TITLE: Can we think about a particle trapped in a potential well in terms of "quantum measurement"?
QUESTION [3 upvotes]: Usually, when I'm thinking about a quantum measurement, I see a sort of particle that is being hit by a photon. The more energy the photon carries, the more the momentum of the particle is disturbed, but at the same time the more precise is the measurement of the particle's position.
But recently, I was thinking about a different scenario. Let's say, that our particle is trapped in a finite/infinite potential well of width $d$ (the distance between the walls). Now, if we would move the walls closer to each other (as a result, reducing the distance $d$), the particle would become "more localized". I guess we wouldn't be able to move the walls infinitely close to each other, because it would probably require an infinite amount of energy, but I think in general there is nothing stopping us from bringing them at least just a little bit closer.
What would happen to the momentum of such a particle? Is this even a reasonable scenario to think about in terms of quantum measurement?

REPLY [1 votes]: The method of measurement has no bearing on the result. The wavefunction for a particle confined at particular position is a delta function, formed from a superposition of momentum states.
$$\delta(x)=\frac 1 {2\pi}\int_{-\infty}^{\infty}e^{ixp}dp=\frac 1 {2\pi}\lim_{l\rightarrow\infty}\int_{-l}^{l}e^{ixp}dp=\frac 1 {\pi}\lim_{l\rightarrow\infty} \frac 1 x \sin{lx}.
$$
In practice, of course, we do not have a true delta function, because as you rightly point out, there is a bound on energy ($l$ does not go to infinity). We can plot the nascent delta function (finite $l$), in which one observes that short wavelengths (large momenta) are needed for a sharp spike (one usually talks of large energies rather than large momenta, to avoid the connotation of movement; the components of momentum are both positive and negative, cancelling out movement)

Incidentally, you can also see the reason for quantum tunneling in the diagram, because a wave function composed of a superposition of momentum states does go to zero outside the central region.


TITLE: Gauss curvature of a scaled metric (by a constant)
QUESTION [0 upvotes]: If I have a metric $g$ and $h=ag$ for some positive constant $a$, how does the Gauss curvature of $h$ ($K_h$) relate to the Gauss curvature of $g$ ($K_g$). I believe it should be $a^{-1}$ but I am not sure why (I'm somewhat new to this). Could someone let me know if this is correct (and maybe a very brief reason why). I'm sorry if this is a simple question, but I would like to use it for a paper I am working on.
If it is helpful, I am working with metrics on $S^2$ and I have conformal metrics $g_2=e^{2w}g_1$ and I would like to conclude that $K_{h}=e^{-2a} K_{g_2}$ where $h=e^{2w+2a}g_1=e^{2a}g_2$.

REPLY [1 votes]: Since the Gaussian curvature is the product of two geodesic curvatures, which scale as $\ell^{-1}$ (where $\ell$ is length as measured by the metric), it scales as $\ell^{-2}$. The metric itself is of scale $\ell^2$, so you're right that $h=ag$ implies $K_h = a^{-1}K_g.$
If you want a computational proof, you could start with one of the explicit formulae for $K_g$ in terms of $g$ and it should fall out pretty quickly.


TITLE: If $\phi$ is an endomorphism of an elliptic curve, and $\phi = \hat{\phi}$ then $\phi = [m]$?
QUESTION [3 upvotes]: I heard a reference to this fact, but I cannot find a reference. (I can find the converse in Silverman, namely that $\hat{[m]} = [m]$.)
Notation: $[m]$ is multiplication by $m$ in the group law, and $\hat{\phi}$ is the dual endomorphism of $\phi$.
Question: If $\phi$ is an endomorphism of an elliptic curve, and $\phi = \hat{\phi}$ then $\phi = [m]$?

REPLY [2 votes]: I think that if $E$ has CM this is false: take an endomorphism $\phi$ such that $\phi^2=[d]$, where $d$ is square-free. Then $\phi\circ\widehat{\phi}=[\deg\phi]=[d]$. Thus $\phi^2=\phi\circ\widehat{\phi}$, which implies $\phi=\widehat{\phi}$, but $\phi$ cannot be multiplication by an integer since it has non-square degree.


TITLE: Are $b$ and $c$ linearly dependent if the inner product $a \cdot b = a \cdot c$?
QUESTION [1 upvotes]: Let $a$ be a vector, but not the zero-vector. If the inner
product $a \cdot  b = a \cdot  c$. Are $b$ and $c$ linear dependent if the vectors $a, b$ and $c$ are 2-dimensional?
I would know how to show this if I would know the actual vector, but now I have no clue what to do. How do you argue whether $b$ and $c$ are linear dependent or not?

REPLY [2 votes]: $a \cdot b = a \cdot c \implies a \cdot (b-c) =0 \implies (b-c) \perp a$, if $a$ or $(b-c)$ is not zero. So, it is possible that both $b, c \in S$, where $S$ is defined as follows-
$$S = \left\{ x\in \mathbb{R}^n: x^Ta = 0, \text{ where } a\in \mathbb{R}^n\right\}.$$
But it is also possible that $b,c \notin S$ and two linearly independent vectors, though $(b-c) \in S$. This is because $S$ is a subspace of $\mathbb{R}^n (\text{if } a\neq \theta_n)$.
One of the examples of such cases can be found in Servaes's answer.


TITLE: facing a small obstacle while trying to proof Wolstenholme's theorem 1 using multiplicative inverse in (mod p)
QUESTION [1 upvotes]: Wolstenholme's theorem states that for a prime $p>3$
  and
  $$ \frac{1}{1^2}+\frac{1}{2^2}+\frac{1}{3^2}+......+\frac{1}{(p-1)^2}=\frac{a}{b}.$$
$ a\equiv 0\pmod p $

I think that I found a reasonable way to prove this theorem but I face a small obstacle at the end of the proof.
My solution steps are as follows.

make common factor for the fractions.
for each term in the numerator take (mod p) and simplify the result using the multiplicative inverse in (mod p)
for example the first term in numerator is
$ 2^2 \times 3^2 \times ..... \times (p-1)^2 $
in this quantity each term has its multiplicative inverse (mod p) multiplied by it so this quantity is only $ \equiv 1\pmod p $.
and after doing the same thing to other terms you get at the end that
the numerator is $ \equiv (1^2+1^2+2^2+3^2+....(p-2)^2)\pmod p $ .
after using the formula for summing n squared numbers and simplifying I got
$$a  \equiv \frac{2p^3-9p^2+13p}{6}\pmod p $$

this is the question :
for p>3 (p is a prime number)
how to prove that this quantity :
 $$ \frac{2p^3-9p^2+13p}{6} $$ is divisible by p .
or equivalently
$$ \frac{2p^2-9p+13}{6} $$ is Integer ( belongs to Z ) .
I appreciate any attempts for help.

REPLY [0 votes]: HINTS

In your numerator, $2p^2$ is always even and the other terms are always odd, and the difference of odd terms is always even, so the numerator is always even.
Since $p$ is a prime, we cannot have $p \equiv 0 \pmod{3}$. Thus, $p = 3n \pm 1$. The middle term clearly is divisible by $3$ and the corner ones yield
$$
\begin{split}
2p^2+13
 &= 2(3n\pm 1)^2+13\\
 &= 2\left(9n^2+6n+1\right)+13\\
 &= 18n^2+12n+15,
\end{split}
$$
and all coefficients are divisible by 3.

Can you finish this?


TITLE: Improper integral
QUESTION [7 upvotes]: I would like to know the value of the following improper integral:
$$\int \limits_{-\infty}^{\infty}\frac{2x}{1+x^2}dx$$
as the function $f(x)=\displaystyle\frac{2x}{1+x^2}$ satisfies $f(-x)=-f(x)$ can I immediately  conclude that the integral is equal to zero?
Thank you in advance for any suggestion.

REPLY [14 votes]: The Cauchy principal value of your integral is defined to be
$$ \lim_{a \to \infty} \int_{-a}^a {2x \over 1+x^2} \: dx $$
and this is zero by symmetry. So there is a sense in which the answer is $0$. However, if we let the limits go to infinity at different rates then we can make this integral "equal" whatever we want; for example, for any positive real constant $k$,
$$ \lim_{a \to \infty} \int_{-a}^{ka} {2x \over 1+x^2} \: dx = \lim_{a \to \infty} \log \left( {1+k^2 a^2 \over 1+a^2} \right) = 2 \log k. $$
Basically, since there is infinitely much area between this curve and the $x$-axis both in the first and third quadrants, we have to be careful about how we arrange things in order to make them "cancel out"; this is why we just throw up our hands and say that the integral doesn't converge.


TITLE: Please Help with Question about Linear Operators
QUESTION [1 upvotes]: The Question: Let $X$ and $Y$ be Banach spaces and $T: X \rightarrow Y$ an injective bounded linear operator. Show that if $R(T)$ is closed in Y, then $T^{-1} : R(T) \rightarrow X $ is bounded.
My attempt: So I was going to show that $T^{-1} $ is a continuous function. Therefore leading to showing $T^{-1} $ is bounded. Since $T$ is a bounded linear operator, $T$ is continuous. Is this the way to go?
Thank you very much!!

REPLY [0 votes]: You're right on track! Note that if we take a closed subset of a complete space, we obtain a complete space again. So, $R(T)$ is another Banach space. Therefore, we see that $T$ is a bijective bounded linear operator on its range. The main trick is to apply the open mapping theorem, which says that for surjective bounded linear operators in $\mathcal{L}(X, Y)$, with $X$, $Y$ Banach, open sets map to open sets.
To show that $T^{-1}$ is continuous, we can show that the pre-image of open sets are again open. So, $(T^{-1})^{-1} (U) = T(U) = O$, with $U$ open. Now, $O$ is also open by the open mapping theorem, so we see that $T^{-1}$ is indeed continuous.


TITLE: does normality ascend along this finite flat morphism?
QUESTION [0 upvotes]: Let $A$ be a Noetherian normal strict Henselian local ring with residue field of characteristic $p\ge 0$. Let $a\in A$ be a uniformizer in $A_P$ for some height 1 prime $P$, and $e$ be an integer coprime to $p$. Then $A[T]/(T^e-a)$ is finite free, hence flat over $A$. Must $A[T]/(T^e-a)$ be normal?

REPLY [0 votes]: The answer is yes! To do this, we use Serre's criteria for normality, which is that a Noetherian ring $R$ is normal if and only if it satisfies:

It has property ($R_1$), which is to say that for every prime ideal $p$ of height $\le 1$, the local ring $R_p$ is regular.
It has property $(S_2$), which is to say that for every prime ideal $p$, the local ring $R_p$ has depth at least min$\{2,dim(R_p)\}$

For $(R_1)$, let $q$ be a prime ideal of $A[T]/(T^e-a)$ be of height $\le 1$, then let $p := q\cap A$. If $a\notin q$, then since $A$ is strictly Henselian, $T^e-a$ splits completely over $A$, and hence $A[T]/(T^e-a)\cong A^e$ which is normal since $A$ is. So we may assume that $a\in q$. Then there is a natural map
$$A_p[T]/(T^e-a)\longrightarrow(A[T]/(T^e-a))_q$$
which realizes the target as a localization of the source. Since both rings are local of the same dimension, we find that this is an isomorphism. The source is a DVR by classical results, and hence the target is regular. This proves that $A[T]/(T^e-a)$ satisfies $(R_1)$.
For $(S_2)$, we use the result of http://stacks.math.columbia.edu/tag/0339 to reduce the problem to proving that the fibers of $A[T]/(T^e-a)$ over $A$ have $(S_2)$, but this is trivial since the fibers have the form $k[T]/(T^e)$ for some residue field $k$, which has dimension 0.


TITLE: Difference in Tikhonov regularization for linear and non-linear case?
QUESTION [0 upvotes]: the Tikhonov regularization for a linear operator $T: X \rightarrow Y, x \mapsto y$ means minimizing the least square problem
$$\begin{align*}
\lVert Tx - y \rVert^2_Y + \alpha \lVert x\rVert_X \rightarrow \min_{x\in X}
\end{align*}$$
This is equivalent to solving the regularized normal equations
$$\begin{align}
(T^*T+\alpha I)x = T^*y,
\end{align}$$
where $T^*$ is the adjoint operator of $T$.
Now I've currently read about the nonlinear case. Apparently, minimizing the above mentioned Tikhonov functional leads to the equation
$$\begin{align}
T'(x)^* (T(x)-y) + \alpha x = 0 \quad
\Rightarrow  \quad T'(x)^*T(x)+\alpha x = T'(x)^*y
\end{align}$$
I compared both expressions and I am wondering why the derivative of $T$ appears in the non-linear case but not in the linear one.
If I plug in a linear operator into the nonlinear expression, I feel like it should be equal to the linear regularized normal equations. But there is definitely a difference between a linear operator and its derivative, which would be a constant operator.
What am I missing here?

REPLY [1 votes]: I think you're confused by the notation. Lets say we have some transformation $f$ acting on $x$, i.e. $f(x)$.
The Problem can be stated as:
$$
\min_x \,\{\lVert f(x)-y\rVert^2 + \lambda\lVert x\rVert^2\}
$$
Taking the gradient w.r.t. $x$ yields:
$$
\frac{d\Phi}{dx} = f'(x)(f(x)-y)+\lambda x
$$
If $f(x)$ is a linear transformation, then
$$
f(x) = Tx
$$
and hence
$$
f'(x) = T^*
$$
Which in turn would again mean:
$$
(T^*T+\lambda I)x = T^*y
$$


TITLE: A quote from Arnold
QUESTION [17 upvotes]: Arnold said the following in a talk on teaching:

Jacobi noted, as mathematics' most fascinating property, that in it one and the same function controls both the presentations of a whole number as a sum of four squares and the real movement of a pendulum.

What's the meaning of this? Is the function he mentioned the theta function or something else? What's the relation with the real movement of a pendulum?

REPLY [4 votes]: OP's guess of "theta functions" isn't really that far off; it is well known (to those who know them) that the Jacobian elliptic function $\mathrm{sn}(u\mid m)$, which turns up in the solution of the DE for the pendulum, is expressible as a ratio of theta functions. (It should probably be emphasized that Jacobi studied both the theta functions and the elliptic functions now named after him, so he was certainly well-aware of this connection.)
As for the sum of four squares: as noted in this review article, Jacobi showed that the theta function $\vartheta_3(0,q)^d$ is the generating function for the number of ways to represent $k$ as $d$ squares $r_d(k)$; that is,
$$\vartheta_3(0,q)^d=\sum_{k=0}^\infty r_d(k) q^k$$
There you have it: two seemingly unrelated applications where the theta functions crop up.


