# This is the title

DISCRETE AND CONTINUOUS WELCH BOUNDS FOR BANACH SPACES WITH APPLICATIONS
K. MAHESH KRISHNA
Post Doctoral Fellow

Statistics and Mathematics Unit

Indian Statistical Institute, Bangalore Centre

Karnataka 560 059, India

Email<EMAIL_ADDRESS>
Date:

Abstract: Let $\\{\tau_{j}\\}_{j=1}^{n}$ be a collection in a finite
dimensional Banach space $\mathcal{X}$ of dimension $d$ and
$\\{f_{j}\\}_{j=1}^{n}$ be a collection in $\mathcal{X}^{*}$ (dual of
$\mathcal{X}$) such that $f_{j}(\tau_{j})=1$, $\forall 1\leq j\leq n$. Let
$n\geq d$ and $\text{Sym}^{m}(\mathcal{X})$ be the Banach space of symmetric
m-tensors. If the operator $\text{Sym}^{m}(\mathcal{X})\ni
x\mapsto\sum_{j=1}^{n}f_{j}^{\otimes m}(x)\tau_{j}^{\otimes
m}\in\text{Sym}^{m}(\mathcal{X})$ is diagonalizable and its eigenvalues are
all non negative, then we prove that

(1) $\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|^{2m}\geq\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{m}\geq\frac{1}{n-1}\left[\frac{n}{{d+m-1\choose
m}}-1\right],\quad\forall m\in\mathbb{N}.$

When $\mathcal{X}=\mathcal{H}$ is a Hilbert space, and $f_{j}$ is defined by
$f_{j}:\mathcal{H}\ni h\mapsto\langle h,\tau_{j}\rangle\in\mathbb{K}$ (where
$\mathbb{K}$ is $\mathbb{R}$ or $\mathbb{C}$), $\forall 1\leq j\leq n$, then
Inequality (1) reduces to Welch bounds. Thus Inequality (1) improves 48 years
old result obtained by Welch [IEEE Transactions on Information Theory, 1974].
We also prove the following continuous version of Inequality (1). Let
$m\in\mathbb{N}$, $(\Omega,\mu)$ be a $\sigma$-finite measure space and
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be a continuous Bessel family for a finite dimensional Banach space
$\mathcal{X}$ of dimension $d$ such that $f_{\alpha}(\tau_{\alpha})=1$,
$\forall\alpha\in\Omega$. If the diagonal
$\Delta\coloneqq\\{(\alpha,\alpha):\alpha\in\Omega\\}$ is measurable in the
measure space $\Omega\times\Omega$,

$\displaystyle\int_{\Omega\times\Omega}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{m}\,d(\mu\times\mu)(\alpha,\beta)<\infty,$

and the operator $\text{Sym}^{m}(\mathcal{X})\ni
x\mapsto\int_{\Omega}f_{\alpha}^{\otimes m}(x)\tau_{\alpha}^{\otimes
m}\,d\mu(\alpha)\in\text{Sym}^{m}(\mathcal{X})$ is diagonalizable and its
eigenvalues are all non negative, then we prove that

(2)
$\displaystyle\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})|^{2m}\geq\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{m}\geq\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{{d+m-1\choose
m}}-(\mu\times\mu)(\Delta)\right].$

Inequality (2) improves the continuous Welch bounds for Hilbert spaces
obtained by the author in the year 2021. Motivated from several high end
Hilbert space results and problems, we formulate several open problems
including Zauner’s conjecture for Banach spaces.

Keywords: Banach space, Welch bound, Approximate Schauder frame, Zauner’s
conjecture.

Mathematics Subject Classification (2020): 42C15.

###### Contents

  1. 1 Introduction
  2. 2 Discrete Welch bounds for Banach spaces
  3. 3 Applications of discrete Welch bounds for Banach spaces
  4. 4 Continuous Welch bounds for Banach spaces
  5. 5 Applications of continuous Welch bounds for Banach spaces

##  1\. Introduction

Given a collection $\\{\tau_{j}\\}_{j=1}^{n}$ of unit vectors in
$\mathbb{C}^{d}$, using Cauchy-Schwarz inequality we get

(3) $\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|\langle\tau_{j},\tau_{k}\rangle|^{2}\leq 1.$

The natural question which comes immediately is that whether there is any
lower bound for the quantity max in Inequality (3). In his celebrated paper
[70], L. Welch proved the following result in 1974.

######  Theorem 1.1.

[70] (Welch bounds) Let $n\geq d$. If $\\{\tau_{j}\\}_{j=1}^{n}$ is any
collection of unit vectors in $\mathbb{C}^{d}$, then

$\displaystyle\sum_{1\leq j,k\leq
n}|\langle\tau_{j},\tau_{k}\rangle|^{2m}=\sum_{j=1}^{n}\sum_{k=1}^{n}|\langle\tau_{j},\tau_{k}\rangle|^{2m}\geq\frac{n^{2}}{{d+m-1\choose
m}},\quad\forall m\in\mathbb{N}.$

In particular,

$\displaystyle\sum_{1\leq j,k\leq
n}|\langle\tau_{j},\tau_{k}\rangle|^{2}=\sum_{j=1}^{n}\sum_{k=1}^{n}|\langle\tau_{j},\tau_{k}\rangle|^{2}\geq\frac{n^{2}}{{d}}.$

Further,

$\displaystyle\text{({Higher order Welch bounds})}\quad\max_{1\leq j,k\leq
n,j\neq
k}|\langle\tau_{j},\tau_{k}\rangle|^{2m}\geq\frac{1}{n-1}\left[\frac{n}{{d+m-1\choose
m}}-1\right],\quad\forall m\in\mathbb{N}.$

In particular,

$\displaystyle\text{({First order Welch bound})}\quad\max_{1\leq j,k\leq
n,j\neq k}|\langle\tau_{j},\tau_{k}\rangle|^{2}\geq\frac{n-d}{d(n-1)}.$

There are several theoretical and practical applications of Theorem 1.1 such
as in the study of root-mean-square (RMS) absolute cross relation of unit
vectors [57], frame potential [7, 15, 12], correlations [56], codebooks [25],
numerical search algorithms [72, 71], quantum measurements [58], coding and
communications [67, 61], code division multiple access (CDMA) systems [41,
42], wireless systems [53], compressed sensing [64], ‘game of Sloanes’ [37],
equiangular tight frames [62], etc.

Some improvements of Theorem 1.1 has been done in [18, 23, 69, 68]. It is in
the paper [22] where the following generalization of Theorem 1.1 has been done
for continuous collections.

######  Theorem 1.2.

[22] Let $\mathbb{C}\mathbb{P}^{n-1}$ be the complex projective space and
$\mu$ be a normalized measure on $\mathbb{C}\mathbb{P}^{n-1}$. If
$\\{\tau_{\alpha}\\}_{\alpha\in\mathbb{C}\mathbb{P}^{n-1}}$ is a continuous
frame for a $d$-dimensional subspace $\mathcal{H}$ of a Hilbert space
$\mathcal{H}_{0}$, then

$\displaystyle\int_{\mathbb{C}\mathbb{P}^{n-1}\times\mathbb{C}\mathbb{P}^{n-1}}|\langle\tau_{\alpha},\tau_{\beta}\rangle|^{2m}\,d(\mu\times\mu)(\alpha,\beta)=\int_{\mathbb{C}\mathbb{P}^{n-1}}\int_{\mathbb{C}\mathbb{P}^{n-1}}|\langle\tau_{\alpha},\tau_{\beta}\rangle|^{2m}\,d\mu(\alpha)\,d\mu(\beta)\geq\frac{1}{{d+m-1\choose
m}},\quad\forall m\in\mathbb{N}.$

Theorem 1.2 has been recently generalized in full generality for Hilbert
spaces and $\sigma$-finite measure spaces by the author in [43].

######  Theorem 1.3.

[43] Let $(\Omega,\mu)$ be a measure space and
$\\{\tau_{\alpha}\\}_{\alpha\in\Omega}$ be a normalized continuous Bessel
family for $\mathcal{H}$ of dimension $d$. If the diagonal
$\Delta\coloneqq\\{(\alpha,\alpha):\alpha\in\Omega\\}$ is measurable in the
measure space $\Omega\times\Omega$, then

$\displaystyle\int_{\Omega\times\Omega}|\langle\tau_{\alpha},\tau_{\beta}\rangle|^{2m}\,d(\mu\times\mu)(\alpha,\beta)=\int_{\Omega}\int_{\Omega}|\langle\tau_{\alpha},\tau_{\beta}\rangle|^{2m}\,d\mu(\alpha)\,d\mu(\beta)\geq\frac{\mu(\Omega)^{2}}{{d+m-1\choose
m}},\quad\forall m\in\mathbb{N}.$

Further, we have the higher order continuous Welch bounds

$\displaystyle\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|\langle\tau_{\alpha},\tau_{\beta}\rangle|^{2m}\geq\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{{d+m-1\choose
m}}-(\mu\times\mu)(\Delta)\right],\quad\forall m\in\mathbb{N}.$

In particular, we have the first order continuous Welch bound

$\displaystyle\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|\langle\tau_{\alpha},\tau_{\beta}\rangle|^{2}\geq\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{d}-(\mu\times\mu)(\Delta)\right].$

Recently Theorem 1.1 and Theorem 1.3 are done in the context of Hilbert
C*-modules by the author in [45].

######  Theorem 1.4.

[45] (Modular Welch bounds) Let $\mathcal{A}$ be a unital commutative
C*-algebra and $\mathcal{A}^{d}$ be the standard Hilbert C*-module over
$\mathcal{A}$. Let $n\geq d$. If $\\{\tau_{j}\\}_{j=1}^{n}$ is any collection
of unit inner product vectors in $\mathcal{A}^{d}$, then

$\displaystyle\sum_{j=1}^{n}\sum_{k=1}^{n}\|\langle\tau_{j},\tau_{k}\rangle\|^{2m}\geq\sum_{j=1}^{n}\sum_{k=1}^{n}\langle\tau_{j},\tau_{k}\rangle^{m}\langle\tau_{k},\tau_{j}\rangle^{m}\geq\frac{n^{2}}{{d+m-1\choose
m}},\quad\forall m\in\mathbb{N}.$

In particular,

$\displaystyle\sum_{j=1}^{n}\sum_{k=1}^{n}\|\langle\tau_{j},\tau_{k}\rangle\|^{2}\geq\sum_{j=1}^{n}\sum_{k=1}^{n}\langle\tau_{j},\tau_{k}\rangle\langle\tau_{k},\tau_{j}\rangle\geq\frac{n^{2}}{{d}}.$

Further,

$\displaystyle\text{({Higher order modular Welch bounds})}\quad\max_{1\leq
j,k\leq n,j\neq
k}\|\langle\tau_{j},\tau_{k}\rangle\|^{2m}\geq\frac{1}{n-1}\left[\frac{n}{{d+m-1\choose
m}}-1\right],\quad\forall m\in\mathbb{N}.$

In particular,

$\displaystyle\text{({First order modular Welch bound})}\quad\max_{1\leq
j,k\leq n,j\neq
k}\|\langle\tau_{j},\tau_{k}\rangle\|^{2}\geq\frac{n-d}{d(n-1)}.$

######  Theorem 1.5.

[45] (Modular Welch bounds) Let $\mathcal{A}$ be a commutative $\sigma$-finite
W*-algebra or a commutative AW*-algebra. Let $\mathcal{E}$ be a Hilbert
C*-module over $\mathcal{A}$ of rank $d$. Let $n\geq d$. If
$\\{\tau_{j}\\}_{j=1}^{n}$ is any collection of unit inner product vectors in
$\mathcal{E}$, then

$\displaystyle\sum_{j=1}^{n}\sum_{k=1}^{n}\|\langle\tau_{j},\tau_{k}\rangle\|^{2m}\geq\sum_{j=1}^{n}\sum_{k=1}^{n}\langle\tau_{j},\tau_{k}\rangle^{m}\langle\tau_{k},\tau_{j}\rangle^{m}\geq\frac{n^{2}}{{d+m-1\choose
m}},\quad\forall m\in\mathbb{N}.$

