TITLE: Are constraint relations valid on friction
QUESTION [1 upvotes]: I was doing a question in Newton's laws of motion where I came across this question .

The question said there was same frictional coefficient between all surfaces and there is no acceleration of the blocks and that they moved with uniform velocity. So we are all good in this case .
If the surfaces were frictionless then we would use constraints ( using constraints would not be necessary , it can be just observed in this question) and get that both accelerate with same acc. in opp. directions.
But I had the following doubts
Doubts:
[ Situation : If the blocks are acc. ]
[ These doubts are not only refering to this question but to all questions of this type]
1) If all the surfaces have friction in between them
Case  1) same frictional co-efficients
Case  2) different friction co-efficients
, then will we able to use constraint relations in them to find acceleration
2) If only some surfaces have friction between them and others are smooth.
I hope to gain better clarity with the help of you guys .
PS: I know some of you may find these questions silly but still getting valuable info. from you all will help greatly

REPLY [0 votes]: Frankly, I think you have far too many variables in your situations and cases. To simplify matters I will assume no friction between the pulley and the cable. You should be able to generalize the situation to include it using the approach given below that assumes friction may or may not be between the blocks and supporting surface.
First, be aware that there are two coefficients of friction, the coefficient of static friction and the coefficient of kinetic or sliding friction. Static friction resists impending motion of an object (resists the start of motion). Kinetic friction resists the relative motion between surfaces when that motion is already under way. The coefficient of kinetic friction involving the same surfaces is generally less than the coefficient of  static friction involving the same surfaces. Since the general situation you present is one in which the blocks are accelerating, motion has already started. Therefore we know that the external force F has caused the maximum static friction forces between surfaces to be exceeded, so that we need only concern ourselves with kinetic friction. The case you first introduced (from the book) is where after acceleration begins the applied force is reduced to exactly equal the kinetic friction force so that there will be no net forces on the blocks and they will move at constant velocity.
The figures below are free body diagrams of blocks A and B. Figure 1 shows the most general case where there is friction between blocks A and B and between block B and the supporting surface of B (friction between all surfaces except the pulley).  Figures 2 and 3 modify the free body diagrams for the cases where there is only friction between one or the other contacting surfaces.
For all the figures, the following are the general constraints that should apply to all of your cases.

If we can assume there is no stretching of the cable and no friction at the pulley then the tension $T$ is the same throughout the cable.
The magnitude of the acceleration of block A will equal block B, or

$$a_{A}=a_{B}$$

If there is friction between A and B the friction force B exerts on A will be equal and opposite to the friction force A exerts upon B, or

$$F_{fBA}=F_{fAB}$$
The other terms in the figures are as follows:

$μ_{AB}$ is the coefficient of kinetic friction, if any, between blocks A and B.
$μ_{BS}$ is the coefficient of kinetic friction, if any, between block B and its supporting surface.
$M_A$ and $M_B$ are the masses of blocks A and B
$g$ is the acceleration due to gravity.

Figure 1- (Friction all surfaces)-
Block A:
$$F-μ_{AB}M_{A}g-T=M_{A}a_A$$
Block B:
$$T-μ_{AB}M_{B}g-μ_{BS}M_{B}g=M_{B}a_B$$
Given the masses and the coefficients of friction, and the general constraints 1-3 above, solve for the acceleration.
Now consider your scenarios

1) If all the surfaces have friction in between them
Case 1) same frictional co-efficients
Case 2) different friction co-efficients
, then will we able to use constraint relations in them to find
  acceleration

There are only two surfaces involving friction. The surfaces of A and B in contact with one another and the surface of B in contact with the supporting surface. If there is friction between both surfaces, whether the coefficients are the same or different, Figure 1 applies and you simply use the applicable coefficients. Given these, knowledge of the masses, and the general constraints 1-3 above, you can solve for the acceleration.

2) If only some surfaces have friction between them and others are
  smooth.

Again, since there are only two surfaces involved, simply remove the friction force for the smooth surface from the free body diagram and keep the friction force for the other surface. There are two possible outcomes:
Figure 2 (no friction between A and B)-
If there is no friction between A and B but friction between B and the supporting surface, then we have the following:
$$T-μ_{BS}M_{B}g=M_{B}a_B$$
and
$$F-T=M_{A}a_A$$
Given the two masses, the coefficient of friction between B and the supporting surface, and the general constraints 1-3 given above, solve for the acceleration.
Figure 3 (friction between A and B)-
There is friction between A and B but no friction between B and its supporting surface.
$$T-μ_{AB}M_{B}g=M_{B}a_B$$
and
$$F-μ_{AB}M_{A}g-T=M_{A}a_A$$
Given the two masses, the coefficient of friction between A and B, and the general constraints 1-3, calculate the acceleration.
Hope this helps.


TITLE: Locating Math Book
QUESTION [3 upvotes]: I am seeking the title of a math book I read several years ago and found interesting. It was intended for a general public audience and covered topics from Arrow's Voting Theorem,  Hilbert's Hotel (and other related set theory concepts), the traveling salesman problem, etc.
I fear that this may not be enough information to identify the book, but I hope it will.

REPLY [3 votes]: Google Books turned up:

James B Stein, "How Math Explains The World: For Car Repair to Modern Physics"
Michael Clark, "Paradoxes from A to Z"
David Foster Wallace, "Everything and More: A compact history of infinity"
Aufmann, Lockwood, Nation, "Mathematical Excursion"

Only the last one showed up when searching:

arrow's voting theorem hilbert's hotel traveling salesman


text:
\begin{document}

 \begin{center}

{\bf SUBGAUSSIAN AND STRICTLY SUBGAUSSIAN \\

\vspace{4mm}

RANDOM VARIABLES} \\

\vspace{4mm}

{\sc Eugene Ostrovsky, Leonid Sirota}\\

\vspace{3mm}

 Bar - Ilan University,  59200, Ramat Gan, ISRAEL; \\

 e-mail: eugostrovsky@list.ru \\
 e-mail: sirota3@bezeqint.net \\

\vspace{5mm}

\hspace{55mm} {\it Devoted to the memory of V.V.Buldygin}\\

 \vspace{5mm}

        {\bf Abstract}

\end{center}

\vspace{3mm}

  We study in this report the so-called Strictly Subgaussian (SSub) random variables (r.v.), which form  a very interest subclass
 of Subgaussian (Sub)  r.v., and obtain the exact exponential bounds for tail of distribution for sums of independent and disjoint such a variables,
 not  necessary to be identical distributed, and give some new examples  of SSub variables  to show  the exactness of our estimates. \par
  We extend also these results on the case of sums of subgaussian martingale differences, and show that the mixture of (Strictly)
  Subgaussian r.v. forms also (Strictly) subgaussian variable. \par

