TITLE: Finding the Volume of a paraboloid using surface integrals via Stokes Theorem.
QUESTION [1 upvotes]: Let, $z = 16-x^2-y^2$ be our paraboloid, and let our surface be bounded above by the paraboloid and bounded bellow by the plane $z=6$.
We can parametrize the surface as, $R(x,y) = (x,y,16-x^2-y^2)$ which gives us $R_{x} = (1,0,-2x)$ and $R_{y} = (0,1,-2y)$, and then we get $ \vec{n} = R_{x} \times R_{y} = (2x,2y,1)$.
We know we can obtain the volume of the solid by using the fact that:
$$V = \frac{1}{3}\int_{S} (x,y,z) \cdot \frac{\vec{n}}{||\vec{n}||} dS.$$
From, $z=16-x^2-y^2$ and $z=6$, we get $x^2+y^2=10$; hence,
$$0 \le \theta \le 2\pi, ~ 0 \le r \le \sqrt{10},$$
gives us a parametrization of the projection $S$ makes on the $x,y$ plane; thus,
$$V = \frac{1}{3}\int^{2\pi}_{0}\int^{\sqrt{10}}_{0} (r^2 + 16)r ~ drd\theta,$$
but the previous expression does not give me the correct answer which is $50\pi$, what am I doing wrong? Thanks.

REPLY [0 votes]: From the work you already have, let's make a few corrections. First, we should use the following vector surface integral instead
$$V = \frac{1}{4}\oint_S(x,y,2z)\cdot d\vec{S}$$
which we can verify has the correct divergence
$$\nabla \cdot \left(\frac{1}{4}(x,y,2z)\right) = \frac{1}{4} + \frac{1}{4} + \frac{1}{2} = 1$$
We will see why this is nice later, but this is a computational simplification choice - your original choice of vector field was correct, too. The only mistake is that you forgot to close your surface and include the integral on the bottom disk. This should look like
$$V = \frac{1}{4}\oint_S(x,y,2z)\cdot d\vec{S} = \frac{1}{4}\int_{S_1}(x,y,2z)\cdot d\vec{S_1} + \frac{1}{4}\int_{S_2}S(x,y,2z)\cdot d\vec{S_2}$$
where $S_1$ denotes the paraboloid and $S_2$ denotes the disk. You already correctly computed the vector Jacobian for the first integral, so let's go ahead and compute that integral
$$\frac{1}{4}\int_{S_1}(x,y,32-2x^2-2y^2)\cdot(2x,2y,1)\:dA = \int_0^{2\pi}\int_0^{\sqrt{10}}8\:r\:dr\:d\theta = 80\pi$$
Next, we have the flat disk. We can compute the vector Jacobian the long way, or note that the disk is flat and horizontal, thus its vector Jacobian will be $1$ times the outward unit normal. In this case, that would be the downward vector $(0,0,-1)$, giving the integral
$$\frac{1}{4}\int_{S_2}(x,y,12)\cdot(0,0,-1)\:dA = \int_0^{2\pi}\int_0^{\sqrt{10}}-3\:r\:dr\:d\theta = -30\pi$$
Therefore the total volume is the sum of the two integrals, $\boxed{50\pi}$


TITLE: Which non-abelian finite groups have the property that every subgroup is normal?
QUESTION [6 upvotes]: If $G$ is an abelian group, every subgroup $H$ of $G$ is normal.
I searched for non-abelian finite groups $G$ , such that every subgroup is normal and
GAP showed only the groups $G'\times Q_8$ , where $Q_8$ is the quaternion-group
of order $8$ and $G'$ is an abelian group with the additional property that $C_4$ is not a subgroup of $G'$. The largest group I found with GAP was $C_{15}\times Q_8$.

Is it true that every non-abelian finite group $G$ with the property that every subgroup of $G$ is normal, is isomorphic to $G'\times Q_8$ with an abelian
  group $G'$ without $C_4$ as a subgroup ?

REPLY [8 votes]: Hall's book "The Theory of Groups" proves (Theorem 12.5.4) that these groups, known as Hamiltonian groups, are all of the form $Q_8 \times A \times B$ where $A$ is an elementary $2$-group, $B$ is an abelian group where every element is of finite odd order, and $Q_8$ is the quaternion group. All groups of this form are Hamiltonian. Note there is no finiteness restriction on these $A$, $B$ groups, so there are also infinite group.


text:
\begin{document}

\title{Positive functionals and Hessenberg matrices}

\author{Jean B. Lasserre}
\address[J-B.~Lasserre]{LAAS-CNRS and Institute of Mathematics, University of Toulouse, France}
\email{\tt lasserre@laas.fr}

\author{Mihai Putinar}
\address[M.~Putinar]{University of California at Santa Barbara, CA,
USA and Newcastle University, Newcastle upon Tyne, UK}
\email{\tt mputinar@math.ucsb.edu, mihai.putinar@ncl.ac.uk}

\date{\today}

\keywords{positive functional, cubature formula, normal matrix, Hessenberg matrix, harmonic polynomial}

\subjclass[2010]{}

\begin{abstract} Not every positive functional defined on bi-variate polynomials of a prescribed degree bound is represented by the integration against a positive measure.
We isolate a couple of conditions filling this gap, either by restricting the class of polynomials to harmonic ones, or imposing the vanishing of a defect indicator. Both criteria offer
constructive cubature formulas and they are obtained via well known matrix analysis techniques involving either the dilation of a contractive matrix to a unitary one or the specific structure of the Hessenberg matrix associated to the multiplier by the underlying complex variable.

\end{abstract}
\maketitle

\section{Introduction} Typically, a positive functional on a space of continuous functions is represented as the integral
against a positive measure. The departure from this paradigm is notable on finite dimensional spaces, turning the existence of a
 Riesz representation theorem into the basic quest of numerical cubature, solutions of the truncated moment problem or algebraic
 certificates for positive elements.

 The present note analyses the structure of positive functionals on spaces of bi-variate real polynomials of a prescribed degree. In spite of the ample
 references and recent progress on this topic, there are still challenging open questions and notorious difficulties, mostly related to the constructive aspects.
 Our approach uses complex variables, hence Hermitian forms over complex variables rather than quadratic forms over the real field. The natural Hilbert space realization of Hermitian forms is
 also prominent in our note. The balance between Hermitian algebra and Hilbert space geometry turns out to be beneficial for entering into the fine structure of positive functional on polynomial spaces.

 Specifically, we import from operator theory the well known Sz.-Nagy unitary dilation of a contractive matrix for constructing numerical cubatures for {\it harmonic} polynomials; second,
 we identify, via a normality criterion in matrix analysis, the numerical obstruction of a positive functional on polynomials bi-variate polynomials of a prescribed degree to possess a cubature on a
 real subspace of roughly half-dimension, but certainly containing all polynomials of half-degree. In this quest we naturally touch the structure of the Hessenberg matrix representing the multiplier by the complex variable, and reveal its normality up to a rank-one additive perturbation. A normal Hessenberg matrix characterizes finite point cubatures for polynomials of a fixed maximum degree in the two variables. For the reader versed in some operator theory, this part of our note deals with the Krylov subspace analysis of truncations of a hyponormal operator.

 A unifying conclusion of our observations is that one has to pay a price for a positive functional to admit numerical quadratures (with positive weights). Namely one has to drop the expectations/identities to some distinguished subspaces of polynomials of approximatively half the original dimension. Undoubtedly there are other paths to achieve such cubature formulas on privileged polynomials, such it is the case of quadrature domains for harmonic functions. A common feature of our study is the numerical and computational accessibility to cubatures, modulo ubiqutiuous matrix analysis techniques, having as solely input data the moments of the original functional.

  While a selected group of operator theory experts may find the main results below not surprising, our aim is to disseminate to a larger audience some possibly novel, but very basic and versatile matrix analysis techniques. The recent authoritative monograph by Schm\"udgen \cite{Sch} contains ample details and references, old and new, on the structure of positive functionals on polynomial subspaces, cubature formulas, orthogonal polynomials and (truncated) moment problems. All presented from the traditional point of view of the functional analyst or function theorist. Hessenberg matrices however are cultivated by rather disjoint groups of mathematicians, notably in numerical analysis \cite{LS} and approximation theory \cite{Simanek}.
 \bigskip

 {\bf Acknowledgement.}
Research of the first author and visit of the second  author at LAAS in Toulouse, funded by the European Research Council (ERC) under the European Union's Horizon 2020 research and innovation program (grant agreement ERC-ADG 666981 TAMING).

\section{Preliminaries}

We denote by $z = x+iy$ the complex variable in $\C$, identified with $\R^2$.
For a fixed positive integer $d$, $\C_d[z]$ is the complex vector space of polynomials
of degree less than or equal to $d$, and similarly for $\C_d[z,\ol{z}]$, respectively $\C_d[x,y]$
where in the case of two variables the total degree is adopted. Similarly $\C_{k, \ell}[z,\ol{z}]$
stands for the vector space of polynomials of bidegree $k$ in $z$, respectively $\ell$ in $\ol{z}$.

The starting point is a linear functional
$$ L : \C_{2d}[z,\ol{z}] \longrightarrow \C$$
which is real $L(\ol{p}) = \ol{L(p)}, \ p  \in  \C_{2d}[z,\ol{z}] ,$
and {\it positive semidefinite}, that is
$$ L( |f(z,\ol{z})|^2) \geq 0,$$ for all $f  \in  \C_{d}[z,\ol{z}] $.
In particular, but not equivalently,  $L$ is {\it hermitian positive semidefinite}, meaning that
$$ L( |f(z)|^2) \geq 0,\  f  \in  \C_{d}[z] .$$
A {\it hermitian square} is by definition a real polynomial of the form $|p(z)|^2$ with $p \in \C[z]$.

The inner product and associated semi-norm
$$ \langle p, q \rangle_L = L(p \ol{q}), \ \ \| p \|_L^2 = L(|p|^2),$$
are considered either on $ \C_{d}[z,\ol{z}] $ or on the subspace $\C_d [z]$.
In either case, Cauchy-Schwarz inequality holds
$$ |\langle p, q \rangle_L| \leq \| p \|_L \|q \|_L,$$
as well as triangle inequality.
An immediate consequence of the latter is that the set of null vectors
$$ N = \{ p \in C_{d}[z,\ol{z}], \ \|p\|_L = 0\},$$
is a vector subspace of $C_{d}[z,\ol{z}]$. The quotient space
$H_d = \C_{d}[z,\ol{z}]/N$ endowed with the induced norm $\| \cdot \|_L$ is therefore a Hilbert space of finite dimension.

For any element $f \in \C_{d-2}[z,\ol{z}] $ Cauchy-Schwarz inequality implies
$$ \| z f \|_L^2 = L( |z|^2 |f|^2) \leq \| |z|^2 f \|_L \|f \|_L.$$ In particular, if
$ \| f \|_L = 0,$ then $\|z f \|_L = 0$. Thus multiplication by the complex variable
is well defined as an induced linear transformation
$$ M_z : \C_{d-2}[z,\ol{z}]/(N \cap \C_{d-2}[z,\ol{z}]) \longrightarrow \C_{d-1}[z,\ol{z}]/(N \cap \C_{d-1}[z,\ol{z}]).$$
The above quotient spaces can be interpreted thanks to the second isomorphism theorem as subspaces of $H_d$.