In particular,

$\displaystyle\sum_{j=1}^{n}\sum_{k=1}^{n}\|\langle\tau_{j},\tau_{k}\rangle\|^{2}\geq\sum_{j=1}^{n}\sum_{k=1}^{n}\langle\tau_{j},\tau_{k}\rangle\langle\tau_{k},\tau_{j}\rangle\geq\frac{n^{2}}{{d}}.$

Further,

$\displaystyle\text{({Higher order modular Welch bounds})}\quad\max_{1\leq
j,k\leq n,j\neq
k}\|\langle\tau_{j},\tau_{k}\rangle\|^{2m}\geq\frac{1}{n-1}\left[\frac{n}{{d+m-1\choose
m}}-1\right],\quad\forall m\in\mathbb{N}.$

In particular,

$\displaystyle\text{({First order modular Welch bound})}\quad\max_{1\leq
j,k\leq n,j\neq
k}\|\langle\tau_{j},\tau_{k}\rangle\|^{2}\geq\frac{n-d}{d(n-1)}.$

######  Theorem 1.6.

[45] (Continuous modular Welch bounds) Let $\mathcal{A}$ be a commutative
$\sigma$-finite W*-algebra or a commutative AW*-algebra. Let $\mathcal{E}$ be
a Hilbert C*-module over $\mathcal{A}$ of rank $d$. Let $(G,\mu_{G})$ be a Lie
group and $\\{\tau_{g}\\}_{g\in G}$ be a continuous modular Bessel family for
$\mathcal{E}$ such that $\langle\tau_{g},\tau_{g}\rangle=1,\forall g\in G$. If
the diagonal $\Delta\coloneqq\\{(g,g):g\in G\\}$ is measurable in $G\times G$,
then

$\displaystyle\int_{G\times
G}\|\langle\tau_{g},\tau_{h}\rangle\|^{2m}\,d(\mu_{G}\times\mu_{G})(g,h)$
$\displaystyle=\int_{G}\int_{G}\|\langle\tau_{g},\tau_{h}\rangle\|^{2m}\,d\mu_{G}(g)\,d\mu_{G}(h)$
$\displaystyle\geq\int_{G\times
G}\langle\tau_{g},\tau_{h}\rangle^{m}\langle\tau_{h},\tau_{g}\rangle^{m}\,d(\mu_{G}\times\mu_{G})(g,h)$
$\displaystyle=\int_{G}\int_{G}\langle\tau_{g},\tau_{h}\rangle^{m}\langle\tau_{h},\tau_{g}\rangle^{m}\,d\mu_{G}(g)\,d\mu_{G}(h)\geq\frac{\mu_{G}(G)^{2}}{{d+m-1\choose
m}},\quad\forall m\in\mathbb{N}.$

In particular,

$\displaystyle\int_{G\times
G}\|\langle\tau_{g},\tau_{h}\rangle\|^{2}\,d(\mu_{G}\times\mu_{G})(g,h)$
$\displaystyle=\int_{G}\int_{G}\|\langle\tau_{g},\tau_{h}\rangle\|^{2}\,d\mu_{G}(g)\,d\mu_{G}(h)$
$\displaystyle\geq\int_{G\times
G}\langle\tau_{g},\tau_{h}\rangle\langle\tau_{h},\tau_{g}\rangle\,d(\mu_{G}\times\mu_{G})(g,h)$
$\displaystyle=\int_{G}\int_{G}\langle\tau_{g},\tau_{h}\rangle\langle\tau_{h},\tau_{g}\rangle\,d\mu_{G}(g)\,d\mu_{G}(h)\geq\frac{\mu_{G}(G)^{2}}{d}.$

Further, we have the higher order continuous modular Welch bounds

$\displaystyle\sup_{g,h\in G,g\neq
h}\|\langle\tau_{g},\tau_{h}\rangle\|^{2m}\geq\frac{1}{(\mu_{G}\times\mu_{G})((G\times
G)\setminus\Delta)}\left[\frac{\mu_{G}(G)^{2}}{{d+m-1\choose
m}}-(\mu_{G}\times\mu_{G})(\Delta)\right],\quad\forall m\in\mathbb{N}.$

In particular, we have the first order continuous modular Welch bound

$\displaystyle\sup_{g,h\in G,g\neq
h}\|\langle\tau_{g},\tau_{h}\rangle\|^{2}\geq\frac{1}{(\mu_{G}\times\mu_{G})((G\times
G)\setminus\Delta)}\left[\frac{\mu_{G}(G)^{2}}{d}-(\mu_{G}\times\mu_{G})(\Delta)\right].$

Even after 48 years of work of Welch, to the best of our knowledge, there are
no Welch bounds or their twins or relatives for Banach spaces. In this paper
we derive both discrete and continuous Welch bounds for Banach spaces. We ask
several open problems for further research.

##  2\. Discrete Welch bounds for Banach spaces

Throughout the paper, $\mathcal{X}$ denotes a finite dimensional Banach space
and $\mathcal{X}^{*}$ denotes its dual. $I_{\mathcal{X}}$ denotes the identity
operator on $\mathcal{X}$. We use $\mathbb{K}$ to denote $\mathbb{R}$ or
$\mathbb{C}$.

######  Definition 2.1.

Let $\mathcal{X}$ be a finite dimensional Banach space. Given a collection
$\\{\tau_{j}\\}_{j=1}^{n}$ in $\mathcal{X}$ and a collection
$\\{f_{j}\\}_{j=1}^{n}$ in $\mathcal{X}^{*}$,

  1. (i)

The frame operator is defined as

$\displaystyle S_{f,\tau}:\mathcal{X}\ni x\mapsto
S_{f,\tau}x\coloneqq\sum_{j=1}^{n}f_{j}(x)\tau_{j}\in\mathcal{X}.$

  2. (ii)

The analysis operator is defined as

$\displaystyle\theta_{f}:\mathcal{X}\ni
x\mapsto\theta_{f}x\coloneqq(f_{j}(x))_{j=1}^{n}\in\mathbb{K}^{n}.$

  3. (iii)

The synthesis operator is defined as

$\displaystyle\theta_{\tau}:\mathbb{K}^{n}\ni(a_{j})_{j=1}^{n}\mapsto\theta_{\tau}(a_{j})_{j=1}^{n}\coloneqq\sum_{j=1}^{n}a_{j}\tau_{j}\in\mathcal{X}.$

Following proposition follows easily.

######  Proposition 2.2.

Given a collection $\\{\tau_{j}\\}_{j=1}^{n}$ in $\mathcal{X}$ and a
collection $\\{f_{j}\\}_{j=1}^{n}$ in $\mathcal{X}^{*}$, the frame operator
factors as $S_{f,\tau}=\theta_{\tau}\theta_{f}$.

######  Definition 2.3.

[29, 65] Let $\\{\tau_{j}\\}_{j=1}^{n}$ be a collection in $\mathcal{X}$ and
$\\{f_{j}\\}_{j=1}^{n}$ be a collection in $\mathcal{X}^{*}.$ The pair
$(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ is said to be an
approximate Schauder frame (ASF) for $\mathcal{X}$ if the map

$\displaystyle S_{f,\tau}:\mathcal{X}\ni x\mapsto
S_{f,\tau}x\coloneqq\sum_{j=1}^{n}f_{j}(x)\tau_{j}\in\mathcal{X}.$

is invertible. If $S_{f,\tau}=\lambda I_{\mathcal{X}}$, for some non zero
scalar $\lambda$, then $(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ is
called as a tight ASF for $\mathcal{X}$.

Following theorem says that we can recover the trace of frame operator using
ASFs.

######  Theorem 2.4.

Given a collection $\\{\tau_{j}\\}_{j=1}^{n}$ in $\mathcal{X}$ and a
collection $\\{f_{j}\\}_{j=1}^{n}$ in $\mathcal{X}^{*}$, we have

$\displaystyle\operatorname{Tra}(S_{f,\tau})=\sum_{j=1}^{n}f_{j}(\tau_{j}),$
$\displaystyle\operatorname{Tra}(S^{2}_{f,\tau})=\sum_{j=1}^{n}\sum_{k=1}^{n}f_{j}(\tau_{k})f_{k}(\tau_{j}).$

###### Proof.

From the expression of $S_{f,\tau}$ and from the definition of trace of
operator in Banach space (see [38]) we get

$\displaystyle\operatorname{Tra}(S_{f,\tau})=\sum_{j=1}^{n}f_{j}(\tau_{j}).$

Now

$\displaystyle
S_{f,\tau}^{2}x=\sum_{j=1}^{n}f_{j}(x)(S_{f,\tau}\tau_{j}),\quad\forall
x\in\mathcal{X}.$

Again using the definition of trace we get

$\displaystyle\operatorname{Tra}(S^{2}_{f,\tau})$
$\displaystyle=\sum_{j=1}^{n}f_{j}(S_{f,\tau}\tau_{j})=\sum_{j=1}^{n}f_{j}\left(\sum_{k=1}^{n}f_{k}(\tau_{j})\tau_{k}\right)$
$\displaystyle=\sum_{j=1}^{n}\sum_{k=1}^{n}f_{j}(\tau_{k})f_{k}(\tau_{j}).$

∎

Now we can derive the first important result of the paper.

######  Theorem 2.5.

(First order Welch bound for Banach spaces) Let $\\{\tau_{j}\\}_{j=1}^{n}$ be
a collection in a finite dimensional Banach space $\mathcal{X}$ of dimension
$d$ and $\\{f_{j}\\}_{j=1}^{n}$ be a collection in $\mathcal{X}^{*}.$ Let
$n\geq d$. If the operator $S_{f,\tau}:\mathcal{X}\ni x\mapsto
S_{f,\tau}x\coloneqq\sum_{j=1}^{n}f_{j}(x)\tau_{j}\in\mathcal{X}$ is
diagonalizable and its eigenvalues are all non negative, then

(4) $\displaystyle\sum_{1\leq j,k\leq
n}f_{j}(\tau_{k})f_{k}(\tau_{j})=\sum_{j=1}^{n}\sum_{k=1}^{n}f_{j}(\tau_{k})f_{k}(\tau_{j})\geq\frac{1}{d}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}=\frac{1}{\text{dim}(\mathcal{X})}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}.$

and

$\displaystyle\max_{1\leq j,k\leq n,j\neq k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|$
$\displaystyle\geq\frac{\frac{1}{d}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2}}{n^{2}-n}$
$\displaystyle=\frac{\frac{1}{\text{dim}(\mathcal{X})}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2}}{n^{2}-n},$
$\displaystyle\max_{1\leq j,k\leq n,j\neq k}|f_{j}(\tau_{k})|$
$\displaystyle\geq\sqrt{\frac{\frac{1}{d}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2}}{n^{2}-n}}$
$\displaystyle=\sqrt{\frac{\frac{1}{\text{dim}(\mathcal{X})}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2}}{n^{2}-n}}.$

Further, equality holds in Inequality (4) if and only if
$(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ is tight ASF for
$\mathcal{X}$. In particular, if $f_{j}(\tau_{j})=1$ for all $1\leq j\leq n$,
then

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|\geq\frac{n-d}{d(n-1)}=\frac{n-\text{dim}(\mathcal{X})}{\text{dim}(\mathcal{X})(n-1)}$

and we have first order (discrete) Welch bound for Banach spaces

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|\geq\sqrt{\frac{n-d}{d(n-1)}}=\sqrt{\frac{n-\text{dim}(\mathcal{X})}{\text{dim}(\mathcal{X})(n-1)}}.$

###### Proof.

Let $\lambda_{1},\dots,\lambda_{d}$ be eigenvalues of $S_{f,\tau}$. Then
$\lambda_{1},\dots,\lambda_{d}\geq 0$ and using the diagonalizability of
$S_{f,\tau}$ we get

$\displaystyle\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}$
$\displaystyle=(\operatorname{Tra}(S_{f,\tau}))^{2}=\left(\sum_{k=1}^{d}\lambda_{k}\right)^{2}\leq
d\sum_{k=1}^{d}\lambda_{k}^{2}$ (5)
$\displaystyle=d\operatorname{Tra}(S^{2}_{f,\tau})=d\sum_{j=1}^{n}\sum_{k=1}^{n}f_{j}(\tau_{k})f_{k}(\tau_{j}).$

