Functional Donoho-Elad-Gribonval-Nielsen-Fuchs Sparsity Theorem

1. Introduction

Let $\mathcal{H}$ be a finite dimensional Hilbert space over $\mathbb{K}$ ($\mathbb{C}$ or $\mathbb{R}$). Recall that [2,28] a finite collection ${\left\{{\tau }_j\right\}}_{j=1}^n$ in $\mathcal{H}$ is said to be a frame (also known as dictionary) for $\mathcal{H}$ if it spans $\mathcal{H}$. A frame ${\left\{{\tau }_j\right\}}_{j=1}^n$ for $\mathcal{H}$ is said to be normalized if $\parallel {\tau }_j\parallel =1$ for all $1\le j\le n$. Given a frame ${\left\{{\tau }_j\right\}}_{j=1}^n$ for $\mathcal{H}$, we define the analysis operator

$$\begin{array}{c}\hfill {\theta }_{\tau }:\mathcal{H}\ni h\mapsto {\theta }_{\tau }h{\left(\langle h,{\tau }_j\rangle \right)}_{j=1}^n\in {\mathbb{K}}^n.\end{array}$$

Adjoint of the analysis operator is known as the synthesis operator whose equation is

$$\begin{array}{c}\hfill {\theta }_{\tau }^{*}:{\mathbb{K}}^n\ni {\left(a_j\right)}_{j=1}^n\mapsto {\theta }_{\tau }^{*}{\left(a_j\right)}_{j=1}^n\sum _{j=1}^n{a}_j{\tau }_j\in \mathcal{H}.\end{array}$$

Given $d\in {\mathbb{K}}^n$, let ${\parallel d\parallel }_0$ be the number of nonzero entries in d. Central problem which occurs in everyday life is the following ${\mathcal{l}}_0$-minimization problem:

Problem 1.1. 

Let ${\left\{{\tau }_j\right\}}_{j=1}^n$ be a frame for $\mathcal{H}$. Given $h\in \mathcal{H}$, solve

$$\begin{array}{c}\hfill \underset{d\in {\mathbb{K}}^n}{minimize}{\parallel d\parallel }_0\mathrm{subject}\mathrm{to}{\theta }_{\tau }^{*}d=h.\end{array}$$

Recall that $c\in {\mathbb{K}}^n$ is said to be a unique solution to Problem 1.1 if it satisfies following two conditions.

(i)

${\theta }_{\tau }^{*}c=h$.

(ii)

If $d\in {\mathbb{K}}^n$ satisfies ${\theta }_{\tau }^{*}d=h$, then

$$\begin{array}{c}\hfill {\parallel d\parallel }_0>{\parallel c\parallel }_0.\end{array}$$

Unfortunately, in 1995, Natarajan showed that Problem 1.1 is NP-Hard [23,33]. Therefore solution to Problem 1.1 has to be obtained using other means. Entire body of work which is built around Problem 1.1 is known as sparseland (term due to Elad [19]) or compressive sensing or compressed sensing [1,4,5,6,7,8,9,10,14,15,16,17,18,19,20,21,22,23,27,32,34,37,38]. We note that as the operator ${\theta }_{\tau }^{*}$ is surjective, for a given $h\in \mathcal{H}$, there is always $d\in {\mathbb{K}}^n$ such that ${\theta }_{\tau }^{*}d=h.$ Thus the central problem is when solution to Problem 1.1 is unique. It is well-known that (see [3,13,18]) following problem is the closest convex relaxation problem to Problem 1.1.

Problem 1.2. 

Let ${\left\{{\tau }_j\right\}}_{j=1}^n$ be a frame for $\mathcal{H}$. Given $h\in \mathcal{H}$, solve

$$\begin{array}{c}\hfill \underset{d\in {\mathbb{K}}^n}{minimize}{\parallel d\parallel }_1\mathrm{subject}\mathrm{to}{\theta }_{\tau }^{*}d=h.\end{array}$$

There are several linear programmings available to obtain solution of Problem 1.2 and it is a P-problem [35,36,39].

Most important result which shows by solving Problem 1.2 we also get a solution to Problem 1.1, obtained independently by Donoho and Elad [17] and Gribonval and Nielsen [27] and Fuchs [25,26], is the following.

Theorem 1.3. 