Similarly we can define the subspace $A_d = C_{d}[z]/(C_{d}[z] \cap N) $ and speak about the well defined operator
$M_z : A_{d-2} \longrightarrow A_{d-1}$. Denote by $\pi_{d-2}$ the orthogonal projection of $A_d$ (or even $H_d$) onto
$A_{d-2}$. When there is no danger of confusion we simply denote $\pi = \pi_{d-2}$. The compressed operator
$$ M = M_{d-2} = \pi M_z|_{A_{d-2}}$$
is therefore well defined and bounded, as an endomorphism of a finite dimensional Hilbert space.

\section{Quadrature formulas}  We start by isolating a direct consequence of Sz-Nagy dilation theorem.

\begin{thm}\label{harmonic} Let $L : \C_{2d+2}[z,\ol{z}] \longrightarrow \C$ be a real functional which is non negative on hermitian squares, and
let $R = \| M_{d}  \|,$ with the notation adopted in the preliminaries. There are at most $N \leq (d+1)^2$ nodes $z_k = R e^{i \theta_k}$ on the circle with center at $z=0$ and radius $R$,
and positive weights $c_k >0, 1 \leq k \leq N$, such that
$$ L(h) = \sum_{k=1}^N c_k h(z_k, \ol{z}_k),$$
for all harmonic polynomials $h  \in \C_{d,d}[z,\ol{z}].$
\end{thm}

\begin{proof} The linear transform $T= M_{d}/R$ is contractive on the space $A_{d}$ of complex polynomials. Counting monomials as generators we find that $\dim A_{d} \leq d+1$.
By the finite dimensional analog of Sz.-Nagy dilation theorem, there exists a Hilbert space $K$ of dimension less than or equal to $(d+1)^2$, containing isometrically $A_d$, and a
unitary transformation $U: K \longrightarrow K$ with the property:
$$ \langle T^k {\bf 1}, {\bf 1} \rangle _L = \langle U^k {\bf 1}, {\bf 1}\rangle_K, \ \ 0 \leq k \leq d.$$
Throughout this note ${\bf 1}$ stands for the constant function equal to $1$.
For a proof see Sz.-Nagy appendix to the Functional Analysis treatise \cite{RN}, or the original construction in \cite{E}.
The spectral resolution of the unitary matrix $U$:
$$ U = \sum_{k=1}^{(d+1)^2} e^{i \theta_k} \langle \cdot, f_k\rangle f_k,$$
implies for any element $p\in \C_d[z]$:
$$ L(p) = \langle p(T) {\bf 1}, {\bf 1} \rangle _L = \langle p(U) {\bf 1}, {\bf 1}\rangle_K = $$
$$ \sum_k p(e^{i \theta_k}) |\langle {\bf 1}, f_k\rangle|^2.$$
Above $f_k$ are the unit, mutually orthogonal eigenvectors of $U$.

Since every real valued harmonic polynomial $h  \in \C_{d,d}[z,\ol{z}]$ can be represented as the real part of a complex polynomial of degree $d$ or less, the proof is complete.
The bound $N \leq (d+1)^2$ counts the non-zero weights $c_k =  |\langle {\bf 1}, f_k\rangle|^2.$

\end{proof}

It is relevant for applications to remind that the construction of the larger unitary matrix $U$ is explicit, on blocks of size $(d+1)\times (d+1)$ involving at most two square roots of
positive matrices, to be more specific, with the notation in the proof, the {\it defect matrices} $\sqrt{I-T^\ast T}$ and $\sqrt{I-TT^\ast}$, cf. \cite{E}.

Note that in the proof of the preceding theorem we have used only the hermitian positivity of the linear form $L$.
If we assume $L$ positive semi-definite, then we may expect more. We elaborate two such consequences.

First, remark that a real polynomial change of variables $X = X(z,\ol{z}), Y = Y(z,\ol{z})$ (specifically $X = \ol{X}$ and $Y = \ol{Y}$) enables to consider the subalgebra $\C[X,Y]$ and the harmonic polynomials
there. That is pull-backs by the base change of harmonic polynomials in the variables $X,Y$. Since we deal with bounded degrees we have to impose the necessary degree reductions. Assume that
$\max( \deg X, \deg Y) \leq n$. Then the functional $L$ defines by pull-back a non-negative functional on $C_{2e+2} [X,Y]$ with $2e+2$ the largest even integer less than or equal to $2d/n$.
The corresponding ``complex" variable is $Z = X(x,y) + i Y(x,y)$ and so on. We leave the details
of adapting Theorem \ref{harmonic} to the reader.

As simple as it might be, the following observation is instrumental for the rest of the article.

\begin{lem}\label{compression}  Assume the real functional $L : \C_{2d+2}[z,\ol{z}] \longrightarrow \C$ is non-negative on hermitian squares and denote
$M = \pi_d M_z \pi_d$.Then
\begin{equation}
L(p\ol{q}) = \langle p(M){\bf 1}, q(M) {\bf 1}\rangle_L,
\end{equation}
provided $\max(\deg(p), \deg(q)) \leq d+1, \deg(p\ol{q}) \leq 2d+1.$
\end{lem}

For the proof note that
$$ M p(M){\bf 1} = M p(z) = \pi_{d} (z p(z))$$
whenever $\deg(p) \leq d.$ In short we do not allow in the Lemma both polynomials $p$ and $q$ to have maximal degree $d+1$.

Second, and more interesting, are some additional constraints on the matrix
$M = M_{d}$, assuming that the functional $L$ is defined on $\C_{2d+2}[z,\ol{z}]$
and it is non-negative on all real squares.

Let $f(z,\ol{z})$ be a polynomial of degree at most $d$ in $z$, respectively $\ol{z}$:
$$ f(z, \ol{z}) = \sum_{j,k=0}^{d} f_{jk} z^j \ol{z}^k.$$
The positivity of $L$ implies in view of Lemma \ref{compression}
$$ 0 \leq L(|f|^2) = \sum_{j,k,r,s} f_{jk} \ol{f_{rs}} L(z^{j+s}\ol{z}^{k+r}) = \sum_{j,k,r,s} f_{jk} \ol{f_{rs}} \langle M^{j+s} {\mathbf 1}, M^{k+r} {\mathbf 1}\rangle.$$
Definitely these are non-trivial constraints on the matrix $M = M_{d}$. If we simplify the form of $f$ to
$$ g(z,\ol{z}) = p(z) + \ol{z} q(z),$$
 Lemma \ref{compression} yields
$$ 0 \leq L(|g|^2) = L(|p(z)|^2 + |z|^2 |q(z)|^2) + 2\Re L(\ol{z} \ol{p(z)}q(z)) =$$
$$ \| p \|^2 + \| M q \|^2 + 2\Re \langle Mp, q\rangle$$
whenever $\deg(p) \leq d$ and $\deg(q) \leq d-1$.
This is reminiscent of Halmos-Bram subnormality condition for the matrix $M$,
except that the quadratic form, defined on $A_d \oplus A_d$:
$$ \sigma_M(p,q) = \| p \|^2 + \| M q \|^2 + 2\Re \langle Mp, q\rangle$$
is positive semi-definite only on the codimension-one subspace $A_d \oplus A_{d-1}$.
For details about the operator theory background see for instance Halmos' problem book \cite{Hal}.
Consequences of such positivity conditions, and in particular relations to moment problems are explicitly
discussed in the monograph \cite{MP} and the article \cite{CP}, where a question raised by Halmos was solved.

If we impose the stronger condition
\begin{equation}\label{hypo}
 \| p \|^2 + \| M q \|^2 + 2 \Re \langle M p, q \rangle \geq 0, \ \ p, q \in A_{d},
 \end{equation}
then the matrix $M$ is what is called hyponormal, that is $[M^\ast,M] \geq 0$. Since the trace of the commutator of two matrices is always zero, we infer that
$M$ is actually normal. The spectral resolution for normal matrices will then provide the desired quadrature formula.

\begin{thm}\label{real-pos} Let $L : \C_{2d+2}[z,\ol{z}] \longrightarrow \C$ be a real functional which is non negative on all real squares.
Assume that $L$ is strictly positive on hermitian squares of bi-degree less than or equal to $(d,d)$.

There are at most $d+1$ points $a_k \in \C$ and weights $c_k>0$, so that
\begin{equation}\label{qd}
 L(p(z) \ol{q(z)}) = \sum_{k=1}^N c_k p(a_k) \overline{q(a_k)}
 \end{equation}
for all $p,q \in \C_{d+1}[z]$, subject to $\deg(p\overline{q}) \leq 2d+1$
if and only if the only non-positive eigenvalue of the quadratic form $\sigma_M$ is equal to zero.
\end{thm}

We have stated the conclusion in terms of one defect number, to stress that a single numerical criterion
certifies the existence of the quadrature (\ref{qd}).

\begin{proof} Assume condition (\ref{hypo}) holds. Then
$$ |\langle Mp, q \rangle| \leq \| p \| \|Mq\|, \ \ p, q \in A_d.$$
On the other hand
$$ |\langle Mp, q\rangle| = |\langle p, M^\ast q\rangle| \leq \|p \| \|M^\ast q\|, \   p, q \in A_d,$$
and the inequality is an equality for $p = M^\ast q$.
Hence
$$ \|M^\ast q\|^2 \leq \|M^\ast q\| \| M q\|, \ q \in A_d,$$
and consequently
$$ \|M^\ast q \| \leq \| M q\|$$
regardless $M^\ast q = 0$ or not.
But the later inequality means
$$ \langle M M^\ast q, q \rangle \leq \langle M^\ast M q, q \rangle, \ \ q \in A_d,$$
or in operator inequality terms, the commutator $[M^\ast, M]$ is non-negative.

Since ${\rm trace} [M^\ast, M] =0$ we deduce that $M$ is a normal matrix, with a spectral decomposition
$$ M = \sum_{k=0}^n a_k \langle \cdot, f_k\rangle f_k,$$
with $n \leq d+1$. Lemma \ref{compression} implies
$$  L(p(z) \ol{q(z)}) = \langle p(M){\bf 1}, q(M){\bf 1}\rangle = \sum_{k=1}^N p(a_k) \overline{q(a_k)} |\langle {\bf 1}, f_k\rangle|^2$$
for all  $p \in \C_{d+1}[z]$ and $q \in \C_{d+1}[z]$, but excluding the top degree corner, that is so that $\deg(p\overline{q}) \leq 2d+1$.