For the second inequality,

$\displaystyle\frac{1}{d}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}$
$\displaystyle\leq\sum_{j=1}^{n}\sum_{k=1}^{n}f_{j}(\tau_{k})f_{k}(\tau_{j})=\sum_{j=1}^{n}f_{j}(\tau_{j})^{2}+\sum_{j,k=1,j\neq
k}^{n}f_{j}(\tau_{k})f_{k}(\tau_{j})$
$\displaystyle\leq\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2}+\sum_{j,k=1,j\neq
k}^{n}|f_{j}(\tau_{k})f_{k}(\tau_{j})|$
$\displaystyle\leq\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2}+(n^{2}-n)\max_{1\leq
j,k\leq n,j\neq k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|$

which gives

(6) $\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|\geq\frac{\frac{1}{d}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2}}{n^{2}-n}.$

Now let $1\leq j,k\leq n,j\neq k$ be fixed. Then

$\displaystyle|f_{j}(\tau_{k})f_{k}(\tau_{j})|$ $\displaystyle\leq\max_{1\leq
j,k\leq n,j\neq k}|f_{j}(\tau_{k})|\max_{1\leq j,k\leq n,j\neq
k}|f_{k}(\tau_{j})|$ $\displaystyle=\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|=\left(\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|\right)^{2}.$

Therefore

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|\leq\left(\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|\right)^{2}$

Using Inequality (6) we now get

(7) $\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|\geq\sqrt{\frac{\frac{1}{d}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2}}{n^{2}-n}}.$

Whenever $f_{j}(\tau_{j})=1$ for all $1\leq j\leq n$, Inequality (6) gives

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|\geq\frac{\frac{n^{2}}{d}-n}{n^{2}-n}=\frac{n-d}{d(n-1)}.$

and Inequality (7) gives

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|\geq\sqrt{\frac{n-d}{d(n-1)}}.$

Equality holds in Inequality (2) if and only if

$\displaystyle\left(\sum_{k=1}^{d}\lambda_{k}\right)^{2}=\left(\sum_{k=1}^{d}1\right)\left(\sum_{k=1}^{d}\lambda_{k}^{2}\right)$

if and only if

$\displaystyle\lambda_{k}=a,\text{ for some }a>0,\forall 1\leq k\leq n$

if and only if $S_{f,\tau}$ is a tight ASF for $\mathcal{X}$. ∎

We next derive higher order Welch bounds for Banach spaces. First we need a
standard result.

######  Theorem 2.6.

[19, 11] If $\mathcal{V}$ is a vector space of dimension $d$ and
$\text{Sym}^{m}(\mathcal{V})$ denotes the vector space of symmetric m-tensors,
then

$\displaystyle\text{dim}(\text{Sym}^{m}(\mathcal{V}))={d+m-1\choose
m},\quad\forall m\in\mathbb{N}.$

######  Theorem 2.7.

(Welch bounds for Banach spaces) Let $\\{\tau_{j}\\}_{j=1}^{n}$ be a
collection in a finite dimensional Banach space $\mathcal{X}$ of dimension $d$
and $\\{f_{j}\\}_{j=1}^{n}$ be a collection in $\mathcal{X}^{*}.$ Let $n\geq
d$ and $m\in\mathbb{N}$. If the operator
$S_{f,\tau}:\text{Sym}^{m}(\mathcal{X})\ni x\mapsto
S_{f,\tau}x\coloneqq\sum_{j=1}^{n}f_{j}^{\otimes m}(x)\tau_{j}^{\otimes
m}\in\text{Sym}^{m}(\mathcal{X})$ is diagonalizable and its eigenvalues are
all non negative, then

$\displaystyle\sum_{1\leq j,k\leq n}f_{j}(\tau_{k})^{m}f_{k}(\tau_{j})^{m}$
$\displaystyle=\sum_{j=1}^{n}\sum_{k=1}^{n}f_{j}(\tau_{k})^{m}f_{k}(\tau_{j})^{m}\geq\frac{1}{{d+m-1\choose
m}}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})^{m}\right)^{2}$ (8)
$\displaystyle=\frac{1}{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})^{m}\right)^{2}.$

and

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{m}$
$\displaystyle\geq\frac{\frac{1}{{d+m-1\choose
m}}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})^{m}\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2m}}{n^{2}-n}$
$\displaystyle=\frac{\frac{1}{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})^{m}\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2m}}{n^{2}-n},$
$\displaystyle\max_{1\leq j,k\leq n,j\neq k}|f_{j}(\tau_{k})|^{m}$
$\displaystyle\geq\sqrt{\frac{\frac{1}{{d+m-1\choose
m}}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})^{m}\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2m}}{n^{2}-n}}$
$\displaystyle=\sqrt{\frac{\frac{1}{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})^{m}\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2m}}{n^{2}-n}}.$

Further, equality holds in Inequality (2.7) if and only if
$(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ is a tight ASF for
$\text{Sym}^{m}(\mathcal{X})$. In particular, if $f_{j}(\tau_{j})=1$ for all
$1\leq j\leq n$, then

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{m}$
$\displaystyle\geq\frac{n-{d+m-1\choose m}}{{d+m-1\choose
m}(n-1)}=\frac{n-\text{dim}(\text{Sym}^{m}(\mathcal{X}))}{\text{dim}(\text{Sym}^{m}(\mathcal{X}))(n-1)}$
$\displaystyle=\frac{1}{n-1}\left[\frac{n}{{d+m-1\choose m}}-1\right]$

and we have  higher order (discrete) Welch bounds for Banach spaces

$\displaystyle\max_{1\leq j,k\leq n,j\neq k}|f_{j}(\tau_{k})|^{m}$
$\displaystyle\geq\sqrt{\frac{n-{d+m-1\choose m}}{{d+m-1\choose
m}(n-1)}}=\sqrt{\frac{n-\text{dim}(\text{Sym}^{m}(\mathcal{X}))}{\text{dim}(\text{Sym}^{m}(\mathcal{X}))(n-1)}}$
$\displaystyle=\sqrt{\frac{1}{n-1}\left[\frac{n}{{d+m-1\choose m}}-1\right]}.$

###### Proof.

We will do the proof of Theorem 2.5 for the space
$\text{Sym}^{m}(\mathcal{X})$ (we refer [55] for the tensor product of Banach
spaces). Let
$\lambda_{1},\dots,\lambda_{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}$ be
eigenvalues of $S_{f,\tau}$. Then

$\displaystyle\left(\sum_{j=1}^{n}f_{j}(\tau_{j})^{m}\right)^{2}$
$\displaystyle=\left(\sum_{j=1}^{n}f_{j}^{\otimes m}(\tau_{j}^{\otimes
m})\right)^{2}=(\operatorname{Tra}(S_{f,\tau}))^{2}=\left(\sum_{l=1}^{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}\lambda_{l}\right)^{2}$
$\displaystyle\leq\text{dim}(\text{Sym}^{m}(\mathcal{X}))\sum_{l=1}^{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}\lambda_{l}^{2}={d+m-1\choose
m}\operatorname{Tra}(S^{2}_{f,\tau})$ $\displaystyle={d+m-1\choose
m}\sum_{j=1}^{n}\sum_{l=1}^{n}f_{j}^{\otimes m}(\tau_{l}^{\otimes
m})f_{l}^{\otimes m}(\tau_{j}^{\otimes m})$ $\displaystyle={d+m-1\choose
m}\sum_{j=1}^{n}\sum_{k=1}^{n}f_{j}(\tau_{k})^{m}f_{k}(\tau_{j})^{m}$

and

$\displaystyle\frac{1}{{d+m-1\choose
m}}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})^{m}\right)^{2}$
$\displaystyle=\sum_{1\leq j,k\leq
n}f_{j}(\tau_{k})^{m}f_{k}(\tau_{j})^{m}=\sum_{1\leq j,k\leq n,j\neq
k}f_{j}(\tau_{k})^{m}f_{k}(\tau_{j})^{m}+\sum_{j=1}^{n}f_{j}(\tau_{j})^{2m}$
$\displaystyle\leq\sum_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{m}+\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2m}$
$\displaystyle\leq(n^{2}-n)\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{m}+\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2m}.$

Other parts are similar to the corresponding part in the proof of Theorem 2.5.
∎

######  Corollary 2.8.

Theorem 1.1 is a corollary of Theorem 2.7.

###### Proof.

Let $\\{\tau_{j}\\}_{j=1}^{n}$ be a finite collection in a finite dimensional
Hilbert space $\mathcal{H}$ of dimension $d$. Define $f_{j}:\mathcal{H}\ni
h\mapsto\langle h,\tau_{j}\rangle\in\mathbb{K}$, $\forall 1\leq j\leq n$. Let
$m\in\mathbb{N}$. Then the operator

$\displaystyle S_{\tau}:\text{Sym}^{m}(\mathcal{H})\ni
h\mapsto\sum_{j=1}^{n}f_{j}^{\otimes m}(h)\tau_{j}^{\otimes
m}=\sum_{j=1}^{n}\langle h,\tau_{j}^{\otimes m}\rangle\tau_{j}^{\otimes
m}\in\text{Sym}^{m}(\mathcal{H})$

is positive and Theorem 2.7 applies. ∎

Theorem 2.7 gives Welch bound for all natural numbers. One can now ask whether
we can replace naturals by positive reals. Following results show that we can
do this. For normalized tight frames for Hilbert spaces, these results are
derived in [33] and [27].

######  Theorem 2.9.

Let $\\{\tau_{j}\\}_{j=1}^{n}$ be a collection in a finite dimensional Banach
space $\mathcal{X}$ of dimension $d$ and $\\{f_{j}\\}_{j=1}^{n}$ be a
collection in $\mathcal{X}^{*}.$ Let $n\geq d$. If the operator
$S_{f,\tau}:\mathcal{X}\ni x\mapsto
S_{f,\tau}x\coloneqq\sum_{j=1}^{n}f_{j}(x)\tau_{j}\in\mathcal{X}$ is
diagonalizable and its eigenvalues are all non negative, then

$\displaystyle\operatorname{Tra}(S^{r}_{f,\tau})\geq\frac{\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{r}}{d^{r-1}},\quad\forall
r\in[1,\infty)$

and

$\displaystyle\operatorname{Tra}(S^{r}_{f,\tau})\leq\frac{\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{r}}{d^{r-1}},\quad\forall
r\in(0,1).$

In particular, if $f_{j}(\tau_{j})=1$ for all $1\leq j\leq n$, then

$\displaystyle\operatorname{Tra}(S^{r}_{f,\tau})\geq\frac{n^{r}}{d^{r-1}},\quad\forall
r\in[1,\infty)$

and

$\displaystyle\operatorname{Tra}(S^{r}_{f,\tau})\leq\frac{n^{r}}{d^{r-1}},\quad\forall
r\in(0,1).$

###### Proof.

Let $\lambda_{1},\dots,\lambda_{d}$ be eigenvalues of $S_{f,\tau}$. Let
$r\in[1,\infty)$. Using Jensen’s inequality, we have

$\displaystyle\left(\frac{\sum_{j=1}^{n}f_{j}(\tau_{j})}{d}\right)^{r}=\left(\frac{\operatorname{Tra}(S_{f,\tau})}{d}\right)^{r}=\left(\frac{\sum_{k=1}^{d}\lambda_{k}}{d}\right)^{r}\leq\frac{\sum_{k=1}^{d}\lambda_{k}^{r}}{d}=\frac{1}{d}\operatorname{Tra}(S^{r}_{f,\tau})$

which gives the first part. Second part again follows from Jensen’s
inequality. ∎

######  Theorem 2.10.

Let $\\{\tau_{j}\\}_{j=1}^{n}$ be a collection in a finite dimensional Banach
space $\mathcal{X}$ of dimension $d$ and $\\{f_{j}\\}_{j=1}^{n}$ be a
collection in $\mathcal{X}^{*}.$ Let $2<p<\infty$. If the operator
$S_{f,\tau}:\mathcal{X}\ni x\mapsto
S_{f,\tau}x\coloneqq\sum_{j=1}^{n}f_{j}(x)\tau_{j}\in\mathcal{X}$ is
diagonalizable and its eigenvalues are all non negative, then

