, so $n$ is exponentially large in $Q$.
Consequently, there is no hope for a polynomial-time algorithm that outputs such a word.  This still leaves open the possibility of a polynomial time algorithm for deciding whether $g$ is in $G$, though.


TITLE: Serret-Frenet for Non-unit "speed" space curves
QUESTION [4 upvotes]: The Serret-Frenet equations form a system of linear, often non-autonomous, ordinary differential equations that recover the local tangent, normal and binormal vectors of a unit "speed" space curve from the curve's curvature and torsion.
The equations are scaled for space curves of non unit "speed." (see sec. 5 of the previous Wikipedia article) This means that in order to find the local frame  one must also supply data about the curve parameterization. Often, one wants to find the curve parameterization for known curvature/torsion variables.
Question 1: So, if one were to solve the SF equations
$
\begin{align}
\dfrac{d\mathbf{T}}{ds} &= & \kappa \mathbf{N}, \\
\dfrac{d\mathbf{N}}{ds} &= - \kappa \mathbf{T} & & + \tau \mathbf{B},\\
\dfrac{d\mathbf{B}}{ds} &= & -\tau \mathbf{N},
\end{align}
$
for a given curvature and torsion, then would the associated space curve be identical, up to its "speed," to the one found by solving
$
\frac{d}{ds} \begin{bmatrix}
\mathbf{T}\\
\mathbf{N}\\
\mathbf{B}
\end{bmatrix}
= \|\mathbf{r}'(s)\|
\begin{bmatrix}
0&\kappa&0\\
-\kappa&0&\tau\\
0&-\tau&0
\end{bmatrix}
\begin{bmatrix}
\mathbf{T}\\
\mathbf{N}\\
\mathbf{B}
\end{bmatrix}?$
It seems that this should not be the case since the $||\textbf{r}'(s)||$ prefactor could greatly complicate the system of equations.
Question 2: Assuming that the answer to question 1 is no, then what is the utility of the SF equations? Knowledge of $||\textbf{r}'(s)||$ likely comes from knowledge of $\textbf{r}'(s)$, which should allow one to find the tangent, normal and binormal vectors bypassing SF.

REPLY [6 votes]: The Serret-Frenet formula in the case where the curve is not parametrised by arc-length are
$$\frac{1}{V_r(t)}\frac{d}{dt}\begin{pmatrix} T\\ N\\ B \end{pmatrix} = \begin{pmatrix} 0 & \kappa & 0\\-\kappa & 0 & \tau \\ 0 &-\tau & 0\end{pmatrix} \begin{pmatrix} T\\N \\B \end{pmatrix}$$
where $V_r(t)=\Vert \dot{r}(t) \Vert$ is the speed of your curve. This is true because in the case of non-arc-legnth parametrisation, the definition of torsion and curvature changes, in order to take into account the speed. Namely, we have that for a $C^3$ biregular curve, the torsion is defined by
$$\tau = \frac{1}{V_r} \langle B, \dot{N} \rangle$$
and the curvature $\kappa$ is defined as the norm of the curvature vector
$$K = \frac{1}{V_r} \dot{T}$$
Assuming that all these quantities are well defined (which requires, as I mentionned earlier, that the curve be at least $C^3$ and biregular), then it holds that the "trace", or the image of the curve, is invariant under reparametrisation (actually we even need a lot less: assuming merely that the curve is $C^1$ and regular is enough, for instance).
Serret-Frenet formulas provide a cinematic interpretation of the curvature and the torsion. Locally, these quantities tell you how the curve is going to evolve. More generally, it is possible to show that (under the usual hypothesis $C^3$ and biregular) if two curves parametrised by arc-length possess same curvature and same torsion, then they are "equal" in the sense that there exists a deplacement of $\mathbb{R}^3$ from one onto the other. Therefore curvature and torsion control all the geometry of the curve, and thus solving Serret-Frenet completely determines the curve.
A further refinement of this theorem would be that if you arbitrarily give two (continuous) functions, one for curvature, and one for torsion, then there exists a unique curve parametrised by arc-length whose curvature and torsion are the one prescribed (up to a deplacement of the space).

REPLY [2 votes]: Observe,
$$ \frac{dT}{dt} = \frac{ds}{dt}\frac{dT}{ds} \ \& \ \frac{dN}{dt} = \frac{ds}{dt}\frac{dN}{ds} \ \& \ \frac{dB}{dt} = \frac{ds}{dt}\frac{dB}{ds}$$
but, the Serret-Frenet equations allow us to write the change in $T,N,B$ in terms of $\kappa, \tau$-weighted sum of $T,N,B$. That is, the equations you quote from Wikipedia. Except, note the distinction between $s$ (arclength) and $t$ (an arbitrary parameter). We define the curvature and torsion with respect to the change of $T,N,B$ in terms of arclength. That choice is important as without it the curvature and torsion would not be well-defined. As the equations above reveal, the magnitude of the rates of change in $T,N,B$ are different for different speed parametrizations of a given curve. That said, it is important to allow for non-unit-speed parametrizations in the theory because often the arclength parametrization is uglier than the $t$-parametrization. In fact, it may not even be possible to calculate the arclength explicitly ! So, we need the $s$-equations for logic, and we need the $t$-equations for computational flexibility. Why do we need the Frenet-Serret equations at all? I suppose it's nice to be able to show congruence of curves which defy highschool geometry contructive geometry. We just have to show the curvature and torsion match then we know the curves are the same curve up to some Euclidean motion. Moreover, this is merely the first chapter in a deeply interesting chapter of math known as Cartan's method of moving frames... much more to see here.


TITLE: Does delta distribution remain continuous with respect to quasinorm?
QUESTION [0 upvotes]: I am thinking the accepted answer which is found here:

When viewing $\delta: \mathbb{S} \to \mathbb{R}$ (linear and
  continuous with respect to the usual  semi-norms on the Schwartz-space
  – or similar on the space of test  functions), it makes sense to say
  that $\delta$ is continuous.

and its extension to quasinorms.
Does the $\delta$ distribution with respect to quasinorm remain remain continuous?
A quasinorm is a nonnegative functional $|| \cdot ||$ on a vector space $X$ that satisfies $||x+y||_{X} \leq K( ||x||_{X} + ||y||_{X})$ for some $K \leq 0$ and all $x,y \in X$ and also $||\lambda x||_{X} = |\lambda| ||x||_{X}$ for all scalars $\lambda$.
When $K=1$, then the quasinorm is called a norm.
(Loukas Grafakos, Classical Fourier Analysis, 2009).

REPLY [1 votes]: The answer is probably no, as the OP doesn't specify a (quasi-)norm. Let $\varphi$ be any non-negative test function with $\varphi(0)=1$ and consider $\varphi_n(x)=\varphi(nx)$.
For every $p<\infty$ you have $||\varphi_n||_p\to 0$, but $\delta(\varphi_n)=1$.


TITLE: Non-separable Hilbert space
QUESTION [0 upvotes]: Let $H$ be a Hilbert space.
Suppose that $H$ has an uncountable orthonormal set.
Why is $H$ is non-separable?

REPLY [2 votes]: Let $S$ be an uncountable orhonormal set. For each $v\in S$, consider the open ball $B_{1/2}(v)$. Then $v\neq w\implies B_{1/2}(v)\cap B_{1/2}(w)=\emptyset$. Therefore, if $D$ is a dense subset of $H$, $D$ must have at least one element in each such ball and, since distinct balls are disjoint and there are uncountably many of them, $D$ must be uncountable. So, no countable subset of $H$ can possibly be dense.


TITLE: How do I go about finding some $f(x)$ that satisfies the following first-order DE?
QUESTION [0 upvotes]: $f'(\sin^2x) = \cos^2x + \tan^2x,\quad$ subject to $\quad0<x<1$
Many thanks.

REPLY [1 votes]: $f'(\sin^2x) = \cos^2x + \tan^2x= (1-sin^2x)+(\frac{sin^2x}{1-sin^2x})$
$\implies f(sin^2x)=(sin^2x-\frac{1}{2}sin^4x)+(-sin^2x-ln(1-sin^2x))+C$
I hope I'm correct


TITLE: Find the tangent plane to the graph of $f:\mathbb R^2\to \mathbb R^2$, $f(x,y)=(\sin(x-y),\cos(x+y))$
QUESTION [1 upvotes]: Let $f:\mathbb R^2\to \mathbb R^2$, $f(x,y)=(\sin(x-y),\cos(x+y))$, find the tangent plane to the graph of $f$ in $\mathbb R^4$ at $({\pi\over 4},{\pi\over 4},0,0)$.

What I did:
The equation of the tangent plane is given by $P(x,y)=f(x_0,y_0)+Df(x_0,y_0)\cdot (x-x_0,y-y_0)$ where $Df(x_0,y_0)$ is the jacobian matrix of $f$ at $(x_0,y_0)$.
Computing the partial derivatives:
$\displaystyle{\partial f_1(x,y)\over \partial x}=\cos(x-y)$
$\displaystyle{\partial f_2(x,y)\over \partial x}=-\sin(x+y)$
$\displaystyle{\partial f_1(x,y)\over \partial y}=-\cos(x-y)$
$\displaystyle{\partial f_2(x,y)\over \partial y}=-\sin(x+y)$
then evaluating at $(\dfrac\pi4,\dfrac\pi4)$ :
$\displaystyle{\partial f_1({\pi\over 4},{\pi\over 4})\over \partial x}=1$; $\displaystyle{\partial f_2({\pi\over 4},{\pi\over 4})\over \partial x}=-1$; $\displaystyle{\partial f_1({\pi\over 4},{\pi\over 4})\over \partial y}=-1$; $\displaystyle{\partial f_2({\pi\over 4},{\pi\over 4})\over \partial y}=-1$
Then we have that:
$$Df({\pi\over 4},{\pi\over 4})=\begin{bmatrix} 1 & -1 \\ -1 & -1\\ \end{bmatrix} \text{ and }f({\pi\over 4},{\pi\over 4})=(0,0)$$
Then we have that the tangent plane to the graph of $f$ at $\displaystyle({\pi\over 4},{\pi\over 4},0,0)$ is
$$P(x,y)=(0,0)+ \begin{pmatrix} 1 & -1 \\ -1 & -1\\ \end{pmatrix}\begin{pmatrix} x-{\pi\over 4} \\ y-{\pi\over 4}\\ \end{pmatrix}= (x-y,-x-y+{\pi\over 2})$$
I would really appreciate if you can tell me if this is the correct approach

REPLY [0 votes]: Here is an alternative method that verifies your result.
First of all, your Jacobian matrix is evaluated correctly—
$$
    \frac{\partial \vec F}{\partial(x,y)} =
    \left[\begin{matrix}
        \partial F_x/\partial x & \partial F_x/\partial y \\
        \partial F_y/\partial x & \partial F_y/\partial y
 \end{matrix}\right] =
 \left[\begin{matrix}
      \cos(x-y) & -\cos(x-y) \\
     -\sin(x+y) & -\sin(x+y)
 \end{matrix}\right]
$$
produces
$$
    \left.\frac{\partial \vec F}{\partial(x,y)}\right|_{(\frac\pi4,\frac\pi4)} =
 \left[\begin{matrix}
       1 & -1 \\
     -1 & -1 \\
 \end{matrix}\right].
$$
In $\mathbb R^4$, the tangent plane is the intersection of 2 hyperplanes.  Let $$ax+by+cu+dv+e=0,$$ be one such hyperplane, where we $u$ and $v$ represent the coordinate along the third and fourth dimension.  A normal vector for the hyperplane would be $\vec n=(a,b,c,d)$, and is perpendicular to all lines that lie in the tangent plane.
From the entries of the Jacobian matrix, we know that the following two vectors are parallel to the tangent plane: $\vec s = (1,0,1,-1)$ and $\vec t = (0,1,-1,-1)$.  Therefore,
$$
    \begin{cases}
    \vec n \cdot \vec s = a + c - d = 0, \\
    \vec n \cdot \vec t = b - c - d = 0.
    \end{cases}.
$$
Furthermore, the hyperplanes must pass through $(\dfrac\pi4,\dfrac\pi4,0,0)$:
$$
    \frac\pi4 a + \frac\pi4 b + e = 0 \implies e = -\frac\pi4 (a+b).
$$
Since there are infinitely many pairs of hyperplanes that intersect at the desired plane, we arbitrary let $(a,b)=(1,1)$ for the first hyperplane and let $(a,b)=(1,-1)$ for the second hyperplane.  This produces the equation of two hyperplanes:
$$
    \begin{cases}
        x + y + v - \frac\pi2 &= 0, \\
        x - y - u &= 0.
    \end{cases}
$$
This is identical to your results of
$$
    P(x,y) = (u,v) = (x-y, -x-y+\frac\pi2).
$$


TITLE: Relative error and roots of significant digits
QUESTION [2 upvotes]: a.) The maximal relative error of the volume of a ball is allowed
  within $1\%$. What is the      maximal relative error measuring the
  radius of the ball?
b.) Given the equation $x^2-40x+1=0$, find its roots to five
  significant digits using $\sqrt{399} = 19.975$, correctly rounded to
  five digits.