\vspace{3mm}

{\it Key words and phrases: } Random variables (r.v.), centering, indicator, binary and Bernoulli's r.v.,  variance, martingales, mixture,
Grand Lebesgue Spaces (GLS), subgaussian norm, subgaussian (Sub) and strictly subgaussian (SSub) r.v., tail or concentrations inequalities,
independence.\\

\vspace{3mm}

 \section { Introduction. Definitions. Notations. Examples.}

\vspace{3mm}

 Let $  \{\Omega, B, {\bf P}  \} $ be some non-trivial probability space with expectation $ {\bf E.} $ \\

 \vspace{2mm}

 {\bf Definition  1.1.}\\

 \vspace{2mm}

  We say that the {\it centered:} $ {\bf E} \xi = 0 $ numerical random variable (r.v.)
 $ \xi = \xi(\omega), \ \omega \in \Omega $ is subgaussian, or equally, belongs to the space $ \Sub(\Omega), $
 if there exists some non-negative constant $ \tau \ge 0 $ such that

$$
\forall \lambda \in R  \ \Rightarrow
{\bf E} \exp(\lambda \xi) \le \exp[ \lambda^2 \ \tau^2 /2]. \eqno(1.1).
$$

\vspace{3mm}

 The minimal value $ \tau $ satisfying (1.1) is called a  {\it subgaussian  norm}
of the variable $ \xi, $ write

 $$
 ||\xi||\Sub = \inf \{ \tau, \ \tau > 0: \ \forall \lambda \in R \ \Rightarrow {\bf E}\exp(\lambda \xi) \le \exp(\lambda^2 \ \tau^2/2) \}.
 $$

 Evidently,

$$
||\xi||\Sub = \sup_{\lambda \ne 0} \left[ \sqrt{ 2 \ln {\bf E}  \exp (  \lambda \xi)  }/|\lambda| \right].  \eqno(1.2)
$$

 This important notion was introduced by  J.P.Kahane \cite{Kahane1}; V.V.Buldygin and Yu.V.Kozachenko in \cite{Buldygin1} proved
that the set $ \Sub(\Omega) $  relative the norm $  ||\cdot|| $ is complete Banach space which is isomorphic to subspace
consisting only from the centered variables of Orlicz's space over $ (\Omega, B,P)  $ with $ N \ -$ Orlicz-Young function
 $ N(u) = \exp(u^2) - 1; $  see also \cite{Kozachenko1}.  \par

  For instance, let us consider the centered indicator (binary) random variable $ \xi_p: $

$$
{\bf P}(\xi_p = 1 - p) = p; \ {\bf P}(\xi_p = -p) = 1 - p, \ p \in (0,1);
$$
then (in our definitions and notations)

$$
||\xi_p||\Sub = \sqrt{ \frac{1 - 2p}{2 \ln(( 1 - p )/p)} }.
$$
 This important result was obtained independently in the articles \cite{Kearns1},  \cite{Buldygin3};  see also
 \cite{Berend1},  \cite{Schlemm1}, \cite{Ostrovsky2}.\par

\vspace{3mm}

  The detail investigation of this class or random variables with very interest  applications into the theory of random fields
  reader may found in the book \cite{Buldygin2};  we reproduce here some main facts from this monograph.\par

   If $ ||\xi||\Sub = \tau \in (0,\infty),  $ then

 $$
 \max [{\bf P}(\xi > x),  {\bf P}(\xi < -x)  ] \le \exp(- x^2/(2 \tau^2)  ), \ x \ge 0; \eqno(1.3)
 $$
 and  the last inequality is in general case non-improvable.  It is sufficient for this to consider the case when
 the r.v. $  \xi  $ has the centered Gaussian non-degenerate distribution.\par

  Conversely, if  $ {\bf E} \xi = 0 $ and if  for some positive finite constant $  K  $

 $$
 \max [{\bf P}(\xi > x),  {\bf P}(\xi < -x)  ] \le \exp(- x^2/K^2  ), \ x \ge 0,
 $$
 then $ \xi \in \Sub(\Omega) $ and $ ||\xi||\Sub < 4 K. $ \par

 The subgaussian norm in the subspace of the centered r.v. is equivalent to the following Grand Lebesgue Space (GLS)
 norm:

 $$
|||\xi||| := \sup_{s \ge 1} \left[ \frac{|\xi|_s}{\sqrt{s}} \right], \hspace{6mm} |\xi|_s =  \left[ {\bf E} |\xi|^s \right]^{1/s}.
 $$

 For the non-centered r.v. $ \xi $  the subgaussian norm may be defined as follows:

 $$
 ||\xi|| \Sub := \left[  \left\{ ||\xi - {\bf E} \xi||\Sub \right\}^2 + ( {\bf E} \xi)^2  \right]^{1/2}.
  $$

 More detail investigation of these spaces see in the monograph \cite{Ostrovsky1}, chapter 1.  \par

 Denote in the sequel for brevity  for any r.v.  $ \eta $

 $$
 \sigma^2(\eta) = \sigma^2 = \Var{\eta} = {\bf E}\eta^2 - ( {\bf E} \eta)^2.
 $$

\vspace{3mm}

{\bf Definition 1.2.} (See \cite{Buldygin2}, chapter 1.) \par

The subgaussian r.v. $ \xi $ is said to be {\it Strictly Subgaussian  (SSub), } iff

$$
\forall \lambda \in R \ \Rightarrow {\bf E} e^{\lambda \xi} \le e^{\lambda^2 \sigma^2(\xi)/2}, \eqno(1.4)
$$
or equally

$$
||\xi ||\Sub \le \sigma(\xi) = ||\xi||L_2(\Omega). \eqno(1.4.a)
$$

 Recall that always $ ||\xi ||\Sub \ge \sigma(\xi) = ||\xi||L_2(\Omega),  $ so that

$$
\xi \in \SSub(\Omega) \Leftrightarrow  {\bf E}\xi = 0, \ ||\xi ||\Sub = \sigma(\xi) = ||\xi||L_2(\Omega).  \eqno(1.4b)
$$

 Many examples of strictly subgaussian distributions may be found in the book of  V.V.Buldygin and Yu.V.Kozachenko \cite{Buldygin2},
chapter 1. For instance, arbitrary mean zero Gaussian distributed r.v. is strictly subgaussian, including the case when
this r.v. is equal to zero  a.e.; the symmetric Rademacher's r.v. $ \rho $ with distribution
$ {\bf P}(\rho = 1) = {\bf P}(\rho = -1) = 1/2  $  belongs to the set  $ \SSub(\Omega). $
The random variable $  \eta $ which has an uniform distribution on the symmetrical interval $ (-b,b), \ b = \const \in (0,\infty) $ is
Strictly Subgaussian. \par
 Consider also following the authors \cite{Buldygin2}  the r.v. $ \zeta $  with the following density:

 $$
 f_{\zeta}(x) =   \frac{\alpha+1}{2 \alpha} \  \left(1 - |x|^{\alpha} \right) \ I(|x| \le 1), \
 \alpha = \const \ge 0, \eqno(1.5)
 $$
 where $ I(A) = I(A,x) = 1, \ x \in A; \ I(A) = I(A,x) = 0, \ x \notin A $ is indicator function;
then $ \zeta \in \SSub(\Omega). $ \par