$\displaystyle\sum_{1\leq j,k\leq
n}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{\frac{p}{2}}\geq
n(n-1)\left(\frac{n-d}{d(n-1)}\right)^{\frac{p}{2}}+n.$

###### Proof.

Define $r\coloneqq 2p/(p-2)$ and $q$ be the conjugate index of $p/2$. Then
$q=r/2$. Using Theorem 2.5 and Holder’s inequality, we have

$\displaystyle\frac{n^{2}}{d}-n$ $\displaystyle\leq\sum_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|\leq\left(\sum_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{\frac{p}{2}}\right)^{\frac{2}{p}}\left(\sum_{1\leq
j,k\leq n,j\neq k}1\right)^{\frac{1}{q}}$ $\displaystyle=\left(\sum_{1\leq
j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{\frac{p}{2}}\right)^{\frac{2}{p}}(n^{2}-n)^{\frac{1}{q}}$
$\displaystyle=\left(\sum_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{\frac{p}{2}}\right)^{\frac{2}{p}}(n^{2}-n)^{\frac{2}{r}}$
$\displaystyle=\left(\sum_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{\frac{p}{2}}\right)^{\frac{2}{p}}(n^{2}-n)^{\frac{p-2}{p}}$

which gives

$\displaystyle\left(\frac{n^{2}}{d}-n\right)^{\frac{p}{2}}\leq\left(\sum_{1\leq
j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{\frac{p}{2}}\right)(n^{2}-n)^{\frac{p}{2}-1}.$

Therefore

$\displaystyle n(n-1)\left(\frac{n-d}{d(n-1)}\right)^{\frac{p}{2}}+n$
$\displaystyle=\frac{1}{(n^{2}-n)^{\frac{p}{2}-1}}\left(\frac{n^{2}}{d}-n\right)^{\frac{p}{2}}+n$
$\displaystyle\leq\sum_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{\frac{p}{2}}+\sum_{j=1}^{n}|f_{j}(\tau_{j})f_{j}(\tau_{j})|^{\frac{p}{2}}$
$\displaystyle=\sum_{1\leq j,k\leq
n}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{\frac{p}{2}}.$

∎

Some of the proofs of Theorem 2.7 (for instance see [54]) use the idea of Gram
matrix and Frobenius norm/Hilbert-Schmidt norm. We now give Welch bound for
Banach spaces using matrices. First we need a definition.

######  Definition 2.11.

Let $\\{\tau_{j}\\}_{j=1}^{n}$ be a collection in a Banach space $\mathcal{X}$
and $\\{f_{j}\\}_{j=1}^{n}$ be a collection in $\mathcal{X}^{*}.$ We define
the Gram matrix $G_{f,\tau}$ of
$(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ as

$\displaystyle G_{f,\tau}\coloneqq[f_{j}(\tau_{k})]_{1\leq j,k\leq
n}=\begin{pmatrix}f_{1}(\tau_{1})&f_{1}(\tau_{2})&\cdots&f_{1}(\tau_{n})\\\
f_{2}(\tau_{1})&f_{2}(\tau_{2})&\cdots&f_{2}(\tau_{n})\\\
\vdots&\vdots&&\vdots\\\
f_{n}(\tau_{1})&f_{n}(\tau_{2})&\cdots&f_{n}(\tau_{n})\\\
\end{pmatrix}_{n\times n}\in\mathbb{M}_{n}(\mathbb{K}).$

In terms of analysis and synthesis operators,
$G_{f,\tau}=\theta_{f}\theta_{\tau}$. We observe that Definition 2.11 reduces
to the definition of Gram matrix in Hilbert spaces. Indeed, let
$\\{\tau_{j}\\}_{j=1}^{n}$ be a collection in a Hilbert space $\mathcal{H}$.
Now define $f_{j}(h)\coloneqq\langle h,\tau_{j}\rangle$ for all
$h\in\mathcal{H}$, for all $1\leq j\leq n$.

######  Theorem 2.12.

Let $\\{\tau_{j}\\}_{j=1}^{n}$ be a collection in a Banach space $\mathcal{X}$
and $\\{f_{j}\\}_{j=1}^{n}$ be a collection in $\mathcal{X}^{*}.$ If the Gram
matrix $G_{f,\tau}=[f_{j}(\tau_{k})]_{1\leq j,k\leq n}$ is diagonalizable and
its eigenvalues are all non negative, then

$\displaystyle\sum_{1\leq j,k\leq
n}f_{j}(\tau_{k})f_{k}(\tau_{j})=\sum_{j=1}^{n}\sum_{k=1}^{n}f_{j}(\tau_{k})f_{k}(\tau_{j})\geq\frac{1}{\text{rank}(G_{f,\tau})}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}.$

and

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|\geq\frac{\frac{1}{\text{rank}(G_{f,\tau})}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2}}{n^{2}-n},$
$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|\geq\sqrt{\frac{\frac{1}{\text{rank}(G_{f,\tau})}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2}}{n^{2}-n}}.$

In particular, if $f_{j}(\tau_{j})=1$ for all $1\leq j\leq n$, then

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|\geq\frac{n-\text{rank}(G_{f,\tau})}{\text{rank}(G_{f,\tau})(n-1)}$

and

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|\geq\sqrt{\frac{n-\text{rank}(G_{f,\tau})}{\text{rank}(G_{f,\tau})(n-1)}}.$

###### Proof.

Let $\lambda_{1},\dots,\lambda_{(\text{rank}G_{f,\tau})}$ be the eigenvalues
of $G_{f,\tau}$. Then

$\displaystyle\left(\sum_{j=1}^{n}f_{j}(\tau_{j})\right)^{2}$
$\displaystyle=(\operatorname{Trace}(G_{f,\tau}))^{2}=\left(\sum_{k=1}^{\text{rank}(G_{f,\tau})}\lambda_{k}\right)^{2}\leq{\text{rank}(G_{f,\tau})}\sum_{k=1}^{\text{rank}(G_{f,\tau})}\lambda_{k}^{2}$
$\displaystyle=\text{rank}(G_{f,\tau})\operatorname{Trace}(G^{2}_{f,\tau})=\text{rank}(G_{f,\tau})\sum_{j=1}^{n}\sum_{k=1}^{n}f_{j}(\tau_{k})f_{k}(\tau_{j})$
$\displaystyle\leq\text{rank}(G_{f,\tau})\left(\sum_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|+\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2}\right)$
$\displaystyle\leq\text{rank}(G_{f,\tau})\left((n^{2}-n)\max_{1\leq j,k\leq
n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|+\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2}\right).$

∎

In the following result, given a matrix $G$, $G^{\circ^{m}}$ denotes the
Hadamard/Schur/pointwise product of $G$ with itself, $m$ times [35].

######  Theorem 2.13.

Let $\\{\tau_{j}\\}_{j=1}^{n}$ be a collection in a Banach space $\mathcal{X}$
and $\\{f_{j}\\}_{j=1}^{n}$ be a collection in $\mathcal{X}^{*}.$ Let
$m\in\mathbb{N}$. If the Hadamard product $G_{f,\tau}^{\circ^{m}}$ is
diagonalizable and its eigenvalues are all non negative, then

$\displaystyle\sum_{1\leq j,k\leq
n}f_{j}(\tau_{k})^{m}f_{k}(\tau_{j})^{m}=\sum_{j=1}^{n}\sum_{k=1}^{n}f_{j}(\tau_{k})^{m}f_{k}(\tau_{j})^{m}\geq\frac{1}{\text{rank}(G_{f,\tau}^{\circ^{m}})}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})^{m}\right)^{2}.$

and

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{m}\geq\frac{\frac{1}{\text{rank}(G_{f,\tau}^{\circ^{m}})}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})^{m}\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2m}}{n^{2}-n},$
$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|^{m}\geq\sqrt{\frac{\frac{1}{\text{rank}(G_{f,\tau}^{\circ^{m}})}\left(\sum_{j=1}^{n}f_{j}(\tau_{j})^{m}\right)^{2}-\sum_{j=1}^{n}|f_{j}(\tau_{j})|^{2m}}{n^{2}-n}}.$

In particular, if $f_{j}(\tau_{j})=1$ for all $1\leq j\leq n$, then

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})f_{k}(\tau_{j})|^{m}\geq\frac{n-\text{rank}(G_{f,\tau}^{\circ^{m}})}{\text{rank}(G_{f,\tau}^{\circ^{m}})(n-1)}$

and

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|^{m}\geq\sqrt{\frac{n-\text{rank}(G_{f,\tau}^{\circ^{m}})}{\text{rank}(G_{f,\tau}^{\circ^{m}})(n-1)}}.$

###### Proof.

We note that

$\displaystyle G_{f,\tau}^{\circ^{m}}=[(f_{j}(\tau_{k}))^{m}]_{1\leq j,k\leq
n}=[(f_{j}^{\otimes m}(\tau^{\otimes m}_{k}))]_{1\leq j,k\leq n}.$

Now the proof is similar to the proof of Theorem 2.12. ∎

We next recall the definition of Gerzon’s bound which allows us to recall the
bounds which are in the same way to discrete Welch bounds in the Hilbert
spaces.

######  Definition 2.14.

[37] Given $d\in\mathbb{N}$, define Gerzon’s bound

$\displaystyle\mathcal{Z}(d,\mathbb{K})\coloneqq\left\\{\begin{array}[]{cc}d^{2}&\quad\text{if}\quad\mathbb{K}=\mathbb{C}\\\
\frac{d(d+1)}{2}&\quad\text{if}\quad\mathbb{K}=\mathbb{R}.\\\
\end{array}\right.$

######  Theorem 2.15.

[37, 71, 49, 60, 13, 20, 32, 51] Define
$m\coloneqq\operatorname{dim}_{\mathbb{R}}(\mathbb{K})/2$. If
$\\{\tau_{j}\\}_{j=1}^{n}$ is any collection of unit vectors in
$\mathbb{K}^{d}$, then

  1. (i)

(Bukh-Cox bound)

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|\langle\tau_{j},\tau_{k}\rangle|\geq\frac{\mathcal{Z}(n-d,\mathbb{K})}{n(1+m(n-d-1)\sqrt{m^{-1}+n-d})-\mathcal{Z}(n-d,\mathbb{K})}\quad\text{if}\quad
n>d.$

  2. (ii)

(Orthoplex/Rankin bound)

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|\langle\tau_{j},\tau_{k}\rangle|\geq\frac{1}{\sqrt{d}}\quad\text{if}\quad
n>\mathcal{Z}(d,\mathbb{K}).$

  3. (iii)

(Levenstein bound)

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|\langle\tau_{j},\tau_{k}\rangle|\geq\sqrt{\frac{n(m+1)-d(md+1)}{(n-d)(md+1)}}\quad\text{if}\quad
n>\mathcal{Z}(d,\mathbb{K}).$

  4. (iv)

(Exponential bound)

$\displaystyle\max_{1\leq j,k\leq n,j\neq
k}|\langle\tau_{j},\tau_{k}\rangle|\geq 1-2n^{\frac{-1}{d-1}}.$

Theorem 2.15 and Theorem 2.5 give the following problem to us.

######  Question 2.16.

Whether there is a version of Theorem 2.15 for Banach spaces?. In particular,
does there exists a (discrete) version of

  1. (i)

Bukh-Cox bound for Banach spaces?

  2. (ii)

Orthoplex/Rankin bound for Banach spaces?

  3. (iii)

Levenstein bound for Banach spaces?

  4. (iv)

Exponential bound for Banach spaces?

##  3\. Applications of discrete Welch bounds for Banach spaces

We begin by defining the RMS of vectors and functionals in Banach spaces.