What I did and know for a is that the volume of a sphere is $V = 4/3\pi * R^3$ and the only term that can have a relative error is the R term. So, I let $dV$ be the relative error of the volume and $dR$ the relative error of the radius then
$dV = dR + dR + dR = 3dR$
$dV/3 = dR$
$1\%/3 = dR$
and I am completely lost in b.

REPLY [1 votes]: Solve for $x$ over the real numbers:
$x^2-40 x+1 = 0$
Solve the quadratic equation by completing the square.
Subtract $1$ from both sides:
$x^2-40 x = -1$
Take one half of the coefficient of $x$ and square it, then add it to both sides.
Add $400$ to both sides:
$x^2-40 x+400 = 399$
Factor the left hand side.
Write the left hand side as a square:
$(x-20)^2 = 399$
Eliminate the exponent on the left hand side.
Take the square root of both sides:
$x-20 = \sqrt{(399)} \text{ or } x-20 = -\sqrt{(399)}$
Look at the first equation:  Solve for $x$.
Add $20$ to both sides:
$x = 20+\sqrt{(399)} \text{ or } x-20 = -\sqrt{(399)}$
Look at the second equation:  Solve for x.
Add $20$ to both sides:
Answer:
$$
  x = 20+\sqrt{(399)} \text{ or } x = 20-\sqrt{(399)}
$$
For question a):
$$
V = \frac{4}{3} \pi R^3 \Rightarrow dV = 4\pi R^2 dR \Rightarrow \frac{dV}{V} = \frac{4\pi R^2dR}{\frac{4}{3} \pi R^3} \le .01 \Rightarrow \frac{3dR}{R} \le .01 \Rightarrow \frac{dR}{R} \le \frac{1}{300} \Rightarrow \max{\left(\frac{dR}{R}\right)} = \frac{1}{300}
$$


TITLE: Solving PDE with Laplace Transform - Non Homogenous ODE
QUESTION [2 upvotes]: Given the following PDE:
$$u_{tt}(x,t)-u_{xx}(x,t)=e^{-t}$$
$$-\infty <x<+ \infty , t>0$$
$$u(x,0)=0,u_t(x,0)=\frac{1}{1+x^2}$$
solve it by using the Laplace transform.
Here's my attempt so far:
Let $$\mathcal L \{u(x,t) \} \equiv U(x,s)\equiv\int_0^\infty u(x,t)e^{-st} \,dt .$$
and the result I've calculated is:
$$U_{xx}(x,s)-s^2 U(x,s) = -\frac{1}{1+x^2}-\frac{1}{s+1}=f(x,s)$$
I've concluded that the relation above is a non-homogenous 2nd order ODE in respect to the variable x. I have tried to solve it but so far it hasn't worked.
Which method do you think would be the best to solve it? Could you provide some guiding steps?

REPLY [1 votes]: So, I solved this the way it asked, by using the Laplace transform.
Continuing from where I left off in the question above, I use the Lagrange method to solve the non homogenous 2nd order linear ODE.
The answer will be in the form:
$$U(x,s)=U_c(x,s)+U_p(x,s)$$
From solving the homogenous part I get the complimentary solution:
$$U_c(x,s)=A(s)e^{sx}+B(s)e^{-sx}$$
But because we need the solution to be finite (since $-\infty<x<\infty$):
$$A(s)=B(s)=0$$
Now, taking the particular solution $U_p(x,s)=C(x,s)e^{sx}+D(x,s)e^{-sx}$, we start by finding the Wronskian and continuing up until we find the derivatives of the factors:
$$C'(x,s)=f(x,s)\frac{e^{-sx}}{2s}$$
$$D'(x,s)=f(x,s)\frac{e^{sx}}{-2s}$$
Then we conclude that $U(x,s)=U_p(x,s)$, while leaving the above in their integral form. Therefore, we can write the following:
$$u(x,t)=\mathcal L^{-1}\{\frac{1}{2s}\int_{-\infty,\infty}(-\frac{1}{1+x'^2}-\frac{1}{1+s})e^{-s|x'-x|}dx'\}$$
By calculating each section of the above we get:
$$\mathcal L^{-1}\{\frac{1}{2s}e^{-s|x'-x|}\}=\frac{1}{2}u_{|x'-x|}(t)$$
and
$$\mathcal L^{-1}\{\frac{1}{2s(s+1)}e^{-s|x'-x|}\}=\frac{1}{2}u_{|x'-x|}(t)-\frac{1}{2}u_{|x'-x|}(t)e^{-t+|x'-x|}$$
where $u_{|x'-x|}(t)$ is the Heaviside function for $c=|x'-x|$. But for $t<|x'-x|=>u_{|x'-x|}(t)=0$, therefore, we limit the solution to where $t\geq|x'-x| => u_{|x'-x|}(t)=1$. That way the new bounds for the integral of $u(x,t)$ are $(x-t,x+t)$.
$$u(x,t)=\frac{1}{2}\int_{(x-t,x+t)}\frac{1}{1+x'^{2}}+1+e^{-t}e^{|x'-x|}dx'$$
and the result is:
$$u(x,t)=\frac{arctan(x+t)-arctan(x-t)}{2}+e^{-t}+t-1$$
which is the same as what @Aleksas Domarkas found by using d'Alambert's formula.
I would say this was solved more easily when tried with a Fourier cosine transform, but since the exercise asked for the Laplace transform I wanted to give the answer just in case anyone else wondered. It's really just a solution to an ODE but I thought the last part was interesting.


TITLE: MAT (Math Advance Tournament) Problem - More Efficient Solutions? - Combinatorics
QUESTION [3 upvotes]: I am a student going into high school next year. Recently, I came across a question from the MAT (Math Advance Tournament) that I spent too long on, but had solved in the end. With the lack of an explained solution on their solutions page, I want to ask the internet for criticism on my solution and possible more efficient solutions for this problem.
The problem is as follows:
Steven has $4$ lit candles and each candle is blown out with a probability $1/2$. After he finishes blowing, he randomly selects a possibly empty subset out of all the candles. The probability his subset has at least one lit candle equals $\frac{m}{n}$ for relatively prime positive integers $m$ and $n$. Find $m + n$.
My Solution:
I first started by writing out all of the possible combinations of blown out and "un-blown out" candles:
a = not blown out
b = blown out
aaaa aaab bbaa bbba bbbb
     aaba baba bbab
     abaa abba babb
     baaa abab abbb
          aabb

A total of $16$ combinations.
I then manually calculated the number of combinations that includes at least one lit candle for every possible size of a subset Steven could have chosen giving me the following:
subset size:  0   1   2   3   4
combinations: 0  32  72  56  15

Then calculating the sum of the combinations: $32 + 72 + 56 + 15 = 175$
And the total: $16 + 4 \cdot 16 + (4 \cdot 3) \cdot 16 + \left(\frac{4 \cdot 3}{2}\right) \cdot 16 + \left(\frac{4 \cdot 3 \cdot 2}{3!}\right) \cdot 16 + 16 = 16 + 64 + 96 + 64 + 16 = 256$
Putting it into a fraction, I get $\frac{175}{256}$. It also happens that $175$ and $256$ are relatively prime, so my answer would be $175+256=431$, which turns out to be the correct answer.

Sorry for the messy work, this is my first problem posted on Stack Exchange.
Thank you in advance!

REPLY [2 votes]: Here's one way to think about this. Since each candle remains lit with probability $1/2$, the set of lit candles is actually a uniformly random subset of the set of $4$ candles. So the problem boils down to:

Let $X_1$ and $X_2$ be uniformly random subsets of $\{1,2,3,4\}$. Find the probability that $X_1 \cap X_2$ is not empty.

Instead, let's find the probability that $X_1 \cap X_2$ is empty, then subtract from $1$. This is mostly straightforward; we have:
$$P(X_1 \cap X_2 = \varnothing) = \frac{1}{16} \sum_{X \subseteq \{1,2,3,4\}} P(X \cap X_2 = \varnothing).$$
Now for a fixed $X \subseteq \{1,2,3,4\}$, the number of subsets of $\{1,2,3,4\}$ disjoint from $X$ is $2^{4-\lvert X \rvert}$, so $P(X \cap X_2 = \varnothing) = 2^{4 - \lvert X \rvert}/2^4 = 2^{-\lvert X \rvert}$. Thus we have
$$P(X_1 \cap X_2 = \varnothing) = \frac{1}{16} \sum_{X \subseteq \{1,2,3,4\}} 2^{-\lvert X \rvert} = \frac{1}{16} \sum_{i=0}^4 \binom{4}{i} 2^{-i}.$$
By the binomial theorem, this becomes
$$P(X_1 \cap X_2 = \varnothing) = \frac{1}{16} \sum_{i=0}^4 \binom{4}{i} 2^{-i} = \frac{1}{16} (1 + 2^{-1})^4 = \frac{1}{16} \cdot \frac{81}{16} = \frac{81}{256}.$$
So the probability that $X_1 \cap X_2$ is not empty is $\frac{175}{256}$. $175$ and $256$ are relatively prime, so the answer is $175+256 = 431$.

Or, even more simply:
$$P(X_1 \cap X_2 = \varnothing) = \prod_{x \in \{1,2,3,4\}} P(x \notin X_1 \cap X_2) = \prod_{x \in \{1,2,3,4\}} (1 - P(x \in X_1) P(x \in X_2)) = \prod_{x \in \{1,2,3,4\}} \frac{3}{4} = \frac{81}{256}.$$


TITLE: Buffon's needle.
QUESTION [1 upvotes]: Suppose we have a
foor made of parallel strips of wood, each the same width,
and we drop a needle onto the
floor. What is the probability that the needle will
lie across a line between two strips?
This is known as the Buffon's needle problem. How can I write a
program that will estimate pi  using the Bufon method?(In Python)

REPLY [0 votes]: If you aren't performing the needle drop experiment physicially, then you need to know $\pi$ in order to do the simulated sampling for drops of the needle (see the other answer by ajotatxe). So you need to already know $\pi$ in order to estimate $\pi$ this way using a computer. That basically defeats the purpose, does it not?


TITLE: Distance between $(-3, 0, 1)$ and the line $(2t, -t, -4t)$
QUESTION [0 upvotes]: Could you please help me with this problem?
How do I calculate the distance between $(-3, 0, 1)$ and the line with the following parametric equation: $(x, y, z) = (2t, -t, -4t)$?
I'm really lost with this one.
Thanks in advance!

REPLY [1 votes]: Let $A=(-3,0,1)$. Note the nearest point on the line is the point  $H$ such that $\overrightarrow{AH}$ is orthogonal to the directing vector $\vec u=(2,-1,-4)$, and the distance is equal to $\;\lVert\overrightarrow{AH}\rVert$.
Hence we have to solve first:
$$\overrightarrow{AH}\cdot\vec u=0\iff 2(2t+3)-(-t)-4(-4t-1)=21t+10=0\iff t=-\frac{10}{21}.$$
The distance is $$\sqrt{\Bigl(\frac{43}{21}\Bigr)^2+\frac{10}{21}\Bigr)^2+\frac{61}{21}\Bigr)^2}=3\sqrt{\frac{10}7}.$$


TITLE: Compute $\cos'(0)$
QUESTION [2 upvotes]: I know that $\cos'(0)$ is $0$, but my work follows:
$$\begin{align}\cos'(0) &= \lim_{\Delta x\to 0}\frac{f(0+\Delta x) - f(0)}{\Delta x}\\
&= \frac{\cos(0+\Delta x) - \cos(0)}{\Delta x}\\
&= \frac{\cos(0) - \cos(0)}{0} = \frac{0}{0}\end{align}$$
Where have I gone wrong? What can I do to show that it equals 0?
(Note: No derivative rules are allowed in my calculus class yet, just difference quotients)

REPLY [3 votes]: Because it is easier to type, I will write $h$ instead of $\Delta x$.  We want to find
$$\lim_{h \to 0} \frac{\cos h -1}{h}.$$
Multiply "top" and "bottom" by $\cos h+1$. We want
$$\lim_{h \to 0} \frac{(\cos h -1)(\cos h+1)}{h(\cos h+1)}.$$
On top we now have $-\cos^2 h +1$, that is, $-\sin^2 h$. So we want
$$\lim_{h \to 0} \frac{-\sin^2 h}{h(\cos h+1)}\qquad\text{that is,}\qquad \lim_{h \to 0} \left(\frac{\sin h}{h}\cdot \frac{-\sin h}{\cos h+1}\right).$$
Finally, let $h \to 0$. We are allowed to use the fact that $\lim_{h\to 0}\frac{\sin h}{h}=1$. And it is clear that $\lim_{h\to 0}\frac{-\sin h}{\cos h +1}=0$.  So our limit is $0$.
Comment: The idea used above is related to the process of "rationalizing the numerator," which you may have seen already, for example in finding the derivative of $\sqrt{x}$ at $x=3$ from the definition of the derivative.