This example is interesting because the kurtosis of the r.v. $ \zeta $ is zero if $ \alpha = \sqrt{10} - 3. $

\vspace{3mm}

 The convenience of these notions is following. Let $  \{ \xi(i ) \}, \ i = 1,2,\ldots,n  $ be (centered) independent
subgaussian r.v. Denote

$$
S(n) = \sum_{i=1}^n \xi(i), \hspace{6mm} \Sigma^2(n) = \sum_{i=1}^n (||\xi(i)||\Sub)^2. \eqno(1.6)
$$

 Then $ ||S(n)||\Sub \le \Sigma(n) $ and following

$$
 \max( {\bf P}(S(n)/\Sigma(n) > x ),  {\bf P}(S(n)/\Sigma(n) < -x )) \le e^{ -x^2/2  }, \ x \ge 0, \eqno(1.7)
$$
 the tail or concentrations inequalities. \par
  If in addition $ \xi(i) $ are identical distributed and $ \beta:= ||\xi(1)||\Sub \in (0,\infty), $ then

 $$
 \sup_n ||S(n)/\sqrt{n}||\Sub = \beta
 $$
  and

$$
\sup_n \max( {\bf P}(S(n)/(\beta \sqrt{n} ) > x ),( {\bf P}(S(n)/(\beta \sqrt{n} ) < - x ) \le e^{ -x^2/2  }, \ x \ge 0, \eqno(1.8)
$$

 If in addition the r.v. $ \xi(i) $  are strictly subgaussian, the estimate (1.8) may be reinforced by lower estimate  used the classical CLT:

$$
\sup_n {\bf P}(S(n)/(\beta \sqrt{n} ) > x ) \ge \lim_{n \to \infty} {\bf P}(S(n)/(\beta \sqrt{n} ) > x ) =
$$

$$
( 2 \pi )^{-1/2} \int_x^{\infty} e^{-y^2/2} \ dy \ge C  \ x^{-1} e^{-x^2/2}, \ x \ge 1. \eqno(1.9)
$$

\vspace{3mm}

{\bf This short report may be considered as a slight addition to the book of V.V.Buldygin and Yu.V.Kozachenko
\cite{Buldygin2}; we give some new examples of Subgaussian and
Strictly Subgaussian random variables, obtain the exponential exact bounds for tails of distribution for
sums of independent Strictly Subgaussian random variables, extend this estimates on the sequence of martingale differences,
 investigate the mixture of ones distributions etc.   }\par

\vspace{3mm}

 Applications of these notions in the non-parametrical statistics may be found in the articles
\cite{Gaivoronsky1}, \cite{Kiefer1}. Another statistical applications is described in \cite{Chen1}, \cite{Ryiabinin1}.
The subgaussian r.v. appears also in the articles \cite{B'enyi1}, \cite{B'enyi2} devoted to the non-linear
Schr\"odinger's equation.  Some applications in the information and coding theory  see  in \cite{Raginsky1}. \par

 \vspace{3mm}

 \section{Mixture of subgaussian random variables}

 \vspace{3mm}

 Let $  Z = \{  z  \}  $ be another set equipped some sigma-algebra and probability measure $ \mu. $ Let also
 $  \xi_z, \ z \in Z $ be a {\it  a family  } of random variables  such that the function

 $$
 z \to {\bf P} (\xi_z \in A), \ A \in B
 $$
is $ \mu \ - $ measurable. By definition, the random variable $  \nu, $ more precisely its distribution is called
{\it mixture  } of individual distributions $ {\bf P} ( \xi_z \in A ) $ relative the (weight) measure  $ \mu, $ if

$$
{\bf P} (\nu \in A) = \int_Z {\bf P} ( \xi_z \in A ) \ \mu(dz).  \eqno(2.1)
$$

Another interpretation-conditional distribution.\par

  If the equality (2.1) there holds, then for all non-negative measurable function $ h: R \to R $

$$
{\bf E} h(\nu) = \int_Z {\bf E} h(\xi_z) \ \mu(dz). \eqno(2.2)
$$

\vspace{3mm}

{\bf Theorem 2.1.}  Let the random variables $  \xi_z $ be Strictly Subgaussian with {\it at the same subgaussian norm}

$$
||\xi_z||\Sub = \sigma = \const \in (0, \infty), \eqno(2.3)
$$
  Then the random variable $ \nu $ is also Strictly Subgaussian with at the same norm. \par

\vspace{3mm}

{\bf Proof.} We derive using the identity (2.2) for the functions $ h(x) = x, x^2, e^{\lambda x} $ correspondingly

$$
{\bf E} \nu = \int_Z {\bf E} \xi_z \ \mu(dz) = 0, \  {\bf E} \nu^2 = \int_Z {\bf E} \xi^2_z \ \mu(dz) = \int_Z  \sigma^2 \ \mu(dz)=
\sigma^2,
$$

$$
{\bf E} e^{\lambda \nu} = \int_Z {\bf E} e^{\lambda \xi_z} \ \mu(dz) \le \int_Z e^{\lambda^2 \sigma^2/2} \ \mu(dz) =  e^{\lambda^2 \sigma^2/2},
$$
Q.E.D.\par

Where the conditions of Theorem 2.1 are not satisfied, the variable $  \nu  $ may be as a Strictly Subgaussian
or no. The correspondent (very spectacular)  example  see in the aforementioned monograph \cite{Buldygin2}, p. 14:

$$
{\bf P} (\nu = 1) = {\bf P}(\nu = -1) = \frac{1-\gamma}{2}, \ {\bf P}(\nu = 0) = \gamma, \ \gamma = \const \in [0,1];
$$
then $  \nu \in \SSub(\Omega) $ iff $  0 \le \gamma \le 2/3 $ or $ \gamma = 1; $ otherwise $ \nu \notin \SSub(\Omega). $\par

 Let now $ \xi_1 $ be a centered Gaussian distributed r.v. with variance $ \sigma^2_1 $ and $  \xi_2 $ be also mean zero Gaussian variable with
other variance $ \sigma_2^2,   $ wherein  $ 0 < \sigma_1^2 < \sigma^2_2 < \infty   $ and $  \mu (\{  1\}) = \mu ( \{ 2 \} )= 1/2. $ It is easy to
verify that the r.v. $ \nu $ is not strictly subgaussian despite both the r.v. $ \xi_1, \ \xi_2  $ are strictly subgaussian.\par