[17,19,25,26,27,31] (Donoho-Elad-Gribonval-Nielsen-Fuchs Sparsity Theorem) Let ${\left\{{\tau }_j\right\}}_{j=1}^n$ be a normalized frame for a Hilbert space $\mathcal{H}$. If $h\in \mathcal{H}$ can be written as $h={\theta }_{\tau }^{*}c$ for some $c\in {\mathbb{K}}^n$ satisfying

$$\begin{array}{c}\hfill {\parallel c\parallel }_0<{\displaystyle \frac{1}{2}}\left(1+{\displaystyle \frac{1}{{\displaystyle \underset{1\le j,k\le n,j\ne k}{max}\left|\langle {\tau }_j,{\tau }_k\rangle \right|}}}\right),\end{array}$$

then c is the unique solution to Problem 1.2 and Problem 1.1.

We naturally ask for (both finite and infinite dimensional) Banach space version Theorem 1.3. More than this natural question, many spaces occurring in functional analysis and in applications are Banach and there is no Hilbert space structure associated with them. As frame theory for Hilbert spaces has been successfully extended to Banach spaces which also found applications, we believe that generalization of Theorem 1.3 will have applications. It is interesting to note that a noncommutative version of Theorem 1.3 has been recently derived [29].

2. Functional Donoho-Elad-Gribonval-Nielsen-Fuchs Sparsity Theorem

In the paper, $\mathbb{K}$ denotes $\mathbb{C}$ or $\mathbb{R}$ and $\mathcal{X}$ denotes a Banach space (need not be finite dimensional) over $\mathbb{K}$. Dual of $\mathcal{X}$ is denoted by ${\mathcal{X}}^{*}$. We need the notion of 1-approximate Schauder frames for Banach spaces which is a subclass of Schauder frames [11,12,24].

Definition 2.1. 

[30] Let $\mathcal{X}$ be a Banach space over $\mathbb{K}$. Let ${\left\{f_n\right\}}_{n=1}^{\infty }$ be a sequence in ${\mathcal{X}}^{*}$ and ${\left\{{\tau }_n\right\}}_{n=1}^{\infty }$ be a sequence in $\mathcal{X}$. The pair $({\left\{f_n\right\}}_{n=1}^{\infty },{\left\{{\tau }_n\right\}}_{n=1}^{\infty })$ is said to be a1-approximate Schauder frame(we write 1-ASF) for $\mathcal{X}$ if the following conditions are satisfied.

(i) 

The map (analysis operator)

$$\begin{array}{c}\hfill {\theta }_f:\mathcal{X}\ni x\mapsto {\theta }_{f}x:={\left\{f_n\left(x\right)\right\}}_{n=1}^{\infty }\in {\mathcal{l}}^1\left(\mathbb{N}\right)\end{array}$$

is a well-defined bounded linear operator.

(ii) 

The map (synthesis operator)

$$\begin{array}{c}\hfill {\theta }_{\tau }:{\mathcal{l}}^1\left(\mathbb{N}\right)\ni {\left\{a\right\}}_{n=1}^{\infty }\mapsto {\theta }_{\tau }{\left\{a\right\}}_{n=1}^{\infty }:=\sum _{n=1}^{\infty }{a}_n{\tau }_n\in \mathcal{X}\end{array}$$

is a well-defined bounded linear operator.

1. 

The map (frame operator)

$$\begin{array}{c}\hfill S_{f,\tau }:\mathcal{X}\ni x\mapsto S_{f,\tau }x:=\sum _{n=1}^{\infty }{f}_n\left(x\right){\tau }_n\in \mathcal{X}\end{array}$$

is a well-defined bounded invertible operator.

We the notion of 1-ASF, we generalize Problems 1.1 and 1.2.

Problem 2.2. 

Let $({\left\{f_n\right\}}_{n=1}^{\infty },{\left\{{\tau }_n\right\}}_{n=1}^{\infty })$ be an 1-ASF for $\mathcal{X}$. Given $x\in \mathcal{X}$, solve

$$\begin{array}{c}\hfill \underset{d\in {\mathcal{l}}^1\left(\mathbb{N}\right)}{minimize}{\parallel d\parallel }_0\mathit{subject}\mathit{to}{\theta }_{\tau }d=x.\end{array}$$

Problem 2.3. 