TITLE: What test would I use to analyze an applications' error rate?
QUESTION [2 upvotes]: I am a software engineer with a stats problem: I want to deploy a new version of a web application to a subset of users, measure the error rates, and if there is not an increase of error rates, direct it to all users. (We call this 'Canary Releasing' - including this term so the problem is searchable)
I'm having trouble determining if the error rate has increased. My naive approach was to just calculate the error rate of each version:
error rate A = # of errors in A / # requests in A
error rate B = # of errors in B / # requests in B
However this is a bit of a coin flip: the rates are never the same and it's possible that error rate B is higher but it is not significant.
Is there a better test I can use?

REPLY [0 votes]: A few years later, Netflix open sourced something in this area and decided to use the The Mann-Whitney U test:
https://github.com/spinnaker/kayenta/blob/master/kayenta-judge/src/main/scala/com/netflix/kayenta/judge/classifiers/metric/MannWhitneyClassifier.scala
https://en.wikipedia.org/wiki/Mann%E2%80%93Whitney_U_test


TITLE: Is it true "If $X \subset Y$ then $\Bbb I(Y) \subset \Bbb I(X)$"(proper inclusion)?
QUESTION [2 upvotes]: Is it true "If $X \subset Y$ then $\Bbb I(Y) \subset \Bbb I(X)$; here I am using proper inclusion. Couldn't prove it though. Trying for long time please help. Actually I saw here "https://people.maths.bris.ac.uk/~mp12500/teaching/Lectures5-7.pdf" in proposition 17 $\Rightarrow$ direction.

REPLY [1 votes]: Looks like it's actually Prop 11 (part 2) in the notes you link that shows inclusion.
To see proper inclusion, suppose $I(Y) = I(X)$. Then $V(I(Y)) = V(I(X))$. But as $X$ and $Y$ are algebraic varieties, $V(I(X)) = X$ and $V(I(Y)) = Y$. So you must have had $X = Y$.


TITLE: Prove that if $A,B$ are closed then, $ \exists\;U,V$ open sets such that $U\cap V= \emptyset$
QUESTION [0 upvotes]: Prove that if $A,B$ are closed sets such that $A\cap B\neq \emptyset$, there exists $U,V$ open sets such that $A\subset U$, $B\subset V$, and $U\cap V= \emptyset$.
I am thinking of going by contradiction, that is: $\forall\; U,V$ open sets such that $A\subset U$, $B\subset V$, and $U\cap V\neq  \emptyset$.
Let $ U,V$ open. Then, $\exists\;r_1,r_2$ such that $B(x,r_1)\subset U$ and $B(x,r_2)\subset V.$ From here, I don't know how to get a contradiction. Can anyone help me out?

REPLY [4 votes]: This is not true in general.
Consider $\Bbb{R}$ with the usual topology, and set $A = [0,1]$ and $B = [1,2]$.
First, we have $A\cap B = \{1\}\ne\emptyset$. Any two open subsets $U\subset A$ and $V\subset B$ cannot contain $1$ since they would have to contain a small neighborhood around $1$, which neither $A$ nor $B$ do. Thus $1\notin U$ and $1\notin V$. Therefore, $U\cap V \subset A\cap B=\{1\}$, so $U\cap V =\emptyset$.

Edit:
I just realized that your question asks for $U\cap V$ to be empty, and thus my counterexample is merely an example. Moreover, the question stipulates that $A\subset U$ and $B\subset V$, whereas I assumed the opposite inclusion.
If you assume that $A\subset U$ and $B\subset V$, then $A\cap B \ne\emptyset$ always implies that $U\cap V\ne\emptyset$ as Tsemo Atistide pointed out in the comments.
If you do in fact assume the opposite inclusion, where you are asking for open sets inside the closed sets, then $U\cap V = \emptyset$ can never hold for closed sets with empty interior, as they contain no open sets by definition.
Are you sure the question is correct as written?


TITLE: Expressing $e^{f(x)}$ as the limit as $x$ goes to infinity of a function.
QUESTION [1 upvotes]: Can someone help me prove that the function $(1+f(a)/x)^x$ as $x $ goes to infinity is equal to $e^{f(a)}$??
I do understand that e = $1+1+(1/2)+(1/3!)+\cdots$
and in general that $e^x = 1+x+(x/2)+(x/3!)+\cdots$
Thank u!!!

REPLY [1 votes]: Another way to do it : consider $$y=\left(1+\frac{f(a)}{x}\right)^x$$ Take logarithms $$\log(y)=x \log\left(1+\frac{f(a)}{x}\right)$$ Since $x\to \infty$, $\frac{f(a)}{x}$ is small. Use now the fact that, using equivalents $\log(1+\epsilon)\sim \epsilon$. Replace $\epsilon$ by $\frac{f(a)}{x}$.
So $$\log(y)\sim x \cdot \frac{f(a)}{x}=f(a)$$ Now, using $y=e^{\log(y)}$ then, for large $x$ $$y\sim e^{f(a)}$$
If you prefer to use Taylor series, since, around $\epsilon=0$ $$\log(1+\epsilon)=\epsilon -\frac{\epsilon ^2}{2}+O\left(\epsilon ^3\right)$$ then $$\log(y)=x \left(\frac{f(a)}{x}-\frac{f(a)^2}{2 x^2}+O\left(\frac{1}{x^3}\right)\right)=f(a)-\frac{f(a)^2}{2 x}+O\left(\frac{1}{x^2}\right)$$ Taylor again leads to $$y=e^{f(a)}-\frac{e^{f(a)} f(a)^2}{2 x}+O\left(\frac{1}{x^2}\right)$$ which shows the limit and how it is approached.


TITLE: Homotopy factorization of morphisms of chain complexes
QUESTION [5 upvotes]: This is a sort of follow up to this MO question.
Let $R$ be a ring (eventually with good properties) and $\mathrm{Chains}(R)$ be the category of chain complexes of $R$-modules (eventually bounded). One has the natural endofunctors $\tau_{\leq n}$ and $\tau_{\geq n}$ associated with the standard t-structure on $\mathrm{Chains}(R)$.
What I'm wondering is the following. Is it true that for every morphism $f\colon A\to B$ of chain complexes there exist morphisms $h:A\to C$ and $g:C\to B$ with $\tau_{\leq -1}(h)\colon \tau_{\leq -1}(A)\to \tau_{\leq -1}(C)$ a weak equivalence (in the standard model structure on chain complexes), $\tau_{\geq 0}(g)\colon \tau_{\geq 0}(C)\to \tau_{\geq 0}(B)$ a weak equivalence, and with $f$ homotopy equivalent to $g\circ h$?

REPLY [6 votes]: Just to explain Eric's construction in more elementary terms (and to point out that you can get something stronger than in your question). Define $C$ as follows:
$$C_n=\left\{
\begin{array}{ll}
B_n,&n>0,\\
A_n&n<0.
\end{array}
\right.$$
Moreover, let $d_n\colon C_n\rightarrow C_{n-1}$ be the differential of $B$ (resp. $A$) if $n>1$ (resp. $n<0$). We must still define $C_0$ and its surrounding differentials.
We define $C_0$ as the push-out
$$\begin{array}{ccc}
Z_0(A)&\rightarrow&Z_0(B)\\
\downarrow&&\downarrow\\
A_0&\rightarrow&C_0
\end{array}$$
The upper arrow is the map induced on $0$-cycles by $f$ and the left arrow is the inclusion of $0$-cycles in $0$-chains, which is injective. Therefore the parallel arrow is also injective. Moreover, $A_0/Z_0(A)=C_0/Z_0(B)$.
The differential $d_0\colon C_0\rightarrow C_{-1}=A_{-1}$ is given by applying the universal property of a push-out to $d_0\colon A_0\rightarrow A_{-1}$ and the trivial map $0\colon Z_0(B)\rightarrow A_{-1}$. This shows that $Z_0(C)=Z_0(B)$ and that the images of $d_0\colon C_0\rightarrow C_{-1}=A_{-1}$ and $d_0\colon A_0\rightarrow A_{-1}$ coincide.
The differential $d_1\colon C_1=B_1\rightarrow C_0$ is the composite $B_1\rightarrow Z_0(B)\hookrightarrow C_0$.
We now take $h\colon A\rightarrow C$ to be $f_n\colon A_n\rightarrow B_n$ for $n>0$, the identity for $n<0$, and the bottom map in the push-out square for $n=0$. The map $g\colon C\rightarrow B$ is the identity for $n>0$, $f_n\colon A_n\rightarrow B_n$ for $n<0$, and for $n=0$, the map induced by applying the universal property of a push-out to $f_0\colon A_0\rightarrow B_0$ and $Z_0(B)\hookrightarrow B_0$.
Clearly $f=gh$. Moreover, $\tau_{\geq 0}g$ and $\tau_{\leq -1}h$ are identity maps, not only weak equivalences, by the previous computations. This construction is actually functorial in $f$.


TITLE: Median vs. Mean
QUESTION [3 upvotes]: Problem:
Consider the following model:
$y_i = \mu + \epsilon_i$, $i = 1,...,n$
Let the mean $\mu$ be estimated by minimizing the criterion $\sum|\mu - y_i|$ over $\mu$.
Show that $m = $median($y_1,y_2,...,y_n$) is optimal for this criterion.
Distinguish the case $n$ is odd and $n$ is even.

My idea to approach the problem:
$1)$ Rewrite the criterion such that no absolute signs are needed (with indicator function for example).
$2)$ Determine the first order condition to recognize the median.
$3)$ Order data to distinguish $n$ odd or $n$ even

However, I dont't know how to to the steps of my apporach in a mathematical/statistical way.
Could anyone please help me?

REPLY [2 votes]: Your approach looks good.
$$
\sum |\mu - y_i| = \sum_{y_i< \mu} [\mu - y_i]+ \sum_{y_i> \mu} [y_i - \mu]
$$
now suppose that $y_1< y_2 <\cdots < y_n$, and let $k(\mu)$ the unique $k$ such as
 $y_k <  \mu < y_{k+1}$. First order conditions are enough to find the solution because the function is convex, and:
$$
\sum |\mu - y_i| = \sum_{i=1}^{k(\mu)} [\mu - y_i]+
\sum_{i=k(\mu)+1}^n [y_i - \mu]\\
0 = \frac{d}{d\mu} \sum |\mu - y_i| = k(\mu) - (n-k(\mu)-1)
$$

if $n$ is odd, any $\mu\in [y_{{n-1}/2}, y_{{n+1}/2}]$ minimizes the function, in particular the median $\frac 12[y_{{n-1}/2}+ y_{{n+1}/2}]$
if $n$ is even, the derivative is never $=0$ but changes its sign when $\mu = y_{n/2}$. The median is the unique minimizer.


text:
\begin{document}
\title{Existence of density for solutions of mixed stochastic
equations}

\author{Taras Shalaiko}
\address{Department of Probability Theory, Statistics and Actuarial Mathematics,\br
Taras Shevchenko National University of Kyiv,\br
Volodymyrska 64, Kyiv 01601, Ukraine\br
and\br
Department of Mathematical Economics II\br
Mathematical Institute of the Manhheim University\br
A5, 6\br
D-68131 Mannheim}
\email{tshalaik@mail.uni-mannheim.de}

\author{Georgiy Shevchenko}

\address{ Department of Probability Theory, Statistics and Actuarial Mathematics,\br
Taras Shevchenko National University of Kyiv,\br
Volodymyrska 64, Kyiv 01601, Ukraine}

\email{zhora@univ.kiev.ua}

\subjclass{60H10, 60H07 , 60G22}

\keywords{Mixed stochastic differential equations, existence of denisty, H\"ormander condition, Malliavin differentiability}

\date{\today}

\begin{abstract}
We consider a mixed stochastic differential equation $\d{X_t}=a(t,X_t)\d{t}+b(t,X_t)\d{W_t}+c(t,X_t)\d{B^H_t}$ driven by independent multidimensional Wiener process and fractional Brownian motion. Under H\"ormander type conditions we show that the distribution of $X_t$ possesses a density with respect to the Lebesgue measure.
\end{abstract}

\maketitle
\section{Introduction}
In this paper we study a so-called mixed stochastic differential equation (SDE) in $\R^d$
\begin{equation}
X_t=X_0+\int_0^t a(s,X_s)\d s+\int_0^t b(s,X_s)\d{W_s}+\int_0^t c(s,X_s)\d{B^H_s}
\label{eqIntr}
\end{equation}
driven by a multidimensional standard Wiener process and a multidimensional fractional Brownian motion (fBm) with Hurst parameter $H\in(1/2,1)$ (see next section for precise definitions). Recently such equations gained a lot of attention thanks to their modeling features. There is already a large literature devoted to them;  the few papers we cite here give an extensive overview of existing results. The unique solvability result in the form suitable for our needs is obtained in the paper \cite{Delay}; although the result is formulated there for equations with delay, it is a fortiori valid for usual equations. The paper \cite{Integr} contains useful estimates of the solution and results on its integrability. Finally, we mention the paper \cite{ShSh}, where the Malliavin differentiability of the solution is obtained.

