Wednesday, November 14, 2012

On a family of symmetric matrices

$\newcommand{\bR}{\mathbb{R}}$ $\newcommand{\eS}{\mathscr{S}}$   Suppose that $w: \bR\to[0,\infty)$ is an integrable function.  Consider its   Fourier transform $\newcommand{\ii}{\boldsymbol{i}}$

$$ \widehat{w}(\theta)=\int_{\bR}  e^{-\ii x\theta}w(\theta) dx. $$



 For any $\vec{\theta}\in\bR^n$ we form the  complex Hermitian  $n\times n$ matrix

$$A_w(\vec{\theta})= \bigl(\; a_{ij}(\vec{\theta})\;)_{1\leq i,j\leq n},\;\; a_{ij}(\vec{\theta})=\widehat{w}(\theta_i-\theta_j) .$$

Observe that  for any $\vec{z}\in\mathbb{C}^n$  we have $\newcommand{\bC}{\mathbb{C}}$

$$ \bigl(\; A_w(\vec{\theta})\vec{z},\vec{z}\;\bigr)=\sum_{i,j} \widehat{w}(\theta_i-\theta_j) z_i\bar{z}_j =\int_{\bR}  | T_{\vec{z}}(x,\vec{\theta})|^2 w(x) dx, $$

where  $T_{\vec{z}}( x)$ is  is the trigonometric polynomial $\newcommand{\vez}{\vec{z}}$

$$T_{\vez}(x,\vec{\theta})= \sum_j z_j e^{\ii \theta_j x}. $$


We denote by $(-,-)_w$ the inner product

$$ (f,g)_w=\int_{\bR} f(x) \bar{g(x)} w(x) dx,\;\;f,g:\bR\to \bC. $$

We see that $A_w(\vec{\theta})$  is the Gramm-Schmidt matrix

$$ a_{ij}(\vec{\theta})= (E_{\theta_i}, E_{\theta_j})_w,\;\;  E_\theta(x)=e^{\ii\theta x}. $$


We see that $\sqrt{\;\det A_w(\vec{\theta})\;}$ is equal to the $n$-dimensional volume   of the parallelepiped   $P(\vec{\theta})=L^2(\bR, wdx)$ spanned  by the  functions $E_{\theta_1},\dotsc, E_{\theta_n}$. We observe  that if these exponentials are  linearly  dependent,  then this volume is  zero.  Here is a first elementary result.


Lemma  1.   The exponentials  $E_{\theta_1},\dotsc, E_{\theta_n}$ are linearly dependent  (over $\bC$) if and only if $\theta_j=\theta_k$ for  some $j\neq k$.


Proof.    Suppose that

$$\sum_{j=1}^n z_j E_{\theta_j}(x)=0,\;\;\forall x\in \bR. $$

Then for any $f\in \eS(\bR)$ we have

$$\sum_{j=1}^n z_j E_{\theta_j}(x)f(x)=0,\;\;\forall x\in \bR. $$


By taking the Fourier Transform of the last equality we deduce

$$ \sum_{j=1}^n z_j \widehat{f}(\theta-\theta_j) =0.  \label{1}\tag{1}$$


If we now choose $\newcommand{\ve}{{\varepsilon}}$ a  family $f_\ve(x)\in\eS(\bR)$ such that, as $\ve\searrow 0$,  $\widehat{f}_\ve(\theta)\to\delta(\theta)=$ the Dirac  delta function concentrated at $0$, we  deduce  from (\ref{1})  that


$$\sum_{j=1}^n z_j\delta(\theta-\theta_j)=0. \tag{2}\label{2} $$

 Clearly this can happen if and only if  $\theta_j=\theta_k$ for  some $j\neq k$.  q.e.d.


If we set

$$ \Delta(\vec{\theta}) :=\prod_{1\leq j<k\leq n} (\theta_k-\theta_j), $$

then we deduce from the above lemma that

$$ \det A_w(\vec{\theta})= 0 \Leftrightarrow  \Delta(\vec{\theta})=0. $$

A more precise statement is true.


Theorem 2.  For any integrable weight $w:\bR\to [0,\infty)$  such that $\int_{\bR} w(x) dx >0$ there  exists a  constant $C=C(w)>0$ such that for any $\theta_1,\dotsc, \theta_n\in [-1,1]$ we have

$$ \frac{1}{C}|\Delta(\vec{\theta})|^2 \leq \det  A_w(\vec{\theta}). \tag{E}\label{E}$$


Proof.          We regard $A_w(\vec{\theta})$ as a hermitian operator

$$ A_w(\vec{\theta}):\bC^n\to \bC^n. $$

We denote by $\lambda_1(\vec{\theta})\leq \cdots \leq \lambda_n(\vec{\theta}) $ its eigenvalues so that

$$\det A_w(\vec{\theta})=\prod_{j=1}^n \lambda_j(\vec{\theta}) \tag{Det}\label{D}. $$