\vspace{3mm}

{\bf Remark 2.1.} If the random variables $ \xi(z) $ in the theorem 2.1  are only subgaussian with variable but uniformly bounded norm

$$
\sigma_+ := \sup_z ||\xi_z||\Sub =  \sup_z \sigma(z) \in (0,\infty), \eqno(2.4)
$$
then the r.v. $ \nu  $ is subgaussian with subgaussian norm less than $ \sigma_+. $\par

 \vspace{3mm}

 \section{Disjoint  subgaussian random variables}

 \vspace{3mm}

 Two r.v. $ \eta_1, \eta_2 $ are called by definition {\it disjoint,} if $ \eta_1 \cdot \eta_2 = 0 $ almost everywhere. The
{\it family } $  \eta_j, \ j = 1,2,\ldots,n; \ n  < \infty $ of r.v.   is named disjoint, if it is pairwise disjoint. \par

 Denote  $ \beta(j) = ||\eta_j||\Sub, \  S(n) = \sum_{j=1}^n \eta_j  $ and suppose $  {\bf E} \eta_j = 0. $\par

\vspace{3mm}

 {\it We intend to obtain in this section the subgaussian and other exponential estimations for the sums of disjoint random variables.} \par

\vspace{3mm}

 The first $  L_p(\Omega) $  estimates for $ S(n) $ was obtained in the famous work of H.P.Rosenthal \cite{Rosenthal1};
the modern results with very interest generalization
see in the articles of  S.V.Astashkin and  F.S.Sukochev \cite{Astashkin1}, \cite{Astashkin2}.\par

\vspace{3mm}

 Let us introduce the following function:

$$
G_n(y_1,y_2, \ldots, y_n) \stackrel{def} =
\inf_{ \mu > 0} \left[ \mu^{-1} \ln \left(  \sum_{j=1}^n e^{\mu y_j} - (n-1)  \right) \right]^{1/2}. \eqno(3.1)
$$

\vspace{3mm}

{\bf Theorem 3.1.} {\it Let $  \eta_j, \ j = 1,2,\ldots,n $ be centered disjoint random variables. Then}

$$
||S(n)||\Sub  \le G_n(\beta_1^2,\beta_2^2, \ldots, \beta_n^2). \eqno(3.2)
$$

\vspace{3mm}

{\bf Proof.} Given:  $  \eta_j = \eta_j \cdot I(A(j)), \ A(j) \cap A(i) = \emptyset, i \ne j, \ A(j) \in B; $

$$
\int_{\Omega} e^{ \lambda \eta_j  } {\bf P}(d \omega) \le e^{\lambda^2 \beta^2_j/2},
$$
therefore

$$
\int_{A(j)}e^{ \lambda \eta_j } {\bf P}(d \omega) \le e^{ \lambda^2 \beta^2_j/2 } - (1 - {\bf P}(A_j)).
$$

 We deduce:

 $$
 {\bf E}e^{\lambda S(n) }= {\bf E} e^{  \lambda \sum_j \eta_j } = {\bf E} e^{  \lambda \sum_j \eta_j  I(A(j))}=
 {\bf E} \prod_j e^{\lambda \eta_j I(A(j)) } =
 $$

$$
\sum_j \int_{A(j)} e^{ \lambda \eta_j } {\bf P}(d \omega) + (1 - \sum_j {\bf P}(A(j))) \le
$$

$$
\sum_j \left[ e^{ \lambda^2 \beta^2_j/2  } -( 1 - {\bf P}(A(j))  ) \right]  + (1 - \sum_j {\bf P}(A(j)))  =
$$

$$
\sum_j e^{ \lambda^2 \beta^2_j/2  } -(n-1) \le e^{ \lambda^2 G^2_n(\beta_1^2, \beta_2^2, \ldots, \beta^2_n) /2 } \eqno(3.3)
$$
on the basis of definition of the function $  G_n(\cdot). $  Theorem 3.1 is proved. \par

 \vspace{3mm}

{\bf  Remark 3.1. } It is interest to note that  our estimation does not depend on the partition

$$
 R = \{ A(j), j = 1,2,\ldots,n; \ \Omega \setminus \cup_j A(j)   \}.
$$

\vspace{3mm}

{\bf  Example 3.1. } Let $ \nu:  \Omega \to R $ be a centered stepwise (simple) r.v. (measurable function):

$$
\nu = \sum_{j=1}^{m} c(j) [ I(A(p(j))) - p(j)  ], \ m = \const \le \infty,  \ c(j) = \const,
$$
and $ \{ A(p(j)) \} $ are pairwise disjoint events. We conclude  using triangle inequality
for the subgaussian norm and the completeness of the space $  \Sub(\Omega) $  in the case when $  m = \infty: $

$$
||\nu||\Sub \le \sum_{j=1}^m |c(j)| Q(p(j)).
$$
 The application of theorem 3.1 gives more exact estimation. \par

\vspace{3mm}

{\it  Another approach to the problem of exponential tail estimates for sums of disjoint random variables.}\par

\vspace{3mm}

 We recall briefly first of all here  for reader conventions some definitions and facts from
the theory of GLS spaces.\par

 Recently, see \cite{Fiorenza1}, \cite{Fiorenza2},\cite{Ivaniec1}, \cite{Ivaniec2}, \cite{Jawerth1},
\cite{Karadzov1}, \cite{Kozachenko1}, \cite{Liflyand1}, \cite{Ostrovsky1}, \cite{Ostrovsky2} etc.
 appear the so-called Grand Lebesgue Spaces (GLS)
 $$
 G(\psi) = G = G(\psi ; B);  \ B = \const \in (1, \infty]
 $$
spaces consisting on all the random variables  (measurable functions) $ f : \Omega \to R  $ with finite norms

$$
||f||G(\psi) \stackrel{def}{=} \sup_{p \in (A;B)} \left[\frac{|f|_p}{\psi(p)} \right], \eqno(3.4)
$$

$$
|f|_p  \stackrel{def}{=} \left[ {\bf E} |f|^p \right]^{1/p}, \ 1 \le p \le \infty.
$$

 Here $ \psi = \psi(p), \ p \in [1,B) $ is some continuous positive on the {\it open} interval $ (1;B) $ function such
that

$$
\inf_{p \in(A;B)} \psi(p) > 0. \eqno(3.5)
$$

We will denote
$$
\supp(\psi) \stackrel{def}{=} [1;B)
$$
 or by abuse of notations $ \supp(\psi) = B.  $ \par

The set of all such a functions with the support $ \supp(\psi) = (1;B) $ will be denoted by  $  \Psi(1;B) = \Psi(B). $  \par

This spaces are rearrangement invariant; and are used, for example, in
the theory of Probability, theory of Partial Differential Equations,
 Functional Analysis, theory of Fourier series,
 Martingales, Mathematical Statistics, theory of Approximation  etc. \par