TITLE: Quantum mechanics of thermal radiation
QUESTION [16 upvotes]: What are the quantum mechanisms behind the emission and absorption of thermal radiation at and below room temperature? If the relevant quantum state transitions are molecular (stretching, flexing and spin changes) how come the thermal spectrum is continuous? What about substances (such as noble gases) which don't form molecules, how do they emit or absorb thermal radiation? Is there a semi-classical mechanism (with the EM field treated classically) and also a deeper explanation using the full apparatus of QFT?

REPLY [3 votes]: Here's a classical (non-quantum-mechanical) answer:
As you probably know, light is an electromagnetic (EM) wave.  EM waves are produced by oscillating electrical charges. This is how an antenna works: connect it to an oscillating voltage source that pushes electrons back and forth in the antenna, and a radio wave is emitted.  This is what Maxwell's field equations are all about.
When you heat an object, its constituent atoms/molecules vibrate, which results in oscillating charges, which in turn emit EM radiation.  The velocities and oscillation periods of the atoms vary in a continuous distribution, and the peak of that distribution increases with temperature.  Hence the EM radiation is also distributed over a range of wavelengths (forming the classic black body curve); the peak wavelength changes (gets shorter) with increasing temperature; and the total power integrated over all wavelengths also increases with temperature.
The above describes how EM radiation is emitted.  As for absorption, when an EM wave hits an object, it is either reflected or absorbed.  The part that is absorbed turns into heat, which is manifested by increasing vibration of its atoms.  Eventually, the object's temperature will rise until emissions exactly equals absorption, at which point it will be in thermal equilibrium with its surroundings.
This is a complicated subject, and I've glossed over a lot of details, but this is a good mental model of what's happening.


TITLE: An example of a finite rank torsion free abelian group which is not finitely generated
QUESTION [0 upvotes]: Let $A$ be an abelian group. A  set $\{ a_1 , \cdots , a_k \}$ of nonzero elements of $A$ is called linearly independent if $\sum_{i=1}^{k}n_i a_i =0$ $(n_i \in  \mathbb{Z})$ implies $n_1 a_1 =\cdots =n_k a_k =0$. More explicitly, this means $n_i =0$ if $o(a_i )=\infty$ and $o(a_i )|n_i $ if $o(a_i )$ is finite.
By the rank $r(A)$ of $A$ is meant the cardinal number of a maximal independent set containing only elements of infinite and prime power orders.
Is there a finite rank torsion-free abelian group $A$ which is not finitely generated?

REPLY [2 votes]: Hint
The rank of the additive group of rationals is $1$.


TITLE: Find constant in system of function equations
QUESTION [4 upvotes]: Question: Let $a \in \mathbb{R}$. Find all possible values of $a$ such that there exists a function $f: \mathbb{R} \to \mathbb{R}$ satisfying $f(x + 2) = -f(x)$ and $f(x + 3) = f(x) + a$ for any real number $x \in \mathbb{R}$.

My (maybe wrong) solution:
Start listing down some values of $f(x)$ with $f(2) = -f(0)$ and $f(3) = f(0) + a$.
$f(2) + f(3) = - f(0) + f(0) + a = a$.
Similarly, you can write $f(1) = -(f(-1))$ and $f(2) = f(-1) + a$.
Hence, $f(1) + f(2) = -(f(-1)) + f(-1) + a = a$.
So, $f(1) + f(2) = f(2) + f(3) \Longrightarrow f(1) = f(3).$
But, according to the problem, $f(x + 2) = -f(x) \Longrightarrow f(3) = -f(1)$.
Since $f(1) = f(3)$ and $f(1) = -f(3) \Longrightarrow f(1) = f(3) = 0.$
Doing similar steps, we find out that $f(0) = f(2) = 0$. Hence, $a = f(2) + f(3) = 0 + 0 = 0$.
The only possible value of $a$ is $\boxed{0}$.
$\\$
If I'm doing something wrong, can someone please explain? Thanks in advance!
(Thanks to the answers, I now know I only need to prove that $a = 0$ can satisfy a valid function :))

REPLY [4 votes]: Your argument is fine - so far.
While looking for necessary conditions for $a$, we might but conclude more about $f$ by morking slightly more generally: We have
$$ f(x+6)=-f(x+4)=f(x+2)=-f(x)$$
and
$$ f(x+6)=f(x+3)+a=f(x)+2a$$
so by equating,
$$-f(x)=f(x)+2a $$
which shows that $f$ is necessarily constant. But then $a=f(x+2)=-f(x)=-a$ shows that in fact $a$ is necessarily $=0$.
What is left (and what is also missing from your attempt!), is to show that this is also sufficient, i.e., that for $a=0$ there actually does exist such a function.
Fortunately, this last step is not hard - all you have to check is that for $a=0$, the constant-zero function fulfills the functional equations.


TITLE: Is parameter free definability consistent with ZF?
QUESTION [1 upvotes]: Is it consistent to add the following axiom to $\sf ZF$?
Definability: $$\forall X \exists \alpha \exists \phi: X=\{y \in V_\alpha \mid V_\alpha \models \phi(y)\}$$
Where $\phi$ is a formula in one free variable only.
This is intended to capture the informal notion of every set being parameter free definable.

REPLY [3 votes]: Yes. In pointwise definable models every $X$ is definable by a formula $\phi$ with one free variable, and by applying reflection we get this schema. You can read more about pointwise definable models in

Hamkins, Joel David; Linetsky, David; Reitz, Jonas, Pointwise definable models of set theory, J. Symb. Log. 78, No. 1, 139-156 (2013). ZBL1270.03101.

On the other hand, your statement is an internal statement, so any model which is elementary equivalent to a pointwise definable model will satisfy it. But that's easy to arrange with an uncountable model which cannot be pointwise definable (since pointwise definable models are necessarily countable).
You can also read What does it really mean for a model to be pointwise definable? for more information.


TITLE: How "fundamental" is quantum information/computation?
QUESTION [1 upvotes]: I am wondering how fundamental the study of quantum information theory and computation is, in the sense of contributing to our understanding of the basic laws of nature. Will quantum information theory give us new insights into fundamental physics, or is it "merely" an application of quantum mechanics to information theory and computation? I find this to be a very interesting question, but can find very few sources that comment on it. I am currently reading Nielsen and Chuang's quantum information textbook, but I haven't seen this point addressed. There's a 2012 article by Aram Harrow (Why now is the right time to study quantum computing) which posits in the abstract that "quantum computing is not merely a recipe for new computing devices, but a new way of looking at the world", but sadly it is behind a paywall. In section 5 of this 1999 article, John Preskill briefly talks about how he thinks quantum information will contribute to fundamental physics, and says "I am anticipating a much broader interface between quantum information science and fundamental physics in the future". I am basically wondering what the current status of this interface is 15 years later, if there is any at all, and if the quantum information community still believes that quantum information may play an important role in fundamental physics.

REPLY [0 votes]: The quantum theory of information and computation sheds light on many issues in fundamental physics.
First, it explains that any physical system can be simulated by a universal computer. This sheds some light on the issue of whether we can expect to be able to understand the laws of physics.
Second, such a computer can be constructed by composing almost any single two-qubit gate:
http://arxiv.org/abs/quant-ph/9505018.
Third, thinking of quantum mechanical processes in terms of information flow in quantum computational networks can help explain experiments that have often been regarded as baffling or paradoxical, see
http://arxiv.org/abs/quant-ph/9906007
http://arxiv.org/abs/1109.6223.
There are also some limits to what we should expect the theory to explain. For example, it can't explain the second law of thermodynamics since it doesn't explain why any process would be irreversible:
http://www.ipod.org.uk/reality/reality_deutsch.pdf.


TITLE: Show that the chord of contact of tangents from the point $(a,-a)$ to the parabola $x^2=4ay$ has length $5a$
QUESTION [0 upvotes]: Show that the chord of contact of tangents from the point $(a,-a)$ to the parabola $x^2=4ay$ has length $5a$. My current method uses the distance formula and takes super long, is there some easier method that i'm missing?
Thanks in advance!

REPLY [0 votes]: The first coordinates of the contact points are easily computed to $x_{1,2}=a\pm a\sqrt5$.  Now the square of the distance of these points is clearly
$$(x_2-x_1)^2+\left(\frac{1}{4a}(x_2^2-x_1^2)\right)^2.$$
Finally notice that $x_2-x_1=2\sqrt5$, $x_2+x_1=2a$ and that $x_2^2-x_1^2=(x_2-x_1)(x_2+x_1)=4\sqrt5a$.


TITLE: Charged conductor in an external electric field
QUESTION [0 upvotes]: If we put a conductor in a place where there is a uniform electric field, then the field will change. Take for example the case of a conducting cylinder, then the field lines would curve to return something like this:

But what would happen if the cylinder was held to some potential $V_0>0$ before "turning on" the external electric field? Stated in other words, how would the resulting electric field change if the cylinder had a uniform surface charge density $\sigma$ before being placed inside the external electric field?

REPLY [0 votes]: Because Maxwell’s equations are linear, and we assume linear media, the principle of superposition holds. Thus, the field of a charged cylinder in an external field is simply the field without the added charge plus the field of the charged cylinder.


TITLE: Leray projector in $L^{\infty}$ and negative order Besov spaces for the Navier-Stokes equations
QUESTION [2 upvotes]: I was reading the paper "Norm inflation for the generalized Navier-Stokes equations" which can be found here: https://arxiv.org/abs/1212.3801.
In Lemma 2.1, the authors said for any $\phi \in L^{\infty}$ and $\alpha>0$ $$\Vert \nabla e^{-t(-\Delta)^{\alpha}}\Vert_{L^\infty} \leq C t^{-\frac{1}{t^{2\alpha}}}\Vert \phi\Vert_{L^\infty} .$$
I am not sure why this is true since the Leray projector $\mathbb P$ is a bounded operator in $L^{2}$. Furthermore, if we want to replace the $L^{\infty}$ norm with some negative Besov norm $\dot B^{-s}_{p,q}$, is the above inequality still true? I appreciate if any reference could be provided.

REPLY [1 votes]: Let $K(x)$ denote the Schwartz kernel of the Fourier multiplier $\nabla e^{-t(-\Delta)^\alpha}$. It's straightforward to check from Fourier inversion and Plancherel's theorem that for $t>0$,
$$K(x) = (2\pi)^{-3}\int_{\mathbb{R}^3} i\xi e^{-t|\xi|^{2\alpha}}e^{i\xi\cdot x}d\xi,\qquad\forall x\in\mathbb{R}^3.$$
I claim that $K\in L^1(\mathbb{R}^3)$. To see this, observe that for any multi-index $\beta=(\beta_1,\beta_2,\beta_3)$ of order $|\beta|=4$ and $x\in\mathbb{R}^3$ such that $x^\beta = x_1^{\beta_1}x_2^{\beta_2}x_3^{\beta_3}\neq0$, we have that
$$K(x)=\frac{1}{x^\beta (2\pi)^3}\int_{\mathbb{R}^3}i\xi e^{-t|\xi|^{2\alpha}} (\partial_{\xi}^\beta e^{i\xi\cdot x})d\xi.$$
Now integrate by parts in $\xi$. The most singular term in the application of the Leibnitz rule comes from repeatedly differentiating $|\xi|^{2\alpha}$. You can show that
$$|K(x)|\lesssim_\alpha \frac{1}{|x^\beta|}\int_{\mathbb{R}^3}|\xi|^{2\alpha-3}e^{-t|\xi|^{2\alpha}}d\xi\lesssim \frac{t^{-1/2\alpha}}{|x^\beta|},$$
where the last expression follows from dilation invariance of Lebesgue measure. Since $|\beta|>3$, the desired conclusion follows.
Regarding negative-order Besov spaces, try playing around with the fact that Littlewood-Paley projectors reduce Fourier multipliers to multiplication by a function of a dyadic frequency.


TITLE: Determining set membership from ordering relationships among disjoint sets
QUESTION [0 upvotes]: Suppose we have a set $P$ (an infinite set), and we have a partition of $P$ into (finitely many) disjoint subsets $P_i$, so that $P = \cup_i P_i$, and $P_i \cap P_j = \emptyset$ for $i \neq j$.
Suppose now that we have a set of partial orderings of $P$, such that for each of the orderings, we know that they strictly order some of the partitions. I'm not sure if that is proper math-speak, but what I mean is, for example, for the first partial ordering given, it guarantees $a < b$ for all $a \in P_1$ and for all $b \in P_2$ (I will write this as $P_1 < P_2$). There may, for example, be another partial ordering given that guarantees $P_2 < P_1$.
My question is, given a finite set of $p_i \in P$ with a finite number of partitions $P_j$, as well as a finite set of partial orderings, is it possible to infer set membership of the $p_i$'s into the partitions ($P_j$'s)? If so, then what is required of the $p_i$ and the partial orderings? How many and which kinds of orderings need to be supplied?
Since I don't know the proper math terminology, I'm making this a wiki.

REPLY [1 votes]: Your problem is a little unclear at the end.  In particular, you need to spell out more clearly what the inputs to the question are.  As written now, it seems to me that part of the initial data you're given are explicit elements $p_i\in P$. But then it seems that you should know right off the bat which partition $P_i$ each element is in.
Maybe the problem you're interested in is trying to determine set membership from a certain number of unspecified elements and the partial ordering on them induced by that on P.  This question then has a similar feel to trying to give necessary conditions for a partially-ordered set to be order-embeddable in another.
Why not work out a simple non-trivial example to make your question more clear?