The main aim of this article is to provide conditions under which the solution to \eqref{eqIntr} has a density with respect to the Lebesgue measure. For It\^o SDEs, such issues were addressed by many authors, see \cite{Nualart} and references therein. Existence and regularity of density for SDEs driven by fBm we proved in \cite{BauHai,NourdinSimon,NualSau} under H\"ormander type conditions. The recent paper \cite{Tindel} contains a generalization of these results to equations driven by Gaussian rough paths, in particular, it allows to deduce the existence of a smooth density of the solution to \eqref{eqIntr} with Stratonovich integral with respect to the Wiener process. However, the machinery used in that article is quite sophisticated, and here we use a more direct approach.

The paper is organised as follows. In Section 2 we introduce our notation, describe the main object and briefly discuss Malliavin calculus of variations for fractional Brownian motion. In Section 3, we prove that the distribution of the solution $X_t, t>0$ possesses density w.r.t. Lebesgue measure under a simplified version of the H\"ormander condition. Section 4 contains the result on existence and smoothness of the density under a strong version of the H\"ormander condition. The Appendix contains some technical lemmas and the Norris lemma for a mixed SDE.
\section{Preliminaries}
\subsection{Definitions and notation}

Throughout the paper, $\abs{\cdot}$ will denote the absolute value of a number, the Euclidean norm of a vector, and the operator norm of a matrix. $\la \cdot, \cdot\ra $ stays for the usual scalar product in the Euclidean space. We will use the symbol $C$ to denote a generic constant, whose value is not important and may change from one line to another. We will write a subscript if a constant is relevant or if its value depends on some parameters.

For a matrix $A=(a_{i,j})$ of arbitrary size, we denote by $a_i$ its $i$-th row and by $a_{\cdot,j}$ its $j$-th column.

The classes of continuous and $\theta$-H\"older continuous functions on $[a,b]$ will be denoted respectively by $C[a,b]$ and $C^\theta[a,b]$. For a function $f\colon [a,b]\to\R$ denote by $\norm{f}_{\infty,[a,b]}$ its supremum norm and by
$$\norm{f}_{\theta,[a,b]}=\sup_{a\leq s<t\leq b}\frac{|f(t)-f(s)|}{|t-s|^\theta}$$
its $\theta$-H\"older seminorm. If there is no ambiguity, we will use the notation $\norm{f}_\infty$ and $\norm{f}_\theta$.

Finally, for a function $h\in C(\R^d)$ denote by $\partial_x h = (\frac{\partial}{\partial x_1}h,\dots, \frac{\partial}{\partial x_d}h)$ its gradient and by $\partial^2_{xx} h = (\frac{\partial^2}{\partial x_i\partial x_j}h)_{i,j=1,\dots,d}$ its second derivative matrix.

\subsection{Main equation and assumptions}
For a fixed time horizon $T>0$, let $\{\Omega, \mathcal F, \mathbf{F}=(\mathcal F_t)_{t\in [0,T]}, \mathsf P\}$ be a standard stochastic basis.  Equation \eqref{eqIntr} is driven by two independent sources of randomness:  an $m$-di\-men\-sional $\mathbf{F}$-Wiener process
$\{W_t=(W^1_t,\ldots,W^m_t), t\in [0,T]\}$ and an $l$-dimensional fBm $\{B^H_t=(B^{H,1}_t,\ldots,B^{H,l}_t), t\ge [0,T]\}$ with Hurst index $H\in(1/2,1)$, i.e. a centered Gaussian process having the covariance
$$\E{B^{H,i}_tB^{H,j}_s}=\frac{\delta_{i,j}}{2}(t^{2H}+s^{2H}-|t-s|^{2H}).$$
It is well known that the fBm $B^H$ has a modification with $\gamma$-H\"older continuous path for any $\gamma<H$, in the following we will assume that the process itself is H\"older continuous.

Equation \eqref{eqIntr} is understood as a system of  SDEs on $[0,T]$
\begin{equation*}

\begin{aligned}
X^i_t=X^i_0  +\int_0^ ta_i(s,X_s)\d s+\sum_{j=1}^m\int_0^tb_{i,j}(s,X_s)\d W^j_s +\sum_{k=1}^l\int_0^tc_{i,k}(s,X_s)\d B^{H,k}_s,
\end{aligned}
\end{equation*}
$i=1,\ldots,d$, with a non-random initial condition $X_0\in\R^d$. In this equation, the integral w.r.t. $W$ is understood in a usual It\^o sense, the one w.r.t. $B^H$ is understood in a pathwise sense, as Young integral. More information on its definition and properties can be found in \cite{friz-victoir}.

The coefficients $a_i,b_{i,j},c_{i,k}\colon [0,T]\times\R^d\to\R^d$, $i=1,\ldots,d,j=1,\ldots, m,k=1,\ldots, l$ are assumed to satisfy  the following conditions.
\begin{itemize}
\item[A1] for all $t\in[0,T]$ \ $a(t,\cdot),b(t,\cdot)\in C^1(\R^d)$, $c(t,\cdot)\in C^2(\R^d)$;
\item[A2] for all
$t\in[0,T]$, $x\in \R^d$
$$\abs{a(t,x)} + \abs{b(t,x)} + \abs{c(t,x)}\le C(1+\abs{x});$$
\item[A3] for all
$t\in[0,T]$, $x\in \R^d$ \ $\abs{\partial_x c(t,x)}\le C $;
\item[A4] there exists $\beta>0$ such that for all $t,s\in[0,T]$, $x\in\R^d$
$$
|c(t,x)-c(s,x)|\leq C|t-s|^\beta(1+|x|),\quad |\partial_x c(t,x)-\partial_x c(s,x)|\leq C|t-s|^\beta.
$$
\end{itemize}
The continuous differentiability implies that $a,b,\partial_x c$ are locally Lipschitz continuous. Therefore, by \cite[Theorem 4.1]{Delay}, equation \eqref{eqIntr} has a unique solution which is H\"older continuous of any order $\theta\in(0,1/2)$.

\subsection{Ad hoc Malliavin calculus}
Here we summarize some facts from the Malliavin calculus of variations, see \cite{Nualart} for a deeper exposition. Denote by $S[0,T]$ the set the of step functions of the form $f(t) = \sum_{k=1}^{n} c_k \ind{[a_k,b_k)}(t)$ defined on $[0,T]$.
Let  $L^2_H[0,T]$ denote the separable Hilbert space obtained by completing $S[0,T]$ w.r.t.\ the scalar product
\begin{gather*}
\la f,g\ra_{L^2_H[0,T]}=\int_0^T\int_0^T f(t)g(s)\phi(t,s)\d t\d s,
\end{gather*}
where $\phi(t,s)=H(2H-1)|t-s|^{2H-2}$.

Consider the product space $$\mathfrak H = \left(L^2_H [0,T]\right)^{ l}\times \left(L^2[0,T]\right)^{m}.$$
It is also a separable Hilbert space with a scalar product
$$\la f,g\ra_{\mathfrak H} = \sum_{i=1}^l \la f_i,g_i\ra_{L^2_H[0,T]} + \sum_{i=l+1}^{l+m} \la f_i,g_i\ra_{L^2[0,T]}.$$

The map
$$
\cI \colon (\ind{[0,t_1)},\dots,\ind{[0,t_l)},\ind{[0,s_1)},\dots,\ind{[0,s_m)})\mapsto (B^{H,1}_{t_1},\dots, B^{H,l}_{t_l},W^1_{s_1},\dots, W^m_{s_m})
$$
can be extended by linearity to $S[0,T]^{l+m}$. It appears that for $f,g\in S[0,T]^{l+m}$
$$\E{\la \cI(f),\cI(g)\ra} = \la f,g\ra_{\mathfrak H},$$
so $\cI$ can be extended to an isometry between $\mathfrak H$ and a subspace of $L^2(\Omega;\R^{m+l})$.

For $\xi = F(\cI(f_1),\dots ,\cI(f_n))$, where $f_1,\dots,f_n\in \mathfrak H$ and $f_i=(f_{i,1},\ldots,f_{i,m+l})$, $i=1,\ldots,n$,  $F\colon \R^{n(m+l)} \to \R$ is a continuously differentiable finitely supported function, define the

Malliavin derivative $\mathsf D \xi$ as an element of $\mathfrak H$, whose  $j$-th coordinate equal to $$ \sum_{i=1}^{n} \partial_{(i-1)(l+m)+j} F (\cI(f_1),\dots, \cI(f_n)) f_{i,j}, j=1,\ldots,l+m.$$
Denote for $p\ge 1$ by $\mathbb D^{1,p}$ the closure of the space of smooth cylindrical random variables with respect to the norm
$$\norm{\xi}^p_{\mathbb D^{1,p}}=\E{|\xi|^{p}+\norm{\md{\xi}}_{\mathfrak H}}^{1/p}.$$
$\md$ is closable in this space and its closure will be denoted likewise. Finally, the Malliavin derivative is a (possibly, generalized) function from $[0,T]$ to $\R^{l+m}$, so we can introduce the notation
$$
\md \xi = \set{\md_t\xi = \big(\md^{H,1}_t \xi,\dots, \md^{H,l}_t\xi,\md^{W,1}_t\xi,\dots, \md^{W,m}_t\xi\big),\  t\in[0,T]}.$$
We say that $\xi\in \mathbb{D}^{1,p}_{\mathit{loc}}$ if exists a sequence $\{\xi_n(\omega),\Omega_n\}_{n\geq 1}$ such that $\Omega_n\subset \Omega_{n+1}$ for $n\geq 1$, $P\left(\Omega\setminus \left(\bigcup_{n\ge 1}\Omega_n\right)\right)=0$;  $\xi_n\in \mathbb D^{1,p}$ and
$\xi|_{\Omega_n} = \xi_n|_{\Omega_n}$ for all $n\geq 1$.

For the reader convenience we state here the theorem concerning the Mallivian differentiability of the solution to \eqref{eqIntr} in the case of SDE with non-homogeneous coefficients. The proof  is similar to that of \cite[Theorem 2]{ShSh}
\begin{theorem}\label{MalliavinDiff}
Suppose that coefficients $a,b,c$ of \eqref{eqIntr} satisfy the assumptions
\begin{itemize}
\item[B1] for all $t\in[0,T]$ \ $a(t,\cdot),b(t,\cdot)\in C^1(\R^d)$, $c(t,\cdot)\in C^2(\R^d)$;
\item[B2] $a,b,\partial_x a, \partial_x b, \partial_x c, \partial^2_{xx}c$ are bounded;
\item[B3] there exists $\beta>0$ such that for all $t,s\in[0,T]$, $x\in\R^d$
$$
|c(t,x)-c(s,x)|\leq C|t-s|^\beta(1+|x|),\quad |\partial_x c(t,x)-\partial_x c(s,x)|\leq C|t-s|^\beta.
$$
\end{itemize}
Then $X_t\in\bigcap_{p\geq 1} \mathbb D^{1,p}$.
\end{theorem}

\section{Existence of density under simplified H\"ormander condition}
In this section we prove that a solution to \eqref{eqIntr} possesses density of a distribution under a quite strong condition, which we call a simplified H\"ormander condition. More precisely, we will assume in this section that the coefficients of \eqref{eqIntr} satisfy
\begin{equation}\label{simpHormander}
\mathsf{span}\{c_{\cdot,k}(0,X_0),b_{\cdot,j}(0,X_0)\mid  1\leq k\leq l,1\leq j\leq m\,\}=
\R^d.
\end{equation}

The first step to establish the existence of density is to show the (local) Malliavin differentiability of the solution to \eqref{eqIntr}.
\begin{theorem}
If the coefficients of \eqref{eqIntr} satisfy the assumptions \emph{A1--A4}, then $X_t\in \bigcap_{p\geq 1} \mathbb{D}^{1,p}_{loc} $.
\end{theorem}
\begin{proof}
Define $\Omega_n=\{\omega: \norm{X_\cdot	(\omega)}_{\infty,[0,t]}<n\}$, $n\ge 1$. Obviously,  $\Omega_n\subset \Omega_{n+1}, n\geq 1$ and, since $\norm{X_\cdot	(\omega)}_{\infty,[0,t]}<\infty$ a.s., $\bigcup_{n\geq 1 }\Omega_n=\Omega$. Consider a smooth function $\psi=\psi(x),\,x\in \R$ such that
\begin{itemize}
\item for all $x\in \R$ \ $0\leq \psi(x)\leq 1$;
\item $\psi(x)=1,x\in[-1,1]$;
\item $\psi(x)=0,\, x\notin[-2,2]$.
\end{itemize}
For $n\ge 1$, put  $\Psi_n=\Psi_n(x_1,\ldots,x_d)=(\int_0^{x_1}\psi(y/n)\d y,\ldots,\int_0^{x_d}\psi(y/n)\d y)$, define $d^{(n)}(s,x) = d(t,\Psi_n(x))$, $d\in \set{a,b,c}$, and let $X^{(n)}$
solve
\begin{equation*}

X^{(n)}_t=X_0  +\int_0^ ta^{(n)}(s,X^{(n)}_s)\d s+\int_0^tb^{(n)}(s,X^{(n)}_s)\d W_s +\int_0^t c^{(n)}(s,X^{(n)}_s)\d B^{H}_s.