Let $({\left\{f_n\right\}}_{n=1}^{\infty },{\left\{{\tau }_n\right\}}_{n=1}^{\infty })$ be an 1-ASF for $\mathcal{X}$. Given $x\in \mathcal{X}$, solve

$$\begin{array}{c}\hfill \underset{d\in {\mathcal{l}}^1\left(\mathbb{N}\right)}{minimize}{\parallel d\parallel }_1\mathit{subject}\mathit{to}{\theta }_{\tau }d=x.\end{array}$$

A very important property used to show Theorem 1.3 is the notion of null space property (see [14,31]). We now define the same property for Banach spaces. We use following notations. Let ${\left\{e_n\right\}}_{n=1}^{\infty }$ be the canonical Schauder basis for ${\mathcal{l}}^1\left(\mathbb{N}\right)$. Given $M\subseteq \mathbb{N}$ and $d={\left\{d_n\right\}}_{n=1}^{\infty }\in {\mathcal{l}}^1\left(\mathbb{N}\right)$, we define

$$\begin{array}{c}\hfill d_M:=\sum _{n\in M}{d}_n{e}_n.\end{array}$$

Definition 2.4. 

An 1-ASF $({\left\{f_n\right\}}_{n=1}^{\infty },{\left\{{\tau }_n\right\}}_{n=1}^{\infty })$ for $\mathcal{X}$ is said to have thenull space property(we write NSP) of order $k\in \mathbb{N}$ if for every $M\subseteq \mathbb{N}$ with $o\left(M\right)\le k$, we have

$$\begin{array}{c}\hfill {\parallel d_M\parallel }_1<{\displaystyle \frac{1}{2}}{\parallel d\parallel }_1,\forall d\in ker\left({\theta }_{\tau }\right),d\ne 0.\end{array}$$

Following characterization relates NSP with Problem 2.3.

Theorem 2.5. 

Let $({\left\{f_n\right\}}_{n=1}^{\infty },{\left\{{\tau }_n\right\}}_{n=1}^{\infty })$ be an 1-ASF for $\mathcal{X}$ and let $k\in \mathbb{N}$. The following are equivalent.

(i) 

If $x\in \mathcal{X}$ can be written as $x={\theta }_{\tau }c$ for some $c\in {\mathcal{l}}^1\left(\mathbb{N}\right)$ satisfying ${\parallel c\parallel }_0\le k$, then c is the unique solution to Problem 2.3.

(ii) 

$({\left\{f_n\right\}}_{n=1}^{\infty },{\left\{{\tau }_n\right\}}_{n=1}^{\infty })$ satisfies the NSP of order k.

Proof. 

(i)

⇒ (ii) Let $M\subseteq \mathbb{N}$ with $o\left(M\right)\le k$ and let $d\in ker\left({\theta }_{\tau }\right),d\ne 0$. Then we have

$$\begin{array}{c}\hfill 0={\theta }_{\tau }d={\theta }_{\tau }(d_M+d_{M^c})={\theta }_{\tau }\left(d_M\right)+{\theta }_{\tau }\left(d_{M^c}\right)\end{array}$$

which gives

$$\begin{array}{c}\hfill {\theta }_{\tau }\left(d_M\right)={\theta }_{\tau }(-d_{M^c}).\end{array}$$

Define $c:=d_M\in {\mathcal{l}}^1\left(\mathbb{N}\right)$ and $x{\theta }_{\tau }\left(d_M\right)$. Then we have ${\parallel c\parallel }_0\le o\left(M\right)\le k$ and

$$\begin{array}{c}\hfill x={\theta }_{\tau }c={\theta }_{\tau }(-d_{M^c}).\end{array}$$

By assumption (i), we then have

$$\begin{array}{c}\hfill {\parallel c\parallel }_1=\parallel d_M{\parallel }_1<\parallel -d_{M^c}{\parallel }_1={\parallel d_{M^c}\parallel }_1.\end{array}$$

Rewriting previous inequality gives

$$\begin{array}{c}\hfill {\parallel d_M\parallel }_1{<\parallel d\parallel }_1-\parallel d_M{\parallel }_1\Rightarrow \parallel d_M{\parallel }_1<{\displaystyle \frac{1}{2}}{\parallel d\parallel }_1.\end{array}$$

Hence $({\left\{f_n\right\}}_{n=1}^{\infty },{\left\{{\tau }_n\right\}}_{n=1}^{\infty })$ satisfies the NSP of order k.