######  Definition 3.1.

Let $\\{\tau_{j}\\}_{j=1}^{n}$ be a collection in a finite dimensional Banach
space $\mathcal{X}$ of dimension $d$ and $\\{f_{j}\\}_{j=1}^{n}$ be a
collection in $\mathcal{X}^{*}$ satisfying $f_{j}(\tau_{j})=1$ for all $1\leq
j\leq n$. Assume that the frame operator $S_{f,\tau}$ is diagonalizable and
its eigenvalues are all non negative. We define the root-mean-square (RMS)
absolute cross relation of $(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$
as

$\displaystyle
I_{\text{RMS}}(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})\coloneqq\left(\frac{1}{n(n-1)}\sum_{1\leq
j,k\leq n,j\neq k}f_{j}(\tau_{k})f_{k}(\tau_{j})\right)^{\frac{1}{2}}.$

Theorem 2.5 gives the following result.

######  Theorem 3.2.

Let $\\{\tau_{j}\\}_{j=1}^{n}$ and $\\{f_{j}\\}_{j=1}^{n}$ be as in Definition
3.1. Then

$\displaystyle\max_{1\leq j,k\leq n,j\neq k}|f_{j}(\tau_{k})|\geq
I_{\text{RMS}}(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})\geq\left(\frac{n-d}{d(n-1)}\right)^{\frac{1}{2}}.$

Here is another notion similar to that of frame potential in Hilbert spaces.

######  Definition 3.3.

Let $\\{\tau_{j}\\}_{j=1}^{n}$ be a collection in a finite dimensional Banach
space $\mathcal{X}$ of dimension $d$ and $\\{f_{j}\\}_{j=1}^{n}$ be a
collection in $\mathcal{X}^{*}$ satisfying $f_{j}(\tau_{j})=1$ for all $1\leq
j\leq n$. Assume that the frame operator $S_{f,\tau}$ is diagonalizable and
its eigenvalues are all non negative. We define the pseudo frame potential of
$(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ as

$\displaystyle
PFP(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})\coloneqq\sum_{j=1}^{n}\sum_{k=1}^{n}f_{j}(\tau_{k})f_{k}(\tau_{j}).$

Note that we defined the notion pseudo frame potential and not frame
potential. The reason is that frame potential for Banach spaces can not be
defined in the way in Definition 3.3. One has to go to the theory of p-summing
operators (see [24, 66]) to define frame potential in Banach spaces, see [17].
Theorem 2.5 again gives the following result.

######  Theorem 3.4.

Let $\\{\tau_{j}\\}_{j=1}^{n}$ and $\\{f_{j}\\}_{j=1}^{n}$ be as in Definition
3.3. Then

(9) $\displaystyle n^{2}\max_{1\leq j,k\leq n}|f_{j}(\tau_{k})|^{2}\geq
PFP(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})\geq\frac{n^{2}}{{d}}.$

Theorem 3.4 gives the following problem.

######  Question 3.5.

Is there a characterization of
$(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ by using equality in
Inequality (9)?

We next introduce the notions of Grassmannian frames and equiangular frames
for Banach spaes. First we need a definition.

######  Definition 3.6.

Let $(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ be an ASF for
$\mathcal{X}$ satisfying $f_{j}(\tau_{j})=1$, $\forall 1\leq j\leq n$. Assume
that the frame operator $S_{f,\tau}$ is diagonalizable and its eigenvalues are
all non negative. We define the frame correlation of
$(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ as

$\displaystyle\mathcal{M}(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})\coloneqq\max_{1\leq
j,k\leq n,j\neq k}|f_{j}(\tau_{k})|.$

######  Definition 3.7.

Let $(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ be an ASF for
$\mathcal{X}$ satisfying $\|f_{j}\|=1,\|\tau_{j}\|=1,f_{j}(\tau_{j})=1$,
$\forall 1\leq j\leq n$. Assume that the frame operator $S_{f,\tau}$ is
diagonalizable and its eigenvalues are all non negative. ASF
$(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ is said to be a
Grassmannian frame for $\mathcal{X}$ if

$\displaystyle\mathcal{M}(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})=\inf\bigg{\\{}$
$\displaystyle\mathcal{M}(\\{g_{j}\\}_{j=1}^{n},\\{\omega_{j}\\}_{j=1}^{n}):(\\{g_{j}\\}_{j=1}^{n},\\{\omega_{j}\\}_{j=1}^{n})\text{
is an ASF for }\mathcal{X}\text{satisfying }$
$\displaystyle\|g_{j}\|=1,\|\omega_{j}\|=1,g_{j}(\omega_{j})=1,\forall 1\leq
j\leq n\text{ and the frame operator }S_{g,\omega}$ $\displaystyle\text{is
diagonalizable and its eigenvalues are all non negative}\bigg{\\}}.$

######  Theorem 3.8.

Grassmannian frames exist in every dimension for every Banach space.

###### Proof.

Our arguments are motivated from the arguments given in [6] for Hilbert spaces
which mainly uses compactness and continuity. We give arguments only for real
Banach spaces and complex case follows by considering real and imaginary parts
(of the linear functionals). Define

$\displaystyle
S_{\mathcal{X}}^{n}\coloneqq\\{(x_{1},\dots,x_{n}):x_{1},\dots,x_{n}\in\mathcal{X},\|x_{1}\|=\cdots=\|x_{n}\|=1\\},$
$\displaystyle
S_{\mathcal{X^{*}}}^{n}\coloneqq\\{(\phi_{1},\dots,\phi_{n}):\phi_{1},\dots,\phi_{n}\in\mathcal{X}^{*},\|\phi_{1}\|=\cdots=\|\phi_{n}\|=1\\}$
$\displaystyle\mathcal{W}\coloneqq\\{((x_{1},\dots,x_{n}),(\phi_{1},\dots,\phi_{n})):((x_{1},\dots,x_{n}),(\phi_{1},\dots,\phi_{n}))\in
S_{\mathcal{X}}^{n}\times S_{\mathcal{X^{*}}}^{n},\phi_{j}(x_{j})=1,\forall
1\leq j\leq n,$
$\displaystyle\quad\quad(\\{\phi_{j}\\}_{j=1}^{n},\\{x_{j}\\}_{j=1}^{n})\text{
is an ASF for }\mathcal{X}\text{ and the frame operator }S_{\phi,x}$
$\displaystyle\quad\quad\text{is diagonalizable and its eigenvalues are all
non negative}\\},$
$\displaystyle\Phi:\mathcal{W}\to[0,1],\quad\Phi((x_{1},\dots,x_{n}),(\phi_{1},\dots,\phi_{n}))\coloneqq\mathcal{M}(\\{\phi_{j}\\}_{j=1}^{n},\\{x_{j}\\}_{j=1}^{n}).$

We re norm $\mathcal{X}^{n}\times\mathcal{X^{*}}^{n}$ by

$\displaystyle\|((x_{1},\dots,x_{n}),(\phi_{1},\dots,\phi_{n}))\|\coloneqq\sum_{j=1}^{n}(\|x_{j}\|+\|\phi_{j}\|)$

and consider $\mathcal{W}$ in this norm. Then $\mathcal{W}$ is compact. We
show that $\Phi$ is continuous on $\mathcal{W}$ which proves the theorem. Let
$(\\{\tau_{j}\\}_{j=1}^{n},\\{f_{j}\\}_{j=1}^{n})$ $\in\mathcal{W}$. Let
$\varepsilon>0$ be given. Define

$\displaystyle R\coloneqq 1+\max_{1\leq j,k\leq
n}\\{\|\tau_{j}\|,\|f_{k}\|\\}$

and let

$\displaystyle 0<\delta<\frac{\sqrt{1+\varepsilon}-1}{R}.$

Now for any given $(\\{\omega_{j}\\}_{j=1}^{n},\\{g_{j}\\}_{j=1}^{n})$
$\in\mathcal{W}$, with

$\displaystyle\|((\omega_{1},\dots,\omega_{n}),(g_{1},\dots,g_{n}))-((\tau_{1},\dots,\tau_{n}),(f_{1},\dots,f_{n}))\|<\delta,$

if we define $h_{j}\coloneqq
g_{j}-f_{j},\rho_{j}\coloneqq\omega_{j}-\tau_{j},\forall 1\leq j\leq n,$ then
$\|h_{j}\|<\delta,\|\rho_{j}\|<\delta$. Then

$\displaystyle|\Phi((\omega_{1},\dots,\omega_{n}),(g_{1},\dots,g_{n}))-\Phi((\tau_{1},\dots,\tau_{n}),(f_{1},\dots,f_{n}))|=\left|\max_{1\leq
j,k\leq n,j\neq k}|g_{j}(\omega_{k})|-\max_{1\leq j,k\leq n,j\neq
k}|f_{j}(\tau_{k})|\right|$ $\displaystyle=\max_{1\leq j,k\leq n,j\neq
k}\big{|}|g_{j}(\omega_{k})|-|f_{j}(\tau_{k})|\big{|}\leq\max_{1\leq j,k\leq
n,j\neq k}\big{|}g_{j}(\omega_{k})-f_{j}(\tau_{k})\big{|}$
$\displaystyle=\max_{1\leq j,k\leq n,j\neq
k}\big{|}(f_{j}+h_{j})(\tau_{k}+\rho_{k})-f_{j}(\tau_{k})\big{|}=\max_{1\leq
j,k\leq n,j\neq
k}\big{|}f_{j}(\tau_{k})+f_{j}(\rho_{k})+h_{j}(\tau_{k})+h_{j}(\rho_{k})-f_{j}(\tau_{k})\big{|}$
$\displaystyle=\max_{1\leq j,k\leq n,j\neq k}|f_{j}(\rho_{k})|+\max_{1\leq
j,k\leq n,j\neq k}|h_{j}(\tau_{k})|+\max_{1\leq j,k\leq n,j\neq
k}|h_{j}(\rho_{k})|$ $\displaystyle\leq\max_{1\leq j,k\leq n,j\neq
k}\|f_{j}\|\|\rho_{k}\|+\max_{1\leq j,k\leq n,j\neq
k}\|h_{j}\|\|\tau_{k}\|+\max_{1\leq j,k\leq n,j\neq k}\|h_{j}\|\|\rho_{k}\|$
$\displaystyle\leq R\cdot\delta+\delta\cdot R+\delta^{2}\leq
R\cdot\delta+\delta\cdot R+R\delta^{2}<\varepsilon.$

∎

######  Definition 3.9.

Let $(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ be an ASF for
$\mathcal{X}$ satisfying $f_{j}(\tau_{j})=1$, $\forall 1\leq j\leq n$. Assume
that the frame operator $S_{f,\tau}$ is diagonalizable and its eigenvalues are
all non negative. ASF $(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ is
said to be $\gamma$-equiangular if there exists $\gamma\geq 0$ such that

$\displaystyle|f_{j}(\tau_{k})|^{2}=\gamma,\quad\forall 1\leq j,k\leq n,j\neq
k.$

Here is the Zauner’s conjecture for Banach spaces (see [5, 3, 73, 59, 30, 52,
2, 9, 4, 40, 31, 10, 36, 8, 48] for a comprehensive look at Zauner’s
conjecture in Hilbert spaces and its connections with Hilbert’s 12-problem and
Stark conjecture).

######  Conjecture 3.10.

(Zauner’s conjecture for Banach spaces) For every $d\in\mathbb{N}$, there
exists a $\frac{1}{d+1}$-equiangular tight ASF
$(\\{f_{j}\\}_{j=1}^{d^{2}},\\{\tau_{j}\\}_{j=1}^{d^{2}})$ for
$\mathbb{C}^{d}$ (w.r.t. any norm) such that
$\|f_{j}\|=\|\tau_{j}\|=|f_{j}(\tau_{j})|=1,\forall 1\leq j\leq n$, i.e.,
there exist a tight ASF for $\mathbb{C}^{d}$ such that
$\|f_{j}\|=\|\tau_{j}\|=|f_{j}(\tau_{j})|=1,\forall 1\leq j\leq n$ and

$\displaystyle|f_{j}(\tau_{k})|^{2}=\frac{1}{d+1},\quad\forall 1\leq j,k\leq
d^{2},j\neq k.$

Following theorem again follows from Theorem 2.5.