\end{equation*}
Since the functions $a_n, b_n, c_n$ satisfy assumptions B1--B3,  in view of Theorem \ref{MalliavinDiff}, $X^{(n)}_t\in \bigcap_{p\geq 1} \mathbb D^{1,p}$. It is not hard to see that that $X^{(n)}_t(\omega)=X_t(\omega)$ for $\omega \in \Omega_n$, which concludes the proof.

\end{proof}
Now we are to prove the main result of this section.
\begin{theorem}
\label{thTOYHorm}
Suppose that the coefficients of \eqref{eqIntr} satisfy assumptions \emph{A1--A4} and the simplified H\"ormander condition \eqref{simpHormander}. Then for all $t>0$ the law of $X_t$ is absolutely continuous with respect to the Lebesgue measure in $\R^d$.
\end{theorem}
\begin{proof}
By the classical condition for existence of density (see e.g. \cite[Theorem 2.1.2]{Nualart}) and thanks to the previous theorem,  it is enough to verify the non-degeneracy of the Mallivain covariation matrix $M(t)=(M_{i,j}(t))_{i,j=1,\ldots,d}$ with $M_{i,j}(t)=\la\md X^i_t,\md X^i_t\ra_{\mathcal H}$. Define the matrix-valued process $J_{t,0}=(J_{t,0}(i,j))_{i,j=1,\ldots,d}$ as the solution to
\begin{equation}
\label{eqJacobian}
\begin{gathered}
J_{t,0}(i,j)=\delta_{i,j}+\sum_{r=1}^d\bigg[\int_0^t \frac{\partial a_i}{\partial x_r}(s,X_s) J_{s,0}(r,j)\d s \\ +\sum_{k=1}^m \int_0^t \frac{\partial b_{i,k}}{\partial x_r}(s,X_s)J_{s,0}(r,j)\d W^{k}_s
+ \sum_{q=1}^l \int_0^t \frac{\partial c_{i,q}}{\partial x_r}(s,X_s)J_{s,0}(r,j) \d B^{H,q}_s\bigg].
\end{gathered}
\end{equation}
where $\delta_{i,j} = \ind{i=j}$ is the Kronecker delta. The system above is linear, hence, possesses a unique solution. In view of  Lemma \ref{lemma1}, $J_{t,0}$ is non-degenerate; denoting $J_{t,s}=J_{t,0}J^{-1}_{s,0}$ and applying Lemma \ref{lemma2} one can write
\begin{equation*}

\begin{gathered}
M(t)=\sum_{k=1}^m \int_0^t (J_{\newcommand{\Rho}{\mathrm{P}}t,s}b_{\cdot,k}(s,X_s))(J_{t,s}b_{\cdot,k}(s,X_s))'\d s\\
+
\sum_{q=1}^l \int_0^t\int_0^t \varphi_H(s,u)(J_{t,s}c_{\cdot,q}(s,X_s))(J_{t,u}c_{\cdot,q}(u,X_u))'\d s\d u=J_{t,0}C_t J'_{t,0},
\end{gathered}
\end{equation*}
where
\begin{equation*}
\begin{gathered}
 C_t=\sum_{k=1}^m \int_0^t (J_{s\newcommand{\Rho}{\mathrm{P}},0}^{-1}b_{\cdot,k}(s,X_s))(J_{s,0}^{-1}b_{\cdot,k}(s,X_s))'\d s \\
+ \sum_{q=1}^l \int_0^t\int_0^t \varphi_H(s,u)(J_{s,0}^{-1}c_{\cdot,q}(s,X_s))(J^{-1}_{u,0}c_{\cdot,q}(u,X_u))'\d s\d u.
\end{gathered}
\end{equation*}
Again, due to the invertibility of $J_{t,0}$,  $M_t$ is invertible if and only if so is $C_t$. Assuming the contrary, there exists a non-zero vector $v\in \R^d$ such that $v'C_tv=0$. Write $$v'C_tv=\sum_{k=1}^m \norm{\la J_{\cdot,0}b_{\cdot,k}(\cdot,X_\cdot),v \ra}^2_{L^2[0,t]}+\sum_{q=1}^l \norm{\la J_{\cdot,0}c_{\cdot,q}(\cdot,X_\cdot),v \ra}^2_{L^2_H[0,t]}.$$
Since the functions
\begin{gather*}
s\mapsto \la J^{-1}_{s,0}b_{\cdot,k}(s,X_s,v \ra,\, k=1,\ldots,m,\\
s\mapsto \la J^{-1}_{s,0}c_{\cdot,q}(s,X_s),v\ra,\, q=1,\ldots,l
\end{gather*}
are continuous, they must be equal zero for all $s\in [0,t]$. For  $s=0$ we get
\begin{align*}
&\sum_{i=1}^d b_{i,k}(0,X_0)v_i=0, k=1,\ldots,m;\\
&\sum_{i=1}^d c_{i,q}(0,X_0)v_i=0, q=1,\ldots,l.
\end{align*}
This, however, contradicts the assumption \eqref{simpHormander}. Consequently, $M_t$ is invertible, as required.
\end{proof}

\section{Existence of density under strong H\"ormander condition}
In this section we consider a homogeneous version of \eqref{eqIntr}:
\begin{gather}
X_t=X_0+\int_0^t a(X_s)\d s+\int_0^t b(X_s)\d{W_s}+\int_0^t c(X_s)\d{B^H_s}.\label{eqHom}
\end{gather}
In this section we assume that Hurst index $H\in (1/2,2/3)$, and some $\theta\in((H-1/2)/(3-4H),1/2)$ is fixed. The role of the restriction $\theta>(H-1/2)/(3-4H)$ will become clear in the proof of the Norris lemma for \eqref{eqHom} (Lemma~\ref{Chuck}). Now we just remark that the expression $(H-1/2)/(3-4H)$ is increasing for $H\in(1/2,3/4)$ and is equal to $1/2$ for $H=2/3$, so the upper bound $H<2/3$ arises naturally.

We impose the following condition on the coefficients of (\ref{eqHom}):
\begin{itemize}
\item[C1] $a,b,c\in C^\infty_b(\R^d)$ with all derivatives bounded.
\end{itemize}
Under this assumption the solution is infinitely differentiable in the Malliavin sense: $X_t\in \bigcap_{k,p=1}^\infty \mathbb D^{k,p}=\mathbb D^{\infty}$, which can be shown similarly to its differentiability under B1--B3.

The aim of this section is to investigate the existence of a density and properties of this density of a distribution of $X_t$ under the strong H\"ormander condition, which reads as follows.

Set $V_0=a$, $V_j(\cdot)=b_{\cdot, j}(\cdot)$ for $j=1,\ldots,m$ and
$V_{j+m}(\cdot)=c_{\cdot,j}(\cdot)$ for $j=1,\ldots,l$.
Using the Lie bracket $[\cdot,\cdot]$, define the set $$\Upsilon_k=\{[V_{i_1},\ldots,[V_{i_{k-1}},V_{i_k}]\ldots], (i_1,\ldots,i_k)\in\{1,\ldots,d\}^k\}.$$
It is said that the vector field $\Upsilon_0=\{V_{j}\}_{j=1,\ldots,m+l}$ satisfies the {H\"ormander condition} at the point $X_0$, if for some positive integer $n_0$ one has
\begin{equation}\label{Hormander}
 \mathsf{span}\set{ V(X_0), V\in \bigcup_{k=1}^{n_0}  \Upsilon_k}=\R^d.
\end{equation}
The main result of this section  is  the following theorem.
\begin{theorem}
 Assume that coefficients of (\ref{eqHom}) satisfy assumption \emph{C1} and the H\"ormander condition \eqref{Hormander}. Then the law of $X_t$ for all $t>0$ possesses  a smooth density with respect to the Lebesgue measure in $\R^d.$
\end{theorem}

\begin{proof}
Using the usual condition for existence of a smooth density (see e.g. \cite[Theorem 2.1.4]{Nualart}) and taking into account that all moments of the Jacobian $J_{t,s}$ are finite, it is enough to show that the matrix inverse to the reduced Malliavin covariance matrix of $X_t$ possesses all moments.

Recall from Theorem \ref{thTOYHorm} that the reduced Malliavin covariance matrix of the solution to (\ref{eqHom}) can be written as
\begin{gather*}
 C(t)=\sum_{k=1}^m \int_0^t(J_{s,0}^{-1} b_{\cdot,k}(X_s))(J_{s,0}^{-1}b_{\cdot,k}(X_s))'\d s\\
+ \sum_{q=1}^l \int_0^t\int_0^t \varphi_H(s,u)(J_{s,0}^{-1}c_{\cdot,q}(X_s))(J_{u,0}^{-1}c_{\cdot,q}(X_u))'\d s\d u.
\end{gather*}
To simplify the notation, we assume from now that $t=1$.
We are to prove that $\E{|\det C_t|^{-p}}<\infty$ for all $p\geq 1$. Due to \cite[Lemma 2.3.1]{Nualart} it suffices to prove that the entries of $C_t$ possess all moments and for any $p\geq 2$ there exists $C_p$ such that for all $\varepsilon>0$ it holds
 $$\sup_{\norm{v}=1}\pr{\la v, C_1 v\ra\leq \varepsilon}\leq C_p \varepsilon^p.$$
Write
\begin{gather*}
\la v,C_1 v\ra=\sum_{k=1}^m \norm{\la J^{-1}_{\cdot,0}b_{\cdot,k}(X_\cdot),v \ra}^2_{L^2[0,1]}+\sum_{q=1}^l
\norm{\la J^{-1}_{\cdot,0}c_{\cdot,q}(X_\cdot),v\ra}_{L^2_H[0,1]}.
\end{gather*}
It is well known that $\norm{f}_{L^2_H[0,1]}\le \norm{f}_{L^2[0,1]}$. Therefore, $$\la v,C_1 v\ra\geq C\sum_{k=1}^{m+l}\norm{G_k}_{L^2_H[0,1]}, \text{ where \ }G_k=\la J^{-1}_{\cdot,0} V_{k}(X_\cdot),v \ra.$$
Applying  \cite[Lemma 4.4]{BauHai} we get that
$$ \la v,C_1 v\ra\ge C\sum_{k=1}^{m+l}\frac{\norm{G_k}^{2(3+1/\theta)}_\infty}{\norm{G_k}^{2(2+1/\theta)}_\theta}$$
for $\theta>H-1/2$.
Thus, $$\pr{\la v,C_1 v\ra \leq \varepsilon}\leq \pr{C\sum_{k=1}^{m+l}\frac{\norm{G_k}^{2(3+1/\theta)}_\infty}{\norm{G_k}^{2(2+1/\theta)}_\theta}\leq \varepsilon}.$$
From \cite[Lemma 4.5]{BauHai} and Theorem \ref{Chuck} we obtain the following estimate
$$\pr{C\sum_{k=1}^{m+l}\frac{\norm{G_k}^{2(3+1/\theta)}_\infty}{\norm{G_k}^{2(2+1/\theta)}_\theta}\leq \varepsilon}\leq C\varepsilon^p+\min_{k=1,\ldots,m+l}\pr{\norm{\la v, J_{\cdot,0}V_k(X_\cdot)}_\infty\leq \varepsilon^\alpha}.$$
Now let $V$ be a bounded vector field with bounded derivatives of all order. The chain rule implies
\begin{gather*}
 J_{t,0}^{-1}V(X_t)=V(X_0)+\int_0^t J_{s,0}^{-1}([V_0,V]+\frac 1 2 \sum_{k=1}^{m+l}[V_k,[V_k,V]])(X_s)\d s
 \\
 + \sum_{k=1}^m \int_0^t J_{s,0}^{-1}[V_k,V](X_s)\d{W_s}+\sum_{k=m+1}^{l+m}\int_0^t J_{s,0}^{-1}[V_k,V](X_s)\d{B^H_s}.
\end{gather*}
Thus, applying Theorem \ref{Chuck} once more, we obtain
\begin{gather*}
 \pr{\norm{\la v,J_{\cdot,0}V(X_\cdot)\ra}_\infty<\varepsilon}\leq C \varepsilon^p+
 \min_{k=1,\ldots,m+l}\pr{\norm{\la v,J^{-1}_{\cdot,0}[V_k,V](X_\cdot) \ra}_\infty\leq \varepsilon^\alpha}.
\end{gather*}
Let  $n_0$ be the integer from the H\"ormander condition. Iterating our consideration above, we obtain
\begin{gather*}
 \pr{\la v, C_1 v\ra\leq \varepsilon}\leq C\varepsilon^p+
 \min_{V\in\bigcup_{k=1}^{n_0} \Upsilon_k} \pr{\norm{\la v,J^{-1}_{\cdot,0}V(X_\cdot)\ra}_\infty\leq \varepsilon^\alpha}
\end{gather*}
for all $\varepsilon$ small enough. Since $\{V(x_0), V\in \bigcup_{k=1}^{n_0} \Upsilon_k\}$ spans $\R^d$, there exists $v$ such
that $\la v, V(x_0)\ra\neq 0$. Hence, there exists $\varepsilon_0(p)$ such that for all $\varepsilon<\varepsilon_0(p)$ the second term vanishes. As a result,
$$\pr{\la v, C_1 v\ra\leq \varepsilon}\leq C_p \varepsilon^p$$
 for all $\varepsilon\leq \varepsilon_0(p)$, as required.
\end{proof}