TITLE: Smallest Prime-Factor of $4^{52} + 52^{2013} + 2013^{52}$
QUESTION [6 upvotes]: I would like to find out the smallest prime factor of $4^{52} + 52^{2013} + 2013^{52}$ by hand.
Thanks in advance

REPLY [2 votes]: As I've commented, the divisibility of $4^{52} + 52^{2013} + 2013^{52}$ by $7$ can be derived very easily as follows
$\displaystyle52\equiv3\pmod7\implies52^{2013}\equiv3^{2013}$
Now as $\displaystyle3^3=27\equiv-1\pmod7$ and $\displaystyle2013=3\cdot\text{ some odd number (namely}, 671 ),$
$\displaystyle 3^{2013}\equiv(-1)^{\text{ some odd number }}\pmod7\equiv-1\ \ \ \ (1)$
Again as $2013\equiv4\pmod 7\implies 2013^{52}\equiv4^{52},$
$\displaystyle 4^{52}+2013^{52}\equiv2\cdot4^{52}\pmod7\equiv2\cdot2^{104}=2^{105}=(2^3)^{35}\equiv1^{35}\pmod7$ as $2^3=8\equiv1\pmod7$
$\displaystyle\implies4^{52}+2013^{52}\equiv1\pmod7\ \ \ \ (2)$


TITLE: If M is a Artinian Module and A submodule of M, show that M/A is Artinian.
QUESTION [1 upvotes]: In the proof we assume that $$A_{1}/A\supseteq A_{2}/A\supseteq A_{3}/A ......$$
is a descending chain of submodules of $M/A$ and so we get that $$A_{1}\supseteq A_{2}\supseteq A_{3} ......$$ is a descending chain of submodules of $M$, but I do not know why, could anyone clarify this form me please?

REPLY [2 votes]: The notation seems a bit unfortunate. Let $\pi: M\to M/A$ be the canonical projection. Then, let us rather assume that $B_1\supseteq B_2 \supseteq B_2\supseteq \cdots$  is a descending chain of submodules of $M/A$ - after all, this is how we should check that it is Artinian. Let then $A_i:=\pi^{-1}(B_i)$. It follows immediately that $A_i\supseteq A_{i+1}$ for all $i$.
In your case, someone simply wrote $A_i/A$ instead of $B_i$. This is a little unprecise with respect to the fact that you might have a choice regarding $A_i$, i.e. there might be several submodules of $M$ that project to $B_i$. Instead of taking just any of these, you can always pick one that completely contains $A$. This way, the inclusion is no longer confusing.


text:
\begin{document}
\maketitle
\centerline{M. RABAOUI\footnote{Institut de Math\'ematiques de Jussieu, Universit\'e Pierre et Marie Curie, 175 rue de chevaleret, 75013 Paris. E-mail : rabaoui@math.jussieu.fr}}

\begin{abstract} In this paper, we determine the spherical functions of positive type on the space $V_\infty= M(\infty, \Bbb C)$ relatively to the action of the product group $K_\infty=U(\infty)\times U(\infty)$. The space $V_\infty$ is the inductive limit of the spaces of square complex matrices $V_n=M(n, \Bbb C)$. The group $K_\infty$ is the inductive limit of the product groups $K_n=U(n)\times U(n)$, where $U(n)$ is the unitary group.
\end{abstract}

\bigskip
\bigskip
\bigskip

\section{Introduction} The work that we present here take place within the framework of the infinite dimensional harmonic analysis on the spherical pairs. We consider in this article the spherical pair $(G_\infty, \ K_\infty)$, which is the inductive limit of the sequence of Gelfand pairs $(G_n, \ K_n)$: $$G_n=K_n\ltimes V_n, \ \ K_n=U(n)\times U(n), \ \ V_n=M(n, \Bbb C), $$ $$G_\infty=K_\infty\ltimes V_\infty, \ \ K_\infty=U(\infty)\times U(\infty).$$ Here $V_\infty=M(\infty,\Bbb C)$ is the space of infinite complex matrices having only a finite number of non-zero entries, and $U(\infty)$ is the group of the infinite unitary matrices $(u_{ij})$ with complex coefficients such that $u_{ij}=\delta_{ij}$ for $i+j$ large enough.

Let $\mathfrak{P}$ be the set of $K_\infty$-biinvariant continuous functions of positive type on $G_\infty$ satisfying $\pp(0)=1$. We are interested in the determination of the extreme points $\pp$ of this convex set which are indeed the spherical functions of postif type relatively to the pair $(G_\infty, \ K_\infty)$. The group $G_\infty$ is equipped with the inductive limit topology. The subgroup $K_\infty$ is closed. The homogeneous space $G_\infty/K_\infty$ is the vector space $$V_\infty=\bigcup_{n=1}^\infty V_n.$$ The law of the inductive limit group $G_\infty=K_\infty\ltimes V_\infty$ is given by : $$(u, \ x)(v, \ y)=\big((u_1v_1, \ u_2v_2), \ x+u_1yu_2^*\big), $$ where $$u=(u_1, u_2), \ v=(v_1, v_2)\in K_\infty \ {\rm and} \ x, \ y\in V_\infty.$$ A function $\pp$ on $G_\infty$ which is right invariant under $K_\infty$ does not depend on the variable $u\in K_\infty$. Therefore it is possible to see it like a function on $V_\infty$ : $$\pp(g)=\pp\big((u, x)\big)=\pp_0(x).$$ Moreover, if the function $\pp$ is $K_\infty$-biinvariant then the function $\pp_0$ which is defined on $V_\infty$ is $K_\infty$-invariant, or $U(\infty)$-biinvariant. Also, the function $\pp$ is of positive type on $G_\infty$ if and only if the function $\pp_0$ is of positive type. Let us note by $D_\infty$ the subspace of diagonal matrices in $V_\infty$. An element of $D_\infty$ can be decomposed as diag$(a_1, a_2, \dots)$, with $a_1,a_2,\dots\in\Bbb R$ and $a_j=0$ for $j$ large enough. Any matrix $x\in V_\infty$ can be diagonalised as $$x=u\ {\rm diag}(a_1, a_2, \dots) \ v^* \ \ \big(u, v\in U(\infty)\big).$$ Consequently, any $K_\infty$-invariant function on $V_\infty$ is uniquely determined by its restriction to the subset $D_\infty$.

For a number of spherical pairs, the spherical functions of positive type (the extreme points of $\mathfrak{P}$) have been determined. One can quote, for example, the work of Schoenberg \cite{Sho1} on $\Bbb R^{(\infty)}$ and those of G. Olshanski and A. Vershik \cite{Ol2} on the space $Herm(\infty,\Bbb C)$ of infinite dimensional Hermitian matrices. In these cases and others the spherical functions of positive type are obtained as limits of the ones of classical Gelfand pairs. In our situation, we obtain the spherical functions in this way. The principal result of this article is the following theorem :\\ \begin{paragraph}{\bfseries{Theorem}} The spherical functions of positive type $\pp$ on $V_\infty$, which are $U(\infty)$-biinvariant and satisfying $\pp(0)=1$, are given by : $$\pp\big({\rm diag}(\xi_1, \dots, \xi_n, 0, \dots)\big)=\Pi(\omega,\xi_1)\dots\Pi(\omega,\xi_n), $$ where $$\Pi(\omega,\lambda):=e^{-\frac{1}{4}\gamma\lambda^2}\prod_{k=1}^\infty\frac{1}{1+\frac{1}{4}\alpha_k\lambda^2}, $$ with $$\omega=(\alpha,\gamma),\ \gamma\in\Bbb R_+, \ \alpha_k\in\Bbb R_+ \ {\rm and} \ \sum_{k=1}^\infty \alpha_k <\infty.$$
\end{paragraph}

\bigskip

\section{Spherical functions of positive type on $(G_n,K_n)$} In this section, we will determine the spherical functions of positive type relatively to the pair $(G_n, K_n)$. An explicit formula of these functions was established for the first time in \cite{Ber}. It was also done in \cite{Mea} by a method using the Abel transform. It was also obtained in \cite{Ben}, by means of a contraction, starting from the spherical functions on $SU(n,n)/S\left(U(n)\times U(n)\right)$. The method that we use here is similar to the one used by J. Faraut in \cite{Far4} for the case of finite dimensional Hermitian matrices. This gives a simple and new proof of the result.\\

 We consider the space $V_n$ of complex matrices on which the group $K_n$ acts as follows : $$T(k):x\mapsto k.x=uxv^* \ \ (u, v\in U(n)).$$ Every matrix $x\in V_n$ admits a polar decomposition $$x=u\,{\rm diag(}\lambda_1, \dots, \lambda_n{\rm)}\,v^* \,, \ \ u, v\in U(n), \, \lambda_j\in\Bbb R. $$ Hence any function $f$ which is $K_n$-invariant on $V_n$ depends only on $\lambda={\rm diag(}\lambda_1, \dots, \lambda_n{\rm)}$ : $$f(x)=F(\lambda_1,\dots,\lambda_n),$$ where $F$ is a function defined on $\Bbb R^n$, invariant under $\mathfrak{S}_n\ltimes\{-1,1\}^n$.

 \bigskip

 Let us consider on $V_n$ the Euclidean structure defined by $ \<x,y\> = \mathfrak{Re}\,{\rm tr}(xy^*).$ The laplacian associated to this structure is : \ba\Delta=\sum_{i,j=1}^n\Big(\diff{}{(\mathfrak{Re}\,x_{jj})}+\diff{}{(\mathfrak{Im}\,x_{jj})}\Big)
         &+&\sum_{j<k}^n\Big(\diff{}{(\mathfrak{Re}\,x_{jk})}+\diff{}{(\mathfrak{Re}\,x_{kj})}\Big)\\
         &+&\sum_{j<k}^n\Big(\diff{}{(\mathfrak{Im}\,x_{jk})}+\diff{}{(\mathfrak{Im}\,x_{kj})}\Big).\ea

\bigskip

The laplacian is invariant under the action of $K_n$ in the following sense: if $f$ is a function of class $\mathscr{C}^2$, then $$\Delta f\big(T(k)\big)=\Delta\Big(f\big(T(k)\big)\Big)\ \ (k\in K_n).$$

Let $f$ be a function of class $\mathscr{C}^2$ on $V_n$ which is invariant under the action of $K_n$: $$f(uxv^*)=f(x)\ \ (k=(u,v)\in K_n).$$ The function $\Delta f$ is also invariant under $K_n$. This leads to the introduction of the operator $L$ defined by : $$\Delta f(x)=LF(\lambda_1,\dots,\lambda_n).$$ The operator $L$ is called the {\it radial part} of the laplacian.

\bigskip

\begin{pro}\label{radprop} {\rm (i)} Let $f$ be a $K_n$-invariant function of class $\mathscr{C}^2$. Then $$\Delta f(x)=LF(\lambda_1,\dots,\lambda_n),$$ where
\ba LF=\sum_{i=1}^n\Big(\diff{F}{{\lambda_i}}+\frac{1}{\lambda_i}\dif{F}{{\lambda_i}}\Big)
  &+&2\sum_{i<j}\frac{1}{\lambda_i-\lambda_j}\Big(\dif{F}{{\lambda_i}}-\dif{F}{{\lambda_j}}\Big)\\
  &+&2\sum_{i<j}\frac{1}{\lambda_i+\lambda_j}\Big(\dif{F}{{\lambda_i}}+\dif{F}{{\lambda_j}}\Big).\ea {\rm (ii)} The preceding formula can also be written as
$$LF=\frac{1}{D(\lambda)}\,\sum_{i=1}^n\Big(\diff{}{\lambda_i}+\frac{1}{\lambda_i}\dif{}{\lambda_i}\Big)\big(D(\lambda)F(\lambda)\big),$$ where $D$ is given by $$D(\lambda)=\prod_{i<j}(\lambda_i^2-\lambda_j^2).$$
\end{pro}

\bigskip

In order to prove the preceding proposition, we will use the following result (see \cite{Far4}, Lemma IX-2.2) :

{{\it {\underline{\bf Lemma :}} Let $f$ be a function of class $\mathscr{C}^2$ on an open set $\mathscr{U}$ of a finite dimensional real vector space $\mathscr{V}$. Let $\mathscr{A}$ be an endomorphism of $\mathscr{V}$, $a\in\mathscr{V}$. Let $\ep>0$ be such that, for $|t|<\ep$, $\exp{t\mathscr{A}}.\,a\in\mathscr{U}$. We assume that, for $|t|<\ep$, $$f(\exp{t\mathscr{A}.\,a)}=f(a).$$
Then \ba&(Df)_a\left(\mathscr{A}.\,a\right)=0,&\\ &(D^2f)_a\left(\mathscr{A}.\,a,\,\mathscr{A}.\,a\right)+\left(Df\right)_a\left(\mathscr{A}^2.\,a\right)=0.&\ea}}

\bigskip

\begin{paragraph}{\bfseries{Proof of Proposition \ref{radprop}.}}{\rm Let $\mathscr{U}$ be an open set in $V_n$ and $\mathscr{A}$ the endomorphism of $V_n$ defined by : $\mathscr{A}.\,a=Xa+aY^*$ where $X,Y\in V_n$. If the matrices $X,\,Y$ are skewHermitian, then for every $t\in\Bbb R$, the matrices $\exp{tX}$, $\exp{tY}$ are unitary and, for every $a\in\mathscr{U}$,\\ $$f(\exp{tX}a\exp{tY^*})=f(a).\\$$

We deduce from (\cite{Far4}, lemma IX-2.2) that\\ $$\left(Df\right)_a\left(Xa+aY^*\right)=0,$$ $$\left(D^2f\right)_a\left(Xa+aY^*,\,Xa+aY^*\right)+\left(Df\right)_a\left(X^2a+2XaY^*+a\left(Y^*\right)^2\right)=0 .$$
\bigskip

(a) Let us put $X=Y=E_{jk}-E_{kj}$ ($j\ne k$), $a={\rm diag(}a_1,\dots,a_n{\rm )}$. We obtain\\ $$Xa+aY^*=\left(a_k-a_j\right)\left(E_{jk}+E_{kj}\right),$$ $$X^2a+2XaY^*+a\left(Y^*\right)^2=2\left(a_k-a_j\right)\left(E_{jj}-E_{kk}\right),$$ and hence $$\left(a_k-a_j\right)^2\left(D^2f\right)_a\left(E_{jk}+E_{kj},\,E_{jk}+E_{kj}\right)+2\left(a_k-a_j\right)\left(Df\right)_a\left(E_{jj}-E_{kk}\right)=0 ,$$
where
 $$\diff{f}{(\mathfrak{Re}\,x_{jk})}(a)+\diff{f}{(\mathfrak{Re}\,x_{kj})}(a)={2\over{\left(a_j-a_k\right)}}\left(\dif{f}{(\mathfrak{Re}\,x_{jj})}(a)- \dif{f}{(\mathfrak{Re}\,x_{kk})}(a)\right).$$
 \bigskip