Conversely, assume that quadrature formula \ref{qd} holds in the given range of bi-degrees. Then we can reverse the preceding computations and find
$$ \langle p(M){\bf 1}, q(M){\bf 1}\rangle = \sum_{k=1}^N c_k p(a_k) \overline{q(a_k)}, $$
whenever $p, q \in \C_{d+1}[z]$ and $\deg(p\overline{q}) \leq 2d+1$. The assumption that the functional $L$
is stricly positive on hermitian squares of bidegree less than or equal to $(d,d)$ is equivalent to the fact that the vectors
${\bf 1}, z = M{\bf 1}, \ldots, z^d = M^d {\bf 1}$ form a basis of the vector space $A_d$. In other terms $M$ is a matrix with a cyclic vector, acting
on a $(d+1)$-dimensional space. Its minimal polynomial $h(\lambda)$ coincides then with the characteristic polynomial $\det(\lambda I - M)$, and in particular
has degree $d+1$. As a matter of fact one can identify $h(\lambda)$ with the orthogonal polynomial in degree $d+1$ of the associated inner product.
Whence
$$ 0 = \langle h(M){\bf 1}, q(M){\bf 1}\rangle = \sum_{k=1}^N c_k h(a_k) \overline{q(a_k)}.$$
By assumption, $N \leq d+1$, so that we can choose the polynomial $q(z)$ of degree at most $d$ to be zero at all points $a_k$ except one.
Consequently $h(a_k) = 0$ for all $k, 1 \leq k \leq N$.

Let $P,Q \in \C[z]$ be arbitrary degree polynomials. The division algorithm yields
$$ P = P_0 h + P_1, \ \ Q = Q_0 h + Q_1,$$
with $\max(\deg(P_1),\deg(Q_1)) \leq d$. Since $P_0(M) h(M) =0$ and $P_0(a_k) h(a_k)=0, 1 \leq k \leq N,$ and similarly for $Q$, we find
the identity
$$ \langle P(M){\bf 1}, Q(M){\bf 1}\rangle = \sum_{k=1}^N c_k P(a_k) \overline{Q(a_k)}$$
valid over the entire polynomial ring $\C[z]$. In other terms the operator $M$ is unitarily equivalent to the multiplication
by the complex variable on Lebesgue space $L^2(\nu)$, with $\nu = \sum_{k=1}^N c_k \delta_{a_k}.$
That is $M$ is normal and the quadratic form $\sigma_M$ is necessarily positive semi-definite.
\end{proof}

A few remarks are in order. One can start with the functional $L$ merely non-negative on hermitian squares, but then
the statement has be adapted to include the positivity of the form $\sigma_M$, and not only the reference to its only possible negative square.
The following section gives more details along this path.
Given the non-degeneracy assumption in the statement of Theorem \ref{real-pos}, we found during the proof that the number $N$ of nodes
is necessarily maximal, that is $N = d+1$. Finally one can compress the multiplier $M_z$ to other subspaces and have cubature formulas derived
from the same normality principle. Some examples collected in the final section will complement these general statements.

\section{The Hessenberg matrix representation} We keep the notation introduced in the previous section and assume that the functional $L : \C_{2d+4}[z,\ol{z}] \longrightarrow \C$ is real and non negative on all real squares.
We adopt the hypothesis of Theorem \ref{real-pos}, namely that $L$ is strictly positive on hermitian squares of bi-degree less than or equal to $(d,d)$. Then one can speak without ambiguity of the associated orthogonal complex polynomials $P_j(z), \ 0 \leq j \leq d+1:$
$$ L(P_j \ol{P_k}) = \delta_{jk}, \ \ 0 \leq j,k \leq d+1.$$
We can also assume that the leading term of $P_j$ is positive:
$$ P_j(z) = \kappa_j z^j + O(z^{j-1}), \ \kappa_j >0, \ \ \ 0 \leq j \leq d+1.$$
In particular the cyclic vector ${\bf 1}$ has the coordinates:
$$ {\bf 1} = (\kappa_0, 0, 0, \ldots, 0)^T.$$
The $(d+1) \times (d+1)$ matrix representation of the compressed multiplier $M = \pi_d M_z \pi_d : \C_d[z] \longrightarrow \C_d[z]$ with respect to the orthonormal basis
$\{ P_0, P_1, \ldots, P_d\}$ has only a first sub-diagonal non-zero:
$$ a_{jk} := \langle M P_k, P_j \rangle = 0, \ \ k < j-1.$$ Such a structure bears the name of a {\it Hessenberg matrix}:
$$ M = \left( \begin{array}{cccccc}
               a_{00}&a_{01}&a_{02}& \ldots& a_{0,d-1}&a_{0d}\\
               a_{10}&a_{11}&a_{12}&\ldots& a_{1,d-1}&a_{1d}\\
               0&a_{21}&a_{22}&\ldots&a_{2,d-1}&a_{2d}\\
               0&0&a_{32}&\ldots& a_{3,d-1}&a_{3d}\\
               \vdots & & \ddots &\ddots & & \vdots\\
               0&0& \ldots & & a_{d,d-1}&a_{dd}\\
               \end{array} \right).$$
Note that the sub-diagonal entries are non-zero:
$$ a_{j+1,j} = \langle M P_j, P_{j+1} \rangle = \langle \kappa_j z^{j+1} + \ldots, P_{j+1} \rangle = \frac{\kappa_j}{\kappa_{j+1}}, \ \ 0 \leq j <d.$$

Returning to the quadratic form $\sigma_M$ defined in the previous section, we turn to its matrix representation:
$$ \sigma_M(p,q) = (\ol{p}, \ol{q}) \left( \begin{array}{cc}
                        I&M^\ast\\
                        M& M^\ast M\\
                        \end{array}\right) \left( \begin{array}{c}
                        p\\
                        q
                        \end{array}\right).$$
         Our assumption implies, as explained in the previous section, that the above block matrix is non-negative on the subspace $C_{d}[z] \oplus C_{d-1}[z]$.

         A notable change of coordinates block-diagonalizes this matrix with self-commutator $[M^\ast, M]$ as one of the blocks:
         $$  \left( \begin{array}{cc}
                        I&0\\
                        -M& I\\
                        \end{array}\right)    \left( \begin{array}{cc}
                        I&M^\ast\\
                        M& M^\ast M\\
                        \end{array}\right)  \left( \begin{array}{cc}
                        I&-M^\ast\\
                        0& I\\
                        \end{array}\right) = \left( \begin{array}{cc}
                        I&0\\
                        0& [M^\ast, M]\\
                        \end{array}\right).$$
                        In virtue of the min-max principle, the self-adjoint matrix $[M^\ast,M]$ has at most one negative eigenvalue.

With some elementary computations, we can say more about the possible negative eigenspace of this self-commutator. To be more precise, start with the one dimension up compression
  $M_{d+1} = \pi_{d+1} M_z \pi_{d+1}$. This is well defined as a linear transform due to the assumption that the original functional is non-negative on $\C_{2d+4}[z,\ol{z}]$. All the above computations hold, except the existence of the orthogonal polynomial $P_{d+2}$, which is not needed in what follows.
  In particular the matrix   $$  \left( \begin{array}{cc}
                        I&M_{d+1}^\ast\\
                        M_{d+1} & M_{d+1} ^\ast M_{d+1}\\
                        \end{array}\right)   $$
                        is non-negative on the subspace $\C_d[z] \oplus \C_d[z]$.
                        That is, the matrix
                        $$ \left( \begin{array}{cc}
                        I&M^\ast\\
                        M& \pi_d M_{d+1} ^\ast M_{d+1}\pi_d \\
                        \end{array}\right)$$
                        is non-negative on   $\C_d[z] \oplus \C_d[z].$
   But   $$ \pi_d M_{d+1} ^\ast M_{d+1}\pi_d   = M^\ast M + \pi_d M_{d+1} ^\ast (\pi_{d+1}-\pi_d) M_{d+1}\pi_d.$$
   It remains to identify along the orthonormal basis given by the complex polynomials $P_j$ the contribution of the last rank-one matrix:
   $$  \pi_d M_{d+1} ^\ast (\pi_{d+1}-\pi_d) M_{d+1} P_j = \pi_d M_{d+1} ^\ast (\pi_{d+1}-\pi_d) (z P_j (z)) = $$ $$\pi_d M_{d+1} ^\ast (  z P_j (z) - z P_j (z)) = 0, \  j < d,$$
   and
$$  \pi_d M_{d+1} ^\ast (\pi_{d+1}-\pi_d) M_{d+1} P_d =  \pi_d M_{d+1}^\ast \langle z P_d, P_{d+1}\rangle P_{d+1} = a_{d, d+1} \pi_d M_{d+1}^\ast P_{d+1}.$$
Note that $K = \pi_d M_{d+1} ^\ast (\pi_{d+1}-\pi_d) M_{d+1}\pi_d$ is a non-negative self-adjoint matrix, with the only non-zero entry on the $d \times d$ diagonal position:
$$ \langle K q, q \rangle = a_{d+1,d}^2 |\langle q, P_d \rangle|^2, \ \ q \in \C_d[z].$$
We infer that the matrix
$$ \left( \begin{array}{cc}
                        I&M^\ast\\
                        M& M^\ast M + K\\
                        \end{array}\right)$$
                        is non-negative on the whole space $\C_d[z]\oplus \C_d[z]$, and via the same elementary transforms we conclude that
                        $[M^\ast,M] + K$ is a non-negative matrix.
  This information gives a lower bound for the only possible negative eigenvalue of the self-commutator:
  $$ \lambda_{-} = \inf \{ \langle [M^\ast,M] q, q \rangle; \ \ \| q \| \leq 1 \}. $$ Indeed, fix a polynomial $q \in \C_d[z]$ of norm less than $1$. Then
  $$ \lambda_{-} + a_{d+1,d}^2 \geq \langle [M^\ast,M] q,q \rangle +    a_{d+1,d}^2   \|q \|^2 \geq    $$ $$
  \langle [M^\ast,M] q,q \rangle + a_{d+1,d}^2 |\langle q, P_d \rangle|^2 = \langle (  [M^\ast,M] + K)q, q \rangle \geq 0.$$

  We collect the above computations into a single statement.

  \begin{thm}\label{Hess} Let $L : \C_{2d+4}[z,\ol{z}] \longrightarrow \C$ be a real functional which is non negative on all real squares.
Assume that $L$ is strictly positive on hermitian squares of bi-degree less than or equal to $(d,d)$. Denote by $(a_{jk})_{j,k=0}^{d+1}$ the Hessenberg matrix associated
to the system of complex orthogonal polynomials induced by $L$.

Then the matrix $M = (a_{jk})_{j,k=0}^{d}$ is normal after a rank-one perturbation, in particular the self-commutator $[M^\ast,M]$ has at most one negative eigenvalue $\lambda_{-} \geq - a_{d+1,d}^2.$
Moreover, the matrix $M$ is normal, that is $[M^\ast,M] =0$ if and only if $a_{d+1,d}=0$.
\end{thm}

Remark that the non-negativity assumption in the statement implies that the $(d+2) \times (d+2)$ Hessenberg matrix $\tilde{M} = (a_{jk})_{j,k=0}^{d+1}$ is well defined, hence the normality of its
submatrix $M$ is equivalent to the fact that $\tilde{M}$ leaves invariant the subspace $\C_d[z]$. More precisely,
$$ \tilde{M} = \left( \begin{array}{cc}
                           M&\ast\\
                           u& a_{d+1,d+1}\\
                           \end{array} \right), \ \ {\rm where} \ \ \ u = (0,0,\ldots, 0, a_{d, d+1}).$$
While there is a simpler, operator theoretic derivation of the conclusion of the above theorem, we preferred to enter minimally into the structure of complex orthogonal polynomials for alerting the
reader about the potential of the Hessenberg matrix approach. For the linear numerical analyst or approximation theory expert it is hardly a surprise to put the Hessenberg matrix representation at work,
see for instance \cite{HJ,Simanek}, to only touch the ample literature devoted to the subject.