\appendix
\section{Technical lemmas}
The following two lemmas concern the Jacobian of the flow generated by the solution $X$ to equation \eqref{eqIntr}. These are quite standard facts, so we just sketch the proofs.
\begin{lemma}
\label{lemma1}
Under assumptions \emph{A1--A4} the matrix valued process $J_{t,0}=(J_{t,0}(i,j))_{i,j=1,\ldots,d}$ given by \eqref{eqJacobian} has an inverse $Z_{t,0}=(Z_{t,0}(i,j))_{i,j=1,\ldots,d}$ for all $t>0$. Moreover, $\set{Z_{t,0},t\ge 0}$ satisfies the following system of equations
\begin{equation}\label{eqInv}
\begin{gathered}
Z_{t,0}(i,j)=\delta(i,j)-\sum_{r=1}^d\bigg[\int_0^t \frac{\partial a_r}{x_j}(s,X_s) Z_{s,0}(r,j)\d s\\
{} - \sum_{k=1}^m \int_0^t \frac{\partial b_{r,k}}{\partial x_j}(s,X_s)Z_{s,0}(i,r)\d W^{k}_s-
\sum_{q=1}^l \int_0^t  \frac{\partial c_{r,q}}{\partial x_j}(s,X_s)Z_{s,0}(i,r) \d B^{H,q}_s\\ + \frac12\sum_{u=1}^d\sum_{v=1}^d \int_0^t \frac{\partial b_{r,v}}{\partial x_u}(s,X_s)\frac{\partial b_{u,v}}{\partial x_r}(s,X_s)Z_{s,0}(i,r)\d s\bigg].
\end{gathered}
\end{equation}
\end{lemma}
\begin{proof}
The equation \eqref{eqInv} is linear, thus possesses a unique solution $Z_{t,0}$. So we need to verify that $Z_{t,0}J_{t,0}=J_{t,0}Z_{t,0}=I_d$, the identity matrix.
The equality clearly holds for $t=0$. To show it for $t>0$, it is enough to show that the differentials of $Z_{t,0}J_{t,0}$ and of $J_{t,0}Z_{t,0}$ vanish.
But this can be routinely checked using the It\^o formula.

\end{proof}
Denote for $t\geq s$ $J_{t,s}=J_{t,0}J^{-1}_{s,0}$.
\begin{lemma}
\label{lemma2}
Under assumptions \emph{A1--A4}, the Malliavin derivatives of the solution to \eqref{eqIntr} are given by
\begin{gather}
\md^{W,k}_s X_t=J_{t,s} b_{\cdot,k}(s,X_s)\ind{s\le t},\,k=1,\ldots,m, \label{eqW}\\
\md^{H,q}_s X_t= J_{t,s} c_{\cdot,q}(s,X_s)\ind{s\le t},\, q=1,\ldots,l.
\end{gather}
\end{lemma}
\begin{proof} The argument is exactly the same for both equations, so we  prove only \eqref{eqW}. Evidently, $\md_s^{W,k} X_t = 0$ for $s>t$, so suppose that $s\le t$. Due to the closedness of the derivative,
we can freely differentiate \eqref{eqIntr} as if the integrals were finite sums, in particular, using the chain rule, we can write for $i=1,\dots,d$
\begin{gather*}
\md_s^{W,k} \int_0^t a_i(u,X_u)\d u =  \int_0^t \md_s^{W,k} a_i(u,X_u)\d u =
\sum_{r=1}^{d}\int_s^t \frac{\partial}{\partial x_r}a_i(u,X_u)\md_s^{W,k}X^r_u \d u
\end{gather*}
and similarly
\begin{gather*}
\md_s^{W,k} \int_0^t c_{i,q}(u,X_u)\d B^{H,q}_u =  \sum_{r=1}^{d}\int_s^t \frac{\partial}{\partial x_r}c_{i,q}(u,X_u)\md_s^{W,k}X^r_u \d B^{H,q}_s, q=1,\dots,l,\\
\md_s^{W,k} \int_0^t b_{i,j}(u,X_u)\d W^{j}_u =  \sum_{r=1}^{d}\int_s^t \frac{\partial}{\partial x_r}b_{i,j}(u,X_u)\md_s^{W,k}X^r_u \d W^j_s, j=1,\dots,m, j\neq k.
\end{gather*}
To differentiate the integral w.r.t. $W^k$, approximate it by an integral sum and note that we will have an extra term corresponding to the derivative of the increment of $W^k$ on the interval containing $s$. Passing to the limit, we get
$$\md_s^{W,k} \int_0^t b_{i,k}(u,X_u)\d W^{k}_u =  b_{i,k}(s,X_s) + \sum_{r=1}^{d}\int_s^t \frac{\partial}{\partial x_r}b_{i,k}(u,X_u)\md_s^{W,k}X^r_u \d W^k_s.$$
Therefore, we have for $s\le t$ the following linear equation on $\md_s^{W,k}X_u$:
\begin{gather*}
\md^{W,k}_s X_t = b_{\cdot,k}(s,X_s) + \sum_{r=1}^{d}\bigg[
\int_s^t \frac{\partial}{\partial x_r}a(u,X_u)\md_s^{W,k}X^r_u \d u
\\ + \sum_{q=1}^l \int_s^t \frac{\partial}{\partial x_r}c_{\cdot,q}(u,X_u)\md_s^{W,k}X^r_u \d B^{H,q}_s
+ \sum_{j=1}^m
\int_s^t \frac{\partial}{\partial x_r}b_{\cdot,j}(u,X_u)\md_s^{W,k}X^r_u \d W^j_s
\bigg].
\end{gather*}
On the other hand, from \eqref{eqJacobian} we can write
\begin{gather*}
J_{t,0}=J_{s,0}+\sum_{r=1}^d\bigg[\int_s^t \frac{\partial a_i}{\partial x_r}(u,X_u) J_{u,0}\d u \\ +\sum_{k=1}^m \int_s^t \frac{\partial b_{\cdot,k}}{\partial x_r}(s,X_s)J_{u,0}\d W^{k}_u
+ \sum_{q=1}^l \int_s^t \frac{\partial c_{\cdot,q}}{\partial x_r}(s,X_s)J_{u,0} \d B^{H,q}_u\bigg],
\end{gather*}
which, upon multiplying by $J_{s,0}^{-1}b_{\cdot,k}(s,X_s)$ on the right leads to the same equation on $J_{u,s} b_{\cdot,k}(s,X_s)$ as that on $\md_s^{W,k}X^r_u$. Hence, by uniqueness, we get the desired result.
\end{proof}
Further we establish a simple estimate on the It\^o integral of a H\"older continuous integrand.
\begin{lemma}
\label{estimInt}
Let $\set{f(t),t\in [0,T]}$ be an $\mathbf{F}$-adapted stochastic process such that $\E{\norm{f}_\theta^p}<\infty $ for all $p\geq 1$, and $0<\delta<\Delta\le T$. Then for all  $s,t,u\in[0,T]$ such that $u<s<t$, $t-s<\delta$, $t-u\le \Delta$ it holds
$$\left|\int_s^t\big(f(v)-f(u)\big)\d{W_v} \right|\leq \Delta^\theta\delta^{1/2}\xi_{\Delta,\delta},$$
where $\E{ \xi^p_{\Delta,\delta}}<C_p\E{\norm{f}_\theta^p}$ for all $p\ge 1$.
\end{lemma}

\begin{proof} It suffices to establish the required result for  $p$ large enough, then one can get deduce it for all $p\ge 1$ with the help of Jensen's inequlaity.

By the Garsia--Rodemich--Rumsey inequality, we get
\begin{gather*}
\left|\int_s^t\big(f(v)-f(u)\big)\d{W_v} \right|\leq C|t-s|^{1/4}\left(\int_s^t\int_s^t \frac{|\int_x^y \big(f(v)-f(u)\big)\d{W_v}|^{8}}{|x-y|^{4}} \d x\d y\right)^{1/8}\\
\le C\Delta^\theta \delta^{1/2	}\xi_{\Delta,\delta},
\end{gather*}
where
$$
\xi_{\Delta,\delta}=\Delta^{-\theta}\delta^{-1/4}\left(\int_s^t\int_s^t \frac{|\int_x^y \big(f(v)-f(u)\big)\d{W_v}|^{8}}{|x-y|^{4}} \d x\d y\right)^{1/8}.
$$
For $p>8$ the  H\"older inequality entails that
 \begin{gather*}
\E{\xi^p_{\Delta,\delta}}\leq \Delta^{-\theta p}\delta^{-p/4}(t-s)^{2(p/8-1)} \int_s^t\int_s^t \E{\dfrac{|\int_x^y \big(f(v)-f(u)\big)\d{W_v}|^p}{|x-y|^{p/2}}}\d x\d y
  \\ \leq C_p\Delta^{-\theta p}\delta^{-2} \int_s^t\int_s^t \E{\dfrac{\left(\int_x^y |f(v)-f(u)|^2\d v\right)^{p/2}}{|x-y|^{p/2}}}\d x\d y\\
  \leq C_p \Delta^{-\theta p}\delta^{-2}\E{\norm{f}_\theta^p}\Delta^{p\theta} \delta^2= C_p\E{\norm{f}_\theta^p}.
 \end{gather*}
Hence, we arrive at the desired statement.
\end{proof}
We also need the result concerning the pathwise regularity property of $X$. It establishes certain exponential integrability of the H\"older seminorm of $X$, so it is an interesting result on its own.
\begin{theorem}
 Let $\{X_t,t\in[0,T]\}$ be the solution to \eqref{eqHom}. Assume that $a,b,c$ satisfy the assumption \emph{C1}. Then $X \in  C^\theta[0,T]$ for $\theta \in (0,1/2)$ and
 $\E{\exp\set{K\norm{X}_\theta^q}}<\infty$ for all $q\in\big(0,q^*\big)$, $K>0$, where
$$
q^* = \frac{4H}{2(H+\theta)+1}\wedge \frac{2H+1}{4H}.
$$
In particular, $\E{\norm{X}^p_\theta}<\infty$ for all $p> 0$.
\end{theorem}
\begin{proof}
Define for $\eps\in(0,T]$
$$
\norm{X}_{\theta,\eps} = \sup_{0\le t-\eps\le s<t\le T}\frac{\abs{X_t-X_s}}{(t-s)^{\theta}}.
$$
Clearly, $\norm{X}_\theta \le \norm{X}_{\theta,\eps} + 2\eps^{-\theta}\norm{X}_\infty$. It follows from \cite[equation (4)]{Integr} that
$$
\norm{X}_{\theta,\eps} \le C_1\left(\big\|I^b\big\|_\theta + \Lambda_\mu\big(1+ \norm{X}_\infty\eps^{\mu-\theta}\big) \right),
$$
for any $\eps\in(0,C_2 \Lambda_\mu^{-1/\mu}] $ where $C_1,C_2$ are some positive constants, $\mu\in(1/2,H)$,
$\Lambda_\mu = \norm{B_H}_\mu+1$, $I^b_t = \int_0^t b(X_s)dW_s.$
Therefore, setting $\eps = C_2 \Lambda_\mu^{-1/\mu}$, we obtain
$$
\norm{X}_{\theta} \le C_1\left(\big\|I^b\big\|_\theta + \Lambda_\mu+ 2 \norm{X}_\infty\Lambda^{\theta/\mu} \right) \le
C\left(\big\|I^b\big\|_\theta + \Lambda_\mu+  \norm{X}_\infty^{p'} + \Lambda^{q'\theta/\mu} \right),
$$
where $p'>1$, and $q' = p'/(p'-1)$ is the exponent conjugate to $p'$.
Therefore,
$$
\norm{X}_{\theta}^q \le
C\left(\big\|I^b\big\|_\theta^q + \Lambda_\mu^q+  \norm{X}_\infty^{p'q}+ \Lambda^{qq'\theta/\mu} \right).
$$
Evidently,  $q^*<1$, so it follows from \cite[Lemma 1]{ShSh} that $\Ex{\exp\set{K\big\|I^b\big\|_\theta^q}}<\infty$ for all $K>0$. Further, $\Lambda_\mu$ is an almost surely finite supremum of a centered Gaussian family, so by Fernique's theorem
$\Ex{\exp\set{K \Lambda_\mu^z}}<\infty$ for any $K>0$, $z\in(0,2)$. Finally, by \cite[Corollary 4]{Integr}, $\Ex{\exp\set{K \norm{X}^{z}_\infty}}<\infty$ for all $K>0, z<4H/(2H+1)$. Now if $p'>1$ is close to $4H q^{-1}(2H+1)^{-1}$ (thanks to the bound on $q$ such choice is possible) and $\mu$ is close to $H$, then $q'$ is close to $4H/(4H-q(2H+1))$, and $qq'\theta/\mu$ is close to
$4q\theta/(4H-q(2H+1))$, which is less than $2$. Indeed, the last statement is equivalent to $q(2\theta+2H+1)<4H$, which is true thanks to the restriction on $q$. Thus, we get the desired integrability.
\end{proof}