 (b) Let us put $X=i\left(E_{jk}+E_{kj}\right)$ et $Y=-X$. We get\\
 $$Xa+aY^*=\left(a_j+a_k\right)\left(iE_{jk}+iE_{kj}\right),$$ $$X^2a+2XaY^*+a\left(Y^*\right)^2=-2\left(a_j+a_k\right)\left(E_{jj}+E_{kk}\right),$$ and hence $$\left(a_j+a_k\right)^2\left(D^2f\right)_a\left(iE_{jk}+iE_{kj},\,iE_{jk}+iE_{kj}\right)-2\left(a_j+a_k\right)\left(Df\right)_a\left(E_{jj}+E_{kk}\right)=0 ,$$
where
 $$\diff{f}{(\mathfrak{Im}\,x_{jk})}(a)+\diff{f}{(\mathfrak{Im}\,x_{kj})}(a)={2\over{\left(a_j+a_k\right)}}\left(\dif{f}{(\mathfrak{Re}\,x_{jj})}(a)+ \dif{f}{(\mathfrak{Re}\,x_{kk})}(a)\right).$$
 \bigskip

 (c) Let us put $X=iE_{jj}$ and $Y=-X$. We obtain\\
 $$Xa+aY^*=i\,2a_jE_{jj},$$ $$X^2a+2XaY^*+a\left(Y^*\right)^2=-4a_jE_{jj},$$ and hence
$$4a_j^2\left(D^2f\right)_a\left(iE_{jj},\,iE_{jj}\right)-4a_j\left(Df\right)_a\left(E_{jj}\right)=0 ,$$
where
 $$\diff{f}{(\mathfrak{Im}\,x_{jj})}(a)={1\over{a_j}}\dif{f}{(\mathfrak{Re}\,x_{jj})}(a).$$

 Finally, \ba\diff{f}{(\mathfrak{Re}\,x_{jj})}(a)&=&\diff{F}{\lambda_j},\\ \diff{f}{(\mathfrak{Im}\,x_{jj})}(a)&=&{1\over{\lambda_j}}\dif{F}{\lambda_j},\\
\diff{f}{(\mathfrak{Re}\,x_{jk})}(a)+\diff{f}{(\mathfrak{Re}\,x_{kj})}(a)&=&{2\over{\left(\lambda_j-\lambda_k\right)}}\left(\dif{F}{\lambda_j}(a)- \dif{F}{\lambda_k}(a)\right),\\
\diff{f}{(\mathfrak{Im}\,x_{jk})}(a)+\diff{f}{(\mathfrak{Im}\,x_{kj})}(a)&=&{2\over{\left(\lambda_j+\lambda_k\right)}}\left(\dif{F}{\lambda_j}(a)+ \dif{F}{\lambda_k}(a)\right).\ea}

This proves (i). In order to prove (ii), we have to use the formula : $$\Delta_0(DF)=\Delta_0F+2(\nabla_0D|\nabla_0F)+\Delta(D),$$ where $\Delta_0$ is the laplacian and $\nabla_0$ the gradiant on $\Bbb R^n$. The polynomial $D$ is harmonic and $$\sum_{j=1}^n{1\over{\lambda_j}}\dif{D}{\lambda_j}=0.$$ We can then conclude that $${1\over D}\Delta_0(DF)+{1\over D}\sum_{j=1}^n{1\over{\lambda_j}}\dif{(DF)}{\lambda_j}=\Delta_0F+ 2{1\over D}(\nabla_0D|\nabla_0F)+\sum_{j=1}^n{1\over{\lambda_j}}\dif{F}{\lambda_j}.$$

Since $${1\over D}\nabla_0D=\nabla_0\log|D|=\sum_{j<k}\frac{1}{\lambda_j-\lambda_k}(e_j-e_k)+\sum_{j<k}\frac{1}{\lambda_j+\lambda_k}(e_j+e_k),$$ where $(e_1,\dots,e_n)$ is the canonical basis of $\Bbb R^n$, we obtain $${1\over D}(\nabla_0D|\nabla_0F)=\sum_{j<k}\frac{1}{\lambda_j-\lambda_k}\left(\dif{F}{\lambda_j}-\dif{F}{\lambda_k}\right)+\sum_{j<k}\frac{1}{\lambda_j+\lambda_k}\left(\dif{F}{\lambda_j}+\dif{F}{\lambda_k}\right).\hspace{1cm}\Box$$
\end{paragraph}

\bigskip

In the preceding Euclidean polar decomposition, the measure $m$ can be written as : \\ $$\alpha(du)\alpha(dv)\prod_{j<k}\big(\lambda_j^2-\lambda_k^2\big)^2\prod_{j=1}^n\lambda_j\,d\lambda_j ,$$ where $\alpha$ is the normalized Haar measure of the unitary group $U(n)$. Moreover, one has the following integration formula :\\

\begin{pro}\label{intprop}{{\rm (see \cite{Far3}, Proposition X.3.4)}} For every integrable function $f$ on $V_n$\\ $$\int_{V_n}f(x)m(dx)=c_n\int_{U(n)\times U(n)}\int_{\Bbb R_+^n}f(u\lambda v)\,\alpha(du)\alpha(dv)\prod_{j<k}\big(\lambda_j^2-\lambda_k^2\big)^2\prod_{j=1}^n\lambda_j\,d\lambda_j .$$ $c_n$ is the following constant : \\ $$c_n=\frac{2^n\pii^{n^2}}{n!\Big(\prod_{j=1}^{n-1}j!\Big)^2}.$$
\end{pro}

\bigskip

By using the preceding results, the resolution of the Cauchy problem for the heat equation on $V_n$ leads to the evaluation of the orbital integral $\mathscr{I}(x,y)$, which is defined for $x,y\in V_n$ by  $$\mathscr{I}(x,y)=\int_{U(n)}\int_{U(n)}e^{\mathfrak{Re}\,{\rm tr}(xuyv^*)}\alpha(du)\alpha(dv).$$

\bigskip

One can remark that the function $\mathscr{I}(x,y)$ is determined by its restriction to the subspace of diagonal matrices because it is invariant under $K_n$ : $$\mathscr{I}(uxv^*,y)=\mathscr{I}(x,uyv^*)=\mathscr{I}(x,y)\ \ (u,v\in U(n)).$$
The Cauchy problem for the heat equation $$\dif{U}{t}=\Delta U,$$ $$U(0,x)=f(x),$$ where $f$ is a bounded continuous function on $V_n$, has a unique solution which is $$ U(t,x)=\frac{1}{(4\pi t)^{\frac{N}{2}}}\int_{V_n}e^{-\frac{1}{4t}|||x-y|||^2}f(y)\,m(dy) \ \ (t>0,\, x\in V_n),$$ where $N=2n^2$ is the dimension of $V_n$, $|||.|||$ is the Hilbert-Schmidt norm on $V_n$ and $m$ is the Euclidean measure.\\

Let us assume that the function $f$ is invariant under the action of $K_n$. Then the solution $U$ will be also $K_n$-invariant. Hence we can write $$f(x)=f_0(\lambda),\ \ U(t,x)=U_0(t,\lambda).$$ By using the Weyl integration formula (Proposition \ref{intprop}), the solution $U_0(t,\lambda)$ is given by:

$$U_0(t,\lambda)=\int_{\Bbb R_+^n} H_0(t,\lambda,\theta)f_0(\theta)D(\theta)\prod_{j=1}^n\theta_j\,d\theta_j,$$
with
\ba H_0(t,\lambda,\theta)&=& c_n\frac{1}{(4\pi t)^{\frac{N}{2}}}\int_{U(n)}\int_{U(n)} e^{-\frac{1}{4t}|||\lambda-u\theta v^*|||^2}f(y)\,\alpha(du)\alpha(dv)\\
&=& c_n\frac{1}{(4\pi t)^{\frac{N}{2}}}e^{-\frac{1}{4t}(||\lambda||^2+||\theta||^2)}\int_{U(n)}\int_{U(n)}e^{\frac{1}{2t}\mathfrak{Re}\,{\rm tr}(\lambda u\theta v^*)}\alpha(du)\alpha(dv)\\
&=&c_n\frac{1}{(4\pi t)^{\frac{N}{2}}}e^{-\frac{1}{4t}(||\lambda||^2+||\theta||^2)}\mathscr{I}(\frac{1}{2t}\lambda,\theta).\ea

\bigskip

\begin{teo} If $\lambda={\rm diag(}\lambda_1,\dots,\lambda_n{\rm)}$ and $\theta={\rm diag(}\theta_1,\dots,\theta_n{\rm)},$ then $$\mathscr{I}(\lambda,\theta) =2^{{n(n-1)}}\big[1!2!\times\dots\times(n-1)!\big]^2\,\frac{1}{D(\lambda)D(\theta)}\,{\rm det\Big(}\big(I_0(\lambda_i\theta_j)\big)_{1\leq i,j\leq n}{\rm \Big)},$$ where $I_0$ is the modified Bessel function: $$I_0(z)=\sum_{k=0}^\infty\frac{1}{2^{2k}(k!)^2}\,z^{2k} \ \ (z\in\Bbb C).$$
\end{teo}
\begin{paragraph}{\bfseries{Proof.}}{\rm From the formula that gives the radial part of the laplacian (Proposition \ref{radprop}) one can deduce that the function $U_0$ is a solution of the equation\\ $$\diff{U_0}{t}=\frac{1}{D(\lambda)}\,\sum_{i=1}^n\Big(\diff{}{\lambda_i}+\frac{1}{\lambda_i}\dif{}{\lambda_i}\Big)\big(D(\lambda)F(\lambda)\big).$$ We put then $$V(t,\lambda)=D(\lambda)U_0(t,\lambda),\ \ g(\lambda)=D(\lambda)f_0(\lambda).$$

\bigskip

 The function $V$ is in consequence a solution of the Cauchy problem\\

  $$\diff{V}{t}=\sum_{i=1}^n\Big(\diff{V}{\lambda_i}+\frac{1}{\lambda_i}\dif{V}{\lambda_i}\Big),$$ $$V(0,\lambda)=g(\lambda).$$

  \bigskip

Let us assume that $f$ belongs to the Schwartz space $\mathscr{S}(V_n)$. By a result similar to (\cite{Far4}, Lemme X-3.1) one can prove that, for every $T>0$, the function $V$ is bounded on $[0,T]\times\Bbb R_+^n$.

 On the other hand, for $n=1$, the preceding problem is equivalent, in cylindrical coordinates, to the Cauchy problem for the heat equation on $\Bbb R^2$ with $f$ radial. The solution of such a problem is expressed using the modified Bessel function $I_0$ and it is given by (see \cite{Far4}, Chapter IX, exercice 3) :

  $$\frac{1}{2t}\int_0^\infty e^{-\frac{r^2+\rho^2}{4t}}g(\rho)\,I_0\Big(\frac{r\rho}{2t}\Big)\,\rho\,d\rho.$$

  \bigskip

  In consequence we can conclude that the solution of our problem for an arbitrary $n$ is given by $$V(t,\lambda)=\frac{1}{(2t)^n}\int_{\Bbb R_+^n}e^{-\frac{1}{4t}(||\lambda||^2+||\theta||^2)}g(\theta)\prod_{i=1}^nI_0\Big(\frac{\lambda_i\theta_i}{2t}\Big)\,\theta_i\,d\theta_i.$$  As the function $g$ is skewsymmetric this last relation can be written $$V(t,\lambda)=\frac{1}{(2t)^n}\int_{\Bbb R_+^n}e^{-\frac{1}{4t}(||\lambda||^2+||\theta||^2)}g(\theta)\frac{1}{n!}\sum_{\sigma\in\mathfrak{S}_n}\varepsilon(\sigma)\prod_{i=1}^nI_0\Big(\frac{\lambda_i\theta_{\sigma(i)}}{2t}\Big)\,\theta_i\,d\theta_i.$$ Hence, for every function $g(\theta)=D(\theta)f_0(\theta)$, where $f_0$ is a symmetric function in $\mathscr{S}(\Bbb R_+^n),$ \ba\lefteqn{\int_{\Bbb R_+^n} H_0(t,\lambda,\theta)g(\theta)\prod_{i=1}^n\theta_i\,d\theta_i}\\& & = \frac{1}{n!(2t)^n}\int_{\Bbb R_+^n}e^{-\frac{1}{4t}(||\lambda||^2+||\theta||^2)}g(\theta)\,{\rm det\bigg(}\Big(I_0\big(\frac{\lambda_i\theta_j}{2t}\big)\Big)_{1\leq i,j\leq n}{\rm \bigg)}\,\prod_{i=1}^n\theta_i\,d\theta_i .\ea This proves that the kernel $H_0$ is equal to $$H_0(t,\lambda,\theta)=\frac{1}{n!(2t)^n}\,\frac{1}{D(\lambda)D(\theta)}\,e^{-\frac{1}{4t}(||\lambda||^2+||\theta||^2)}\,{\rm det\bigg(}\Big(I_0\big(\frac{\lambda_i\theta_j}{2t}\big)\Big)_{1\leq i,j\leq n}{\rm \bigg)}.$$ We obtain the result by comparing, for $t=\frac{1}{2}$, the two expressions we obtained for $H_0$.$\hspace{1cm}\Box$}
\end{paragraph}

\bigskip

For $x\in V_n$ the orbital measure $\m_x$ is defined on $V_n$ by\\ $$\int_{V_n}f(y)\,\m_x(dy)=\int_{U(n)}\int_{U(n)}f(uxv^*)\,\alpha(du)\alpha(dv),$$ where $\alpha$ is the normalized Haar measure on $U(n)$ and $f$ is a continuous function on $V_n$. The Fourier transform of $\m_x$ is the following function $\widehat{\m_{x}}$ :\\ \ba \widehat{\m_x}(\xi)&=&\int_{V_n}e^{i\<\xi,y\>}\m_x(dy)\\
&=&\int_{U(n)}\int_{U(n)}e^{i\mathfrak{Re}\,{\rm tr}(\xi uxv^*)}\alpha(du)\alpha(dv)\\
&=& \mathscr{I}(x,i\xi).\ea