Observe that $\newcommand{\Lra}{\Leftrightarrow}$ $\newcommand{\eO}{\mathscr{O}}$

$$\vec{z}\in \ker A(\vec{\theta}) \Lra  \sum_{j=1}^n z_j E_{\theta_j}(x) =0,\;\;\forall x\in{\rm supp}\; w \Lra   \sum_{j=1}^n z_j E_{\theta_j}(x) =0,\;\;\forall x\in\bR. \tag{Ker}\label{K}$$

We want to give a  more precise description    of $\ker A_w(\vec{\theta})$.    Set

$$ I_n:=\{1,\dotsc, n\},\;\; \Phi_{\vec{\theta}}=\{ \theta_1,\dotsc,\theta_n\}\subset \bR. $$


We want to emphasize that $\Phi{\vec{\theta}}$ is not a multi-set so that $\#\Phi(\vec{\theta})\leq n.$  $\newcommand{\vet}{{\vec{\theta}}}$.

 Example 3.  For example  with $n=6$ and $\vet=(1,2,3,2,2,4)$ we have

$$ \Phi_\vet=\Phi_{(1,2,3,2,2,4)}=\{1,2,3,4\}. $$






For $\newcommand{\vfi}{{\varphi}}$ $\vfi\in\Phi_\vet$ we set


$$ J_\vfi=\bigl\{ j\in I_n;\;\; \theta_j=\vfi\;\bigr\}. $$

In the example above for $ \vet=(1,2,3,2,2,4)$ and $\vfi=2$ we have $J_\vfi=\{2,4,5\}$.   $\newcommand{\vez}{\vec{z}}$  For $J\subset I_n$ we set


$$S_J:\bC^n\to \bC,\;\;S_J(\vez)=\sum_{j\in J} z_. $$

In particular, for  any $\vfi\in\Phi_\vet$ we define

$$S_\vfi:\bC^n\to \bC,\;\; S_{\vfi}(\vec{z})=S_{J_\vfi}(\vez)=\sum_{j\in J_\vfi} z_j. $$

We deduce

$$  \sum_{j\in I_n} z_jE_{\theta_j}=\sum_{\vfi\in \Phi_\vet} S_\vfi(\vec{z}) E_\vfi. $$

Using (\ref{K}) we deduce

$$ \vez\in\ker A(\vet)\Lra \sum_{\vfi\in \Phi_\vet} S_\vfi(\vec{z}) E_\vfi\Lra S_\vfi(\vez)=0,\;\;\forall \vfi\in \Phi_\vet . \tag{3}\label{3}$$

In particular we deduce

$$\dim \ker A(\vet)=n-\#\Phi_\vet. $$


Step 1.   Assume  that $w$ has compact support so that $\widehat{w}(\theta)$  is real analytic over $\bR$.   We will show  that we have the two-sided estimate

$$ \frac{1}{C}|\Delta(\vec{\theta})|^2 \leq \det  A_w(\vec{\theta}) \leq C |\Delta(\vec{\theta})|^2. \tag{$E_*$}\label{Es} $$

 In this case  $\det A_w(\vet)$ is real analytic and symmetric in the variables $\theta_1,\dotsc, \theta_n$  and vanishes   if and only if $\theta_j=\theta_k$ for some $j=k$.  Thus $\det A_w(\vet)$ has a  Taylor series expansion (near $\vet=0$)


$$\det A_w(\vet)= \sum_{\ell\geq 0} P_\ell(\vet), $$

where $P_\ell(\vet)$ is a  symmetric polynomial  in $\vet$ that vanishes   when $\theta_j=\theta_k$ for some $j\neq k$. Symmetric  polynomials of this type  have the form,


$$\Delta(\vet)^{2N} \cdot Q(\vet) $$

where $N$ is some positive integer  and $Q$ is a symmetric polynomial.  We deduce from  the \Lojasewicz inequality  for  subanalytic functions that there exists  $C=C(w)>0$, a positive integer $N$  and a rational number and $r>0$ such that


$$ \frac{1}{C} |\Delta(\vet)|^{r}\leq   \det A_w(\vet) \leq C \Delta(\vet)^{2N},\;\;\forall |\vet|\leq 2\pi. \tag{4} \label{4} $$

We want to show  that in (\ref{4})  we have $2N=r=2$.   We argue by contradiction, namely we assume that $r\neq 2$ or $N\neq 1$.      Let

$$\vet(t)= (0, t, \theta_3, \dotsc, \theta_n), \;\; 0\leq |t| < \theta_3<\cdots < \theta_n. $$