 Notice that the classical Lebesgue - Riesz spaces $ L_p $  are extremal case of Grand Lebesgue Spaces, see
 \cite{Ostrovsky2}. \par

 Let a function $  \xi:  \Omega \to R  $ be such that

 $$
 \exists B > 1  \Rightarrow  \forall p \in [1, B)  \ |\xi|_p < \infty.
 $$
Then the function $  \psi = \psi_{\xi}(p) $ may be naturally defined by the following way:

$$
\psi_{\xi}(p) := |\xi|_p, \ p \in [1,B). \eqno(3.6)
$$

 The finiteness of the $ G\psi \ - $ norm for some r.v. $  \xi $ allows to obtain the exact exponential tail inequalities for the
distribution  $  \xi; $ for instance,

$$
\sup_{p \ge 1} \left[ \frac{|\xi|_p}{p^m} \right]  < \infty \ \Leftrightarrow \exists C > 0, \ \forall x \ge 0 \ \Rightarrow
{\bf P} (|\xi| > x) \le e^{ - C x^{1/m} }, \ m = \const > 0, \eqno(3.7)
$$
see \cite{Kozachenko1}, \cite{Ostrovsky1}, chapter 1, section 3.\par

\vspace{3mm}

 Let  us return to the the problem of exponential estimations for sums of disjoint variables.\par

\vspace{3mm}

{\bf  Proposition 3.1.  } Let $ \xi, \eta  $ be two disjoint r.v. belonging to some space $ G\psi, \ \psi \in G\Psi $ with
$ \supp \psi = B \in (1, \infty]. $ Then

$$
||\xi + \eta||G\psi \le \left[ (||\xi||G\psi)^B +   ( ||\eta||G\psi)^B   \right]^{1/B}, \eqno(3.8)
$$
where at $  B = \infty $

$$
||\xi + \eta||G\psi \le \max [ ||\xi||G\psi,  ||\eta||G\psi ]. \eqno(3.8a)
$$

{\bf Proof.} Denote for brevity $ a = ||\xi||G\psi, \    b = ||\eta||G\psi. $ Then

$$
|\xi|_p \le a \psi(p), \ |\eta|_p \le b \psi(p); \  |\xi|_p^p \le a^p \psi^p(p), \ |\eta|_p^p \le b^p \psi^p(p).
$$
Since the r.v.  $  \xi, \eta $ are disjoint,

$$
|\xi + \eta|_p^p = |\xi|_p^p + |\eta|_p^p = (a^p + b^p) \cdot \psi^p(p).
$$

 Therefore,

$$
|\xi + \eta|_p \le \psi(p) \cdot (a^p + b^p)^{1/p} \le \psi(p) \cdot (a^B + b^B)^{1/B}.
$$
 It remains to divide on the $ \psi(p) $ and take maximum over $ p; 1\le p < B. $\par

 The generalization on the sum of $  n  $ disjoint variables is clear.\par

\vspace{3mm}

 \section{Martingale case}

 \vspace{3mm}

 Let $  F $ be non-trivial sigma subalgebra of the source sigma-algebra $  B. $  The r.v. $ \eta $ is said to be
{\it conditional subgaussian,}  if there is a {\it non-random } non-negative constant $  \tau  $ for which

$$
\forall \lambda \in R \ \Rightarrow   {\bf E} e^{ \lambda \eta  }/F \le e^{ \lambda^2 \tau^2/2  }. \eqno(4.1)
$$

 The minimal value of the constant $  \tau $ from the inequality (4.1) is called {\it conditional subgaussian norm}
of the r.v. $  \eta $  relative the sigma-algebra $  F, $ write

$$
||\eta||\Sub(F):=  \inf \{ \tau, \ \tau > 0: \
 \forall \lambda \in R \ \Rightarrow   {\bf E} e^{ \lambda \eta  }/F \le e^{ \lambda^2 \tau^2/2  }  \}. \eqno(4.2)
$$

 The set of all r.v. with $ ||\eta||\Sub(F) < \infty   $  relative this norm and ordinary algebraic operations
 forms by definition the complete Banach space $  \Sub(\Omega,F). $ \par

For example, arbitrary centered bounded r.v.  $ \eta = \eta(\omega), \ \omega \in \Omega  $ belongs to any space $ \Sub(\Omega,F).  $\par

Obviously, $  ||\eta||\Sub(\Omega) \le  ||\eta||\Sub(\Omega,F). $\par

 If in the equality the value $ \tau $ may  be selected such that

 $$
\tau = ||\eta||\Sub(F) = \sqrt{ \Var(\eta)/F } = \sqrt{ {\bf E} \eta^2/F  },
 $$
 then as before the r.v. $  \eta  $ may be named Strictly  conditional subgaussian:  $  \eta \in \SSub(\Omega,F). $\par

\vspace{4mm}

 Recall that sequence $ (X(i), F(i)), \ i = 0,1,2,\ldots,n  $ where the $  X(i) $ are random
variables and the $ F(i) $ are sigma-algebras, is a martingale if the following
conditions are satisfied: \\

1. The sequence of sigma-algebras $ \{ F(i) \} $ forms a filtration, i.e. $ F(0) \subset F(1) \subset F(2) \ldots \subset F(n); $ usually, $ F(0) $
is the trivial sigma-algebra $  (\emptyset, \Omega)  $  and $ F(n) $ is sigma-subalgebra  of  source sigma-algebra $ B.$ \\

2. $  X(i) \in L_1(\Omega,P) $ and

$$
X_{i-1} \stackrel{a.e.}{=} {\bf E} X(i)/F(i-1),  \ i = 1,2,\ldots,n-1.
$$

 We suppose in the sequel $ {\bf E} X(i)  = 0 $ and introduce the correspondent sequence of martingale-differences $ \xi(i) $ as follows:

$$
\xi(0) = 0, \ \xi(i) = X(i+1) - X(i), \ i = 1,2,\ldots, n-1.
$$

\vspace{3mm}

 Denote
 $$
  \theta(j) = ||\xi(j)||(\Sub(\Omega, F(j-1)), \ j = 1,,2,\ldots,n, \ \Delta(n) = \sqrt{\sum_{j=1}^n \theta^2(j)  }, \eqno(4.3)
 $$
if there exists.\par

\vspace{3mm}

{\bf Theorem 4.1.}

$$
||X(n)||\Sub(\Omega) \le \Delta(n); \eqno(4.4)
$$

$$
\max( {\bf P}(X(n)/\Delta(n)> x ),  {\bf P}(X(n)/\Delta(n) < - x ) \le e^{-x^2/2}, \ x \ge 0. \eqno(4.4a)
$$