With the above understanding of the normality of the compressed multiplier $M$, we can return to the main result of the previous section, with a quantitative criterion.

\begin{cor} Let $L : \C_{2d+4}[z,\ol{z}] \longrightarrow \C$ be a real functional which is non negative on all real squares.
Assume that $L$ is strictly positive on hermitian squares of bi-degree less than or equal to $(d,d)$.

There are at most $d+1$ points $a_k \in \C$ and weights $c_k>0$, so that
$$
 L(p(z) \ol{q(z)}) = \sum_{k=1}^N c_k p(a_k) \overline{q(a_k)}
$$
for all $p,q \in \C_{d+1}[z]$, subject to $\deg(p\overline{q}) \leq 2d+1$
if and only if the Hessenberg matrix entry $a_{d, d+1}$ vanishes.
\end{cor}

  Summing up, the main observation supporting the preceding Corollary is stated in the following rather surprising property of the associated Hessenberg matrix.

  \begin{prop} Let $L : \C_{2d+4}[z,\ol{z}] \longrightarrow \C$ be a real functional which is non negative on all real squares.
Assume that $L$ is strictly positive on hermitian squares of bi-degree less than or equal to $(d,d)$ and denote by $M_k = \pi_k M_z \pi_k$
the associated Hessenberg matrix of order $k, \  k \leq d+1$.

The following are equivalent:

\begin{enumerate}
\item $M_d$ is a normal matrix,

\item $\det ([M_d^\ast,M_d]+ \epsilon I) \geq 0$ for $\epsilon >0$,

\item The entry $a_{d, d+1}$ in $M_{d+1}$ vanishes,

\item$M_{d+1}$ leaves invariant the subspace $\C_d[z]$.

\end{enumerate}

\end{prop}

To complete the proof, it suffices to recall that the self-commutator $[M^\ast,M]$ can possess at most one negative eigenvalue. Condition $(2)$ simply states that $[M^\ast,M]$ cannot
have a negative eigenvalue, hence $[M^\ast,M] \geq 0$, hence $M$ is normal. We stress that in general condition (2) offers a numerical certificate for a symmetric matrix with at most one
negative eigenvalue to be non-negative.

So far, we have only used the positivity of the functional $L$ on hermitian squares and on squares of the form $|f(z) + \ol{z} g(z)|^2$. A whole hierarchy of stronger non-equivalent conditions, for instance on the
Hessenberg matrix $M$, can be derived from the positivity of $L$ on elements $|P(z,\ol{z})|^2$ where $\deg_{\ol{z}}P \leq k$. This later property, called $k$-hyponormality, is analyzed in detail and put at work in \cite{CP}.

\section{Examples}

In this section we gather a few simple examples which support the general theme of this note: that combining complex variables with positivity of linear functionals
is leading to useful, sometimes unexpected, consequences.

\begin{example} Contrary to the tacit convention of this note, we start with a scalar product rather than a potential integration functional with positivity
assumption. Specifically, let $[-a,a]$ be a compact interval of the real line and consider the Dirichlet type inner product:
$$ \langle p, q \rangle = \int_{-a}^a [p(x)\overline{q(x)} + p'(x)\overline{q'(x)}] dx,$$
defined for all polynomials $p,q \in \C[x]$ with complex coefficients. Apparently there is no linear functional $L$, so that
$$ L(p\overline{q}) = \langle p, q \rangle.$$
One can remedy this by passing to complex coordinates. Namely, define for $p, q \in \C[z]$ the sesquilinear form
$$ \Lambda(p\overline{q}) = \int (1+\frac{1}{4}\Delta)(p\overline{q}) d\mu,$$
where $\mu$ is any rapidly decreasing at infinity positive measure. Above $\Delta$ is Laplace operator, and its factorization via complex variables
$$ \Delta = 4 \frac{\partial}{\partial z} \frac{\partial}{\partial \ol{z}}$$ implies
$$  \Lambda(p\overline{q}) = \int (p\overline{q} + p'\overline{q'}) d\mu,$$
when this time $p'(z), q'(z)$ denote complex derivatives. By choosing $d\mu = \chi_{[-a,a]} dx$ one finds unexpectedly
$$ \Lambda(p\ol{q}) =  \langle p, q \rangle, \ \ p,q \in \C[z].$$
The multiplier $M_z$ is well defined on the whole algebra $\C[z]$, but its compressions to prescribed degrees are not normal.
However, a familiar computation borrowed from Sturm-Liouville theory shows that one can restrict $M$ to a space of polynomials where it is. To be precise,
$$ \langle M_z p(z), q(z) \rangle + \langle p(z), M_z q(z) \rangle = $$ $$ \int_{-a}^a [ xp(x)\ol{q(x)} + p(x) \ol{xq(x)} + (xp'(x) + p(x))\ol{q'(x)} +p'(x) \ol{(x q'(x) + q(x))}] dx=
$$ $$ 2 \int_{-a}^a (1+\frac{1}{4}\Delta)(p\overline{q}) x dx + \int_{-a}^a [p(x)\ol{q'(x)} +p'(x) \ol{q(x)}] dx.$$
Fix a dimension $d$ and consider the subspace $V(e)_d$ of even polynomials vanishing at $\pm a$, of degree less or equal than $d$, or the subspace $V(o)_d$ of odd polynomials
vanishing at $\pm a$. Along every one of these two subspaces the last to integrals vanish. Hence the compressions $M_e$, respectively $M_o$ of $M_z$ to these subspaces are skew symmetric,
i.e. $M_e + M_e^\ast = 0,$ respectively $ \ M_o + M_o^\ast = 0$. Hence $M_o$ and $M_e$ are normal with purely imaginary spectra.

In conclusion, there are quadrature formulas of the form
$$ \int_{-a}^a [p(x)\overline{q(x)} + p'(x)\overline{q'(x)}] dx = \sum_{k=1}^N c_k p(i a_k) \ol{q(ia_k)},$$
with $c_k >0$ and $a_k \in \R$ for all $k, 1 \leq k \leq N$, valid for both polynomials $p,q$ either of the form
$ (z^2-a^2) f(z^2)$, or $(z^2-a^2) z f(z^2)$ and $f \in \C_n[z]$ with a prescribed $n$. Counting degrees we can specify $N \leq n+1$.
The complex location of the quadrature nodes is another strong indication that passing from real to complex variables is very natural in this case.

\end{example}

\begin{example} To prove that the minimal node number (producing a so called Gaussian type quadrature) in Theorem \ref{real-pos} is needed, we simply look at a counting measure supported by the
vertices of a regular $n$-gon.

Indeed, let $n>2$ be a positive integer and let $\epsilon = e^{i 2\pi/n}$ be the primitive root of order $n$ of unity. The measure
$d\nu = \frac{1}{n}(\delta_{1} + \delta_{\epsilon}+ \ldots + \delta_{\epsilon^{n-1}})$ has complex moments concentrated on the diagonal, up to degree $n$:
$$ \int z^k \ol{z}^\ell d\nu = 0, \ \ 0 \leq k,\ell \leq n, \ k \neq \ell.$$
The associated complex orthonormal polynomials are ${\bf 1}, z, \ldots, z^n$. Let $d<n$ and consider the orthogonal projection $\pi_d$ of the finite Hilbert space
$L^2(\nu)$ onto the space $A_d$ of complex polynomials of degree less than or equal to $d$. The compressed matrix $M_d = \pi_d M_z \pi_d$ of the multiplier $M_z$
is the Jordan block of size $(d+1)\times (d+1)$, hence far from being normal. However, a finite point quadrature exists (with more than $d+1$ nodes)
$$ \langle p(M_d){\bf 1}, q(M_d){\bf 1} \rangle = \int p \ol{q} d\nu,$$
whenever $\deg(p) \leq d+1$ and $\deg(q) \leq d$.

\end{example}

\begin{example}

A non-negative real functional which is positive on hermitian squares is offered by the integration with respect to arc length on the unit circle:
$$ L( z^j \ol{z}^k) = \delta_{jk}, \ \ j,k \geq 0.$$
The associated Hessenberg matrix of order $(d+1) \times (d+1)$ is the Jordan block, that is the Toeplitz matrix with equal entries to $1$ on the sub diagonal
and zero elsewhere. The complex orthogonal polynomials are $P_j(z)=z^j, \ \ j \geq 0.$

Note that
$$ L(|z|^2 q(z,\ol{z})) = L( q(z,\ol{z})) $$
for any polynomial $q \in \C[z, \ol{z}]$
and in particular
$$ L(|1-\ol{z} z|^2) =0.$$
In other terms the form $\sigma_M$ and the self-commutator $[M^\ast,M]$ may be degenerate, in spite of the fact that $L$ is positive definite on hermitian squares, that is:
$$ L(|f(z)|^2) = 0 \ \ \Rightarrow f =0.$$
\end{example}

config: arxiv


TITLE: Convergence in Probability and Convergent almost surely
QUESTION [2 upvotes]: I am new to probability theory. Pardon me if question is too lame.
I came across the following theorem:
Suppose $X_n \xrightarrow P X_\infty$ and $\sum_{n=1}^\infty P( \left| X_n-X_\infty \right| > \epsilon) < \infty ~~~~~~~     \forall \epsilon >0$
Then $X_n \xrightarrow {a.s} X_\infty$
My doubts:

Doesn't $\sum_{n=1}^\infty P( \left| X_n-X_\infty \right| > \epsilon) < \infty$ implies that $ \lim_{n \to \infty} P( \left| X_n-X_\infty \right| > \epsilon)=0$ and hence $X_n \xrightarrow P X_\infty$? I feel that $X_n \xrightarrow P X_\infty$ is redundant given the summation condition. Am I missing out on something? Is it to imply the existence of a limit point, $X_\infty$ ?
The proof given with the theorem, uses Borel-Cantelli's lemma and then states that $ P(|X_n-X_\infty|>\epsilon, i.o.) = 0$, which implies almost sure convergence. I tried to break it down into following smaller steps.

Let $A_n = \{|X_n-X_\infty|>\epsilon\}$ be a sequence of sets.
Using Borel-Cantelli's lemma, $P( \limsup_{n \to \infty} A_n) =0$
$\implies P (\bigcap_{m=1}^ \infty \bigcup_{n=m}^\infty \{|X_n-X_\infty|>\epsilon\}) =0$
Let $B_m = \bigcup_{n=m}^\infty \{|X_n-X_\infty|>\epsilon\}$. This is a decreasing sequence of sets. Using continuity of probability measure,
$ P( \lim_{m \to \infty} B_m) = \lim_{m \to \infty} P(B_m)$
$\implies P( \lim_{m \to \infty} \bigcup_{n=m}^\infty \{|X_n-X_\infty|>\epsilon\}) = \lim_{m \to \infty} P(\bigcup_{n=m}^\infty \{|X_n-X_\infty|>\epsilon\})$
$ \implies P( \lim_{n \to \infty} \{|X_n-X_\infty|>\epsilon\}) =  \lim_{n \to \infty} P(\{|X_n-X_\infty|>\epsilon\})$
This implies equivalence in convergence in probability and almost sure convergence under the given condition.
I am not sure if all the steps are right especially the equivalence of last two implied statement. Even if they are equivalent I am not very clear about them. It would be very helpful if you verify the proof. Thank you.