The following result is a Norris type lemma for mixed SDEs. It is a crucial result to prove existence of density under the H\"ormander condition. Loosely speaking, this statement says that if
\begin{equation}
Y_t=Y_0+\int_0^t a(s)\d{s}+\int_0^t b(s)\d{W_s}+\int_0^tc(s)\d{B^H_s},\label{NorrisEq}
\end{equation}
 $\norm{Y}_\infty  = \norm{Y}_{\infty; [0,T]}$ is small, then $\norm{b}_\infty$ and $\norm{c}_\infty$ can not be large. This means that the integral w.r.t. $W$ and w.r.t. $B^H$ can not compensate each other well. The rigorous formulation is as follows.
\begin{theorem}
\label{Chuck}
Assume that $H\in(1/2,2/3)$, $\theta\in\big(\theta_*,1/2\big)$,
where
$$
\theta_* =  \frac{H-\frac 12}{3-4H},
$$
 and that $a,b,c$ in \eqref{NorrisEq}  are $\mathbf F$-adapted processes  satisfying $\E{\norm{a}^p_\infty+\norm{b}^p_\theta+\norm{c}^p_\theta}<\infty$ for all $p\geq 1$. Then exists $q>0$ such that for all $p\geq 1, \varepsilon>0$
$$\pr{\norm{Y}_\infty<\varepsilon \textrm{ and } \norm{b}_\infty+\norm{c}_\infty>\varepsilon^q}\leq C_p\varepsilon^p.$$
\end{theorem}
\begin{proof}
Here we imitate the proof of in \cite[Proposition 3.4]{BauHai}.
For notational simplicity, we assume that $T=1$. For some positive integers $M$ and $r$ denote $\Delta = 1/M$, $\delta = \Delta/r$ and
define the following uniform partitions of $[0,1]$: $T_N = N\delta$, $N=0,\dots,M$; $t_n=\delta n$, $n=0,\ldots, Mr$. Further, fix some $\breve H\in (1/2,H)$ and write for $N=0,\dots,M-1$, $n= Nr,\dots, (N+1)r-1$ (so that $t_n\in [T_N,T_{N+1})$),  $i=1,\ldots, d$
\begin{equation}\label{eqS}
\begin{gathered}
\big\la c_i(T_N), B^H_{t_{n+1}}-B^H_{t_n} \big\ra+
\big\la b_{i}(T_N), W_{t_{n+1}}-W_{t_n}\big\ra
\leq |Y^i_{t_{n+1}}-Y^i_{t_n}|+ \delta\norm{a}_\infty \\
+ \abs{\int_{t_n}^{t_{n+1}}\la b_i(s)-b_i(T_N),\d{W_s}\ra}+\abs{\int_{t_n}^{t_{n+1}}\la c_i(s)-c_i(T_N), \d{B^H_s}\ra }\\ \leq 2\norm{Y}_\infty+\delta\norm{a}_\infty +C\Delta^\theta\delta^{\breve{H}}\norm{c}_\theta\norm{B^H}_{\breve{H}}+\Delta^{\theta}\delta^{1/2}\xi_{\Delta,\delta}=:S,
\end{gathered}
\end{equation}where in the last step we have used the Young--Love inequality (see e.g. \cite[Proposition 1]{NualSau}) and {Lemma \ref{estimInt}}.

 For processes $\xi,\zeta$ denote

\begin{equation}
\label{V_N}
V_N(\xi,\zeta) = \sum_{n=Nr}^{(N+1)r-1} \left(\xi_{t_{n+1}}-\xi_{t_n}\right)\left(\zeta_{t_{n+1}}-\zeta_{t_n}\right);
\end{equation}
we remind that the summation is in fact over $t_n\in[T_N,T_{N+1})$.
Squaring the both sides of \eqref{eqS}, summing  over $n= Nr,\dots, (N+1)r-1$ and then taking the square root we get
\begin{gather}
 \biggl(\sum_{u,v=1}^m \sum_{v=1}^m b_{i,u}(T_N)b_{i,v}(T_N)V_N(W^u,W^v) + \sum_{u,v=1}^l  c_{i,u}(T_N)c_{i,v}(T_N)V_N(B^{H,u},B^{H,v})
\\
+ \sum_{u=1}^m \sum_{v=1}^l b_{i,u}(T_N)c_{i,v}(T_N)V_N(W^u,B^{H,v}) \biggr)^{1/2}\leq C \Delta^{1/2}\delta^{-1/2}\mathit{S}.
\end{gather}
Therefore,
\begin{equation}\label{sumincrements}
\begin{gathered}
\sum_{u=1}^m \abs{b_{i,u}(T_N)}V_N(W^u,W^u)^{1/2}+\sum_{v=1}^l \abs{c_{i,v}(T_N)}V_N(B^{H,v},B^{H,v}) \\
\le  C\biggl ( \sum_{1\le u<v\le m}|b_{i,u}(T_N)|^{1/2}|b_{i,v}(T_N)|^{1/2}|V_N(W^u,W^v)|^{1/2}\\ +\sum_{1\le u<v\le l}|c_{i,u}(T_N)|^{1/2}|c_{i,v}(T_N)|^{1/2}|V_N(B^{H,u},B^{H,v})|^{1/2}\\
+\sum_{u=1}^m \sum_{v=1}^l |b_{i,u}(T_N)|^{1/2}|c_{i,v}(T_N)|^{1/2}|V_N(W^u,B^{H,v}|^{1/2}+\Delta^{1/2}\delta^{-1/2}\mathit S \biggr).
\end{gathered}
\end{equation}
Further, for arbitrary $f\in C^\theta[0,1]$,
$$\abs{\Delta\sum_{N=0}^{M-1}|f(T_N)|-\norm{f}_{L^1[0,1]}}\leq \norm{f}_\theta\Delta^{\theta},$$
which yields
\begin{gather*}
\sum_{u=1}^m\norm{b_{i,u}}_{L^1[0,1]}
+\sum_{v=1}^l\norm{c_{i,v}}_{L^1[0,1]}
\\
\leq \sum_{u=1}^m\bigg(\Delta^{\theta}\norm{b_{i,u}}_\theta+
\Delta\sum_{N=0}^{M-1} \abs{b_{i,u}(T_N)}\bigg) + \sum_{v=1}^l\bigg(\Delta^{\theta}\norm{c_{i,v}}_\theta+
\Delta\sum_{N=0}^{M-1} \abs{c_{i,v}(T_N)}\bigg)\\
\leq \sum_{u=1}^m\bigg(\Delta^{\theta}\norm{b}_\theta+
\Delta^{1/2}\norm{b}_\infty\sum_{N=0}^{M-1} \abs{\Delta^{1/2} - V_N(W^u,W^u)^{1/2}}
\\+
\Delta^{1/2}\delta^{1/2-H}\sum_{N=0}^{M-1} \abs{b_{i,u}(T_N)} V_N(W^u,W^u)^{1/2}\bigg)
\\
+ \sum_{v=1}^l\bigg(\Delta^{\theta}\norm{c}_\theta+
\Delta^{1/2}\delta^{1/2-H}\norm{c}_\infty\sum_{N=0}^{M-1} \abs{\Delta^{1/2}\delta^{H-1/2} - V_N(B^{H,v},B^{H,v})^{1/2}}\\
  +
 \Delta^{1/2}\delta^{1/2-H}\sum_{N=0}^{M-1}\abs{c_{i,v}(T_N)}
 V_N(B^{H,v},B^{H,v})^{1/2}\bigg).
\end{gather*}
Therefore, using \eqref{sumincrements}, we arrive at
\begin{gather*}
\norm{b}_{L^1[0,1]}+\norm{c}_{L^1[0,1]}\leq C\bigg ( \Delta^{\theta}\big(\norm{b}_\theta+ \norm{c}_\theta\big)\\
+\Delta^{1/2}\delta^{1/2-H} \norm{b}_\infty\sum_{N=0}^{M-1}\sum_{u,v=1}^m \abs{\Delta^{1/2}\delta_{u,v} - \abs{V_N(W^u,W^v)}^{1/2}}\\ +\Delta^{1/2}\delta^{1/2-H} \norm{c}_\infty\sum_{N=0}^{M-1}\sum_{u,v=1}^l \abs{\Delta^{1/2}\delta^{H-1/2}\delta_{u,v} - \abs{V_N(B^{H,u},B^{H,v})}^{1/2}}\\
+ \Delta^{1/2}\delta^{1/2-H}\norm{c}^{1/2}_\infty
\norm{b}^{1/2}_\infty
\sum_{N=0}^{M-1}\sum_{u=1}^m\sum_{v=1}^l  \abs{V_N(W^u,B^{H,v})}^{1/2}
+\delta^{-H}S\bigg)\\
\le C\bigg ( \norm{b}_\theta+ \norm{c}_\theta+
\Delta^{-1/4}\delta^{3/4-H} \norm{b}_\infty R^W\\ +\Delta^{H-1}\delta^{1-H} \norm{c}_\infty R^B
+ \Delta^{-1/4}\delta^{(1-H)/2}\left(\norm{c}_\infty +
\norm{b}_\infty\right)R^{W,B}
+\delta^{-H}S\bigg),
\end{gather*}
where
\begin{equation}\label{RWB}
\begin{gathered}
R^W = \Delta^{3/4}\delta^{-1/4} \sum_{N=0}^{M-1}\sum_{u,v=1}^m \abs{\Delta^{1/2}\delta_{u,v} - \abs{V_N(W^u,W^v)}^{1/2}},\\
R^B = \Delta^{H-3/2}\delta^{1/2}\sum_{N=0}^{M-1}\sum_{u,v=1}^l  \abs{\Delta^{1/2}\delta^{H-1/2}\delta_{u,v} - \abs{V_N(B^{H,u},B^{H,v})}^{1/2}},\\
R^{W,B} = \Delta^{3/4}\delta^{-H/2} \sum_{N=0}^{M-1}\sum_{u=1}^m
\sum_{v=1}^{l} \abs{V_N(W^u,B^{H,v})}^{1/2}.
\end{gathered}
\end{equation}
Further we use the following interpolation inequality, valid for any $f\in C^\theta[0,1]$ and $\gamma<1$:
$$\norm{f}_\infty\leq C\big(\gamma \norm{f}_\theta+\gamma^{-1/\theta}\norm{f}_{L^1[0,1]}\big).$$
for any $\gamma\leq1$. Thus,
\begin{equation}\label{bcinfty}
\begin{gathered}
\norm{b}_\infty+\norm{c}_\infty\leq C\big(\norm{b}_\theta + \norm{c}_\theta\big)\gamma +
C\gamma^{-1/\theta}\Big(\big(\norm{b}_\theta + \norm{c}_\theta\big)\Delta^\theta  \\
+ (\norm{b}_\infty + \norm{c}_\infty)\Big[
\Delta^{-1/4}\delta^{3/4-H}R^W + \Delta^{H-1}\delta^{1-H}R^B + \Delta^{-1/4}\delta^{(1-H)/2} R^{W,B}\Big] \\
+ \delta^{-H}\norm{Y}_\infty + \delta^{1-H}  \norm{a}_\infty +  \Delta^\theta \delta^{\breve H-H}\norm{c}_\theta\norm{B^H}_{\breve H} + \Delta^\theta \delta^{1/2-H} \xi_{\Delta,\delta}\Big).
\end{gathered}
\end{equation}
Now we want to put
\begin{gather*}
\Delta^{\beta}\sim \varepsilon^\beta, \delta\sim \varepsilon^{\alpha}, \gamma\sim \varepsilon^\eta, \alpha>\beta>0, \eta>0,
\end{gather*}
so that in the right-hand side of \eqref{bcinfty}, the exponents of $\varepsilon$ are positive for all terms except $\norm{Y}_\infty$. Since $(H-1/2)/\theta\le (3-4H)<1$, it is possible to take $\beta/\alpha\in \big((H-1/2)/\theta, (3-4H)\big)$ so that both  $\theta \beta + (1/2-H)\alpha$ and  $-\beta/4 + (3/4-H)\alpha$ are positive. Also $(H-1)\beta +(1-H)\alpha = (1-H)(\alpha-\beta)>0$,
$-\beta/4 + (1-H)\alpha/2 > -\beta/4 + (3/4-H)\alpha>0$, $\theta \beta + (\breve H - H)\alpha > \theta \beta + (1/2-H)\alpha>0$. Therefore, by choosing $\eta$ small enough we can make all needed exponents positive.