\vspace{3mm}

{\bf Proof }  is alike to the one in the article of K.Azuma \cite{Azuma1}; see also \cite{Raginsky1}. Namely, let $  \lambda = \const \in R; $ then

$$
{\bf E} e^{\lambda X(n)} = {\bf E} e^{ \lambda \sum_{j=1}^{n-1} \xi(j) } =
{\bf E} \left[ {\bf E} \left[ e^{ \lambda \sum_{j=1}^{n-1} \lambda \xi(j) } \right] /F(n-2)  \right] =
$$

$$
{\bf E} \left[  \left[ e^{ \lambda \sum_{j=1}^{n-2} \lambda \xi(j) } \right]  \ \cdot {\bf E} e^{ \lambda \xi(n-1) }/F(n-2)  \right] \le
$$

$$
{\bf E} \left[  \left[ e^{ \lambda \sum_{j=1}^{n-2} \lambda \xi(j) } \right]  \ \cdot  e^{ \lambda^2 \theta^2(n)/2 } \right] \le \ldots \le
$$

$$
e^{ \lambda^2 \sum_{j=1}^n \theta^2(j)/2  } = e^{\lambda^2 \Delta^2(n)/2 }. \eqno(4.5)
$$
Thus,

$$
||X(n)||\Sub(\Omega) \le \Delta(n). \eqno(4.6)
$$
 This completes the proof of theorem 4.1.\par

 Note that this estimate is exponential exact if for instance $ \{ \xi(j) \} $ are independent and strictly subgaussian.\par

\vspace{3mm}

 \section{Another examples of subgaussian random variables}

 \vspace{3mm}

{\bf A. Symmetrized beta distribution.}\par

\vspace{3mm}

 Let us consider a symmetrical r.v. $  \xi = \xi_{\alpha, \beta} $ with the density

$$
f(x) = f_{\alpha,\beta}(x) = 0.5 \frac{ |x|^{\alpha - 1} \ ( 1 - |x|)^{\beta - 1}}{ B(\alpha, \beta)} \ I(|x| < 1), \ \alpha, \beta = \const > 0,
\eqno(5.1)
$$

where as usually $  B(\alpha, \beta) $ denotes the beta-function. \par

\vspace{3mm}

{\bf Theorem 5.1.}  {\it If }

$$
B(\alpha,\beta) \le 1,  \eqno(5.2)
$$
{\it then the r.v. $ \xi = \xi_{\alpha, \beta}  $ is strictly subgaussian. } \par

\vspace{3mm}

Note that the condition (5.2) of theorem 5.1 is satisfied if for instance $ \alpha \ge 1, \ \beta \ge 1.  $ \par

\vspace{3mm}

{\bf Proof.}  We have:

$$
{\bf E} \xi^{2k + 1} = 0, \ {\bf E} \xi^{2k} = \frac{B(2k + \alpha, \beta)}{B(\alpha, \beta )}, \ k = 0,1, \ldots,
$$
so that

$$
\sigma^2 := \Var(\xi) = \frac{B(2 + \alpha, \beta)}{B(\alpha, \beta )}= \frac{\alpha(\alpha + 1)}{(\alpha + \beta)( \alpha + \beta + 1) };
$$

$$
{\bf E} e^{\lambda \xi} = \sum_{k=0}^{\infty} \frac{\lambda^{2k}}{(2k)!} \cdot {\bf E} \xi^{2k} = \sum_{k=0}^{\infty}
\frac{B(2k + \alpha, \beta)}{B(\alpha, \beta )} \cdot \frac{\lambda^{2k}}{(2k)!};
$$

$$
e^{\lambda^2 \sigma^2/2} = \sum_{k=0}^{\infty} \frac{\lambda^{2k} \sigma^{2k}}{2^k \ k!}.
$$

 It is sufficient to prove that

$$
\frac{B(2k + \alpha, \beta)}{B(\alpha, \beta )} \cdot \frac{1}{(2k)!} \le \frac{\sigma^{2k}}{2^k \ k!}, \ k = 1,2,\ldots; \eqno(5.3)
$$
see \cite{Buldygin2}, p.8; the case $  k = 0 $ is trivial.\par

 The inequality (5.3) is equivalent to the following:

$$
\frac{B(2k + \alpha, \beta)}{B(\alpha, \beta )} \cdot \frac{1}{(2k)!}  \le  \frac{B^{2k}(\alpha + 2, \beta)}{B^{2k}(\alpha, \beta) \cdot   2^k \cdot k!}.
\eqno(5.4)
$$

 We denote

$$
\theta(k) = \left[\frac{B(2k + \alpha, \beta)}{B(\alpha, \beta )} \cdot \frac{1}{(2k)!} \right] :
\left[ \frac{B^{2k}(\alpha + 2, \beta)}{B^{2k}(\alpha, \beta) \cdot   2^k \cdot k!} \right], \ k = 0,1,\ldots;
\eqno(5.5)
$$
then $ \theta(0) = 1. $ Further, the expression for the fraction $ \theta(k+1)/\theta(k), \ k = 0,1,2, \ldots  $ has a form:

$$
\theta(k+1)/\theta(k) = \frac{\Gamma^2(\alpha) \Gamma^2(\beta)}{\Gamma^2(\alpha + \beta)} \cdot \frac{k+1}{k(2k+1)} \times
$$

$$
\frac{(2k + \alpha)(2k+ \alpha + 1)}{(2k + \alpha + \beta + 1)(2k + \alpha + \beta)} < B^2(\alpha,\beta) < 1;
$$
therefore $  \theta(k) \le 1, \ k = 0,1,2, \ldots, $ Q.E.D.\par

 \vspace{3mm}

{\bf B. Symmetrized gamma distribution.}\par

\vspace{3mm}

 We consider here the r.v. $ \gamma = \gamma_{\alpha, \beta} $ with the (symmetrical) density of distribution

$$
g_{\alpha, \beta}(x) = 0.5 \ \frac{\beta \ |x|^{\alpha} \ e^{- |x|^{\beta}} }{ \Gamma( (\alpha + 1)/\beta )}, \ x \in R, \ \alpha = \const > -1,
\beta = \const > 0. \eqno(5.6)
$$
 We have:   $   {\bf E} \gamma^{2k+1} = 0, \ k = 0, 1, 2,\ldots;   $

$$
{\bf E} \gamma^{2k} = \frac{\Gamma( (\alpha + 2k +1)/\beta  )}{\Gamma( (\alpha + 1)/\beta)}. \eqno(5.7)
$$
 In particular,

$$
\Var(\gamma) = \frac{\Gamma( (\alpha + 3)/\beta  )}{\Gamma( (\alpha + 1)/\beta)}. \eqno(5.7a)
$$

 We suppose in the sequel $  \beta > 2; $ obviously, when $  \beta < 2, $ the r.v. $ \gamma $ is even not subgaussian.\par
The r.v.   $ \gamma  $ is subgaussian iff $ \beta \ge 2. $

\vspace{3mm}

 In order  to formulate the next result, we need to introduce some preliminary notations.