REPLY [1 votes]: Your observation in 1) is correct.
For 2) let $A_k=\{\omega: |X_n(\omega) -X_{\infty}(\omega))| >\frac 1 k\}$. Then $P(A_k)=0$ for each $k$. If $A =\cup_k A_k$ then $P(A)=0$. If $\omega \notin A$ then $\omega \notin A_k$ for any $k$. Hence, for each $k$ we have $|X_n(\omega) -X_{\infty}(\omega))| \leq \frac 1 k$ whenever $n$ is sufficiently large. By deifinition of limit this says that $X_n(\omega) \to X_{\infty}(\omega)$.
The main step in 2) is to vary $\epsilon$ and get one set of probability $0$. So the equivalence you stated in your argument is not correct.


TITLE: How do I compute the limit of a radical outside of a fraction?
QUESTION [0 upvotes]: How do I compute this?
$$\lim_{x\to\infty}\left(\sqrt[3]{n^2+5}-\sqrt[3]{n^2+3}\right)$$
I tried multiplying it by the conjugate exactly as if it were in a fraction, but it didn’t quite work out, because you end up with a way bigger expression.
Arriving at the solution likely uses the sandwich theorem, but with two separate expressions. I find it to be quite the challenge.

REPLY [0 votes]: $$\lim_{x\to\infty}\left(\frac{\sqrt[3]{n² +5} \ - \ \sqrt[3]{n² + 3}}{1}\right)$$
Multiply numerator and denominator by
$$\left(\sqrt[3]{n² +5}\right)^2+\left(\sqrt[3]{n² +5}\right)\left(\sqrt[3]{n² +3}\right)+\left(\sqrt[3]{n² +3}\right)^2$$
the fraction becomes
$$\frac{\left(\sqrt[3]{n² +5}-\sqrt[3]{n² + 3}\right)\left[\left(\sqrt[3]{n² +5}\right)^2+\left(\sqrt[3]{n² +5}\right)\left(\sqrt[3]{n² +3}\right)+\left(\sqrt[3]{n² +3}\right)^2\right]}{\left(\sqrt[3]{n² +5}\right)^2+\left(\sqrt[3]{n² +5}\right)\left(\sqrt[3]{n² +3}\right)+\left(\sqrt[3]{n² +3}\right)^2}$$
Recalling the identity $(a-b)(a^2+ab+b^2)=a^3+b^3$ we get
$$\lim_{x\to\infty}\frac{(n^2+5)-(n^2+3)}{\sqrt[3]{(n² +5)^2}+\sqrt[3]{(n² +5)(n^2+3)}+\sqrt[3]{(n² +3)^2}}=$$
$$=\lim_{x\to\infty}\frac{2}{\sqrt[3]{(n² +5)^2}+\sqrt[3]{(n² +5)(n^2+3)}+\sqrt[3]{(n² +3)^2}}=0$$
Hope this helps


TITLE: When finding the minimal polynomials in GF(16), Why the 4th conjugate of α³ becomes α^9?
QUESTION [1 upvotes]: Shouldn't it be α^18, which in GF(16) simplifies to α^3 since α^15 = 1 in GF16?
Same is also confusing for the 3rd and 4th conjugate of α^7 while finding the minimal polynomial in GF (16), how they get to have α^13 and α^11?
To make matter easy I am also attaching. an image for conjugacy classes different roots of GF16

REPLY [4 votes]: The conjugates, apart from $\newcommand{\al}{\alpha}\al^3$ are $\al^6$, $\al^{12}$ and $\al^{24}=\al^9$. I got these by repeatedly applying the
Frobenius map $x\mapsto x^2$ to $\al^3$.


TITLE: Prove there's $x_0$ such that $f'(x_0)=0$
QUESTION [1 upvotes]: Let $f:\mathbb{R}\rightarrow \mathbb{R}$ differentiable at $\mathbb{R}$ and:
$$\lim_{x\rightarrow \infty}\left(  f(x)-f(-x) \right) = 0$$
Show there's $x_0$ such that $f'(x_0) = 0$.
I tried to use the limit definition, but couldn't get much further.
I'll be glad for guidance.
Thanks.

REPLY [3 votes]: If $f$ is a constant function, the result is easy.
Here is an outline of one way to prove such an $x_0$ exists if $f$ is not a constant function:

Choose $a, b$ with $f(a)< f(b)$ for some $a,b$.

Show that there is an $N>\max\{|a|,|b|\}$ so that either $\max\{f(N), f(-N)\}<f(b)$ or   $\min\{f(N), f(-N)\}>f(a)$ (choose $N$ so that $f(N)$ is within $|f(b)-f(a)|/2$ of $f(-N)$; drawing a picture will help here).

Use the result of 2. and the Intermediate Value Theorem to show $f(c)=f(d)$ for some $c\ne d$.

Wrap things up with Rolle's Theorem.

Alternatively:
If $f'(x)$ is never $0$, then by Darboux's Theorem,  $f'$ is either positive everywhere or negative everywhere). In this case $f$ is either strictly increasing or strictly decreasing. But then, for $x>1$, $|f(x)-f(-x)|> |f(1)-f(-1)|>0$; whence  $\lim\limits_{x\rightarrow\infty} \bigl(f(x)-f(-x)\bigr)\ne0$.


TITLE: How is $F=ma$ a differential equation? (Confirmation)
QUESTION [1 upvotes]: Just wanna make sure I have this right. The derivative of position is velocity, and  the derivative of velocity is acceleration
so if $$F = ma$$
where
$$a=\frac{d}{dx} v(t)\\
v(t)=\frac{d}{dx} s(t)$$  where $s(t)$ is the position in time.
This means that $$a = s''(t)$$
which means $$F = ms''(t)$$
and its an ordinary differential equation because it contains ordinary derivatives and it doesn't have the weird partial derivative symbol anywhere?
Is everything above right? Thanks.

REPLY [1 votes]: Just expand a little on what everyone has already said: In general, assuming the mass $m$ and velocity $\mathbf{v}$ are functions of time only:
$$ \mathbf{F}= \frac{d}{dt}\left(m\mathbf{v}\right) = m\frac{d \mathbf{v}}{dt}+
 \frac{dm}{dt}\mathbf{v}$$
The above results follows from the fact that force is the rate of change of momentum with time, and momentum is the product of mass and velocity.
Now, if the mass $m$ is constant, then $\mathbf{v} =\frac{d \mathbf{s}}{dt}$ and
$$\mathbf{F}= m\frac{d \mathbf{v}}{dt} = m\mathbf{a}$$
Finally (again assuming that $m$ is constant), if $\mathbf{s}$ is the displacement vector as a function of time, then $\mathbf{v} =\frac{d \mathbf{s}}{dt}$, which implies that
$$\mathbf{F}= m\frac{d}{dt}\left(\frac{d \mathbf{s}}{dt}\right) = m\left[\frac{d^2}{{dt}^2}\mathbf{s}(t)\right] = m\mathbf{a}$$


TITLE: non homogeneous heat equation?
QUESTION [0 upvotes]: consider the non homogeneous heat equation :
let $f:\mathbb{R}^n\times \mathbb{R}^+ \rightarrow \mathbb{R}$ be a smooth function
$\frac{\partial u}{\partial t}-\Delta u = f(x,t)$ , $x\in\mathbb{R}^n,t\in\mathbb{R}$
$u(x,0)=0$
so can someone tell me how to verify that the solution of this equation is given by :
$u(x,t)=\int_0^t(\int_{\mathbb{R}^n}\phi_n(x-y,t-s)f(y,s)dy)ds$
where $\phi_n$ is the fundamentale solution of the homogeneous heat equation.
thank you very much.

REPLY [0 votes]: See Evans PDE book, Theorem 2, p. 50. or see page 18 of PDF.


TITLE: Prove that the linear mapping $f: V \to V'$ is injective if and only if $g: V' \to V$ exists such that $gf = 1_V$
QUESTION [0 upvotes]: Prove that the linear mapping $f: V \to V'$ is injective if and only
  if $g: V' \to V$ exists such that $gf = 1_V$

If $f$ is injective, $Kerf=0$, but I actually don't understand what $(1_V)$ means, I know - elementary knowledge. Because our mapping is mapped back into $V$?

REPLY [1 votes]: You already showed in the comments that if $f$ is not injective then no such $g$ exists. So assume $f$ is injective. Let $M$ be a basis of $V$. Then since $f$ is injective, $f(M)$ is linearly independent in $V'$ (check this). Complete $f(M)$ to a basis $M'$ of $V'$. Define $g$ on $M'$ by saying $g(x)=y$ if $x\in f(M)$ and $x=f(y)$, and define $g(x)=0$ if $x\notin f(M)$. Then extend $g$ linearly. This gives the required left inverse.


TITLE: Is this a known learning problem?
QUESTION [4 upvotes]: Let $(\mathcal{X},\rho)$ be a metric space (say, $\mathcal{X}=[0,1]$ with the Euclidean metric). Let $\alpha:\mathcal{X}\to[0,1]$ be unknown. Suppose that $\mathcal{X}$ is endowed with a distribution $\mu$ from which $(X_1,\ldots,X_n)$ is sampled iid. Further, for each $X_i$, we observe $Y_i\sim\mathrm{Bernoulli}(\alpha(X_i))$.
Our goal is to recover $\alpha$ from the finite sample, under some loss function (I tend to favor the pointwise loss $\ell(\hat\alpha,\alpha)=$
$$KL(\hat\alpha||\alpha)
=
\hat\alpha\log(\hat\alpha/\alpha)
+
(1-\hat\alpha)\log[(1-\hat\alpha)/(1-\alpha)]
,$$
but other loss functions can be considered also.)
Obviously, there is no hope of recovering $\alpha$ without some regularity condition, such as Lipschitz continuity.
Does this problem fall under some known framework?
Edit: Following Clement's comment, a clarification. The loss I defined above is pointwise, at each $x$ --- so it should really be $\ell(\hat\alpha(x),\alpha(x))$. The overall risk is then
$$R(\alpha)=E_X[\ell(\hat\alpha(X),\alpha(X))].$$

REPLY [3 votes]: Well, we wrote a paper on it, so now it's definitely known:
https://arxiv.org/abs/2010.09886


TITLE: Average of all 3 digit numbers with distinct digits
QUESTION [0 upvotes]: I tried adding up all the digits in each digits place and average them, but I got some decimals.
I don't know if I am doing it correctly
Can someone please explain this to me

REPLY [1 votes]: The only thing that makes this hard is that the numbers can't start with $0.$  If they were allowed to start with $0,$ it would be easy, so let's do the problem on the assumption that numbers are allowed to start with $0,$ and then adjust the result.
If numbers can start with $0$, there are $10\cdot9\cdot8=720$ nunbers with distinct digits.  By symmetry, each of the digits appears the same number of times $(72)$ in each column, and since the sum of the numbers from $0$ t $9$ is $45,$ the sum of the numbers is $$72\cdot45\cdot111=359640,$$ where the $111$ comes from the fact that we are summing the ones, tens, and hundreds columns.
Now we just have to subtract out the sum of the numbers that started with $0$.  There are $9\cdot8=72$ of these, since the last two digits must be distinct, and neither can be $0.$  So, to finish this off, we need to subtract the sum of the two-digit numbers with distinct non-zero digits, and divide by $648(720-72).$
I'll leave that to you.