\bigskip

The spherical functions of positive type for the Gelfand pair $(G_n,K_n)$ are the functions $\pp_x=\widehat{\m_{x}}, \ (x\in\Bbb R^n)$, Fourier transforms of the orbital measures $\m_x$.\\

\begin{cor}\label{spherfin} If $x={\rm diag(}x_1,\dots,x_n{\rm)}$ and $\xi={\rm diag(}\xi_1,\dots,\xi_n{\rm)},$ $$\pp_x(\xi):=\widehat{\m_x}(\xi)=\big({\bf \delta!}\big)^2\,\frac{(-4)^{\frac{n(n-1)}{2}}}{D(x)D(\xi)}\,{\rm det\Big(}\big(J_0(x_j\xi_k)\big)_{1\leq j,k\leq n}{\rm \Big)},$$ where  $${\bf \delta}=(\delta_1,\delta_2,\dots,\delta_n):=(n-1,n-2,\dots,0),$$ $${\bf \delta!}=\delta_1!\times\dots\times\delta_n!,$$ and $J_0$ is the classical Bessel function.
\end{cor}

\bigskip

\section{Multiplicativity property of spherical functions}

\bigskip

A spherical function, for the spherical pair $(G_\infty,K_\infty)$, is a continuous function $\pp$ on $G_\infty$ satisfying $$\lim_{n\rightarrow\infty}\int_{K_n}\pp(xky)dk=\pp(x)\pp(y),$$ where $dk$ is the normalized Haar measure of the product group $K_n=U(n)\times U(n)$ (see \cite{Far2}, Theorem 5.1). In our case, the function $\pp$ can be seen as a function on $V_\infty$ and hence  $$\lim_{n\rightarrow\infty}\int_{U(n)\times U(n)}\pp(x+k_1yk_2^*)\alpha_n(dk_1)\alpha_n(dk_2)=\pp(x)\pp(y),$$ where $\alpha_n$ is the normalized Haar measure of the unitary group $U(n)$.\\

\begin{teo}\label{multi}{{\rm (The multiplicativity property)}} Let $\pp\in\mathfrak{P}$. The function $\pp$ is spherical, if and only if, there exists a continuous function $\Phi$ on $\Bbb R$, with $\Phi(0)=1$ such that $$\pp\big({\rm diag}(a_1,\dots,a_n,0,\dots)\big)=\Phi(a_1)\dots\Phi(a_n).$$
\end{teo}

\bigskip

Let us put, for $m\leq n$,$$K_m(n)=\Biggl\{\left(\begin{array}{cc}       I_m & 0  \\       0  & v \end{array}\right)\,\Big|\, v\in U(n-m)\Biggl\}\simeq U(n-m), $$ and $$K_m(\infty)=\bigcup_{n=1}^\infty K_m(n)\subset U(\infty).$$ Also, let us put $$K(m,n)=\Biggl\{\left(\begin{array}{cc}       u & 0  \\       0  & v \end{array}\right)\,\Big|\, u\in U(m),\, v\in U(n-m)\Biggl\}\subset U(n).$$ We introduce in addition, for $n\geq 2m$, a Cartan subgroup for the symmetric pair $\big(U(n),\,K(m,n)\big)$:$$a(\theta)=\left(\begin{array}{ccccccc}  \cos\theta_1 & & & -\sin\theta_1 & & & \\
 & \ddots & & & \ddots & & \\
 & & \cos\theta_m & & & -\sin\theta_m & \\
\sin\theta_1 & & & \cos\theta_1 & & & \\
 & \ddots & & & \ddots & & \\
 & & \sin\theta_m & & & \cos\theta_m & \\
 & & & & & & I_{n-2m}\end{array}\right).$$ Hence, for every $k\in K(n)$, $$k=h_1a(\theta)h_2,$$ with $h=(h_1,h_2)\in K(m,n)\times K(m,n)=:K^2(m,n)$, and the Weyl integration formula that corresponds to this last decomposition is given, for every integrable function $f$ on $U(n)\times U(n)$, by  \ba\lefteqn{\int_{U(n)\times U(n)} f(k_1,\,k_2)\,\alpha_n(dk_1)\alpha_n(dk_2)}\\& & =\int_{[0,\pi]^m\times [0,\pi]^m}\int_{K^2(m,n)}\int_{K^2(m,n)}f\big(h_1a(\theta)h_2,\,g_1a(\zeta)g_2\big)\,\kappa(dh)\kappa(dg)D_{m,n}(\theta)D_{m,n}(\zeta)\,d\theta\,d\zeta,\ea where $\kappa=\beta\otimes\beta$ and $\beta$ is the standardized Haar measure of $K(m,n)$, and $$D_{m,n}(\theta)=c_{m,n}\Big|\prod_{1\leq i<j\leq m}\big(\sin(\theta_i+\theta_j)\big)^2\big(\sin(\theta_i-\theta_j)\big)^2\prod_{i=1}^m\big(\sin2\theta_i\big)\big(\sin\theta_i\big)^{2(n-2m)}\Big|,$$ where $c_{m,n}$ is a constant such that $$\int_{[0,\pi]^m}D_{m,n}(\theta)\,d\theta_1\dots\theta_m=1.$$

\bigskip

 \begin{pro}\label{convkey} Let $f$ be a continuous function on $K_\infty$ which is $K_m(\infty)\times K_m(\infty)$-invariant. Then \ba\lefteqn{\lim_{n\rightarrow\infty}\int_{U(n)\times U(n)}f(k_1,\,k_2)\,\alpha_n(dk_1)\alpha_n(dk_2)}\\& & =\int_{K_m^2}\int_{K_m^2}f(h_1w_mh_2,\,g_1w_mg_2)\,\alpha_m(dh_1)\alpha_m(dh_2)\alpha_m(dg_1)\alpha_m(dg_2),\ea
  where $$w_m=a(\frac{\pi}{2},\dots,\frac{\pi}{2})=\left(\begin{array}{ccc}  0 & -I_m & 0 \\
I_m &  0   & 0 \\
 0 & 0 & I \end{array}\right).$$
\end{pro}

\bigskip

\begin{lem}\label{convmult} Let $X$ be a compact topological space and $\m$ a positive measure on $X$ such that for every non-null open set $B$ we have $\m(B)\geq0$. Let $\delta$ be a continuous positive function on $X$, which reaches its maximum at a unique point $x_0$. Let us put $$\frac{1}{c_n}=\int_X\delta(x)^n\,\m(dx).$$ Then, if $f$ is a continuous function on $X\times X$, $$\lim_{n\rightarrow\infty}c_n^2\int_{X\times X}f(x,y)\delta(x)^n\delta(y)^n\,\m(dx)\m(dy)=f(x_0,x_0).$$
\end{lem}
\begin{paragraph}{\bfseries{Proof.}}{\rm We apply Lemma 5.4 in \cite{Far2} to the function $\delta(x,y)=\delta(x)\delta(y)$.$\hspace{1cm}\Box$}
\end{paragraph}

\bigskip

By using Lemma \ref{convmult}, for every fixed $m$, and for every continuous function $f$ on $[0,\pi]^m\times[0,\pi]^m$,$$\lim_{n\rightarrow\infty}\int_{[0,\pi]^m\times[0,\pi]^m}f(\theta,\,\zeta)D_{m,n}(\theta)D_{m,n}(\zeta)\,d\theta \,d\zeta=f\Big(\frac{\pi}{2},\dots,\frac{\pi}{2};\frac{\pi}{2},\dots,\frac{\pi}{2}\Big).$$

\bigskip

\begin{paragraph}{\bfseries{Proof of Proposition \ref{convkey}.\\}}{\rm By using the integration formula and the invariance under $K_m(\infty)\times K_m(\infty)$, we obtain $$\int_{U(n)\times U(n)} f(k_1,\,k_2)\,\alpha_n(dk_1)\alpha_n(dk_2)=\int_{[0,\pi]^m\times [0,\pi]^m}F(\theta,\,\zeta)D_{m,n}(\theta)D_{m,n}(\zeta)\,d\theta\,d\zeta,$$ with $$F(\theta,\,\zeta)=\int_{K_m^2}\int_{K_m^2}f\big(h_1a(\theta)h_2,\,g_1a(\zeta)g_2\big)\,\alpha_m(dh_1)\alpha_m(dh_2)\alpha_m(dg_1)\alpha_m(dg_2).$$ As a result, by Lemma \ref{convmult}  \ba\lefteqn{\lim_{n\rightarrow\infty}\int_{U(n)\times U(n)} f(k_1,\,k_2)\,\alpha_n(dk_1)\alpha_n(dk_2)}\\& &=\int_{K_m^2}\int_{K_m^2}f(h_1w_mh_2,\,g_1w_mg_2)\,\alpha_m(dh_1)\alpha_m(dh_2)\alpha_m(dg_1)\alpha_m(dg_2).\hspace{1cm}\Box\ea}

\end{paragraph}

\bigskip

\begin{cor}\label{cormult} Let $\pp$ be a $K_\infty$-invariant continuous function on $V_\infty$. Then, for\\ $x={\rm diag(}a_1,\dots,a_m,0,\dots{\rm )}$ and $y={\rm diag(}b_1,\dots,b_m,0,\dots{\rm )}$, $$\lim_{n\rightarrow\infty}\int_{U(n)\times U(n)}\pp(x+k_1yk_2^*)\,\alpha_n(dk_1)\alpha_n(dk_2)=\pp\big({\rm diag(}a_1,\dots,a_m,b_1,\dots,b_m,0\dots{\rm )}\big).$$
\end{cor}
\begin{paragraph}{\bfseries{Proof.}}{\rm The function $(k_1,\,k_2)\mapsto\pp(x+k_1yk_2^*)$ is $K_m(\infty)\times K_m(\infty)$-invariant. Hence, we can apply Proposition \ref{convkey}: \ba\lefteqn{\lim_{n\rightarrow\infty}\int_{U(n)\times U(n)}\pp(x+k_1yk_2^*)\,\alpha_n(dk_1)\alpha_n(dk_2)}\\ & &=\int_{K_m^2}\int_{K_m^2}\pp(x+h_1w_mh_2yg_2^*w_m^{-1}g_1^*)\,\alpha_m(dh_1)\alpha_m(dh_2)\alpha_m(dg_1)\alpha_m(dg_2).\ea Finally, we obtain the result by using the fact that $$x+h_1w_mh_2yg_2^*w_m^{-1}g_1^*\in U(\infty){\rm diag(}a_1,\dots,a_m,b_1,\dots,b_m{\rm )}U(\infty).\hspace{1cm}\Box$$}
\end{paragraph}

\bigskip

\begin{paragraph}{\bfseries{Proof of Theorem \ref{multi}.}}{\rm Let $\pp\in\mathfrak{P}$. If $\pp$ is spherical then for $$x={\rm diag(}a_1,\dots,a_m,0,\dots{\rm )},\  y={\rm diag(}b_1,\dots,b_m,0,\dots{\rm )},$$ $$\lim_{n\rightarrow\infty}\int_{U(n)\times U(n)}\pp(x+k_1yk_2^*)\,\alpha_n(dk_1)\alpha_n(dk_2)=\pp(x)\pp(y).$$ By Corollary \ref{cormult}, $$\pp\big({\rm diag(}a_1,\dots,a_m,0,\dots{\rm )}\big)\big({\rm diag(}b_1,\dots,b_m,0,\dots{\rm )}\big)=\pp\big({\rm diag(}a_1,\dots,a_m,b_1,\dots,b_m,0,\dots{\rm )}\big).$$ By applying Corollary \ref{cormult} many times as necessary, one obtains $$\pp\big({\rm diag(}a_1,\dots,a_n,0,\dots{\rm )}\big)=\Phi(a_1)\dots\Phi(a_n),$$ where $$\Phi(\lambda)=\pp\big({\rm diag(}\lambda,0,\dots{\rm )}\big).$$

Inversely, let us assume that there exists a continuous function $\Phi$ on $\Bbb R$ such that $$\pp\big({\rm diag(}a_1,\dots,a_n,0,\dots{\rm )}\big)=\Phi(a_1)\dots\Phi(a_n).$$ Then, by Corollary \ref{cormult}, $$\lim_{n\rightarrow\infty}\int_{U(n)\times U(n)}\pp(x+k_1yk_2^*)\,\alpha_n(dk_1)\alpha_n(dk_2)=\pp(x)\pp(y).$$ Hence the function $\pp$ is spherical.$\hspace{1cm}\Box$ \\}
\end{paragraph}

\bigskip

\bigskip

\section{Modified P{\'o}lya functions : definition and convergence}

\bigskip

\begin{defi}\label{dif}\rm{
{\it The modified P{\'o}lya function} of parameter $\omega=(\alpha,\gamma)$ with $\alpha=\{\alpha_j\}_{j\geq1}\in\ell^{1}(\Bbb N)$, $\alpha_j\in\Bbb R_+$ and $\gamma\in\Bbb R_+$ is defined on $\Bbb R$ by :\\
$${\Pi}(\omega,\lambda):=e^{-\frac{1}{4}\gamma\lambda^2}\prod_{j=1}^{\infty}\frac{1}{1+\frac{1}{4}\alpha_j
  \lambda^2}.$$}
\end{defi}

\bigskip

We consider on the set $\mathcal P$ of modified P{\'o}lya functions the topology of uniform convergence on compact sets of $\Bbb R$. The topological space $\mathcal P$ is metrizable and complete. This topology can be expressed in terms of the set of parameters : $$\Omega=\left\{\omega=(\alpha,\gamma)\,\Big|\, \alpha=(\alpha_j)_{j\geq1},\, \alpha_j\geq0,\, \sum_{j=1}^\infty\alpha_j<\infty,\, \gamma\geq0\right\}.$$ For a continuous function $f$ on $\Bbb R$, we define the function $L_f$ on $\Omega$ by \bq\label{log}L_f(\omega)=\int_{\mathbb R} f(t)\sigma_\omega(dt)=\gamma f(0)+\sum\limits_{j=1}^{\infty}
{\alpha}_j f(\alpha_j).\eq Let us remark that the moments of the measure $\sigma_\omega$ are given by $${\mathcal M}_0(\sigma_\omega)=\int_{\Bbb
  R}\sigma_\omega(dt)=\gamma+\sum_{k=1}^{\infty}\alpha_k=\gamma+p_1(\alpha),$$
and for $m\geq 1$,
$${\mathcal M}_m(\sigma_\omega)=\int_{\Bbb
  R}t^m\sigma_\omega(dt)=\sum_{k=1}^{\infty}\alpha_k^{m+1}=p_{m+1}(\alpha),$$
where $p_m$ is the Newton power sum function : for $x=(x_1,x_2,\ldots)\in\ell^{1}(\Bbb N)$ and $m\geq 1$, $$p_m(x)=\sum_{k=1}^{\infty}x_k^m.$$