 $$
 \theta = \max(\alpha - \beta + 1,1),  \hspace{6mm} k_0 = \max(1, (\beta - \alpha - 1)/2),
 $$

 $$
 G_1 = \sqrt{\pi}
 \cdot \Gamma \left( \frac{\alpha+1}{\beta}  \right) \cdot e^{- 3/8 } \cdot e^{ -\theta/(2 \beta)  },
 $$

$$
G_2 = 0.25 \ e  \cdot \frac{\Gamma ( ( \alpha + 3  )/\beta )}{ \Gamma((\alpha + 1)/\beta)} \cdot
\left[ 2 \beta^{-1} (1 + \theta/(2 k_0) \right]^{2/\beta},
$$

$$
G = \left[( \max(1, G_1)  ) \cdot G_2 \right]^{\beta/(\beta - 2) },
$$

$$
\zeta(k)= \zeta_{\alpha,\beta}(k) := \Gamma( ( \alpha + 2k + 1 )/\beta ) \cdot
 \frac{2^k \ k!}{(2k)!} \cdot  \frac{\Gamma^{ k-1 }( ( \alpha +1)/\beta )}{\Gamma( (\alpha + 3)/\beta) }.
$$

Recall that we consider the values $ \beta $ greatest than 2. \par

\vspace{3mm}

 {\bf  Theorem 5.2.} {\it If the following conditions }

$$
\forall k < G \ \Rightarrow \zeta_{\alpha,\beta}(k) \le 1 \eqno(5.9)
$$
 {\it there holds, then the r.v. $ \gamma = \gamma_{\alpha,\beta} $ is strictly subgaussian.} \par

\vspace{3mm}

 {\bf Proof.}\par

It is sufficient to prove as before the following inequality for all the values $ k = 0,1,2, \ldots $

$$
\zeta(k)= \zeta_{\alpha,\beta}(k) := \Gamma( ( \alpha + 2k + 1 )/\beta ) \cdot
 \frac{2^k \ k!}{(2k)!} \cdot  \frac{\Gamma^{ k-1 }( ( \alpha +1)/\beta )}{\Gamma( (\alpha + 3)/\beta) } \le 1. \eqno(5.10)
$$

 Since $  \beta > 2, \ \lim_{k \to \infty} \zeta(k) = 0. $  We  conclude omitting complicated computation more exact
estimate  using  Stirling's formula:

$$
k \ge G  \ \Rightarrow  \zeta_{\alpha, \beta}(k) \le 1.
$$
 We derive taking into account the condition (5.9) that (5.10) is true for all the integer values $ k = 1,2,\ldots,  $ Q.E.D.\par

\vspace{3mm}

{\bf Remark 5.1. Example.} Let $  \alpha = \beta \to \infty;  $ then condition (5.9) is satisfied; moreover,

$$
\lim_{\beta \to \infty} \sup_{k \ge 1} \zeta_{\beta, \beta}(k)  = 0.
$$
 Therefore, there exists a value  $ Z   $ such that for all the values $  \alpha = \beta > Z $ the r.v. $  \gamma_{\beta, \beta}  $
is strictly subgaussian.

\vspace{3mm}

{\bf Remark 5.2.} Recall that the condition $  \gamma(1) \le 3 $ or in detail the inequality

$$
\Gamma((\alpha + 5)/\beta) \Gamma( ( \alpha + 1  )/\beta ) \le 3 \Gamma^2( (\alpha + 3   )/\beta ) \eqno(5.11)
$$
is necessary for strictly subgausianness.  Denote $ \epsilon = 1/\beta $ and assume $ \alpha = \const, \
 \epsilon \to 0+ \ \Leftrightarrow \beta \to \infty. $
 Taking into account the behavior of the Gamma function near to the value $ 0+: $

$$
\Gamma(\epsilon) \sim 1/\epsilon, \ \epsilon \to 0+,
$$
we deduce from (5.11) as $  \beta \to \infty $

$$
\frac{1}{\alpha + 5} \cdot  \frac{1}{\alpha+1} \le  \frac{3}{( \alpha + 3 )^2},
$$
which is not true as $ \alpha \to -1 + 0. $\par

 Moreover, the solution of the last inequality subject to the limitation $ \alpha > -1 $ has a form
 $ \alpha \ge 3(\sqrt{3} - 1) \approx 2.19.   $ \par
  Thus, for all the values $  \alpha $ from the interval $ -1 < \alpha < 3(\sqrt{3} - 1)  $ there exists a positive the value $ \beta_0 $
such that for all the values $ \beta > \beta_0 $  the random value $  \gamma = \gamma_{\alpha,\beta} $ is subgaussian but not strictly subgaussian.

\vspace{4mm}

Another (but more simple) {\bf example } of strictly subgaussian r.v. with {\it unbounded } support may be constructed by means of section 2.
Namely, let the r.v. $ \xi_1 $ has an uniform distribution on the symmetric interval  $ [-b,b], \ b = \const > 0 $ and let $  \xi_2 $ has a normal
(Gaussian) mean zero distribution  with at the same variation $ \Var \xi_2 = b^2/3. $ The arbitrary non - trivial mixture of these distributions
has an unbounded support, is not Gaussian and is strictly subgaussian by theorem 2.1. \par

 \vspace{3mm}

\section{ Acknowledgement.}\par

 Authors are very grateful to prof. S.V.Astashkin  and L.Maligranda for sending Your remarkable articles and comments. \\

\vspace{4mm}

config: arxiv


TITLE: Does gas pressure impact the rate of evaporation?
QUESTION [0 upvotes]: As we know, evaporation occurs when a liquid molecule has a kinetic energy that is greater than the minimum amount of energy needed to overcome the intermolecular attractive forces holding it in the liquid, allowing it to escape into the vapor phase.
We also know that the pressure of a gas is a measure of the average linear momentum of the moving molecules of a gas.
Does pressure of the gas above the liquid have any impact on the number of liquid molecules escaping into the vapor phase? Is there a gas pressure (linear momentum of gas molecules) high enough that a direct collision of a gas molecule with a liquid molecule would prevent the liquid molecule from vaporizing?

REPLY [0 votes]: Does pressure of the gas above the liquid have any impact on the number of liquid molecules escaping into the vapor phase?

Yes. If the pressure is low enough, the net rate at which molecules in the liquid will leave the liquid phase will be so great the liquid will boil.
Above that pressure, the rapidity with which gaseous molecules of the liquid species are moved away from near the surface will affect the rate of evaporation.
Molecules of the liquid species that do NOT move away from the surface may be reabsorbed into the liquid phase, thus reducing the rate of evaporation.
The two means by which molecules move away from the surface are convection (bulk flow) and diffusion. The greater the number of non-liquid-species molecules per volume, the lower the rate of diffusion.