######  Theorem 3.11.

Let $(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})$ be as in Definition
3.7. Then

(10)
$\displaystyle\mathcal{M}(\\{f_{j}\\}_{j=1}^{n},\\{\tau_{j}\\}_{j=1}^{n})\geq\sqrt{\frac{n-d}{d(n-1)}}\eqqcolon\gamma.$

If the ASF is $\gamma$-equiangular, then we have equality in Inequality (10).

######  Question 3.12.

  1. (i)

Whether the equality in Inequality (10) implies that the ASF
$(\\{f_{j}\\}_{j=1}^{n},$ $\\{\tau_{j}\\}_{j=1}^{n})$ is $\gamma$-equiangular?

  2. (ii)

Whether the validity of Inequality (10) implies there is a relation between
the number of elements in the ASF and dimension of the space (like Theorem 2.3
in [61])?

##  4\. Continuous Welch bounds for Banach spaces

Unlike the discrete setting, we need some concepts to get continuous version
of Theorem 2.7 for Banach spaces. Following definition is motivated from the
notion of continuous frames for Hilbert spaces [1, 39], continuous framings
for Banach spaces [47], continuous Schauder frames for Banach spaces [28] and
approximate Schauder frames for Banach spaces [29, 65] (which is motivated
from the notion of Schauder frame [14] which is motivated from the notion of
framing [16] which is motivated from the Han-Larson-Naimark dilation theorem
[21] for Hilbert space frames [34]).

######  Definition 4.1.

Let $(\Omega,\mu)$ be a measure space. Let
$\\{\tau_{\alpha}\\}_{\alpha\in\Omega}$ be a collection in a Banach space
$\mathcal{X}$ and $\\{f_{\alpha}\\}_{\alpha\in\Omega}$ be a collection in
$\mathcal{X}^{*}$. The pair
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
is said to be a continuous approximate Schauder frame for $\mathcal{X}$ if the
following holds.

  1. (i)

For every $x\in\mathcal{X}$ and for every $\phi\in\mathcal{X}^{*}$, the map

$\displaystyle\Omega\ni\alpha\mapsto
f_{\alpha}(x)\phi(\tau_{\alpha})\in\mathbb{K}$

is measurable and integrable.

  2. (ii)

The frame operator

$\displaystyle S_{f,\tau}:\mathcal{X}\ni x\mapsto
S_{f,\tau}x\coloneqq\int_{\Omega}f_{\alpha}(x)\tau_{\alpha}\,d\mu(\alpha)\in\mathcal{X}$

is a well-defined invertible bounded linear operator, where the integral is
weak integral (Pettis integrals [63]).

If $S_{f,\tau}=\lambda I_{\mathcal{X}}$, for some non zero scalar $\lambda$,
then
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
is called as a tight continuous ASF for $\mathcal{X}$. If we do not demand the
invertibility of $S_{f,\tau}$, then we say that
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
is a continuous approximate Bessel family for $\mathcal{X}$.

Our first observation is that there is a large supply of continuous frames for
finite dimensional Banach spaces. Here is a result of existence of them for
Banach spaces. Our result is motivated from the work in [50].

######  Theorem 4.2.

Let $\mathcal{X}$ be a finite dimensional Banach space of dimension $d$. Let
$(\Omega,\mu)$ be a finite measure space such that there are measurable
subsets $\Omega_{1},\dots,\Omega_{n}$ satisfying

$\displaystyle\Omega=\Omega_{1}\cup\cdots\cup\Omega_{n},\quad\Omega_{j}\cap\Omega_{k}=\emptyset,\quad\forall
1\leq j,k\leq n,j\neq n.$

If $n\geq d$, then there exists a continuous ASF
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
for $\mathcal{X}$.

###### Proof.

Let $(\\{g_{j}\\}_{j=1}^{n},\\{\omega_{j}\\}_{j=1}^{n})$ be an ASF for
$\mathcal{X}$ (they exist. Infact, any spanning collection can be turned to an
ASF or a pair of basis for the space and its dual works, see [46, 44]). Define

$\displaystyle
f_{\alpha}\coloneqq\frac{g_{j}}{\sqrt{\mu(\Omega_{j})}},\quad\forall\alpha\in\Omega_{j},\forall
1\leq j\leq
n,\quad\tau_{\alpha}\coloneqq\frac{\omega_{j}}{\sqrt{\mu(\Omega_{j})}},\quad\forall\alpha\in\Omega_{j},\forall
1\leq j\leq n.$

Then

$\displaystyle\int_{\Omega}f_{\alpha}(x)\tau_{\alpha}\,d\mu(\alpha)$
$\displaystyle=\sum_{j=1}^{n}\int_{\Omega_{j}}f_{\alpha}(x)\tau_{\alpha}\,d\mu(\alpha)=\sum_{j=1}^{n}\int_{\Omega_{j}}\frac{g_{j}(x)}{\sqrt{\mu(\Omega_{j})}}\frac{\omega_{j}}{\sqrt{\mu(\Omega_{j})}}\,d\mu(\alpha)$
$\displaystyle=\sum_{j=1}^{n}g_{j}(x)\omega_{j},\quad\forall x\in\mathcal{X}.$

Hence
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
is a continuous ASF for $\mathcal{X}$. ∎

Like discrete case, we can get trace of frame operator using continuous ASFs.

######  Theorem 4.3.

Let
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be a continuous approximate Bessel family for $\mathcal{X}$. Then

$\displaystyle\text{Tra}(S_{f,\tau})=\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha),$
$\displaystyle\text{Tra}(S_{f,\tau}^{2})=\int_{\Omega}\int_{\Omega}f_{\beta}(\tau_{\alpha})f_{\alpha}(\tau_{\beta})\,d\mu(\alpha)\,d\mu(\beta).$

###### Proof.

Let $\\{\omega_{j}\\}_{j=1}^{d}$ be any basis for $\mathcal{X}$, where $d$ is
the dimension of $\mathcal{X}$. Let $\\{\zeta_{j}\\}_{j=1}^{d}$ be the dual
basis associated with $\\{\omega_{j}\\}_{j=1}^{d}$. Then
$S_{f,\tau}x=\sum_{j=1}^{n}\zeta_{j}(x)(S_{f,\tau}\omega_{j})$ which gives

$\displaystyle\text{Tra}(S_{f,\tau})$
$\displaystyle=\sum_{j=1}^{n}\zeta_{j}(S_{f,\tau}\omega_{j})=\sum_{j=1}^{n}\zeta_{j}\left(\int_{\Omega}f_{\alpha}(\omega_{j})\tau_{\alpha}\,d\mu(\alpha)\right)$
$\displaystyle=\sum_{j=1}^{n}\int_{\Omega}f_{\alpha}(\omega_{j})\zeta_{j}(\tau_{\alpha})\,d\mu(\alpha)=\int_{\Omega}f_{\alpha}\left(\sum_{j=1}^{n}\zeta_{j}(\tau_{\alpha})\omega_{j}\right)\,d\mu(\alpha)$
$\displaystyle=\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha)$

and

$\displaystyle\text{Tra}(S^{2}_{f,\tau})$
$\displaystyle=\sum_{j=1}^{n}\zeta_{j}(S^{2}_{f,\tau}\omega_{j})=\sum_{j=1}^{n}\zeta_{j}\left(\int_{\Omega}f_{\alpha}(S_{f,\tau}\omega_{j})\tau_{\alpha}\,d\mu(\alpha)\right)$
$\displaystyle=\sum_{j=1}^{n}\int_{\Omega}f_{\alpha}(S_{f,\tau}\omega_{j})\zeta_{j}(\tau_{\alpha})\,d\mu(\alpha)=\int_{\Omega}f_{\alpha}\left(\sum_{j=1}^{n}\zeta_{j}(\tau_{\alpha})S_{f,\tau}\omega_{j}\right)\,d\mu(\alpha)$
$\displaystyle=\int_{\Omega}f_{\alpha}(S_{f,\tau}\tau_{\alpha})\,d\mu(\alpha)=\int_{\Omega}f_{\alpha}\left(\int_{\Omega}f_{\beta}(\tau_{\alpha})\tau_{\beta}\,d\mu(\beta)\right)\,d\mu(\alpha)$
$\displaystyle=\int_{\Omega}\int_{\Omega}f_{\beta}(\tau_{\alpha})f_{\alpha}(\tau_{\beta})\,d\mu(\alpha)\,d\mu(\beta).$

∎

Note that we did not assume any condition on set of vectors and functionals to
derive Theorem 2.4. The reason is that frame operator always exists in
discrete case. To get the existence of frame operator we assumed Besselness in
Theorem 4.3. We now derive continuous versions of Theorem 2.5 and Theorem 2.7.

######  Theorem 4.4.

(First order continuous Welch bound for Banach spaces) Let $(\Omega,\mu)$ be a
$\sigma$-finite measure space and
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be a continuous Bessel family for finite dimensional Banach space
$\mathcal{X}$ of dimension $d$. If the diagonal
$\Delta\coloneqq\\{(\alpha,\alpha):\alpha\in\Omega\\}$ is measurable in the
measure space $\Omega\times\Omega$,

$\displaystyle\int_{\Omega\times\Omega}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|\,d(\mu\times\mu)(\alpha,\beta)<\infty,$

and the operator $S_{f,\tau}:\mathcal{X}\ni x\mapsto
S_{f,\tau}x\coloneqq\int_{\Omega}f_{\alpha}(x)\tau_{\alpha}\,d\mu(\alpha)\in\mathcal{X}$
is diagonalizable and its eigenvalues are all non negative, then

$\displaystyle\int_{\Omega\times\Omega}f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})\,d(\mu\times\mu)(\alpha,\beta)$
$\displaystyle=\int_{\Omega}\int_{\Omega}f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})\,d\mu(\alpha)\,d\mu(\beta)$
(11)
$\displaystyle\geq\frac{1}{d}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha)\right)^{2}=\frac{1}{\text{dim}(\mathcal{X})}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha)\right)^{2}.$

and

$\displaystyle\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|$
$\displaystyle\geq\frac{\frac{1}{d}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha)\right)^{2}-\int_{\Delta}|f_{\alpha}(\tau_{\alpha})|^{2}\,d(\mu\times\mu)(\alpha,\alpha)}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}$
$\displaystyle=\frac{\frac{1}{\text{dim}(\mathcal{X})}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha)\right)^{2}-\int_{\Delta}|f_{\alpha}(\tau_{\alpha})|^{2}\,d(\mu\times\mu)(\alpha,\alpha)}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)},$
$\displaystyle\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})|$
$\displaystyle\geq\sqrt{\frac{\frac{1}{d}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha)\right)^{2}-\int_{\Delta}|f_{\alpha}(\tau_{\alpha})|^{2}\,d(\mu\times\mu)(\alpha,\alpha)}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}}$
$\displaystyle=\sqrt{\frac{\frac{1}{\text{dim}(\mathcal{X})}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha)\right)^{2}-\int_{\Delta}|f_{\alpha}(\tau_{\alpha})|^{2}\,d(\mu\times\mu)(\alpha,\alpha)}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}}.$

Further, equality holds in Inequality (4.4) if and only if
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
is a tight continuous ASF for $\mathcal{X}$. In particular, if
$f_{\alpha}(\tau_{\alpha})=1$ for all $\alpha\in\Omega$, then

$\displaystyle\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|$
$\displaystyle\geq\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{d}-(\mu\times\mu)(\Delta)\right]$
$\displaystyle=\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{\text{dim}(\mathcal{X})}-(\mu\times\mu)(\Delta)\right].$

and we have first order continuous Welch bound for Banach spaces

$\displaystyle\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})|$
$\displaystyle\geq\sqrt{\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{d}-(\mu\times\mu)(\Delta)\right]}$
$\displaystyle=\sqrt{\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{\text{dim}(\mathcal{X})}-(\mu\times\mu)(\Delta)\right]}.$

###### Proof.

Let $\lambda_{1},\dots,\lambda_{d}$ be eigenvalues of the frame operator
$S_{f,\tau}$. Then $\lambda_{1},\dots,\lambda_{d}\geq 0$. Now using Theorem
4.3 we get