(ii)

⇒ (i) Let $x\in \mathcal{X}$ can be written as $x={\theta }_{\tau }c$ for some $c\in {\mathcal{l}}^1\left(\mathbb{N}\right)$ satisfying ${\parallel c\parallel }_0\le k$. Define $M:=supp\left(c\right)$. Then $o\left(M\right)={\parallel c\parallel }_0\le k$. By assumption (ii), we then have

$$\begin{array}{c}\hfill {\parallel d_M\parallel }_1<{\displaystyle \frac{1}{2}}{\parallel d\parallel }_1,\forall d\in ker\left({\theta }_{\tau }\right),d\ne 0.\end{array}$$

(1)

Let $b\in {\mathcal{l}}^1\left(\mathbb{N}\right)$ be such that $x={\theta }_{\tau }b$ and $b\ne c$. Define $a:=b-c\in {\mathcal{l}}^1\left(\mathbb{N}\right)$. Then ${\theta }_{\tau }a={\theta }_{\tau }b-{\theta }_{\tau }c=x-x=0$ and hence $a\in ker\left({\theta }_{\tau }\right),a\ne 0$. Using Inequality (1), we get

$$\begin{array}{c}\hfill {\parallel a_M\parallel }_1<{\displaystyle \frac{1}{2}}{\parallel a\parallel }_1\Rightarrow \parallel a_M{\parallel }_1<{\displaystyle \frac{1}{2}}(\parallel a_M{\parallel }_1+\parallel a_{M^c}{\parallel }_1)\Rightarrow \parallel a_M{\parallel }_1<{\parallel a_{M^c}\parallel }_1.\end{array}$$

(2)

Using Inequality (2) and the information that c is supported on M, we get

$$\begin{array}{cc}\hfill {\parallel b\parallel }_1-{\parallel c\parallel }_1& =\parallel b_M{\parallel }_1+\parallel b_{M^c}{\parallel }_1-\parallel c_M{\parallel }_1-\parallel c_{M^c}{\parallel }_1=\parallel b_M{\parallel }_1+\parallel b_{M^c}{\parallel }_1-{\parallel c_M\parallel }_1\hfill \\ \hfill & =\parallel b_M{\parallel }_1{+\parallel (b-c)}_{M^c}{\parallel }_1-\parallel c_M{\parallel }_1=\parallel b_M{\parallel }_1+\parallel a_{M^c}{\parallel }_1-{\parallel c_M\parallel }_1\hfill \\ \hfill & >\parallel b_M{\parallel }_1+\parallel a_M{\parallel }_1-\parallel c_M{\parallel }_1\ge \parallel b_M{\parallel }_1{+\parallel (b-c)}_M{\parallel }_1-{\parallel c_M\parallel }_1\hfill \\ \hfill & \ge \parallel b_M{\parallel }_1-\parallel b_M{\parallel }_1+\parallel c_M{\parallel }_1-{\parallel c_M\parallel }_1=0.\hfill \end{array}$$

Hence c is the unique solution to Problem 2.3.

□

Using Theorem 2.5 we obtain Banach space version of Theorem 1.3. We do this by relating Problem 2.3 to Theorem 2.5 and then Problem 2.2 to Theorem 2.5.

Theorem 2.6. 

Let $({\left\{f_n\right\}}_{n=1}^{\infty },{\left\{{\tau }_n\right\}}_{n=1}^{\infty })$ be an 1-ASF for $\mathcal{X}$ such that

$$\begin{array}{c}\hfill |f_n\left({\tau }_n\right)|\ge 1,\forall n\in \mathbb{N}.\end{array}$$

(3)

If $x\in \mathcal{X}$ can be written as $x={\theta }_{\tau }c$ for some $c\in {\mathcal{l}}^1\left(\mathbb{N}\right)$ satisfying

$$\begin{array}{c}\hfill {\parallel c\parallel }_0<{\displaystyle \frac{1}{2}}\left(1+{\displaystyle \frac{1}{{\displaystyle \underset{n,m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}}}\right),\end{array}$$

(4)

then c is the unique solution to Problem 2.3.

Proof. 