TITLE: Second-order polynomial with cosine as a factor: Numeric vs. Analytic solution
QUESTION [0 upvotes]: I want to do the following seemingly simple thing:
Solve $$E = x + x^2\lambda\cos(3\phi)$$ with $x$ being positively defined number and $\lambda$ some small and irrelevant number (for dimensional reasons).
In essence, for a fixed constant value of $E$, I want to map each angles $\phi \in [0, 2\pi)$ to a real and positive $x$ value. (I am aware that I only need $[0, 2\pi/3)$ but for simplicity let's use $2\pi$)
The problem is the following though: Solving this numerically and analytically just doesn't give the same results.
I know that the solution $x$ oscillates around the value $E$ with frequency $2\pi/3$. I have also proven this numerically, just by calculating the solution of the equation numerically (while having the constraint of $x>0$). What I do is simply start from a small value of $x>0$ and increase it until the equation is satisfied (for a given $\phi$).
My problem is that when I try to solve this expression analytically, I run intro trouble! Let $a = \lambda\cos(3\phi)$, and
$$ \chi = \frac{-1 \pm \sqrt{1 + 4\alpha E}}{2\alpha}$$
be the solution. This is of course valid only for $a\neq0$. But this is not the problem. The problem is that solutions get arbitrary large when $a$ gets close to zero. This does not happen for the numeric solution. If necessary I can also upload figures which show the analytic and numeric solutions plotted together.
The numeric solution is simply: figure, where I have used actual values for $E, \lambda$.
So the question is: How can I get the numerical solution analytically, given that I know that $x>0$?

REPLY [0 votes]: Of course you will get one very large solution if the quadratic coefficient is close to zero, this simply because by the product part of Vietes formula the product of both roots is $E/α$.
You just need to select the small root, if possible with a numerically stable formula. As it happens, this exists for quadratic equations by way of the binomial formulas in that the solutions can also be written as
$$
\chi=\frac{2E}{1\pm\sqrt{1+4αE}}
$$
where then the "$+$" solution close to $E$ or one order better $\dfrac{E}{1+αE}+O(α^2)= E-αE^2+O(α^2)$ is the one you are looking for.


TITLE: An operator exponential/commutator question
QUESTION [1 upvotes]: There is "an important lemma" related to the Baker-Campbell-Haussdorff theorem which says that
$$
e^XYe^{-X} = Y + [X,Y] + \frac{1}{2!}[X,[X,Y]]+\ldots
$$
Clearly if $[X,Y]=0$ we get (noting that $e^{-X}e^{X} = \mathbf{1}$) that
$$
e^X Y-Ye^X = [e^X,Y]=0
$$
Thus $[X,Y]=0$ implies $[e^X,Y]=0$. Is the converse of this true? If $[e^X,Y]=0$ then does that imply $[X,Y]=0$? By the above formula this would imply that if
$$
[X,Y] + \frac{1}{2!}[X,[X,Y]]+\ldots =0
$$
then $[X,Y]=0$. It is not immediately obvious to me why this might be the case.

REPLY [2 votes]: If $[X,Y]=0$ all terms in your sum are zero, so it's no surprise that the sum is zero.
I will use the holomorphic functional calculus in a Banach algebra; this allows us to define $f(A)$ for any function $f$ holomorphic on an open domain that contains that contains the spectrum of the matrix $A$. There is  probably a less technical way of doing this.
For $s\in[0,1]$, consider the function $f_s(z)=z^s$ (this requires choosing a branch for the logarithm, but that's no issue). Since the spectrum $\sigma(e^X)$ of $e^X$ is discrete, we can choose an domain $D$ with smooth boundary, $0\not\in D$, and $\sigma(e^X)\subset D$. Note that
$$
f_s(e^X)=(f_s\circ\exp)(X)=e^{sX},
$$
since we have $f_s\circ\exp (z)=e^{sz}$. Then, since products are continuous, and integrals are limits of sums,
$$
f_s(e^X)Y=\left(\frac1{2\pi i}\int_{\Gamma} f(z)\,(zI-e^X)^{-1}\,dz\right)Y
=\left(\frac1{2\pi i}\int_{\Gamma} f(z)\,(zI-e^X)^{-1}Y\,dz\right)
=Y\left(\frac1{2\pi i}\int_{\Gamma} f(z)\,(zI-e^X)^{-1}\,dz\right)=Yf_s(e^X).
$$
(note that $(zI -e^X)Y=Y(z-e^X)$, so $(zI -e^X)^{-1}Y=Y(z-e^X)^{-1}$). So we have that
$$
e^{sX}Y=Ye^{sX},\ \ s\in[0,1].
$$
Now differentiate both sides and evaluate at $s=0$, to get
$$
XY=YX.
$$


TITLE: Fourier transform of the dilation of a distribution
QUESTION [1 upvotes]: Let $F$ a tempered distribution,  in my textbook the transform of $F$ is defined by $\hat{f}[ \phi]=F[\hat{\phi}]$ , having said that l'm interested in prove the following property of the Fourier transform:
$\mathbf{F}(F(tx))=|t|^{-n} \hat{f}(t^{-1} \xi)$
My attempt:
$(\mathbf{F}[F(tx)])[\phi]=F[\hat{(t\phi)}]=F[|t|^{-n} \hat{\phi }(t^{-1} \xi)]=|t|^{-n}F[\hat{\phi} (t^{-1} \xi)]=|t|^{-n} \hat{f}(t^{-1} \xi )$
This is correct? I am not sure that I have applied the definitions correctly. Best regards.
Added: n is the number of dimensions

REPLY [1 votes]: How I would write it
I here use the notation $\langle u, \varphi \rangle$ for the application of the distribution $u$ on the test function $\varphi.$
Introduce the operator $\lambda_t$ defined by $(\lambda_t f)(x) = f(tx).$ For a distribution $u$ we defined $\lambda_t u$ by $\langle \lambda_t u, \varphi \rangle = |t|^{-n} \langle u, \lambda_{1/t} \varphi \rangle.$
For a test function we have $\lambda_t\hat\varphi = |t|^{-n}\widehat{\lambda_{1/t}\varphi}.$ Using the above formulas, for a distribution we get
$$
\langle \widehat{\lambda_t u}, \varphi \rangle
= \langle \lambda_t u, \hat\varphi \rangle
= |t|^{-n} \langle u, \lambda_{1/t}\hat\varphi \rangle
= |t|^{-n} \langle u, |1/t|^{-n} \widehat{\lambda_t\varphi} \rangle \\
= \langle u, \widehat{\lambda_t\varphi} \rangle
= \langle \hat u, \lambda_t\varphi \rangle
= |t|^{-n} \langle \lambda_{1/t}\hat u, \varphi \rangle
,
$$
i.e. $\widehat{\lambda_t u} = |t|^{-n} \lambda_{1/t}\hat u.$