\bigskip

We consider on $\Omega$ the initial topology associated to the functions $L_f$. A point $\omega\in \Omega$ is seen as a point configuration, i.e. a permutation of the numbers $\{\alpha_k\}$, $\gamma$ does not change $\omega$. For $\lambda$ fixed, the function $\omega\mapsto \Pi(\omega,\lambda)$ is injective and continuous on $\Omega$. This can be seen by looking at the logarithmic derivative of $\Pi(\omega,\lambda)$ : \bq\label{logg}\begin{aligned}{{\Pi^{'}(\omega,\lambda)}\over
  {\Pi(\omega,\lambda)}}&=-\dfrac{1}{2}(\gamma+p_{1}(\alpha))\lambda+i\sum\limits_{m=2}^{\infty}p_m(\alpha)\left(\dfrac{i\lambda}{2}\right)^{2m-1}.\end{aligned}\eq

\bigskip

\begin{lem}\label{reff} Let $\mathcal I$ be the set of positive measures $\m$ on $\Bbb R_+^*$ such that $\m\left([a,+\infty[\right)\in\Bbb N$, for all $a>0$. Then, the following properties hold :\\

 {\rm (i)} For all $\m\in\mathcal I$, there exists a sequence of positive reals $\{\alpha_k\}_k$ such that $$\m=\sum_{k=1}^N\delta_{\alpha_k},$$ with $N\leq\infty$. If $N=\infty$, then the sequence $\{\alpha_k\}$ converges to $0$.\\

 {\rm (ii)} Let $\m_n$ be a sequence of measures in $\mathcal I$. Assume that there exists a measure $\m$ on $\Bbb R_+^*$ such that, for every function $f$ in the set $\mathcal C_b^0(\Bbb R_+^*)$ of bounded continuous functions on $\Bbb R_+^*$, vanishing near $0$, \bq\label{us}\lim_{n\to\infty}\int_{\Bbb R_+^*}f(x)\;\m_n(dx)=\int_{\Bbb R_+^*}f(x)\;\m(dx).\eq Then $\m\in\mathcal I$.\\

 \end{lem}

 \begin{paragraph}{\bfseries{Proof.}}{\rm (i) For $t>0$, let $F$ be the function defined on $\Bbb R_+^*$ by $F(t)=\m\left([t,\infty[\right)$. We can observe that $F$ is integer-valued, decreasing, and left-continuous. In consequence, the set $D$ of discontinuity points of $F$ is countable : $D=\left\{\alpha_k,\ k\in I\right\}$. Furthermore, we can observe that $D=\{\ t>0 \ |\ \m\left(\{t\}\right)\neq0\}$. We can also remark that the jump at each discontinuity point $\alpha_k$ is an integer $m_k$. Hence, if the number of discontinuity points $N$ is finite, then the measure $\m$ is given by : $\m=\sum_{k=1}^N m_k\delta_{\alpha_k}$ and, provided that we repeat the $\alpha_k$ as many times as their multiplicity $m_k$, we get $\m=\sum_{k=1}^N\delta_{\alpha_k}$. Finally, if $D$ is not finite, i.e. $N=\infty$, then the sequence $\{\alpha_k\}$ converges necessarily to $0$.

 (ii) Let $F_n$ be the function defined on $\Bbb R_+^*$ by $F_n(t)=\m_n\left([t,\infty[\right)$, $F(t)=\m\left([t,\infty[\right)$, and $D$ the set of discontinuity points of $F$. For $t\in D^c$ fixed, the positive measures $\m_n$ and $\m$ are bounded on $[t,\infty[$. In addition, we have $\m(\{t\})=\m(\{\infty\})=0$. Furthermore, it follows by (\ref{us}) that $\m_n$ converges weakly to $\m$ on $[t,\infty[$. As a result $$\lim_{n\to\infty}F_n(t)=F(t)\quad (t\in D^c).$$ Since $F_n(t)$ is a sequence of integers, it follows that, if $t\in D^c$, then $F(t)\in\Bbb N$. On the other hand, because $D$ is countable, its complementary set $D^c$ is dense in $\Bbb R_+^*$. This implies, since $F$ is left-continuous, that $F(t)\in\Bbb N$ for all $t>0$. Hence $\m\in\mathcal I$. $\hspace{1cm}\Box$\\ }

\end{paragraph}

\bigskip

\begin{teo}\label{fermx} $\Omega$ is weakly closed in the set of bounded measures on $\Bbb R_+$.
\end{teo}

\begin{paragraph}{\bfseries{Proof.}}{\rm Let $\omega^{(n)}=(\alpha^{(n)},\gamma^{(n)})$ be a sequence in $\Omega$. Assume that there is a bounded positive measure $\sigma$ on $\Bbb R_+$ such that, for every bounded continuous function $f$ on $\Bbb R_+$, \bq\label{cutis}\lim_{n\to\infty}\gamma^{(n)}f(0)+\sum_{k=1}^\infty\alpha_k^{(n)}f(\alpha_k^{(n)})=\int_{\Bbb R_+}f(t)\sigma(dt) .\eq For every $a>0$, there is a finite number of $\alpha_k^{(n)}$ which are greater than $a$. As a result, the measure $\m_n$ on $\Bbb R_+^*$ given by $$\int_{\Bbb R_+^*}f(t)\;\m_n(dt)=\sum_{k=1}^\infty f(\alpha_k^{(n)}) $$ belongs to $\mathcal I$. Furthermore, by (\ref{cutis}), we can observe that $\m_n$ converges weakly on $\Bbb R_+^*$ to $\dfrac{\tilde{\sigma}}{t}$, where $\tilde{\sigma}$ is the restriction of the measure $\sigma$ to $\Bbb R_+^*$. By (ii) of Lemma \ref{reff}, there exists a sequence $\alpha=\{\alpha_k\}_k$ such that, for every function $f\in\mathcal C_b^0(\Bbb R_+^*)$, $$\lim_{n\to\infty}\int_{\Bbb R_+^*}f(t)\;\m_n(dt)=\sum_{k=1}^\infty f(\alpha_k).$$ Since thet set $\mathcal C_c(\Bbb R_+^*)$ of continuous functions with compact support in $\Bbb R_+^*$ is included in $\mathcal C_b^0(\Bbb R_+^*)$, it follows that $\tilde{\sigma}$ is given by $$\int_{\Bbb R_+^*}f(t)\;\tilde{\sigma}(dt)=\sum_{k=1}^\infty\alpha_kf(\alpha_k).$$ Since the measure $\sigma$ is bounded, $\sum_{k=1}^\infty\alpha_k<\infty$. Finally $$\sigma=\gamma\delta_0+\sum_{k=1}^\infty\alpha_k\delta_{\alpha_k},$$ with $\gamma=\sigma(\{0\})$. Hence $\sigma=\sigma_\omega$ with $\omega=(\alpha,\gamma)\in\Omega$. $\hspace{1cm}\Box$ \\}

\end{paragraph}

\bigskip

A modified P{\'o}lya function of parameter $\omega$ is the Fourier transform of a probability measure $\m_\omega$. Let $\mathfrak M_\Omega$ be the set of these measures : $$\mathfrak M_\Omega=\left\{\m_\omega \ |\ \Pi(\omega,\lambda)=\widehat{\m_\omega}(\lambda)\right\}.$$ We consider on $\mathfrak M_\Omega$ the weak topology of measures. We will prove that the topology defined on $\Omega$ is equivalent to the weak topology of $\mathfrak M_\Omega$. We will need the following proposition (see \cite{Far8}, Proposition 3.11) :\\

\begin{pro}\label{proofc} Let $\psi_n$ be a sequence of $\mathcal C^\infty$-functions on $\Bbb R^d$ of positive type with $\psi_n(0)=1$, and $\psi$ an analytic function on a neighborhood of $0$. Assume that, for every $\alpha=(\alpha_1,\dots,\alpha_d)$, $$\lim_{n\to\infty}\partial^\alpha\psi_n(0)=\partial^\alpha\psi(0).$$ Then $\psi$ has an analytic extension to $\Bbb R^d$, and $\psi_n$ converges to $\psi$ uniformly on compact sets in $\Bbb R^d$. \\\end{pro}

\bigskip

\begin{pro}\label{irman} The topology of $\Omega$ is equivalent to the topology of $\mathfrak M_\Omega$.\\
\end{pro}

\begin{paragraph}{\bfseries{Proof.}}\rm{ (i) Assume that $\omega^{(n)}$ converges to $\omega$ in the topology of $\Omega$. Then the modified P{\'o}lya functions $\Pi(\omega^{(n)},\lambda)$ and $\Pi(\omega,\lambda)$ are holomorphic in $D(0,R)$ where $\displaystyle\frac{1}{R}=\sup\limits_{m,
  n}\left|\alpha_m^{(n)}\right|$. Therefore, for every $\lambda\in D(0,R)$, their logarithmic derivatives are given by $${{\Pi^{'}(\omega^{(n)},\lambda)}\over {\Pi(\omega^{(n)},\lambda)}}=-\dfrac{1}{2}(\gamma^{(n)}+p_1(\alpha^{(n)}))\lambda+i\sum\limits_{m=2}^{\infty}p_m(\alpha^{(n)})\left(\dfrac{i\lambda}{2}\right)^{2m-1},$$
$${{\Pi^{'}(\omega,\lambda)}\over
  {\Pi(\omega,\lambda)}}=-\dfrac{1}{2}(\gamma+p_1(\alpha))\lambda+i\sum\limits_{m=2}^{\infty}p_m(\alpha)\left(\dfrac{i\lambda}{2}\right)^{2m-1}.$$
For every bounded continuous function $f$ on $\Bbb R$ $$\lim_{n\to\infty}\int_{\Bbb R} f(t)\sigma_{\omega^{(n)}}(dt)=\int_{\Bbb R}
f(t)\sigma_\omega(dt),$$
where $\sigma_{\omega^{(n)}}$ is the bounded positive measure on $\Bbb R$
  associated to $\omega^{(n)}=(\alpha^{(n)},\gamma^{(n)})$.
Hence, for $f=1$, the sequence $\gamma^{(n)}+p_1(\alpha^{(n)})$ converges. In consequence, there exists a constant $A>0$ such that, for every $n$,
$$ 0\leq \gamma^{(n)}+p_1(\alpha^{(n)})\leq
A.$$\\
Therefore, the sequence $\{\alpha_m^{(n)}\}$ is bounded by $A$ for every $m$ and every $n$.\\ Hence,
$$\mbox{supp}(\sigma_{\omega^{(n)}})\subset [-A,A]\,\mbox{ and}\ \
\mbox{supp}(\sigma_\omega)\subset [-A,A].$$
It follows that the sequence of measures $\sigma_{\omega^{(n)}}$ converges to $\sigma_\omega$ for every continuous function on $\Bbb R$. We can then deduce that, for every $m\geq
2$,$$\begin{aligned}\lim\limits_{n\to {\infty}}p_m({{\alpha}^{(n)}})&=\lim\limits_{n\to
  {\infty}}\int_{\mathbb R}t^{m-1}\sigma_{\omega^{(n)}}(dt)\\&=\int_{\mathbb
  R}t^{m-1}\sigma_\omega(dt)=p_m(\alpha),\end{aligned}$$ and $$p_{m}({{\alpha}^{(n)}})\leq
\big(p_{1}{({\alpha}^{(n)})}\big)^{m}\leq A^{m}.$$\\
Furthermore, we have
$$\lim_{n\to\infty}\gamma^{(n)}+p_1(\alpha^{(n)})= \gamma+p_1(\alpha).$$
 In consequence,
  $$\lim_{n\to\infty}{{\Pi^{'}(\omega^{(n)},\lambda)}\over {\Pi(\omega^{(n)},\lambda)}}={{\Pi^{'}(\omega,\lambda)}\over {\Pi(\omega,\lambda)}},$$
 the convergence being uniform on compact sets in $D(0,R)$. Hence, for $|\lambda|< R$,
  $$\lim_{n\to\infty}\Pi(\omega^{(n)},\lambda)=\Pi(\omega,\lambda),$$ since
  $\Pi(\omega^{(n)},0)=1$ and $\Pi(\omega,0)=1$.\\

 The functions
  $\Pi(\omega^{(n)},\lambda)$ and $\Pi(\omega,\lambda)$ being of positive type, by Proposition \ref{proofc}, $\Pi(\omega^{(n)},\lambda)$ converges uniformly on compact sets in $\Bbb R$ to $\Pi(\omega,\lambda)$. Finally, by applying the L\'evy-Cramer theorem, one can prove that $\m_{\omega^{(n)}}$ converges weakly to $\m_\omega$.\\

(ii) Assume that $\m_{\omega^{(n)}}$ converges weakly to $\m_\omega$. This implies that $\Pi(\omega^{(n)},\lambda)$ converges uniformly on compact sets in $\Bbb R$ to $\Pi(\omega,\lambda)$. Let $\lambda_0>0$. Since the modified P{\'o}lya function $\Pi(\omega^{(n)},\lambda)$ is continuous, non-zero on $\Bbb R$ and satisfies $\Pi(\omega^{(n)},0)=1$, there exists $M>0$ such that, for every $n$, $\Pi(\omega^{(n)},\lambda_0)\geq M.$ In consequence $$\dfrac{1}{4}\gamma^{(n)}\lambda_0^2+\dfrac{1}{4}\lambda_0^2\sum_{k=1}^\infty\alpha_k^{(n)}\leq e^{\frac{1}{4}\gamma^{(n)}\lambda_0^2}\prod_{k=1}^\infty\left(1+\dfrac{1}{4}\alpha_k^{(n)}\lambda_0^2\right)\leq {1\over M}.$$ Therefore $$-2\Pi^{''}(\omega^{(n)},0)=p_1(\alpha^{(n)})+\gamma^{(n)}\leq{4\over{\lambda_0^2\,M}}=:R.$$