$\displaystyle\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha)\right)^{2}$
$\displaystyle=(\operatorname{Tra}(S_{f,\tau}))^{2}=\left(\sum_{k=1}^{d}\lambda_{k}\right)^{2}\leq
d\sum_{k=1}^{d}\lambda_{k}^{2}$
$\displaystyle=d\operatorname{Tra}(S^{2}_{f,\tau})=d\int_{\Omega}\int_{\Omega}f_{\beta}(\tau_{\alpha})f_{\alpha}(\tau_{\beta})\,d\mu(\alpha)\,d\mu(\beta).$

which gives the first inequality. Using Fubini’s theorem, now we get

$\displaystyle\frac{1}{d}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha)\right)^{2}$
$\displaystyle\leq\int_{\Omega}\int_{\Omega}f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})\,d\mu(\alpha)\,d\mu(\beta)=\int_{\Omega\times\Omega}f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})\,d(\mu\times\mu)(\alpha,\beta)$
$\displaystyle=\int_{\Delta}f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})\,d(\mu\times\mu)(\alpha,\beta)+\int_{(\Omega\times\Omega)\setminus\Delta}f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})\,d(\mu\times\mu)(\alpha,\beta)$
$\displaystyle=\int_{\Delta}|f_{\alpha}(\tau_{\alpha})|^{2}\,d(\mu\times\mu)(\alpha,\alpha)+\int_{(\Omega\times\Omega)\setminus\Delta}f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})\,d(\mu\times\mu)(\alpha,\beta)$
$\displaystyle\leq\int_{\Delta}|f_{\alpha}(\tau_{\alpha})|^{2}\,d(\mu\times\mu)(\alpha,\alpha)+(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|.$

∎

######  Theorem 4.5.

(Continuous Welch bounds for Banach spaces) Let $m\in\mathbb{N}$,
$(\Omega,\mu)$ be a $\sigma$-finite measure space and
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be a continuous Bessel family for finite dimensional Banach space
$\mathcal{X}$ of dimension $d$. If the diagonal
$\Delta\coloneqq\\{(\alpha,\alpha):\alpha\in\Omega\\}$ is measurable in the
measure space $\Omega\times\Omega$,

$\displaystyle\int_{\Omega\times\Omega}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{m}\,d(\mu\times\mu)(\alpha,\beta)<\infty,$

and the operator $S_{f,\tau}:\text{Sym}^{m}(\mathcal{X})\ni
x\mapsto\int_{\Omega}f_{\alpha}^{\otimes m}(x)\tau_{\alpha}^{\otimes
m}\,d\mu(\alpha)\in\text{Sym}^{m}(\mathcal{X})$ is diagonalizable and its
eigenvalues are all non negative, then

$\displaystyle\int_{\Omega\times\Omega}f_{\alpha}(\tau_{\beta})^{m}f_{\beta}(\tau_{\alpha})^{m}\,d(\mu\times\mu)(\alpha,\beta)$
$\displaystyle=\int_{\Omega}\int_{\Omega}f_{\alpha}(\tau_{\beta})^{m}f_{\beta}(\tau_{\alpha})^{m}\,d\mu(\alpha)\,d\mu(\beta)$
$\displaystyle\geq\frac{1}{{d+m-1\choose
m}}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})^{m}\,d\mu(\alpha)\right)^{2}$
(12)
$\displaystyle=\frac{1}{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})^{m}\,d\mu(\alpha)\right)^{2}.$

and

$\displaystyle\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{m}$
$\displaystyle\geq\frac{\frac{1}{{d+m-1\choose
m}}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})^{m}\,d\mu(\alpha)\right)^{2}-\int_{\Delta}|f_{\alpha}(\tau_{\alpha})|^{2m}\,d(\mu\times\mu)(\alpha,\alpha)}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}$
(13)
$\displaystyle=\frac{\frac{1}{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})^{m}\,d\mu(\alpha)\right)^{2}-\int_{\Delta}|f_{\alpha}(\tau_{\alpha})|^{2m}\,d(\mu\times\mu)(\alpha,\alpha)}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)},$
$\displaystyle\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})|^{m}$
$\displaystyle\geq\sqrt{\frac{\frac{1}{{d+m-1\choose
m}}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})^{m}\,d\mu(\alpha)\right)^{2}-\int_{\Delta}|f_{\alpha}(\tau_{\alpha})|^{2m}\,d(\mu\times\mu)(\alpha,\alpha)}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}}$
(14)
$\displaystyle=\sqrt{\frac{\frac{1}{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})^{m}\,d\mu(\alpha)\right)^{2}-\int_{\Delta}|f_{\alpha}(\tau_{\alpha})|^{2m}\,d(\mu\times\mu)(\alpha,\alpha)}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}}.$

Further, equality holds in Inequality (4.5) if and only if
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
is a tight continuous ASF for $\text{Sym}^{m}(\mathcal{X})$. In particular, if
$f_{\alpha}(\tau_{\alpha})=1$ for all $\alpha\in\Omega$, then

$\displaystyle\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{m}$
$\displaystyle\geq\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{{d+m-1\choose
m}}-(\mu\times\mu)(\Delta)\right]$
$\displaystyle=\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}-(\mu\times\mu)(\Delta)\right].$

and we have higher order continuous Welch bound for Banach spaces

$\displaystyle\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})|^{m}$
$\displaystyle\geq\sqrt{\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{{d+m-1\choose
m}}-(\mu\times\mu)(\Delta)\right]}$
$\displaystyle=\sqrt{\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}-(\mu\times\mu)(\Delta)\right]}.$

###### Proof.

Let $\lambda_{1},\dots,\lambda_{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}$ be
eigenvalues of $S_{f,\tau}$. Then

$\displaystyle\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})^{m}\,d\mu(\alpha)\right)^{2}$
$\displaystyle=\left(\int_{\Omega}f_{\alpha}^{\otimes
m}(\tau_{\alpha}^{\otimes
m})\,d\mu(\alpha)\right)^{2}=(\operatorname{Tra}(S_{f,\tau}))^{2}=\left(\sum_{l=1}^{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}\lambda_{l}\right)^{2}$
$\displaystyle\leq\text{dim}(\text{Sym}^{m}(\mathcal{X}))\sum_{l=1}^{\text{dim}(\text{Sym}^{m}(\mathcal{X}))}\lambda_{l}^{2}={d+m-1\choose
m}\operatorname{Tra}(S^{2}_{f,\tau})$ $\displaystyle={d+m-1\choose
m}\int_{\Omega}\int_{\Omega}f_{\alpha}^{\otimes m}(\tau_{\beta}^{\otimes
m})f_{\beta}^{\otimes m}(\tau_{\alpha}^{\otimes
m})\,d\mu(\alpha)\,d\mu(\beta)$ $\displaystyle={d+m-1\choose
m}\int_{\Omega}\int_{\Omega}f_{\alpha}(\tau_{\beta})^{m}f_{\beta}(\tau_{\alpha})^{m}\,d\mu(\alpha)\,d\mu(\beta).$

and

$\displaystyle\frac{1}{{d+m-1\choose
m}}\left(\int_{\Omega}f_{\alpha}(\tau_{\alpha})^{m}\,d\mu(\alpha)\right)^{2}=\int_{\Omega}\int_{\Omega}f_{\alpha}(\tau_{\beta})^{m}f_{\beta}(\tau_{\alpha})^{m}\,d\mu(\alpha)\,d\mu(\beta)$
$\displaystyle=\int_{\Omega\times\Omega}f_{\alpha}(\tau_{\beta})^{m}f_{\beta}(\tau_{\alpha})^{m}\,d(\mu\times\mu)(\alpha,\beta)$
$\displaystyle=\int_{(\Omega\times\Omega)\setminus\Delta}f_{\alpha}(\tau_{\beta})^{m}f_{\beta}(\tau_{\alpha})^{m}\,d(\mu\times\mu)(\alpha,\beta)+\int_{\Delta}f_{\alpha}(\tau_{\alpha})^{2m}\,d(\mu\times\mu)(\alpha,\alpha)$
$\displaystyle\leq\int_{(\Omega\times\Omega)\setminus\Delta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{m}\,d(\mu\times\mu)(\alpha,\beta)+\int_{\Delta}|f_{\alpha}(\tau_{\alpha})|^{2m}\,d(\mu\times\mu)(\alpha,\alpha)$
$\displaystyle\leq(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{m}+\int_{\Delta}|f_{\alpha}(\tau_{\alpha})|^{2m}\,d(\mu\times\mu)(\alpha,\alpha).$

∎

######  Corollary 4.6.

Theorem 2.7 is a corollary of Theorem 4.5.

###### Proof.

Take $\Omega=\\{1,\dots,n\\}$ and $\mu$ as the counting measure. ∎

######  Theorem 4.7.

Theorem 1.3 is a corollary of Theorem 4.5.

###### Proof.

Let $\\{\tau_{\alpha}\\}_{\alpha\in\Omega}$ be a normalized continuous Bessel
family for $\mathcal{H}$ of dimension $d$. We define

$\displaystyle f_{\alpha}:\mathcal{H}\ni h\mapsto\langle
h,\tau_{\alpha}\rangle\in\mathbb{K},\quad\forall\alpha\in\Omega$

and apply Theorem 4.5. ∎

As observed in [43], we again observe that given a measure space $\Omega$, the
diagonal $\Delta$ need not be measurable (see [26]).

######  Question 4.8.

Classify measure spaces $(\Omega,\mu)$ such that Theorem 4.5 holds? In other
words, given a measure space $(\Omega,\mu)$, does the validity of Inequality
(4.5) or Inequality (4.5) or Inequality (4.5) implies conditions on meausre
space $(\Omega,\mu)$, say $\sigma$-finite?

Now we derive continuous versions of Theorem 2.9 and Theorem 2.10.

######  Theorem 4.9.

Let $(\Omega,\mu)$ be a $\sigma$-finite measure space and
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be a continuous Bessel family for finite dimensional Banach space
$\mathcal{X}$ of dimension $d$. If the diagonal
$\Delta\coloneqq\\{(\alpha,\alpha):\alpha\in\Omega\\}$ is measurable in the
measure space $\Omega\times\Omega$ and the operator $S_{f,\tau}:\mathcal{X}\ni
x\mapsto
S_{f,\tau}x\coloneqq\int_{\Omega}f_{\alpha}(x)\tau_{\alpha}\,d\mu(\alpha)\in\mathcal{X}$
is diagonalizable and its eigenvalues are all non negative, then

$\displaystyle\frac{1}{d}\operatorname{Tra}(S_{f,\tau})^{r}\geq\left(\frac{1}{d}\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha)\right)^{r},\quad\forall
r\in[1,\infty)$

and

$\displaystyle\frac{1}{d}\operatorname{Tra}(S_{f,\tau})^{r}\leq\left(\frac{1}{d}\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha)\right)^{r},\quad\forall
r\in(0,1).$

In particular, if $f_{\alpha}(\tau_{\alpha})=1$ for all $\alpha\in\Omega$,
then

$\displaystyle\frac{1}{\mu(\Omega)}\operatorname{Tra}(S_{f,\tau})^{r}\geq\left(\frac{\mu(\Omega)}{d}\right)^{r-1},\quad\forall
r\in[1,\infty)$

and

$\displaystyle\frac{1}{\mu(\Omega)}\operatorname{Tra}(S_{f,\tau})^{r}\leq\left(\frac{\mu(\Omega)}{d}\right)^{r-1},\quad\forall
r\in(0,1).$

###### Proof.

Let $\lambda_{1},\dots,\lambda_{d}$ be eigenvalues of $S_{f,\tau}$. Let
$r\in[1,\infty)$. Jensen’s inequality gives

$\displaystyle\left(\frac{1}{d}\sum_{k=1}^{d}\lambda_{k}\right)^{r}\leq\frac{1}{d}\sum_{k=1}^{d}\lambda_{k}^{r}.$

Therefore

$\displaystyle\left(\frac{1}{d}\int_{\Omega}f_{\alpha}(\tau_{\alpha})\,d\mu(\alpha)\right)^{r}=\left(\frac{1}{d}\operatorname{Tra}(S_{f,\tau})\right)^{r}\leq\frac{1}{d}\operatorname{Tra}(S_{f,\tau})^{r}.$