Set $A_w(t)=A_w\bigl(\,\vet(t)\;\bigr)$.   Denote its eigenvalues by

$$0\leq \lambda_1(t)\leq \lambda_2(t)\cdots \leq \lambda_n(t). $$

The eigenvalues are so arranged so that the functions $\lambda_k(t)$ are real analytic for $t$ in a neighborhood of $0$. We deduce from (\ref{3}) that $\ker A_w(0)$ is one dimensional so that $\lambda_1(0) =0$,  $\lambda_k(0)>0$, $\forall  k>1$. Hence

$$ \det  A_w(t) \sim \lambda_1(t) \prod_{k=2}^n \lambda_k(0)\;\;\mbox{as $t\searrow 0$}.  \tag{5}\label{5} $$

On the other hand

$$\Delta(\vet(t))^2 \sim Zt^2  \;\;\mbox{as}\;\; t\searrow 0 $$

for some positive constant $Z$.  Using this estimate in (\ref{4}) we deduce $r=2N$.  On the other hand, using the above estimate in (\ref{5}) we deduce

$$ \lambda_1(t) \sim Z_1 t^{2N} \;\;\mbox{as}\;\;t\searrow 0. \tag{6}\label{6}, $$

for another positive constant $Z_1$.

The   kernel of $A_w(0)$ is spanned by the  unit vector

$$ \vez(0)= (\frac{1}{\sqrt{2}}, -\frac{1}{\sqrt{2}}, 0,\dotsc 0). $$

We can find a real analytic family of vectors $t\mapsto \vec{z}(t)$ satisfying

$$|\vez(t)|=1,\;\; A_w(t) \vez(t)=\lambda_1(t)\vez(t),\;\;\lim_{t\to 0}\vez(t)=\vez(0). $$

In particular, we deduce

$$ \dot{A}_w(0)\vez(0)+A_w(0)\dot{\vez}(0)=\dot{\lambda}_1(0)\vez(0)+\lambda_1(0)\dot{\vez}(0)=0. $$

A simple computation  shows that $\dot{A}_w(0) \vez(0)=0$ so we deduce  $A_w(0)\dot{\vez}(0)=0$.  This shows that

$$\dot{z}_1(0)+\dot{z}_2(0)=0,\;\;\dot{z}_k(0)=0,\;\;\forall k>2. $$

$$ \lambda_1(t)= (A_w(t) \vez(t),\vez(t))= \int_{\bR}   \Bigl| \;\underbrace{\sum_{j=1}^n z_j(t) e^{\theta_j(t) x}}_{=:f_t(x)}\;\Bigr|^2 w(x) dx. $$


Observe that

$$f_t(x):= \sum_{j=1}^n z_j(t) e^{\theta_j(t) x}= \frac{1}{\sqrt{2}}(1-e^{\ii t x}) +\sum_{j=1}^k \ve_j(t) e^{\ii\theta_j(t) x},\;\;\ve_j(t)=z_j(t)-z_j(0). $$

We   deduce that

$$  \lim_{t\to 0} \frac{1}{t}f_t(x) = -\frac{\ii x}{\sqrt{2}} + \sum_{k=1}^n \dot{z}_k(0)= -\frac{\ii x}{\sqrt{2}}\tag{7}\label{7}$$

uniformly  for  $x$   on compacts. Since  $w$ has compact support  we deduce that (\ref{7}) holds for uniformly for $x$ in the support of $w$.  We deduce that

$$ \lambda_1(t)\sim \frac{1}{2}\;\underbrace{\left(\int_{\bR}  x^2 w(x)dx \right)}_{=\widehat{w}''(0)}\;t^2\;\;\mbox{as $t\to 0$}. $$

Using the last equality in (\ref{6}) we obtain $2N=2$ which proves  (\ref{Es}) .


Step 2.    We will show that if (\ref{E}) holds for $w_0$ and $w_1(x) \geq w_0(x)$,  $\forall x$,  then (\ref{E}) holds for  $w_1$ as well.       For any weight $w$  and any $\vet$ such that the $\Delta(\vet)\neq 0$ consider the ellipsoid

$$ \Sigma_w:=\bigl\{\vez\in\bC^n;\;\; (A_w\vez,\vez)\leq 1\bigr\}. $$

Then


$$ {\rm vol}\, \bigl(\;\Sigma_w(\vet)\;\bigr)=\frac{\pi^n}{n!\det A_w(\vet)}. $$

Observe that if $ w_0\leq w_1$ then $\Sigma_{w_0}(\vet)\subset \Sigma_{w_1}(\vet)$ and we  deduce

$$ \det A_{w_0}(\vet) \leq \det A_{w_1}(\vet). $$

This proves our claim.

Step 3. We show that (\ref{E}) holds for any integrable weight.   At  least one of the level sets $\{w\geq \ve\}$, $\ve>0$ is nonempty.  We can find a compact set of nonzero measure  $K \subset \{w\geq \ve \}$. Now define $w_0=I_{K}$. Clearly $I_K\leq w$.   From  Step 1 we know that (\ref{E}) holds for $w_0$. Invoking Step 2 we deduce that (\ref{E}) holds for  $w$.  Q.E.D.






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