Thus, for some $\kappa>0$ and $C_1>0$ we have
\begin{gather*}
\norm{b}_\infty+\norm{c}_\infty\leq C_1\norm{Y}_\infty \eps^{-\lambda} +
C_1\eps^\kappa\Big(  (\norm{b}_\infty + \norm{c}_\infty)\big[
R^W + R^B + R^{W,B}]\\ + \norm{b}_\theta + \norm{c}_\theta
+   \norm{a}_\infty +  \norm{c}_\theta\norm{B^H}_{\breve H} +  \xi_{\Delta,\delta}\Big),
\end{gather*}
where $\lambda= H\alpha+\eta/\theta$. Consequently, for $\eps$ small enough
\begin{gather*}
\pr{\norm{b}_\infty+\norm{c}_\infty>\varepsilon^{\kappa/2} \text{ and } \norm{Y}_\infty<\varepsilon^{\lambda+\kappa}} \\
\le \pr{R^W \ge \eps^{-\kappa/3}} + \pr{R^B \ge \eps^{-\kappa/3}} +  \pr{R^{W,B} \ge \eps^{-\kappa/3}} \\+ \pr{\norm{b}_\theta + \norm{c}_\theta + \norm{c}_\theta\norm{B^H}_{\breve H} +  \xi_{\Delta,\delta}\ge \eps^{-\kappa/3}}.
\end{gather*}
Now the statement follows by applying Lemmas \ref{estimInt} and \ref{concentrat} and the Chebyshev inequality.
\end{proof}

\begin{lemma}\label{concentrat}
Let $R^W,R^B$ and $R^{W,B}$ be given by \eqref{RWB} and \eqref{V_N}. Then we have for any $h>1$ the following concentration inequalities
\begin{align}
\pr{R^{W}\geq h}&\leq \dfrac{C}{\Delta}\exp(-Ch^2),\label{conc1}\\
\pr{R^B\geq h}&\leq \dfrac{C}{\Delta}\exp(-Ch^2),\label{conc2}\\
\pr{R^{W,B}\geq h}&\leq \dfrac{C}{\Delta}\exp(-Ch^2)\label{conc3}.
\end{align}
\end{lemma}
\begin{proof}

By \cite[Lemma 3.1]{BauHai} we have for $h>0$
\begin{equation}\label{WvWv}
\pr{\big|\Delta^{1/2}-V_N(W^u,W^u)^{1/2}\big|\Delta^{-1/4}\delta^{-1/4}\geq h}\leq C\exp\left(-C {h^2}\right).
\end{equation}
Further, let $u\neq v$. Since  $W^u$ and $W^v$ are independent, and $W^v$ has independent increments, then conditional on $W^v$,  $V_N(W^w,W^u)\Delta^{-1/2}\delta^{-1/2}$ has a centered Gaussian distribution with the variance $ V_N(W^v,W^v)\Delta^{-1} $. Therefore,
\begin{gather*}
\pr{|V_N(W^u,W^v)|^{1/2}\Delta^{-1/4}\delta^{-1/4}\geq h}\\ = \E{\pr{|V_N(W^u,W^v)|\Delta^{-1/2}\delta^{-1/2}\geq h^2}\big|\ W^v}
\le C\,\E{\exp\set{-\frac{h^4\Delta}{4 V_N(W^v,W^v)}}}\\
\le C\exp\set{-\frac{h^4\Delta}{4(\Delta^{1/2}+v)^2}} + \pr{\big|\Delta^{1/2}-V_N(W^v,W^v)^{1/2}\big|\ge v}\\
\le C\exp\set{-\frac{h^4\Delta }{8(\Delta+v^2)}} + C\exp\set{-C\frac{v^2}{\Delta^{1/2}\delta^{1/2}}},
\end{gather*}
where we have used \eqref{WvWv}.
Setting $v^2 = h^2\Delta$ and recalling that $\Delta\ge\delta$ we get
$$ \pr{|V_N(W^u,W^v)|^{1/2}\Delta^{-1/4}\delta^{-1/4}\geq h}\leq C\exp\left(-C {h^2}\right).
$$
Combining this with \eqref{WvWv}, we get
\begin{gather*}
\pr{R^W\ge h}\le \sum_{N=0}^{M-1} \sum_{u,v=1}^m \pr{\Delta^{-1/4}\delta^{-1/4} \big|\Delta^{1/2}\delta_{u,v}-\abs{V_N(W^u,W^v)}^{1/2}\big|\ge hm^2}\\
\le \frac{C}{\Delta} \exp\set{-Ch^2}.
\end{gather*}
Using the inequalities from \cite[Lemma 3.2]{BauHai} and repeating the last step, we get \eqref{conc2}.

The estimate \eqref{conc3} is proved similarly to \eqref{conc1}, so we omit some details. Write
\begin{gather*}
\pr{|V_N(W^u,B^{H,v})|^{1/2}\Delta^{-1/4}\delta^{-H/2}\geq h}\\ = \E{\pr{|V_N(W^u,B^{H,v})|\Delta^{-1/2}\delta^{-H}\geq h^2}\big|\ B^{H,v}}
\le C\,\E{\exp\set{-\frac{h^4\Delta\delta^{2H-1}}{4 V_N(B^{H,v},B^{H,v})}}}\\
\le C\exp\set{-\frac{h^4\Delta\delta^{2H-1}}{4(\Delta^{1/2}\delta^{H-1/2}+v)^2}} + \pr{\big|\Delta^{1/2}\delta^{H-1/2}-V_N(B^{H,v},B^{H,v})^{1/2}\big|\ge v}\\
\le C\exp\set{-\frac{h^4\Delta\delta^{2H-1} }{8(\Delta\delta^{2H-1}+v^2)}} + C\exp\set{-C\frac{v^2}{4\Delta^{2H-1}\delta}}.
\end{gather*}
Setting $v^2 = h^2 \Delta \delta^{2H-1}$ and taking into account that $\Delta\ge \delta$, we arrive at
$$
\pr{|V_N(W^u,B^{H,v})|^{1/2}\Delta^{-1/4}\delta^{-H/2}\geq h}\le C\exp\set{-Ch^2}.
$$
From here \eqref{conc3} is deduced similarly to \eqref{conc1}.
\end{proof}

config: arxiv


TITLE: Representation theory and simultaneous block diagonalization of matrices with common symmetries
QUESTION [6 upvotes]: Group representation theory has applications in a lot of areas and problems in mathematics, one of such an example is the finest simultaneous block diagonalization of the set of matrices with common symmetries. According to Page 7 of this paper, let $\{T(g)\}$ be an orthogonal matrix representation of a group $G$, if a set of matrices $\{A_p\}$ all share the symmetry described by $G$, that is,
$T(g)^\intercal A_p T(g) = A_p, \ \forall g\in G \ \text{and} \ \forall p,$
then a simultaneous block diagonalization of $A_p$ can be obtained through the decomposition of the representation $\{T(g)\}$ into irreducible representations. In particular, if $\{P^\intercal T(g) P\}$ is the direct sum of irreducible representations with dimensions $n_i$ and multiplicities $m_i$, then each element in $\{P^\intercal A_p P\}$ has a common block structure as the direct sum of blocks of dimensions $m_i$ and multiplicities $n_i$.
Can anyone point me to references/theorems in group representation theory from which this claim is based? Thanks!

REPLY [3 votes]: [I am sure that there is a slicker way of serving this up, and will happily delete this when it emerges. This is just the old-fashioned matrix way of looking at things.]
Suppose $T$ is a representation of the finite groups $G$ on the finite-dimensional space $V$. That is the same as saying that $V$ is a $\mathbb{C}G$-module. By Mashke's Theorem we can write $V=V_1\oplus\dots\oplus V_m$ where the $V_i$ are irreducibles. Let $\alpha:V\to V$ commute with all the $T(g)$.
Choose a basis of $V$ compatible with this decomposition. Then in terms of block matrices we have that
$$
T(g)=\left[\begin{matrix}
T_1(g) & 0 & \dots &0\\
0    &T_2(g) &\dots& 0\\
0   & 0 & \ddots & 0\\
0 & 0& \dots &T_m(g)\\
\end{matrix}
\right],
\ \ \
A=\left[\begin{matrix}
A_{11} & A_{12} & \dots &A_{1m}\\
A_{21} & A_{22} & \dots &A_{2m}\\
\vdots &\vdots& \ddots & \vdots\\
A_{m1} & A_{m2}& \dots &A_{mm}\\
\end{matrix}
\right].
$$
The hypothesis of commutativity now gives $T_{r}(g)A_{rs}=A_{rs}T_{s}(g)$. As the $V_i$ are irreducible, Schur's Lemma yields that $A_{rs}=O$ whenever $V_r$ and $V_s$ are not isomorphic, and $A_{rs}=\lambda_{rs} I$ ($\lambda_{rs}\in\mathbb{C}$) when they are isomorphic.
If we now group the $V_i$ into "clumps" of isomorphic modules, we have that both $T(g)$ and $A$ are in diagonal block form; both havezeros where the row and column correspond to non-isomorphic modules.
That means that we need only  consider the case when $V=V_1\oplus\dots\oplus V_m$ is the direct sum of isomorphic irreducibles. What we have proved in this case is that $A=\Lambda \otimes I_n$, where $n$ is the vector space dimension of $V_1$, and $\Lambda$ is the $m\times m$ matrix of the $\lambda_{rs}$.


TITLE: Question about dimension of a subspace
QUESTION [2 upvotes]: Let $K$ be a field and define the following subspaces
$$V=\textrm{span}(e_1,e_2,e_3),\;\; V^\bot = \textrm{span}(e_4,e_5,e_6)$$
inside $K^6$. Let $\dim L=4$ and assume that $\dim L\cap V\leq 1$. Can I conclude that $\dim L\cap V^\bot > 1$? I know of course that $\dim L\cap V^\bot \geq 1$, but I'm wondering if the fact that $V^\bot$ is the orthogonal complement of $V$ would force the dimension to be strictly larger than $1$?

REPLY [2 votes]: Take $V=\mathbb{R}^{6}$ and $\{e_{i}\}_{i=1}^{6}$ as the standard
basis.
Define
$$
L=sp\{e_{1},e_{4},e_{2}+e_{5},e_{3}+e_{6}\}
$$
then since
$$
\begin{pmatrix}1 & 0 & 0 & 0 & 0 & 0\\
0 & 0 & 0 & 1 & 0 & 0\\
0 & 1 & 0 & 0 & 1 & 0\\
0 & 0 & 1 & 0 & 0 & 1
\end{pmatrix}
$$
is of rank $4$ we conclude $dim(L)=4$.
Now, a general element in $L$ is of the form
$$
ae_{1}+be_{4}+c(e_{2}+e_{5})+d(e_{3}+e_{6})=(a,c,d,b,c,d)
$$
for $a,b,c,d,e,f\in\mathbb{R}$
a general element in $V$ is of the form
$$
(x,y,z,0,0,0)
$$
and thus the intersection is
$$
\{(a,c,d,b,c,d)\in\mathbb{R}^{6}\mid b=c=d=0\}=\{(a,0,0,0,0,0)\mid a\in\mathbb{R}\}
$$
is of dimension $1$.
But now calculate the intersection with $V^{\perp}=sp\{e_{4},e_{5},e_{6}\}$
: the intersection is
$$
\{(a,c,d,b,c,d)\in\mathbb{R}^{6}\mid a=c=d=0\}=\{(0,0,0,b,0,0)\mid b\in\mathbb{R}\}
$$
is also of dimension $1$.
Since
$$
6\geq dim(L\cup V^{\perp})=dim(L)+dim(V^{\perp})-dim(L\cap V^{\perp})=7-dim(L\cap V^{\perp})
$$
We conclude that in general, we can only say that
$$
dim(L\cap V^{\perp})\geq1
$$
and not $dim(L\cap V^{\perp})>1$.