\bigskip

It follows that the function $\Pi(\omega^{(n)},\lambda)$ is holomorphic for $|\lambda|<R$ and then also in the strip $\Sigma_R=\{x+iy\,|\,|y|<R\}$ (see \cite{Far2}, Lemma 6.5). Furthermore, for $r<R$, there exists a constant $M(r)>0$ such that $|\Pi(\omega^{(n)},\lambda)|\leq M(r) \ {\rm for}\ \lambda\in\Sigma_r.$ From the theorem of Montel, it follows that there is a subsequence $\Pi(\omega^{(n_j)},\lambda)$ which converges uniformly on compact sets in $\Sigma_R$. Since the sequence itself converges to $\Pi(\omega,\lambda)$ on $\Bbb R$, we get that $\Pi(\omega^{(n)},\lambda)$ converges to $\Pi(\omega,\lambda)$ uniformly on compact sets in $\Sigma_R$. As a result, the logarithmic derivatives $\frac{\Pi^{'}(\omega^{(n)},\lambda)}{\Pi(\omega^{(n)},\lambda)}$ converge uniformly in a neighborhood of $0$ which implies the convergence of the coefficients of their Taylor expansions at $0$ : $$\lim_{n\to\infty}\gamma^{(n)}+p_1(\alpha^{(n)})=\gamma+p_1(\alpha),$$ $$\lim_{n\to\infty}p_m(\alpha^{(n)})=p_m(\alpha), \ \ {\rm for} \ m\geq2.$$ This proves that $\omega^{(n)}$ converges to $\omega$ in the topology of $\Omega$. $\hspace{1cm} \Box$\\ }
\end{paragraph}

\bigskip

Let us put for every modified P{\'o}lya function of parameter $\omega=(\alpha,\gamma)$,\\ $$\widetilde{p}_1(\omega)=p_1(\alpha)+\gamma,$$ and, for every $R\geq0$,\\ $$\Omega_R=\{\omega\in\Omega \ |-2\Pi^{''}(\omega,0)=\widetilde{p}_1(\omega)\leq R\}.$$

\bigskip

\begin{cor}\label{compact} $\Omega_R$ is a compact subset of $\Omega$. \end{cor}
\begin{paragraph}{\bfseries{Proof.}}\rm{ Since the modified P{\'o}lya function is the Fourier transform of a probability measure $\m$, we have  $$-\Pi^{''}(\omega,0)=\int_{\Bbb R}t^2\m_\omega(dt).$$ We also know that the set of probability measures such that $\int_{\Bbb R}t^2\m_\omega(dt)\leq {R\over2}$ is relatively compact. By using Theorem \ref{fermx} and Proposition \ref{irman}, we can conclude that the set $\Omega_R$ is relatively compact in $\Omega$. Moreover, the convergence of a sequence $\omega^{(n)}$ to $\omega$ in $\Omega$ implies that $\widetilde{p}_1(\omega^{(n)})$ converges to $\widetilde{p}_1(\omega)$. Hence, the map $\omega\mapsto \widetilde{p}_1(\omega)$ is continuous. In consequence, the set $\Omega_R$ is closed and relatively compact in $\Omega$, therefore compact. $\hspace{1cm} \Box$ }
\end{paragraph}

\bigskip

\bigskip

\section{Convergence of orbital measures and spherical functions}

\bigskip

Let $(X, {\mathcal B})$ be a measurable space on which a group $G$ acts by measurable transformations. Let $\nu$ be a $G$-invariant probability measure on $X$. A set $E\in\mathcal B$ is said to be $G$-invariant relatively to $\nu$ if, for every $g\in G$, $\nu((gE)\Delta E)=0,$ where $\Delta$ is the symmetric difference. The measure $\nu$ is said to be ergodic relatively to the action of $G$ if, for every $E\in\mathcal
B$ which is $G$-invariant : $\nu(E)=0\ \mbox{or}\ 1.$ If $X$ is locally compact and $K$ a compact group acting on $X$ by homeomorphisms, then the ergodic measures are exactly the orbital ones. It is the case, if one takes $X=V_n=M(n,\Bbb C)$ and $K=K_n=U(n)\times U(n)$.\\

Let $V^{\infty}$ be the space of infinite complex matrices. It is the dual space of $V_\infty$. The space $V^{\infty}$ is defined as the projective limit of the sequence $V_n$ relatively to the orthogonal projections $$p_{m,n}:V_n\rightarrow V_m\ \ (n>m).$$ The space $V^\infty$ can be identified to $\Bbb R^\infty$. The group $K_\infty$ acts on it in the same way as on $V_\infty$. Let $\mathfrak M$ be the set of probability measures on $V^\infty$ which are invariant by $K_\infty$. The Fourier transform of a measure $\nu$ in $\mathfrak M$ is the $K_\infty$-invariant function of positive type $\varphi$ defined on $V_\infty$ by\\
$$\varphi(\xi)=\int_{V^\infty}e^{i\<x,\xi\>}\nu(dx).$$ The Fourier transform establishes a bijection from $\mathfrak{M}$ onto $\mathfrak{P}$ and also between their corresponding subsets of extreme points ext($\mathfrak{M}$) and ext($\mathfrak{P}$). A function $\pp\in\mathfrak{P}$ is spherical if and only if the measure $\nu$ is ergodic.\\

Let $\lambda^{(n)}$ be a sequence of diagonal matrices \\
$$\lambda^{(n)}={\rm diag}({\lambda_1^{(n)}},\ldots,{\lambda_n^{(n)}}).$$ We associate to it the sequence of orbital measures $\nu^{(n)}$ such that, for every continuous function $f$ on $V^\infty$ :\\
$$\int_{V^\infty}f(x)\nu^{(n)}(dx)=\int_{U(n)\times U(n)}f(u\lambda^{(n)}v^*)\alpha_n(du)\alpha_n(dv).$$

\bigskip

The Fourier transform $\pp_n$ of $\nu^{(n)}$ is defined on $V_\infty$ by\\
$$\begin{aligned}\pp_n(\xi)=\pp_n(\xi,\lambda^{(n)})&=\int_{V^\infty}e^{i\<\xi,x\>}\nu^{(n)}(dx)\\
&=\int_{U(n)\times U(n)}e^{i\<\xi, u \lambda^{(n)}v^*\>}\,\alpha_n(du)\alpha_n(dv).\end{aligned}$$

\bigskip

We calculated the preceding integral in Corollary \ref{spherfin}. Let us put $$\ F(z)=\sum_{k=0}^\infty \frac{1}{(k!)^2}\,z^k,\ (z\in\Bbb C),$$ $$\Xi=\big(\Xi_1,\Xi_2,\dots,\Xi_n,0,\dots\big):=\big(-\frac{\xi_1^2}{4},-\frac{\xi_2^2}{4},\dots,-\frac{\xi_n^2}{4},0,\dots\big),$$ $$\Lambda=\big(\Lambda_1,\Lambda_2,\dots,\Lambda_n\big):=\big({\lambda_1^{(n)}}^2,{\lambda_2^{(n)}}^2,\dots,{\lambda_n^{(n)}}^2\big).$$

\bigskip

For $\xi={\rm diag}(\xi_1,\dots,\xi_n,0,\dots)$, we can rewrite the spherical function $\pp_n$ as

\bq\label{formule}\pp_n(\xi,\lambda^{(n)})=\Psi(\Lambda,\ \Xi)=\big({\bf \delta}!\big)^2\,\frac{{\rm det}\bigg(\Big(F\big(\Lambda_i\Xi_j\big)\Big)_{1\leq i,j\leq n}\bigg)}{D(\Lambda)D(\Xi)},\eq where   $$D\big(\Xi\big)=\prod_{i<j}\big(\Xi_i-\Xi_j\big), \ D\big(\Lambda\big)=\prod_{i<j}\big(\Lambda_i-\Lambda_j\big).$$

\bigskip

By a result in (\cite{Far2}, section 2.3), we can write (\ref{formule}) as \bq\label{nes3}\pp_n(\xi,\lambda^{(n)})=\sum_{m_1\geq\dots\geq m_n\geq0} \left(\frac{{\bf \delta!}}{({\bf m+\delta})!}\right)^2\ s_{\bf m}(\Lambda)s_{\bf m}(\Xi).\eq

\bigskip

The generating function of the complete symmetric function $$h_m(x)=\sum_{|\alpha|=m}x^\alpha$$ is given by : $$H(x,t):=\sum_{m=0}^\infty h_m(x)t^m=\prod_{j=1}^n\dfrac{1}{1-x_jt}.$$ The logarithmic derivative of $H(x,-\frac{t^2}{4})$ with respect to $t$ is given by : \bq\label{log1}\dfrac{H^{'}(x,-\frac{t^2}{4})}{H(x,-\frac{t^2}{4})}=-\sum_{j=1}^n\dfrac{\frac{1}{2}x_jt}{1+\frac{1}{4}x_jt^2}=i\sum_{m=1}^\infty p_{m}(x)\left(\dfrac{it}{2}\right)^{2m-1}.\eq

\bigskip

Let $\Gamma$ be the algebra of symmetric functions on $${\Bbb C}^{(\infty)}=\{z_1,z_2,\dots)\, |\, z_i\in\Bbb C \ \textrm{are zero for {\it i} large enough}\}.$$ The set $\{s_{\bf m}\}$ where ${\bf m}$ runs over all partitions is a system of linear generators of $\Gamma$. Furthermore, the sets $\{h_m\}_{m\geq1}$ and $\{p_m\}_{m\geq1}$ are systems of generators of $\Gamma$ : $$\Gamma=\Bbb C [h_1,h_2,\dots]=\Bbb C [p_1,p_2,\dots].$$

 Let us consider the algebra morphism : \bq\label{log2}\Gamma \rightarrow \mathscr{C}(\Omega),\, g\mapsto\widetilde{g},\eq which is uniquely determined by : \bq\label{log3}\widetilde{p}_1(\omega)=\gamma+\sum_{j=1}^\infty\alpha_j, \quad \quad \widetilde{p}_m(\omega)=\sum_{j=1}^\infty\alpha_j^m \quad (m\geq2).\eq

The functions $\widetilde{p}_m$ are continuous on $\Omega$. This can be shown by taking in (\ref{log}) $f\equiv1$ for $m=1$, or $f(t)=t^{2m-2}$ for $m\geq2$.

\bigskip

\begin{pro}\label{nes1} It holds that $$\widetilde{H}(\omega,-\frac{\lambda^2}{4})=\Pi(\omega,\lambda).$$ Furthermore, the Taylor expansion of $\Pi(\omega,\lambda) $ is : $$\Pi(\omega,\lambda)=\sum_{m=0}^\infty \widetilde{h}_m(\omega)\left(-\frac{\lambda^2}{4}\right)^m.$$ Here $\widetilde{H}(\omega,-\frac{\lambda^2}{4})$ is the image of $H(.,-\frac{\lambda^2}{4})$ under the morphism {\rm(\ref{log2})} and $\Pi(\omega,\lambda)$ is the modified P{\'o}lya function.\end{pro}
\begin{paragraph}{\bfseries{Proof.}} {\rm  By (\ref{logg}), (\ref{log3}) and (\ref{log1}), we have $$\dfrac{\Pi^{'}(\omega,\lambda)}{\Pi(\omega,\lambda)}=i\sum_{m=1}^\infty\widetilde{p}_{m}(\omega)\left(\dfrac{i\lambda}{2}\right)^{2m-1}=\dfrac{\widetilde{H}^{'}(\omega,-\frac{\lambda^2}{4})}{\widetilde{H}(\omega,-\frac{\lambda^2}{4})}.$$ Since $\Pi(\omega,0)=1$ and $\widetilde{H}(\omega,0)=1$, the statement follows.$\hspace{1cm}\Box$\\}
\end{paragraph}

Let us consider now the map $$T_n\; : {\Bbb R}^n\rightarrow\Omega;\; \big(\lambda_1,\lambda_2,\dots,\lambda_n\big)\mapsto\omega=(\alpha,\gamma)$$ given by

$$\alpha_j=\left(\frac{\lambda_j}{n}\right)^2,\quad \quad \gamma=0.$$

\begin{teo}\label{nes2}Let $\lambda^{(n)}\in\Bbb R^n$ be a sequence such that the following limit exists for the topology of $\Omega$ : $$\lim_{n\to\infty}T_n\left(\lambda^{(n)}\right)=\omega .$$ Then, for every $g\in\Gamma$, homogeneous of degree $m$, $$\lim_{n\to\infty}\dfrac{1}{n^{2m}}\;g\left(\left(\lambda^{(n)}\right)^2\right)=\widetilde{g}(\omega).$$
\end{teo}
\begin{paragraph}{\bfseries{Proof.}} {\rm It is enough to prove the result for $g=p_m$ since the Newton power sums generate $\Gamma$. Let $m=1$, then $$p_1(\lambda^{(n)})=\sum_{j=1}^n\lambda_j^{(n)}, \quad \quad \widetilde{p}_1(\omega)=\gamma+\sum_{j=1}^\infty\alpha_j.$$  By assumption, for every continuous function $f$ on $\Bbb R$, $$\lim_{n\to\infty}\sum_{k=1}^n\left(\dfrac{\lambda_k^{(n)}}{n}\right)^2 f\left(\dfrac{\lambda_k^{(n)}}{n}\right)= \gamma f(0)+\sum_{k=1}^\infty \alpha_k f(\alpha_k) .$$ In particular, by taking $f\equiv1$ we get  $$\lim_{n\to\infty}\sum_{k=1}^n\left(\dfrac{\lambda_k^{(n)}}{n}\right)^2= \gamma+\sum_{k=1}^\infty \alpha_k,$$ $$\lim_{n\to\infty}\dfrac{1}{n^2}\;p_1\left(\left(\lambda^{(n)}\right)^2\right)=\widetilde{p}_1(\omega) .$$    For $m\geq2$, by taking $f(t)=t^{2m-2}$, one obtains $$\lim_{n\to\infty}\dfrac{1}{n^{2m}}\;p_m\left(\left(\lambda^{(n)}\right)^2\right)=\widetilde{p_m}(\omega).\hspace{1cm}\Box$$\\}