TITLE: Does UFD imply ACC?
QUESTION [0 upvotes]: UFD imply ACCP (ascending chain condition for principal ideals). But is it necessarily UFD implies ACC also (ascending chain condition for ideals)?

REPLY [6 votes]: No. If $K$ is a field, $K[X_1,X_2,\dots, X_n,\dots]$  is an example of a non-noetherian UFD.


text:
In this section the main estimates and proofs of positivity of the commutator are discussed. First we introduce the two terms $A_0$ and $\CQ{Q}$ which we add to $C_1$ as already mentioned in the overview of the proof. Then we show how the main theorem is proven given that we know that the sum of all three terms is positive (\thref{th:positive}). Then, in \Autoref{subsec:error} we prove how we estimate the three terms separately and which error terms occur. With that we conclude by proving \thref{th:positive}.

In this section the conditions of \autoref{th:main} are assumed, that is, that \ref{ass:h1},  \ref{ass:h2},  \ref{assumption:I1}, \ref{assumption:I2}, \ref{assumption:I3} and \ref{fgrc} are satisfied.

\subsection{Putting Things together, Proof of the Main Theorem}
\label{subsec:main proof}
In the following we use  for $m \in \IN_0$ the short-hand notation
\[
\Nfh := \Id_\p \otimes \Id_\p \otimes \Nf, \qquad \Poh := \Id_\p \otimes \Id_\p \otimes P_\Omega.
\]
Remember that we have on $\Ds$,
\begin{equation} \label{eq:calccomm}
C_1 = \Vc^* \q^2 \Vc  \otimes \Id_\p \otimes \Id_\f +  \Id_\p \otimes  \Vc^* \q^2 \Vc    \otimes \Id_\f + \Nfh + \lambda W_1 ,
\end{equation}
where $W_1$ was the commutator with the interaction, cf. \eqref{eq:cn_formula 2}. Obviously, $C_1$ is strictly positive on the orthogonal complement of the vacuum subspace for $\lambda = 0$ and its first two terms are positive on $(\ran (\Pd \otimes \Pd))^\perp \otimes \FF$.

On the space $\ran \Pi$ we use the Fermi Golden Rule and introduce the corresponding conjugation operator $A_0$ in order to obtain a positive expression in \thref{th:fgr} as a commutator with $L_\lambda$. It was first developed for zero temperature systems in \cite{a0firsttime} and later adapted to the positive temperature case in \cite{merkli_positive}.
It is a bounded self-adjoint operator on $\H$, given by
\muu
{
A_0 := \i \lambda (\Pi W R_\epsilon^2 \oPi - \oPi R_\epsilon^2 W \Pi),
}
where  $R_\epsilon^2 := (L_0^2 + \epsilon^2)^{-1}$ and $\epsilon > 0$. We will choose $\epsilon > 0$ such that \ref{fgrc} holds.
One can show that $\ran A_0 \subseteq \Def(L_\lambda)$ since $\ran A_0 \subseteq \H_\p \otimes \H_\p \otimes \FF_\fin$ (cf. \autoref{th:a0 welldefined}). Furthermore,
\muun
{
\label{eq:L A0}
\i [L_\lambda, A_0] = - \lambda [L_\lambda, \Pi W R_\epsilon^2 \oPi - \oPi R_\epsilon^2 W \Pi]
}
extends to a bounded self-adjoint operator as well and we have $ \epsilon \Pi \i[  L_\lambda, A_0] \Pi > 0$ for small $\epsilon > 0$, see \thref{th:fgr}.

To obtain positivity on the remaining space $(\ker L_\p)^\perp \otimes \ran P_\Omega$ we introduce the bounded operator
\begin{align}
\label{eq:CQ}
\CQ{Q} := Q \otimes P_\Omega    +   \frac{\lambda}{2}    W    ( L_\p^{-1}  Q \otimes  P_\Omega )   + \frac{\lambda}{2}    ( W ( L_\p^{-1}  Q \otimes P_\Omega)   )^*
\end{align}
on $\H$, where $L_\p^{-1}$ is to be understood in the sense of functional calculus as an unbounded operator, and
\begin{align}
\label{eq:Q defn}
Q := L_\p^2 \ind_{[-1,1]}(L_\p) + \ind_{(-\infty,-1) \cup (1,\infty)}(L_\p).
\end{align}
Notice that by construction $\ran Q \subseteq \Def(L_\p^{-1})$, $L_\p^{-1} Q$ is bounded and self-adjoint, so the definition of $\CQ{Q}$ makes sense. Furthermore, the first summand $ Q \otimes P_\Omega $  is indeed positive on $(\ker L_\p)^\perp \otimes \ran P_\Omega$.

The idea will be to show that the sum of the three operators \eqref{eq:calccomm}, \eqref{eq:L A0} and \eqref{eq:CQ} is  positive and has zero expectation with any element of the kernel of $L_\lambda$.
To this end we define for $\psi \in \Def(\qce) \cap \Def(L_\lambda)$ and some $\theta > 0$,
\[
\qtot(\psi) := \qce(\psi) +  \theta \sc{\psi,\i [L_\lambda, A_0] \psi} + \sc{\psi,\CQ{Q} \psi}.
\]
By the virial theorem for $C_1$ and by construction of the other two terms, this form is actually non-positive for any $\psi \in \ker L_\lambda$. This is the content of the following proposition.
\begin{prop}
\label{th:virial}
For arbitrary $\lambda$, $\epsilon$, $\delta$, and $\psi \in \ker L_\lambda$ we have $\psi \in \Def(\qce)$ and $\qtot(\psi) \leq 0$.
\end{prop}
\begin{proof}
Let $\psi \in \ker L_\lambda$. By \thref{th:result virial} we know that $\psi \in \Def(\qce)$, and
\[
\qce(\psi) \leq 0.
\]
It is clear that  $\sc{\psi,\i [L_\lambda, A_0] \psi} = 0$. Furthermore, by construction, the last term vanishes:
\begin{align*}
0 &= \sc{ (L_\p^{-1} Q \otimes P_\Omega)  L_\lambda \psi, \psi} \\
&= \sc{ \psi, L_\lambda  (L_\p^{-1} Q \otimes P_\Omega)  \psi}\\
&= \sc{ \psi, (L_\p \otimes \Id_\f + \lambda W) (L_\p^{-1} Q \otimes P_\Omega)  \psi} \\
&= \sc{ \psi,  (Q \otimes P_\Omega) \psi  + \lambda W  (L_\p^{-1} Q \otimes P_\Omega )  \psi }.
\end{align*}
Then $\sc{\psi,\CQ{Q} \psi} = 0$ follows by adding the complex conjugate term. Thus,
\[
\qtot(\psi) = \qce(\psi) \leq 0. \qedhere
\]
\end{proof}
We prove in the next subsection the following proposition which states that $\qtot$ is in fact positive. Remember that $\gb$ is the constant appearing in \ref{fgrc} which might depend on $\beta$.
\begin{prop}
\label{th:positive}
Let $\beta_0 > 0$. Then there exists constant $\lambda_0,C > 0$ such that for all $0 < \abs \lambda < \lambda_0$ with $\abs \lambda < C\gb^2$ and all $\beta \geq \beta_0$, we have $\qtot > 0$. That is, $\qtot \geq 0$ and $\qtot(\psi) = 0$ for some $\psi \in \Def(\qce) \cap \Def(L_\lambda)$ implies $\psi = 0$.
\end{prop}
This positivity result together with \autoref{th:virial} now implies the main theorem \thref{th:main}.
\subsection{Error Estimates}
\label{subsec:error}

In the following proposition we prove separate estimates from below  of the three operators \eqref{eq:calccomm}, \eqref{eq:L A0} and \eqref{eq:CQ}.
\begin{prop}
\label{th:total estimate steps} \thmenum
\begin{enumerate}[label=(\alph*)]
\item
\label{it:C1 estimate}
There exist constants $c_1$, $c_2 > 0$ such that, for all $\lambda \in \IR$, we have in the sense of quadratic forms on $\Def(\qce)$,
\begin{align}
\label{eq:C1 estimate}
C_1 &\geq   \big[\Vc^* \q^2 \Vc  \otimes \Id_\p  + \Id_\p \otimes   \Vc^* \q^2 \Vc   \\ &- c_1 (1+ \beta^{-1}) \lambda^2 \big( \Vc^* \sc{\po}^{-2} \Vc \otimes \Id_\p +  \Id_\p  \otimes  \Vc^* \sc{\po}^{-2} \Vc + \PP{(\Pc \otimes \Pc)} \big)  \big]    \nonumber
\\ &\qquad  \otimes \Id_\f  + c_2  \PP{\Poh}. \nonumber
\end{align}

\item
For all $\epsilon >0$ there exist constants $c_1, c_2, c_3 > 0$ (depending on $\epsilon$) such that for $\abs \lambda < 1$,
\muu
{
\i [L_\lambda, A_0] &\geq (1- c_1 \abs \lambda) 2  \lambda^2 \Pi W R_\epsilon^2 \oPi W \Pi  - c_2 \abs \lambda \Id_\f \otimes \Id_\f \otimes  \PP{P_\Omega}   \\  &\qquad -  c_3 (1+\beta^{-1}) \lambda^2    \ind_{L_\p \not= 0} \big( \Vc^* \sc{\po}^{-2} \Vc     \otimes \Id_\p +  \Id_\p  \otimes \Vc^* \sc{\po}^{-2} \Vc \\ &\qquad + \PP{ (\Pc \otimes \Pc)}   \big) \otimes P_\Omega.
}