Similarly the case $r\in(0,1)$ follows by using Jensen’s inequality. ∎

######  Theorem 4.10.

Let $2<p<\infty$, $(\Omega,\mu)$ be a $\sigma$-finite measure space and
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be a continuous Bessel family for finite dimensional Banach space
$\mathcal{X}$ of dimension $d$ such that $f_{\alpha}(\tau_{\alpha})=1$ for all
$\alpha\in\Omega$. If the diagonal
$\Delta\coloneqq\\{(\alpha,\alpha):\alpha\in\Omega\\}$ is measurable in the
measure space $\Omega\times\Omega$,

$\displaystyle\int_{\Omega\times\Omega}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|\,d(\mu\times\mu)(\alpha,\beta)<\infty,$

and the operator $S_{f,\tau}:\mathcal{X}\ni x\mapsto
S_{f,\tau}x\coloneqq\int_{\Omega}f_{\alpha}(x)\tau_{\alpha}\,d\mu(\alpha)\in\mathcal{X}$
is diagonalizable and its eigenvalues are all non negative, then

$\displaystyle\int_{\Omega\times\Omega}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{\frac{p}{2}}\,d(\mu\times\mu)(\alpha,\beta)$
$\displaystyle=\int_{\Omega}\int_{\Omega}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{\frac{p}{2}}\,d\mu(\alpha)\,d\mu(\beta)$
$\displaystyle\geq\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)^{\frac{p}{2}-1}}\left(\frac{\mu(\Omega)^{2}}{d}-(\mu\times\mu)(\Delta)\right)^{\frac{p}{2}}+(\mu\times\mu)(\Delta).$

###### Proof.

Define $r\coloneqq 2p/(p-2)$ and $q$ be the conjugate index of $p/2$. Then
$q=r/2$. Using Theorem 4.4 and Holder’s inequality, we have

$\displaystyle\frac{\mu(\Omega)^{2}}{d}-(\mu\times\mu)(\Delta)$
$\displaystyle\leq\int_{(\Omega\times\Omega)\setminus\Delta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|\,d(\mu\times\mu)(\alpha,\beta)$
$\displaystyle\leq\left(\int_{(\Omega\times\Omega)\setminus\Delta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{\frac{p}{2}}\,d(\mu\times\mu)(\alpha,\beta)\right)^{\frac{2}{p}}\left(\int_{(\Omega\times\Omega)\setminus\Delta}d(\mu\times\mu)(\alpha,\beta)\right)^{\frac{1}{q}}$
$\displaystyle=\left(\int_{(\Omega\times\Omega)\setminus\Delta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{\frac{p}{2}}\,d(\mu\times\mu)(\alpha,\beta)\right)^{\frac{2}{p}}(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)^{\frac{1}{q}}$
$\displaystyle=\left(\int_{(\Omega\times\Omega)\setminus\Delta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{\frac{p}{2}}\,d(\mu\times\mu)(\alpha,\beta)\right)^{\frac{2}{p}}(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)^{\frac{2}{r}}$
$\displaystyle=\left(\int_{(\Omega\times\Omega)\setminus\Delta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{\frac{p}{2}}\,d(\mu\times\mu)(\alpha,\beta)\right)^{\frac{2}{p}}(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)^{\frac{p-2}{p}}$

which gives

$\displaystyle\left(\frac{\mu(\Omega)^{2}}{d}-(\mu\times\mu)(\Delta)\right)^{\frac{p}{2}}\leq\left(\int_{(\Omega\times\Omega)\setminus\Delta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{\frac{p}{2}}\,d(\mu\times\mu)(\alpha,\beta)\right)(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)^{\frac{p}{2}-1}.$

Therefore

$\displaystyle\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)^{\frac{p}{2}-1}}\left(\frac{\mu(\Omega)^{2}}{d}-(\mu\times\mu)(\Delta)\right)^{\frac{p}{2}}+(\mu\times\mu)(\Delta)$
$\displaystyle~{}=\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)^{\frac{p}{2}-1}}\left(\frac{\mu(\Omega)^{2}}{d}-(\mu\times\mu)(\Delta)\right)^{\frac{p}{2}}+\int_{\Delta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{\frac{p}{2}}\,d(\mu\times\mu)(\alpha,\beta)$
$\displaystyle~{}\leq\int_{(\Omega\times\Omega)\setminus\Delta}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{\frac{p}{2}}\,d(\mu\times\mu)(\alpha,\beta)+\int_{\Delta}|f_{\alpha}(\tau_{\alpha})f_{\alpha}(\tau_{\alpha})|^{\frac{p}{2}}\,d(\mu\times\mu)(\alpha,\alpha)$
$\displaystyle~{}=\int_{\Omega\times\Omega}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|^{\frac{p}{2}}\,d(\mu\times\mu)(\alpha,\beta).$

∎

In the continuous case, Theorem 2.15 leads to the following problem.

######  Question 4.11.

Whether there is a continuous version of Theorem 2.15 for Banach spaces?. In
particular, does there exists a continuous version of

  1. (i)

Bukh-Cox bound for Banach spaces?

  2. (ii)

Orthoplex/Rankin bound for Banach spaces?

  3. (iii)

Levenstein bound for Banach spaces?

  4. (iv)

Exponential bound for Banach spaces?

##  5\. Applications of continuous Welch bounds for Banach spaces

Here we list continuous versions of corresponding concepts, results and open
problems stated in Section 3. Throughout this section, $(\Omega,\mu)$ is a
$\sigma$-finite measure space. We further assume that the diagonal $\Delta$ is
measurable.

######  Definition 5.1.

Let
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be a continuous Bessel family for a finite dimensional Banach space
$\mathcal{X}$ of dimension $d$ satisfying $f_{\alpha}(\tau_{\alpha})=1$ for
all $\alpha\in\Omega$. Assume that the frame operator $S_{f,\tau}$ is
diagonalizable and its eigenvalues are all non negative. Also assume that

$\displaystyle\int_{\Omega\times\Omega}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|\,d(\mu\times\mu)(\alpha,\beta)<\infty.$

We define the  continuous root-mean-square (RMS) absolute cross relation of
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
as

$\displaystyle
I_{\text{CRMS}}(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})\coloneqq\left(\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\int_{(\Omega\times\Omega)\setminus\Delta}f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})\,d(\mu\times\mu)(\alpha,\beta)\right)^{\frac{1}{2}}.$

######  Theorem 5.2.

Let
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be as in Definition 5.1. Then

$\displaystyle\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})|\geq
I_{\text{CRMS}}(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})\geq\left(\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{d}-(\mu\times\mu)(\Delta)\right]\right)^{\frac{1}{2}}.$

######  Definition 5.3.

Let
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be a continuous Bessel family for a finite dimensional Banach space
$\mathcal{X}$ of dimension $d$ satisfying $f_{\alpha}(\tau_{\alpha})=1$ for
all $\alpha\in\Omega$. Assume that the frame operator $S_{f,\tau}$ is
diagonalizable and its eigenvalues are all non negative. Also assume that

$\displaystyle\int_{\Omega\times\Omega}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|\,d(\mu\times\mu)(\alpha,\beta)<\infty.$

We define the continuous pseudo frame potential of
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
as

$\displaystyle
CPFP(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})\coloneqq\int_{\Omega\times\Omega}f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})\,d(\mu\times\mu)(\alpha,\beta).$

######  Theorem 5.4.

Let
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be as in Definition 5.3. Then

(15)
$\displaystyle\mu(\Omega)^{2}\sup_{\alpha,\beta\in\Omega}|f_{\alpha}(\tau_{\beta})|^{2}\geq
CPFP(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})\geq\frac{\mu(\Omega)^{2}}{d}.$

######  Question 5.5.

Is there a characterization of
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
by using equality in Inequality (15)?

######  Definition 5.6.

Let
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be a continuous ASF for $\mathcal{X}$ satisfying $f_{\alpha}(\tau_{\alpha})=1$
for all $\alpha\in\Omega$. Assume that the frame operator $S_{f,\tau}$ is
diagonalizable and its eigenvalues are all non negative. Also assume that

$\displaystyle\int_{\Omega\times\Omega}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|\,d(\mu\times\mu)(\alpha,\beta)<\infty.$

We define the  continuous frame correlation of
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
as

$\displaystyle\mathcal{M}(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})\coloneqq\sup_{\alpha,\beta\in\Omega,\alpha\neq\beta}|f_{\alpha}(\tau_{\beta})|.$

######  Definition 5.7.

Let
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be a continuous ASF for $\mathcal{X}$ satisfying
$\|f_{\alpha}\|=1,\|\tau_{\alpha}\|=1,f_{\alpha}(\tau_{\alpha})=1$ for all
$\alpha\in\Omega$. Assume that the frame operator $S_{f,\tau}$ is
diagonalizable and its eigenvalues are all non negative. Also assume that

$\displaystyle\int_{\Omega\times\Omega}|f_{\alpha}(\tau_{\beta})f_{\beta}(\tau_{\alpha})|\,d(\mu\times\mu)(\alpha,\beta)<\infty.$

Continuous ASF
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
is said to be a continuous Grassmannian frame for $\mathcal{X}$ if

$\displaystyle\mathcal{M}(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})=\inf\bigg{\\{}$
$\displaystyle\mathcal{M}(\\{g_{\alpha}\\}_{\alpha\in\Omega},\\{\omega_{\alpha}\\}_{\alpha\in\Omega}):(\\{g_{\alpha}\\}_{\alpha\in\Omega},\\{\omega_{\alpha}\\}_{\alpha\in\Omega})\text{
is a continuous ASF for }\mathcal{X}$ $\displaystyle\text{satisfying
}\|g_{\alpha}\|=1,\|\omega_{\alpha}\|=1,g_{\alpha}(\omega_{\alpha})=1,\forall\alpha\in\Omega\text{,
the frame operator }S_{g,\omega}$ is diagonalizable and its eigenvalues are
all non negative and
$\displaystyle\int_{\Omega\times\Omega}|g_{\alpha}(\omega_{\beta})g_{\beta}(\omega_{\alpha})|\,d(\mu\times\mu)(\alpha,\beta)<\infty\bigg{\\}}.$

######  Question 5.8.

Classify measure spaces and (finite dimensional) Banach spaces so that
continuous Grassmannian ASFs exist.

######  Definition 5.9.

Let
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be a continuous ASF for $\mathcal{X}$ satisfying $f_{\alpha}(\tau_{\alpha})=1$
for all $\alpha\in\Omega$. Assume that the frame operator $S_{f,\tau}$ is
diagonalizable and its eigenvalues are all non negative. Continuous ASF
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
is said to be $\gamma$-equiangular if there exists $\gamma\geq 0$ such that

$\displaystyle|f_{\alpha}(\tau_{\beta})|=\gamma,\quad\forall\alpha,\beta\in\Omega,\alpha\neq\beta.$

######  Conjecture 5.10.

(Continuous Zauner’s conjecture for Banach spaces) For a given measure space
$(\Omega,\mu)$ and for every $d\in\mathbb{N}$, there exists a
$\gamma$-equiangular continuous ASF
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
for $\mathbb{C}^{d}$ such that $\mu(\Omega)=d^{2}$.

######  Theorem 5.11.

Let
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
be as in Definition 5.7. Then

(16)
$\displaystyle\mathcal{M}(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})\geq\left(\frac{1}{(\mu\times\mu)((\Omega\times\Omega)\setminus\Delta)}\left[\frac{\mu(\Omega)^{2}}{d}-(\mu\times\mu)(\Delta)\right]\right)^{\frac{1}{2}}\eqqcolon\gamma.$

If the continuous ASF is $\gamma$-equiangular, then we have equality in
Inequality (16).

######  Question 5.12.

  1. (i)

Whether the equality in Inequality (16) implies that the continuous ASF
$(\\{f_{\alpha}\\}_{\alpha\in\Omega},\\{\tau_{\alpha}\\}_{\alpha\in\Omega})$
is $\gamma$-equiangular?

  2. (ii)

Whether the validity of Inequality (16) implies there is a relation between
the measure of $\Omega$ and dimension of the space (like Theorem 2.3 in [61])?

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