\end{paragraph}

\begin{pro}\label{nes4} For any $\omega\in\Omega$ and $\xi_1,\dots,\xi_k\in\Bbb C$, it holds that $$\sum_{{\bf m}:{\rm partitions}}\widetilde{s}_{\bf m}(\omega)s_{\bf m}\left(-\dfrac{\xi_1^2}{4},\dots,-\dfrac{\xi_k^2}{4}\right)=\prod_{j=1}^k\Pi(\omega,\xi_j).$$  \end{pro}
\begin{paragraph}{\bfseries{Proof.}} {\rm  Recall the Cauchy formula (see \cite{Far2}, Proposition 2.5) : $$\sum_{\bf m}s_{\bf m}(x_1,x_2,\dots)s_{\bf m}(y_1,\dots,y_k)=\prod_{i=1}^\infty\prod_{j=1}^k\frac{1}{1-x_iy_j}=\prod_{j=1}^k H(x,y_j).$$ We apply the morphism (\ref{log2}) to both sides of the preceding equality with $y_j=-\dfrac{\xi_j^2}{4}$. The result follows by Proposition \ref{nes1}.$\hspace{1cm}\Box$\\        }
\end{paragraph}

\bigskip

\begin{teo}\label{nes5} As in Theorem \ref{nes2}, we assume $$\lim_{n\to\infty}T_n\left(\lambda^{(n)}\right)=\omega .$$ Then, for a fixed diagonal matrix $\xi={\rm diag}(\xi_1,\dots,\xi_k,0,\dots)$, the sequence $\pp_n(\xi,\lambda^{(n)})$ converges uniformly on compact sets in $\Bbb R^k$ : $$\lim_{n\to\infty}\pp_n(\xi,\lambda^{(n)})=\prod_{j=1}^k\Pi(\omega,\xi_j).$$

\end{teo}
\begin{paragraph}{\bfseries{Proof.}} {\rm  Consider first $k=1$. This case corresponds to a single variable $\xi=(\xi,0,\dots)$ and the unique non-zero terms in the Schur function expansion are those for which ${\bf m}=(m,0,\ldots,0)$. Hence by (\ref{nes3}) we have $$\pp_n(\xi,\lambda^{(n)})=\sum_{m=0}^\infty \left(\frac{(n-1)!}{(m+n-1)!}\right)^2\ h_m({\lambda^{(n)}}^2)\left(-\dfrac{\xi^2}{4}\right)^m.$$  Since $$\dfrac{(n-1)!}{(m+n-1)!}\sim\dfrac{1}{n^m} \quad {\rm as}\ n\to\infty ,$$ by Theorem \ref{nes2} $$\lim_{n\to\infty}\left(\frac{(n-1)!}{(m+n-1)!}\right)^2\ h_m({\lambda^{(n)}}^2)=\widetilde{h}_m(\omega).$$ Now by applying Proposition \ref{proofc} about the convergence of $\mathcal C^\infty$-functions of positive type on $\Bbb R^d$, we obtain  $$\lim_{n\to\infty}\pp_n(\xi,\lambda^{(n)})=\sum_{m=0}^\infty \widetilde{h}_m(\omega)\left(-\dfrac{\xi^2}{4}\right)^m.$$ Finally, by Proposition \ref{nes1}, $$\sum_{m=0}^\infty \widetilde{h}_m(\omega)\left(-\frac{\lambda^2}{4}\right)^m=\Pi(\omega,\lambda).$$ Now, let us consider the multivariable case $\xi=(\xi_1,\dots,\xi_k,0,\dots)$ for which $k>1$. If $m_{k+1}> 0$, then $s_{\bf m}(\xi_1,\dots,\xi_k,0,\dots)=0$. In consequence $$\pp_n(\xi,\lambda^{(n)})=\sum_{m_1\geq\dots\geq m_k\geq0}\left(\dfrac{{\bf \delta!}}{(\bf m+\bf \delta)!}\right)^2s_{\bf m}\left({\lambda^{(n)}}^2\right)s_{\bf m}\left(-\dfrac{\xi_1^2}{4},\dots,-\dfrac{\xi_k^2}{4},0,\dots\right).$$ But, for ${\bf m}$ fixed, $n\to\infty$,  $$\dfrac{{\bf \delta!}}{(\bf m+\bf \delta)!}\sim\dfrac{1}{n^{|{\bf m}|}},\quad |{\bf m}|=m_1+m_2+\dots.$$ Hence by Theorem \ref{nes2} $$\lim_{n\to\infty}\left(\dfrac{{\bf \delta!}}{(\bf m+\bf \delta)!}\right)^2s_{\bf m}\left({\lambda^{(n)}}^2\right)=\widetilde{s}_{\bf m}(\omega).$$ Similarly, by Proposition \ref{proofc},   $$\lim_{n\to\infty}\pp_n(\xi,\lambda^{(n)})=\sum_{{\bf m}:{\rm partitions}}\widetilde{s}_{\bf m}(\omega)s_{\bf m}\left(-\dfrac{\xi_1^2}{4},\dots,-\dfrac{\xi_k^2}{4},0,\dots\right).$$ Finally, by Proposition \ref{nes4}, $$\sum_{{\bf m}:{\rm partitions}}\widetilde{s}_{\bf m}(\omega)s_{\bf m}\left(-\dfrac{\xi_1^2}{4},\dots,-\dfrac{\xi_k^2}{4},0,\dots\right)=\prod_{j=1}^k\Pi(\omega,\xi_j).\hspace{1cm}\Box$$      }
\end{paragraph}

\bigskip

The preceding theorem shows that the limit of a spherical function of positive type on $V_n$ is a spherical function of positive type on $V_\infty$ which is given as a finite product of modified P{\'o}lya functions. In order to prove that all spherical functions of positive type on $V_\infty$ are obtained in the same way, we need to prove the converse of Theorem \ref{nes5}. This will be done in Theorem \ref{bi} using the following lemma :\\

\begin{lem}\label{keyrev} For every $n\geq1$, let $\omega^{(n)}$ be the point in $\Omega$ associated to an orbital measure $\n^{(n)}$ that weakly converges to an ergodic measure $\nu$ on $V^\infty$. If \bq\label{big}\sup_{n} \Big(p_1(\alpha^{(n)})\Big)=R<\infty,\eq then the sequence of parameters $\omega^{(n)}$ converges in $\Omega$.\\
\end{lem}

\begin{paragraph}{\bfseries{Proof.}}{\rm By condition (\ref{big}), the sequence $\omega^{(n)}$ belongs to the compact set $\Omega_R$ (see Corollary \ref{compact}). We can then extract a subsequence $\left(\omega^{(n_k)}\right)_k$ which converges in $\Omega$ to $\omega=(\alpha,\gamma)$. Hence, by Theorem \ref{nes5}, the Fourier transform $\pp^{(n_k)}$ of $\nu^{(n_k)}$ uniformally converges on compact sets : $$\lim_{k\rightarrow\infty}\pp^{(n_k)}\left({\rm diag}\left(\xi,0,\dots\right)\right)=\Pi(\omega,\xi).$$ Therefore, the Fourier transform $\pp$ of $\nu$ is identically equal to $\Pi(\omega,.).$\\

If another sub-sequence such that $\omega^{({n_k}^{'})}$ converges in $\Omega$ to $\omega^{'}$, then $\pp\equiv\Pi(\omega^{'},.)$. By uniqueness, we get $\omega=\omega^{'}$, and $\omega^{(n)}$ do have in consequence a unique accumulation point. Therefore, it necessarily converges to $\omega$.$\hspace{1cm}\Box$\\}
\end{paragraph}

\begin{teo}\label{bi}
The spherical functions of positive type on $V_\infty$ (i.e. the extreme points of $\mathfrak P$) are the functions $\varphi_\omega$ defined, for every $k\geq1$, by : $$\pp_\omega\big({\rm diag}(\xi_1,\dots,\xi_k,0,\dots)\big)=\Pi(\omega,\xi_1)\dots\Pi(\omega,\xi_k),$$  where $\Pi(\omega,.)$ is the modified P{\'o}lya function associated to $\omega\in\Omega$.\\
\end{teo}

\begin{paragraph}{\bfseries{Proof.}}{\rm
(a) The function $\pp_{\omega}$, which is of positive type, is spherical since it is multiplicative (Theorem \ref{multi}). Therefore, it is an extreme point in $\mathfrak P$.

(b) Let $\varphi\in {\rm ext}(\mathfrak P)$. It is the Fourier transform of an ergodic measure $\nu$ on $V^\infty$ relatively to the action of $K_\infty$. By using a theorem due to A. Vershik (\cite{Ol2}, Theorem 3.2), the measure $\nu$ is the weak limit of $\nu^{(n)}$, where $\nu^{(n)}$ is a sequence of orbital measures relatively to $K_n$. Hence $\varphi$
 is the uniform limit on compact sets of the sequence $\pp_n$, where $\pp_n$ is the Fourier transform of $\nu^{(n)}.$ In particular we have $$\lim_{n\rightarrow\infty}\pp_n\left({\rm diag}\left(\xi,0,\dots\right),\lambda^{(n)}\right)=\pp\left({\rm diag}\left(\xi,0,\dots\right)\right).$$ We will prove that $\omega^{(n)}=T_n(\lambda^{(n)})$ converges in $\Omega$, which gives the converse of Theorem \ref{nes5}. Let us suppose now that the condition (\ref{big}) is not satisfied :

 $$\sup_n\left(p_1\left(\left(\dfrac{\lambda^{(n)}}{n}\right)^2\right)\right)=\infty.$$

There exists a positive sequence $\ep_n$ such that $$\lim_{n\rightarrow\infty}\ep_n=0\ {\rm and}\ \lim_{n\rightarrow\infty}p_1\left(\ep_n\left(\dfrac{\lambda^{(n)}}{n}\right)^2\right)=1.$$ One can remark that multiplying $(\dfrac{\lambda^{(n)}}{n})^2$ by $\ep_n$ is the same as multiplying $\xi$ by $\ep_n$. Hence, by Lemma \ref{keyrev}, there exists $\omega\in\Omega$ such that :

 $$\lim_{n\rightarrow\infty}\pp_n\left({\rm diag}\left(\ep_n\xi,0,\dots\right),\lambda^{(n)}\right)=\Pi(\omega,\xi).$$

 The modified P{\'o}lya function $\Pi(\omega,.)$ is not identically equal to $1$, because $\omega\ne0$. Therefore, in a neighborhood $D(0,R)$ of $0$, the function $\Pi(\omega,.)$ is not identically equal to $1$. But this leads to a contradiction, because, by our assumption, \ba\lefteqn{\lim_{n\rightarrow\infty}\pp_n\left({\rm diag}\left(\ep_n\xi,0,\dots\right),\lambda^{(n)}\right)}\\& &=\lim_{n\rightarrow\infty}\sum_{m=0}^\infty \left(\frac{n^m\ep_n^m}{n(n+1)\dots(n+m-1)}\right)^2h_m\left(\left(\dfrac{\lambda^{(n)}}{n}\right)^2\right)\;\left(-\dfrac{\xi^2}{4}\right)^m=1,\ea and then $\Pi(\omega,.)\equiv1.$ Hence, there exists $\omega\in\Omega$ such that : $\pp(\xi)=\pp_{\omega}(\xi)\ \ (\xi\in V_\infty).\hspace{1cm}\Box$}
\end{paragraph}

\bigskip

\begin{nota} {\rm  A measurable space is said to be standard if it is isomorphic to a Borel subset in a polish space which is equipped with the $\sigma$-algebra induced by the Borel one. One can prove that the correspondence $\Omega\leftrightarrow$ {\rm ext(}$\mathfrak P${\rm)} is an isomorphism between two standard spaces. This enables us to prove a parameterized version of the generalized Bochner theorem (\cite{Rab}, Theorem 7) : let $\varphi$ be a $K_\infty$-invariant continuous function of positive type on $V_\infty$ with $\varphi(e)=1$. Then, there exists a unique probability measure $\m$ defined on $\Omega$ such that, for every $g\in V_\infty$,
$$\varphi(g)=\int_{\Omega}\varphi_{\omega}(g)\, \m(d\omega).$$}

\end{nota}

\begin{paragraph}{\bfseries {Acknowledgement}}{\rm I would like to express my gratitude to Professor Jacques Faraut for numerous suggestions during the preparation of this paper.}
\end{paragraph}

\bibliographystyle{amsalpha}

config: arxiv


TITLE: Finding a function that best fits another in an inner product
QUESTION [0 upvotes]: I'm really confused about what this question is looking for:
Let $V$ be the $\mathbb R$-vector space of continuous functions over $[−1, 1]$ with inner product $\langle f,g \rangle = \int_{-1}^1 f(t)g(t)dt$. Find the quartic (i.e. degree 4) polynomial $g(t)$ that best fits
the function $f(t) = cos(πt)$ over $[−1, 1]$.
What does it mean by a function that best fits the other?
Thanks

REPLY [0 votes]: Going one step further, find an orthonormal basis on $[-1,1]$ using Gram Schmidt, which will give you Legendre Polynomials. Then find the inner products (coefficients) upto degree 4.


TITLE: Module of vector bundles
QUESTION [1 upvotes]: If E→M and F→M are two vector bundles over manifold M, then $C^{\infty}$(E,F) is a $C^{\infty}$(M) module. I am confused about this claim for the operation of module.
To f$\in$$C^{\infty}$(E,F) and g$\in$$C^{\infty}$(M), how could g operate on f? f is from E to F while g is from M to R.

REPLY [1 votes]: Let $p:E\to M$ and $q:F\to M$ be the projections, so $f$ is a smooth map $E\to F$ such that $qf=p$ and $f$ is linear on each fiber.  Define $g\cdot f$ by $g\cdot f(x)=g(p(x))f(x)$ for $x\in E$ (here $g(p(x))\in\mathbb{R}$ and we are multiplying $f(x)$ by it using the vector space structure of $F_{p(x)}$).  It is then straightforward to verify that $g\cdot f\in C^\infty(E,F)$, and this makes $C^\infty(E,F)$ a module over $C^\infty(M)$.