\item
There exists a constant $c_1 > 0$ such that, for all $\lambda \in \IR$,
\muu
{
\CQ{Q} \geq (1-c_1 \abs \lambda (1+\beta^{-1}) ) Q \otimes P_\Omega - \abs \lambda \PP{\Poh}.
}
\end{enumerate}
\end{prop}
Before we can give the proof of \thref{th:total estimate steps} we need some preparatory lemmas.
It is convenient to introduce some further notation for the interaction and the commuted interaction. We separate them into parts which act on the left and right of the particle space tensor product, respectively,
\begin{align*}
I_1(u,\Sigma) &:= I_1^{(\l )}(u,\Sigma) \otimes \Id_\p + \Id_\p \otimes I_1^{(\r )}(u,\Sigma),   \\
I^{(\l )}_1(u,\Sigma)  &:= (-\i \partial_u) \tau_\beta(G)(u,\Sigma)  +  \tau_\beta( \ad^{(1)}_{\Aps}(G))(u,\Sigma)  , \\
I^{(\r )}_1(u,\Sigma) &:=  (-\i \partial_u)  e^{-\beta u  /2}   \tau_\beta(\G^*)(u,\Sigma)    -   e^{-\beta u  /2}   \tau_\beta( \ad^{(1)}_{\Aps} (\G^*))(u,\Sigma).
\end{align*}
We introduce also integrated versions which will be used in the further estimates.
\begin{align*}
W_{\p} &:=  \int I(u,\Sigma)^* I(u,\Sigma) \d(u, \Sigma), \\
W_{1,\p} &:=  \int I_1 (u,\Sigma)^* I_1(u,\Sigma) \d(u, \Sigma),
\end{align*}
and the left and right parts, for  $\alpha = \l,\r$,
\begin{align*}
W^{(\alpha)}_{\p} &:=  \int I^ {(\alpha)}(u,\Sigma)^* I^ {(\alpha)}(u,\Sigma) \d(u, \Sigma), \\
W^{(\alpha)}_{1,\p} &:=  \int I_1^ {(\alpha)}(u,\Sigma)^* I_1^ {(\alpha)}(u,\Sigma) \d(u, \Sigma).
\end{align*}
By construction $W_1 = \Phi(I_1)$ and we have the decomposition
\muun
{
\label{eq:I1_decomposition}
I_1(u,\Sigma) &= I^{(\l)}_1(u,\Sigma) \otimes \Id_\p + \Id_\p  \otimes  I^{(\r)}_1(u,\Sigma).
}
First, we estimate the commuted interaction term $W_1$ appearing in \eqref{eq:C1 estimate}.
\begin{lemma}
\label{lemma:W1_estimate}
For any $\delta > 0$  we have
\begin{align} \label{estoncommint}
\pm  \lambda  W_1    &\leq  \delta  \Nfh    + \frac{1}{\delta}  \lambda^2  W_{1,\p} \otimes \Id_\f
\end{align}
in the sense of forms.
\end{lemma}
\begin{proof}
By \ref{assumption:I1} and \ref{assumption:I2} we know that $I_1 \in L^2(\IR \times \IS^2, \L(H_\p))$. The standard estimates
of the creation and annihilation operators lead to the inequality above.
\end{proof}
Next, we prove a bound for $W_\p$ and $W_{1,\p}$.
\begin{lemma}
\label{th:I1_q}
There exists constants $C$ independent of $\beta$ such that, for  $\alpha = \l,\r$,
\begin{enumerate}[label=(\alph*)]
\item $W^{(\alpha)}_{\p} \leq C ( 1+ \beta^{-1})$,
\item $W^{(\alpha)}_{1, \p} \leq C ( 1+ \beta^{-1})$,
\item $\Pc W^{(\alpha)}_{\p} \Pc \leq C ( 1+ \beta^{-1}) \Vc^* \sc{\po}^{-2} \Vc$,
\item $ \Pc W^{(\alpha)}_{1, \p} \Pc \leq C ( 1+ \beta^{-1}) \Vc^*  \sc{\po}^{-2} \Vc$.
\end{enumerate}
\end{lemma}
\begin{proof}
By \thref{th:G commutator conditions}

we have that, for all $j \in \{1,2,3\}$, $\alpha = \l, \r$,
\[
I^{(\alpha)},~
I^{(\alpha)}_1,~
I^{(\alpha)} \Vc^* \po_j \Vc, ~
I^{(\alpha)}_1 \Vc^* \po_j \Vc  \in L^2(\IR \times \IS^2, \H_\p ),
\]
 and there is a constant $C$ independent of $\beta$ such that we can estimate the norm of these expressions by
\[
C (1+\beta^{-1}).
\]
Thus, the same applies for $ I^{(\alpha)} \Vc^* \sc{\po}  \Vc, ~ I^{(\alpha)}_1 \Vc^* \sc{\po} \Vc \in L^2(\IR \times \IS^2, \H_\p )$.
Consequently, we obtain, for $\alpha = \l, \r$, and some constant $C >0$ not depending on $\beta$,
\muu
{
\Pc &W_\p^{(\alpha)}  \Pc  \\ &=   \Vc^* \sc{\po}^ {-1} \Vc  \int  (I^{(\alpha)}(u,\Sigma)  \Vc^* \sc{\po} \Vc  )^* I^{(\alpha)}(u,\Sigma)  \Vc^* \sc{\po} \Vc \d( u, \Sigma) \Vc^* \sc{\po}^ {-1} \Vc \\ &\geq C ( 1+ \beta^{-1}) \Vc^* \sc{\po}^{-2 } \Vc.
}
The proof for $W^{(\alpha)}_{1, \p}$ is analogous.
\end{proof}
Now, we estimate the second term involving $A_0$. We have
\muu
{
\i [L_\lambda, A_0] = - \lambda [L_\lambda, \Pi W R_\epsilon^2 \oPi - \oPi R_\epsilon^2 W \Pi].
}
With respect to the different subspaces we obtain
\muun
{
\Pi \i [ L_\lambda, A_0] \Pi &= 2  \lambda^2 \Pi W R_\epsilon^2 \oPi W \Pi, \label{eq:pipi_normal}  \\
\oPi \i [ L_\lambda, A_0] \oPi &= -  \lambda^2 ( \oPi W \Pi W R_\epsilon^2 \oPi + \oPi R_\epsilon^2 W \Pi W \oPi), \label{eq:pipi} \\
\Pi \i [ L_\lambda, A_0] \oPi &=  \lambda \Pi W R_\epsilon^2 \oPi L_\lambda \oPi.
}
The Fermi Golden Rule condition implies strict positivity of  $\Pi \i [ L_\lambda, A_0] \Pi$, see \thref{th:fgr}. The other expressions can be potentially negative and are estimated in the following lemma. It contains sharper estimates than \cite{merkli2}  which we use later for a Birman-Schwinger argument.
\begin{lemma}
\label{lemma:pi_estimate_new}
For all $\epsilon, $ and all $\lambda$ the following holds.
\begin{enumerate}[label = (\alph*)]
\item
We have  $ \oPi \i [L_\lambda, A_0] \oPi =  \ind_{\Nfh=1}  \oPi \i [L_\lambda, A_0] \oPi \ind_{\Nfh=1}   $.  Moreover,
\[
\nn{ \oPi \i [L_\lambda, A_0] \oPi }  \leq 2  \frac{\lambda^2}{\epsilon^2} \nn{W_\p } .
\]
\item  For arbitrary $\delta_1, \delta_2 > 0 . $
\muu
{
 \oPi  &\i [L_\lambda, A_0] \Pi + \Pi  \i [L_\lambda, A_0] \oPi  \leq
  (\abs\lambda \delta_1 + \delta_2 \lambda^2)  \Pi W R_\epsilon^2 \oPi W \Pi  + \frac{  \abs\lambda}{ \delta_1}  \PP{\Poh} \\
  &\qquad  + \frac{ \lambda^2} {\delta_2}   \bigg( \PP{\Poh}  \oPi a^*(I)  \oPi R_\epsilon^2 a(I) \oPi \PP{\Poh}  \\ &\qquad \qquad +   \oPi  \int \frac{ I(u,\Sigma)^*  \oPi  I(u,\Sigma)}{u^2 + \epsilon^2}  \d(u, \Sigma)  \oPi \Poh  \bigg).
}
\end{enumerate}
\end{lemma}
\begin{proof}
\begin{enumerate}[label = (\alph*)]
\item
Consider the first term in \eqref{eq:pipi},
\muu
{
 \oPi W \Pi W R_\epsilon^2 \oPi = a^*(I) \Pi  a(I) R_\epsilon^2 \oPi.
}
Clearly, this operator vanishes everywhere except on $\ran \ind_{\Nfh=1}$.
By standard estimates of creation annihilation operators we obtain
 \[\nn{  a^*(I) \Pi  a(I)} \leq  \nn{ \int I^*(k) I(k) \d k },
\] thus
\muu
{
\nn{  \oPi W \Pi W R_\epsilon^2 \oPi } \leq \frac{ \nn{ \int I^*(k) I(k) \d k }  }{\epsilon^2},
}
which proves the first inequality, as the second term in \eqref{eq:pipi} is just the adjoint of the first one.
\item
Using the formulas above, and the operator inequality
\begin{align}
\label{eq:op estimate}
A^*B + B^*A \leq A^* A + B^* B,
\end{align}
we get
\muun
{
\oPi & \i [L_\lambda, A_0] \Pi + \Pi  \i [L_\lambda, A_0] \oPi \nonumber  \\ &=  \lambda( \Pi W R_\epsilon^2 \oPi L_0 \oPi \ind_{\Nfh=1} + \ind_{\Nfh=1} \oPi L_0 R_\epsilon^2 W \Pi) + \nonumber \\ &\qquad \lambda^2 ( \Pi W R_\epsilon^2 \oPi \ind_{\Nfh=1} W \oPi + \oPi W \ind_{\Nfh=1} \oPi R_\epsilon^2 W \Pi)  \nonumber \\
&\leq   \abs\lambda ( \delta_1 \Pi W R_\epsilon^2 \oPi W \Pi  + \delta_1^{-1} \ind_{\Nfh=1} \oPi L_0 \oPi R_\epsilon^2 L_0 \oPi \ind_{\Nfh=1} ) \label{eq:firstpiop} \\ &\qquad + \lambda^2( \delta_2 \Pi W R_\epsilon^2 \oPi W \Pi + \delta_2^{-1} \oPi W \ind_{\Nfh=1} \oPi R_\epsilon^2 \ind_{\Nfh=1} W \oPi).   \label{eq:secondpiop}
}
We have
\[
\nn{ \ind_{\Nfh=1} \oPi L_0 \oPi R_\epsilon^2 L_0 \oPi \ind_{\Nfh=1} } \leq 1,
\]
which yields a bound for the second operator in \eqref{eq:firstpiop}.  The second one \eqref{eq:secondpiop} only operates on the space $\ran (\Poh + \ind_{\Nfh = 2})$, so we can write
\begin{align}
\oPi W \ind_{\Nfh=1} &\oPi R_\epsilon^2 \ind_{\Nfh=1} W \oPi \nonumber  \\ &=  (\Poh + \ind_{\Nfh = 2}  ) \oPi W \ind_{\Nfh=1} \oPi R_\epsilon^2 \ind_{\Nfh=1} W \oPi (\Poh + \ind_{\Nfh = 2}  ). \label{eq:local term}
\end{align}
Now, we can use the same operator inequality as above to bound \eqref{eq:local term} by
\[
2  ( \oPi a^*(I)  \oPi R_\epsilon^2 a(I) \oPi \ind_{\Nfh=2}  +  \oPi  \int  \frac{ I^*(u,\Sigma)  \oPi   I(u,\Sigma)}{u^2 + \epsilon^2}  \d(u,\Sigma)  \oPi \Poh ). \qedhere
\]
\end{enumerate}
\end{proof}
Having all this we can give the proof for the concrete error estimates we stated.
\begin{proof}[Proof of \thref{th:total estimate steps}]
\begin{enumerate}[label=(\alph*)]
\item First, we use \thref{lemma:W1_estimate} and the explicit form of $C_1$ to obtain on $\D$, for any $\delta_1 > 0$,
\muu
{
C_1 &= (\Vc^* \q^2 \Vc   \otimes \Id_\p + \Id_\p \otimes  \Vc^* \q^2 \Vc)    \otimes \Id_\f +  \Nfh + \lambda W_1 \\
&\geq ( \Vc^* \q^2 \Vc  \otimes \Id_\p + \Id_\p \otimes  \Vc^* \q^2 \Vc  )  \otimes \Id_\f +  (1 - \delta_1)  \PP{\Poh}  - \frac{\lambda^2}{\delta_1} W_{1,\p} .
}
Next, note that the operator inequality \eqref{eq:op estimate} yields
\[
W_{1,\p} \leq 2 (  W^{(\l)}_{1,\p} \otimes \Id_\p +  \Id_\p  \otimes  W^{(\r)}_{1,\p}  ).
\]
Then, using \eqref{eq:op estimate} again, and subsequently \thref{th:I1_q},  a decomposition into $\ran \Pc$ and $\ran \Pd$ gives, for $\alpha = \l, \r$,
\begin{align*}
W^{(\alpha)}_{1,\p}  &\leq 2(\Pc W^{(\alpha)}_{1,\p}  \Pc +  \Pd W^{(\alpha)}_{1,\p}  \Pd ) \\
&\leq C (1+\beta^{-1}) ( \Vc^* \sc{\po}^{-2} \Vc + \Pd),
\end{align*}
where $C > 0$ is a constant not depending on $\beta$. Choosing any $0< \delta_1 < 1$ yields \eqref{eq:C1 estimate} on $\D$.  As $\D$ is a core for $C_1$, it is also a form core for $q_{C_1}$, so the operator inequality can be extended to the corresponding forms in the form sense on $\Def(q_{C_1})$.