We show that $({\left\{f_n\right\}}_{n=1}^{\infty },{\left\{{\tau }_n\right\}}_{n=1}^{\infty })$ satisfies the NSP of order ${k:=\parallel c\parallel }_0$. Then Theorem 2.5 says that c is the unique solution to Problem 2.3. Let $x\in \mathcal{X}$ can be written as $x={\theta }_{\tau }c$ for some $c\in {\mathcal{l}}^1\left(\mathbb{N}\right)$ satisfying ${\parallel c\parallel }_0\le k$. Let $M\subseteq \mathbb{N}$ with $o\left(M\right)\le k$ and let $d\in ker\left({\theta }_{\tau }\right),d\ne 0$. Then we have

$$\begin{array}{c}\hfill {\theta }_f{\theta }_{\tau }d=0.\end{array}$$

By writing $d={\left\{d_n\right\}}_{n=1}^{\infty }\in {\mathcal{l}}^1\left(\mathbb{N}\right)$, above equation gives

$$\begin{array}{cc}\hfill 0& ={\theta }_f{\theta }_{\tau }{\left\{d_m\right\}}_{m=1}^{\infty }={\theta }_f\left(\sum _{m=1}^{\infty }{d}_m{\theta }_{\tau }{e}_m\right)\hfill \\ \hfill & ={\theta }_f\left(\sum _{m=1}^{\infty }{d}_m{\tau }_m\right)=\sum _{m=1}^{\infty }{d}_m{\theta }_f\left({\tau }_m\right)=\sum _{m=1}^{\infty }{d}_m\sum _{k=1}^{\infty }{f}_k\left({\tau }_m\right)e_k.\hfill \end{array}$$

Let ${\left\{{\zeta }_n\right\}}_{n=1}^{\infty }$ be the coordinate functionals associated with the canonical Schauder basis ${\left\{e_n\right\}}_{n=1}^{\infty }$ for ${\mathcal{l}}^1\left(\mathbb{N}\right)$. Let $n\in \mathbb{N}$. By evaluating previous equation at ${\zeta }_n$, we get

$$\begin{array}{cc}\hfill 0& ={\zeta }_n\left(\sum _{m=1}^{\infty }{d}_m\sum _{k=1}^{\infty }{f}_k\left({\tau }_m\right)e_k\right)=\sum _{m=1}^{\infty }{d}_m\sum _{k=1}^{\infty }{f}_k\left({\tau }_m\right){\zeta }_n\left(e_k\right)\hfill \\ \hfill & =\sum _{m=1}^{\infty }{d}_m{f}_n\left({\tau }_m\right)=d_n{f}_n\left({\tau }_n\right)+\sum _{m=1,m\ne n}^{\infty }{d}_m{f}_n\left({\tau }_m\right).\hfill \end{array}$$

Therefore

$$\begin{array}{c}\hfill d_n{f}_n\left({\tau }_n\right)=-\sum _{m=1,m\ne n}^{\infty }{d}_m{f}_n\left({\tau }_m\right),\forall n\in \mathbb{N}.\end{array}$$

By using Inequality (3),

$$\begin{array}{cc}\hfill |d_n|& \le |d_n\left|\right|f_n\left({\tau }_n\right)|=\left|-\sum _{m=1,m\ne n}^{\infty }{d}_m{f}_n\left({\tau }_m\right)\right|\hfill \\ \hfill & \le \sum _{m=1,m\ne n}^{\infty }|d_m{f}_n\left({\tau }_m\right)|\le \left({\displaystyle \underset{m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}\right)\sum _{m=1,m\ne n}^{\infty }\left|d_m\right|\hfill \\ \hfill & \le \left({\displaystyle \underset{n,m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}\right)\sum _{m=1,m\ne n}^{\infty }\left|d_m\right|=\left({\displaystyle \underset{n,m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}\right)\left(\sum _{m=1}^{\infty }|d_m|-|d_n|\right)\hfill \\ \hfill & =\left({\displaystyle \underset{n,m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}\right){(\parallel d\parallel }_1-|d_n\left|\right),\forall n\in \mathbb{N}.\hfill \end{array}$$

By rewriting above inequality we get

$$\begin{array}{c}\hfill \left(1+{\displaystyle \frac{1}{{\displaystyle \underset{n,m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}}}\right)|d_n{|\le \parallel d\parallel }_1,\forall n\in \mathbb{N}.\end{array}$$

(5)

Summing Inequality (5) over M leads to

$$\begin{array}{cc}\hfill \left(1+{\displaystyle \frac{1}{{\displaystyle \underset{n,m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}}}\right){\parallel d_M\parallel }_1& =\left(1+{\displaystyle \frac{1}{{\displaystyle \underset{n,m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}}}\right)\sum _{n\in M}\left|d_n\right|\hfill \\ \hfill & \le {\parallel d\parallel }_1\sum _{n\in M}1={\parallel d\parallel }_{1}o\left(M\right).\hfill \end{array}$$

Finally using Inequality (4)

$$\begin{array}{cc}\hfill \parallel d_M{\parallel }_1& \le {\left(1+{\displaystyle \frac{1}{{\displaystyle \underset{n,m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}}}\right)}^{-1}{\parallel d\parallel }_{1}o\left(M\right)\le {\left(1+{\displaystyle \frac{1}{{\displaystyle \underset{n,m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}}}\right)}^{-1}{\parallel d\parallel }_{1}k\hfill \\ \hfill & ={\left(1+{\displaystyle \frac{1}{{\displaystyle \underset{n,m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}}}\right)}^{-1}{\parallel d\parallel }_1{\parallel c\parallel }_0<{\displaystyle \frac{1}{2}}{\parallel d\parallel }_1.\hfill \end{array}$$

Hence $({\left\{f_n\right\}}_{n=1}^{\infty },{\left\{{\tau }_n\right\}}_{n=1}^{\infty })$ satisfies the NSP of order k which completes the proof. □

Theorem 2.7.(Functional Donoho-Elad-Gribonval-Nielsen-Fuchs Sparsity Theorem)

Let $({\left\{f_n\right\}}_{n=1}^{\infty },{\left\{{\tau }_n\right\}}_{n=1}^{\infty })$ be an 1-ASF for $\mathcal{X}$ such that

$$\begin{array}{c}\hfill |f_n\left({\tau }_n\right)|\ge 1,\forall n\in \mathbb{N}.\end{array}$$

If $x\in \mathcal{X}$ can be written as $x={\theta }_{\tau }c$ for some $c\in {\mathcal{l}}^1\left(\mathbb{N}\right)$ satisfying

$$\begin{array}{c}\hfill {\parallel c\parallel }_0<{\displaystyle \frac{1}{2}}\left(1+{\displaystyle \frac{1}{{\displaystyle \underset{n,m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}}}\right),\end{array}$$

then c is the unique solution to Problem 2.2.

Proof. 

Theorem 2.6 says that c is the unique solution to Problem 2.3. Let $d\in {\mathcal{l}}^1\left(\mathbb{N}\right)$ be such that $x={\theta }_{\tau }d$. We claim that ${\parallel d\parallel }_0>{\parallel c\parallel }_0$. If this fails, we must have ${\parallel d\parallel }_0\le {\parallel c\parallel }_0$. We then have

$$\begin{array}{c}\hfill {\parallel d\parallel }_0<{\displaystyle \frac{1}{2}}\left(1+{\displaystyle \frac{1}{{\displaystyle \underset{n,m\in \mathbb{N},n\ne m}{sup}\left|f_n\left({\tau }_m\right)\right|}}}\right).\end{array}$$

Theorem 2.6 again says that d is also the unique solution to Problem 2.3. Therefore we must have ${\parallel c\parallel }_1<{\parallel d\parallel }_1$ and ${\parallel c\parallel }_1>{\parallel d\parallel }_1$ which is a contradiction. Therefore claim holds and we have ${\parallel d\parallel }_0>{\parallel c\parallel }_0$. □

Corollary 2.8. 

Theorem 1.3 follows from Theorems 2.6 and 2.7.

Proof. 

Let ${\left\{{\tau }_j\right\}}_{j=1}^n$ be a normalized frame for a Hilbert space $\mathcal{H}$. For each $1\le j\le n$, define

$$\begin{array}{c}\hfill f_j:\mathcal{H}\ni h\mapsto f_j\left(h\right):=\langle h,{\tau }_j\rangle \in \mathbb{K}.\end{array}$$

Then

$$\begin{array}{c}\hfill |f_j\left({\tau }_j\right)|=1,\forall 1\le j\le n\end{array}$$

and

$$\begin{array}{c}\hfill f_k\left({\tau }_j\right)=\langle {\tau }_j,{\tau }_k\rangle ,\forall 1\le j,k\le n.\end{array}$$

□