\item
We have for all $\delta_1,  \delta_2 > 0$, by \thref{lemma:pi_estimate_new},
\muu
{
\i [L_\lambda, A_0] &= \Pi i [ L_\lambda , A_0 ]   \Pi +   \oPi i [ L_\lambda , A_0 ]   \oPi +   \oPi i [ L_\lambda , A_0 ]   \Pi +   \Pi i [ L_\lambda , A_0 ]   \oPi  \\

 &\geq (1-  (\abs \lambda \delta_1 + \delta_2 \lambda^2)) \Pi i [ L_\lambda , A_0 ]   \Pi  -  2 \frac{{\lambda}^2}{\epsilon^2} \nn{ W_\p }  \ind_{\Nfh =2}   \\
& \qquad -  \frac{ \abs\lambda}{\delta_1} \ind_{\Nfh =1}  -  2  \frac{ \lambda^2}{\delta_2}  \oPi a^*(I)  \oPi R_\epsilon^2 a(I) \oPi \ind_{\Nfh =2} \\ &\qquad  - 2 \frac{ \lambda^2}{\delta_2 \epsilon^2}  \oPi  W_\p  \oPi \Poh.
}
Notice that the operator $ \oPi a^*(I)  \oPi R_\epsilon^2 a(I) \oPi \ind_{\Nfh =2}$ is in fact bounded.  The last term can be decomposed and estimated as above in the proof of \ref{it:C1 estimate}.
\item
We have
\muu
{
W (L_\p^{-1} Q \otimes P_\Omega ) = a^*(I)  (L_\p^{-1} Q \otimes P_\Omega)  = a^*(I L_\p^{-1} Q   ) \Poh.
}
Thus, by the standard estimates for creation and annihilation operators, we obtain for all  $\delta > 0$ on $\Def(\Nfh)$,
\begin{align*}
&  W ( L_\p^{-1}  Q \otimes P_\Omega )    +   ( W ( L_\p^{-1}  Q \otimes P_\Omega )  )^*  \\
& \quad  \leq \delta \Nfh + \delta^{-1} \left(   \int
L_\p^{-1} Q  I(u,\Sigma)^* I(u,\Sigma) L_\p^{-1} Q   \d(u,\Sigma) \right) \otimes  P_\Omega.
\end{align*}
Hence,  we get
\muun
{
\CQ{Q} &\geq Q \otimes  P_\Omega  - \abs \lambda \delta  \Nfh -  \abs \lambda \nn{ W_\p }  \delta^{-1}
( L_\p^{-1} Q )^2 \otimes P_\Omega   \nonumber  \\
 &\geq  (1 - 2 \lambda (1+\beta^{-1}) \delta^{-1}) Q \otimes  P_\Omega  - \abs \lambda \delta  \Nfh,  \label{eq:final3}
}
where we used the that the concrete choice of $Q$ as in \eqref{eq:Q defn} yields
\[
( L_\p^{-1} Q )^2 =  ( L_p^2 \ind_{[-1,1]}(L_\p) + L_\p^{-2} \ind_{(-\infty,-1) \cup (1, \infty)}(L_\p)) \leq 2\Id_\p \otimes \Id_\p. \qedhere
\]
\end{enumerate}
\end{proof}


TITLE: Why my cosine interpolation of a cube's face doesn't work?
QUESTION [2 upvotes]: NB please : executable use cases are available at the end of this question. I begin this question by showing you the problems of my program, then I explain how the latter works, and finally I end up with these both executable use cases.
What I want to do
I want to draw the bottom face of a cube. To do that, I interpolate between the 4 vertices of this face. I use two kinds of interpolation :

Linear
Cosine

I know it's typically the aim of GPUs, but I wanted to implement it myself to learn and try the "deep" mechanisms of 3D rendering.
Results
Each point of the following both pictures are interpolated between the 4 vertices of the bottom face (except the 8 vertices of my cube).
With linear interpolation
It's OK, no problem.

With cosine interpolation
Error. As you can see, there is some weird problem : face's points are inclined towards left.

How do I interpolate between 4 points ?
I test all the combinations of interpolation's weights and, if the sum of the weights equals 1, I draw the resulting point. I know this method is not very good (a lot of consumption of CPU and RAM), and there are P duplicates. But this problem is (1) secondary and (2) out of the range of this StackOverflow question.
In other words : the definition of the interpolation between 4 points is : P = aA + bB + cC + dD. a + b + c + d must equals 1 to draw P, otherwise it's not drawn. (a;b;c;d) \in [0;1]^4 by the way.
A concrete example is (step = 1) :
P = 0a + 0b + 0c + 0d is NOT drawn
P = 0a + 0b + 0c + 1d is drawn
P = 0a + 0b + 1c + 0d is drawn
P = 0a + 0b + 1c + 1d is NOT drawn
etc. ("I test all the combinations"). (Last is : P = 1a + 1b + 1c + 1d, which is not drawn).
To test all these combinations, I iterates on the weights, with a step, using recursion.
Case of the linear interpolation
The above definition doesn't change : P = aA + bB + cC + dD.
Case of the cosine interpolation
The above definition does change : P = aA + bB + cC + dD becomes P = a'A + b'B + c'C + d'D, with : a' = 0.5(1 - cos(a * \pi)) and the idea is the same for the three other weights.
This new definition is not random, I could explain how I found it, but it's out of the range of this topic (in résumé : a simple remap of the cosine function).
Question
Why does it works well with the linear interpolation, and not the cosine one ? Indeed, the cosine face's points are inclined towards left.
By the way : interpolations between two points work well
Linear and cosine interpolations both work very well when used between two points, as you can see below.
Linear interpolation
It's OK, no problem.

Cosine interpolation
It's OK, no problem.

Implementation and Executable
I wrote the program in Scala. First I show you the functions, then two use cases you can execute along with the given functions.
Tests all the combinations of interpolations between n points of k coordinates
def computeAllPossibleInterpolatedPoints(step : Double, points : Seq[Seq[Double]], transform: (Double) => Double) : Seq[Seq[Double]] = {
  var returned_object : Seq[Seq[Double]] = Seq.empty[Seq[Double]]
  recursiveInterpolation(0, Seq.empty[Double])

  def recursiveInterpolation(current_weight_id : Int, building_weights : Seq[Double]) : Unit = {

    (.0 to 1.0 by step).foreach(current_step => {
      if (current_weight_id < points.size - 1) {
        recursiveInterpolation(current_weight_id + 1, building_weights :+ current_step)
      } else {
        val found_solution = (building_weights :+ current_step).map(transform)
        if(BigDecimal(found_solution.sum).setScale(5, BigDecimal.RoundingMode.HALF_UP).toDouble == 1.0) {
          returned_object = returned_object :+ interpolation(found_solution, points)
        }
      }
    })
  }

  returned_object
}

Interpolates between n points of k coordinates
def interpolation(weights: Seq[Double], points: Seq[Seq[Double]], transform: (Double) => Double = null) : Seq[Double] = {
  var transformed_weights = weights
  if(transform != null) {
    transformed_weights = weights.map(transform)
  }
  if(BigDecimal(transformed_weights.sum).setScale(5, BigDecimal.RoundingMode.HALF_UP).toDouble != 1.0) {
    println("ERROR : `SUM(weights) != 1`. Returning `null`.")
    return null
  }
  if(transformed_weights.exists(weight => BigDecimal(weight).setScale(5, BigDecimal.RoundingMode.HALF_UP).toDouble < 0)
    ||
    transformed_weights.exists(weight => BigDecimal(weight).setScale(5, BigDecimal.RoundingMode.HALF_UP).toDouble > 1)) {
    println("ERROR : `EXISTS(weight) / weight < -1 OR weight > 1`. Returning `null`.")
    return null
  }

  transformed_weights.zip(points).map(
    weight_point => weight_point._2.map(coordinate => weight_point._1 * coordinate)
  ).reduce((point_a : Seq[Double], point_b : Seq[Double]) => point_a.zip(point_b).map(coordinate_points => coordinate_points._1 + coordinate_points._2))
}

Interpolation functions
def linear(weight : Double) : Double = {
  weight
}

def cosine(weight : Double) : Double = {
  (1 - Math.cos(weight * Math.PI)) * 0.5
}

Two use cases : between 2 points, each made of 2 coordinates (first use case : linear interpolation, second use case : cosine interpolation)
Note that you can add 2 new points to each of the following functions, to interpolate a face instead of a segment (for now, the below code indeed interpolates points of a segment).
val interpolated_points : Seq[Seq[Int]] = interpolator.computeAllPossibleInterpolatedPoints(0.05, Seq(Seq(22, 22), Seq(33, 33)), interpolator.linear).map(_.map(_.intValue()))

val interpolated_points : Seq[Seq[Int]] = interpolator.computeAllPossibleInterpolatedPoints(0.05, Seq(Seq(22, 22), Seq(33, 33)), interpolator.cosine).map(_.map(_.intValue()))

REPLY [4 votes]: I only remember after struggling to read through code, but I'd recommend posting pseudo-code rather than code here. Regardless, I tried to understand your implementation, and as far as I can tell, you test weights $a,b,c,d$ such that they sum to 1, and for these $a+b+c+d=1$ you directly plug them into your cosine interpolation to compute $a',b',c',d'$. In general, this will not give you a point in the face because you actually want the constraint $a'+b'+c'+d'=1$ to perform an interpolation.
Now, why does this work with a segment interpolation and not a face interpolation? The tl;dr is that $a+b=1$ will ensure $a'+b'=1$.
Indeed $$\cos(a\pi)=\cos(\pi-b\pi)=-\cos(-b\pi)=-\cos(b\pi)$$
and this yields
$$a'+b' = \frac 12(1-\cos(a\pi)) +\frac 12(1-\cos(b\pi))=1$$
When you have more coefficients however, this is unlikely to hold, so your interpolation is not really an interpolation.

To solve this, you can use the "transform" (linear or cosine functions) before testing the validity of the weights.