On some equalities and inequalities for $K$-frames

TL;DR

This paper extends known equalities and inequalities from ordinary frames to K-frames in Hilbert spaces, providing new mathematical tools relevant for sampling theory and operator analysis.

Contribution

It introduces several new equalities and inequalities for K-frames, expanding the theoretical framework and applying Jensen's operator inequality for further results.

Findings

New equalities for K-frames established

Novel inequalities derived using Jensen's operator inequality

Enhanced understanding of K-frame properties in Hilbert spaces

Abstract

K-frame theory was recently introduced to reconstruct elements from the range of a bounded linear operator K in a separable Hilbert space. This significant property is worthwhile especially in some problems arising in sampling theory. Some equalities and inequalities have been established for ordinary frames and their duals. In this paper, we continue and extend these results to obtain several important equalities and inequalities for K-frames. Moreover, by applying Jensen's operator inequality we obtain some new inequalities for $K$-frames.

Equations

$$A\Vert K^{\ast }f{\Vert }^2\le \sum _{i\in I}|⟨f,f_i⟩{|}^2\le B\Vert f{\Vert }^2,(f\in \mathcal{H}).$$

$$Kf=\sum _{i\in I}⟨f,g_i⟩f_i,(f\in \mathcal{H}).$$

$$T_{F}X=K,$$

$$\Vert X_F{\Vert }^2=\inf \{\alpha >0,K{K}^{\ast }\le \alpha T_F{T}_F^{\ast }\}.$$

$$A$$

$$\left(\sum _{i\in J}⟨f,g_i⟩\overline{⟨Kf,f_i⟩}\right)-{\Vert \sum _{i\in J}⟨f,g_i⟩f_i\Vert }^2=\left(\sum _{i\in J^c}\overline{⟨f,g_i}⟩⟨Kf,f_i⟩\right)-{\Vert \sum _{i\in J^c}⟨f,g_i⟩f_i\Vert }^2.$$

$$M_{J}f=\sum _{i\in J}⟨f,g_i⟩f_i,(f\in \mathcal{H}).$$

$$\left(\sum _{i\in J}⟨f,g_i⟩\overline{⟨Kf,f_i⟩}\right)-{\Vert \sum _{i\in J}⟨f,g_i⟩f_i\Vert }^2$$

$$⟨T_J\{(X_{F}f{)}_i{\}}_{i\in J},Kf⟩-\Vert T_J\{(X_{F}f{)}_i{\}}_{i\in J}{\Vert }^2$$

$$\left(\sum _{i\in I}{\alpha }_i⟨f,g_i⟩\overline{⟨Kf,f_i⟩}\right)-{\Vert \sum _{i\in I}{\alpha }_i⟨f,g_i⟩f_i\Vert }^2$$

$${\Vert \sum _{i\in J\cup E}⟨f,f_i⟩f_i\Vert }^2-{\Vert \sum _{i\in J^c\setminus E}⟨f,f_i⟩f_i\Vert }^2$$

$$S_J^2-S_{J^c}^2$$

$$\Vert S_{J}f{\Vert }^2-\Vert S_{J^c}f{\Vert }^2=⟨K{K}^{\ast }{S}_{J}f,f⟩-⟨S_{J^c}K{K}^{\ast }f,f⟩,$$

$${\Vert \sum _{i\in J\cup E}⟨f,f_i⟩f_i\Vert }^2-{\Vert \sum _{i\in J^c\setminus E}⟨f,f_i⟩f_i\Vert }^2$$

$$Re\left(\sum _{i\in J^c}⟨f,f_i⟩\overline{⟨K{K}^{\ast }f,f_i⟩}\right)+{\Vert \sum _{i\in J}⟨f,f_i⟩f_i\Vert }^2$$

$$S_J^2+S_{J^c}^2=2{\left(\frac{K{K}^{\ast }}{2}-S_J\right)}^2+\frac{(K{K}^{\ast }{)}^2}{2}\ge \frac{(K{K}^{\ast }{)}^2}{2},$$

$$K{K}^{\ast }{S}_J+S_{J^c}^2+(K{K}^{\ast }{S}_J+S_{J^c}^2{)}^{\ast }$$

$$Re\left(\sum _{i\in J^c}⟨f,f_i⟩\overline{⟨K{K}^{\ast }f,f_i⟩}\right)+{\Vert \sum _{i\in J}⟨f,f_i⟩f_i\Vert }^2$$

$$v_{+}(F,K,J)=\underset{f\ne 0}{\sup }\frac{Re\left({\sum }_{i\in J^c}\overline{⟨f,f_i⟩}⟨K{K}^{\ast }f,f_i⟩\right)+{\Vert {\sum }_{i\in J}⟨f,f_i⟩f_i\Vert }^2}{\Vert K{K}^{\ast }f{\Vert }^2},$$

$$v_{-}(F,K,J)=\underset{f\ne 0}{\inf }\frac{Re\left({\sum }_{i\in J^c}\overline{⟨f,f_i⟩}⟨K{K}^{\ast }f,f_i⟩\right)+{\Vert {\sum }_{i\in J}⟨f,f_i⟩f_i\Vert }^2}{\Vert K{K}^{\ast }f{\Vert }^2}.$$

$${\left(\sum _{i\in I}|⟨f,f_i⟩{|}^2\right)}^2$$

$${\Vert \sum _{i\in J}⟨f,f_i⟩f_i\Vert }^2$$

$$Re\left(\sum _{i\in J^c}⟨f,f_i⟩\overline{⟨K{K}^{\ast }f,f_i⟩}\right)$$

$$v_{-}(F,K,J)\le v_{+}(F,K,J)\le \Vert K\Vert \Vert K^{†}\Vert (1+\Vert K\Vert \Vert K^{†}\Vert ).$$

$$K{K}^{\ast }{S}_J-S_{J^c}K{K}^{\ast }=S_J^2-S_{J^c}^2.$$

$$⟨S_J^{2}f,f⟩+⟨S_{J^c}K{K}^{\ast }f,f⟩=⟨S_{J^c}^{2}f,f⟩+⟨K{K}^{\ast }{S}_{J}f,f⟩,$$

$${\Vert \sum _{i\in J}⟨f,f_i⟩f_i\Vert }^2+\overline{\sum _{i\in J^c}⟨f,f_i⟩\overline{⟨K{K}^{\ast }f,f_i⟩}}$$

$$\sum _{i\in J^c}\overline{⟨f,f_i⟩}⟨K{K}^{\ast }f,f_i⟩+{\Vert \sum _{i\in J}⟨f,f_i⟩f_i\Vert }^2$$

$$\Vert S_{J}f{\Vert }^2-⟨K{K}^{\ast }{S}_{J}f,f⟩=\Vert S_{J^c}f{\Vert }^2-⟨K{K}^{\ast }{S}_{J^c}f,f⟩.$$

$$⟨S_{J^c}f,S_{J}f⟩=⟨f,S_{J^c}{S}_{J}f⟩=⟨(K{K}^{\ast }{S}_J-S_J^2)f,f⟩,(f\in \mathcal{H}),$$

$$h(\lambda A+(1-\lambda )B)\le \lambda h(A)+(1-\lambda )h(B),$$

Full text

Key words: $K$-frames; Parseval $K$-frames; $K$-duals, Jensen’s operator inequality.

1. Introduction and preliminaries

Frames are redundant systems in separable Hilbert spaces, which provide non-unique representations of vectors and are applied in a wide range of applications [4, 5, 6, 7]. Working on efficient algorithms for computing reconstruction of signals without noisy phase, Balan et al. [3] discover some new identities for Parseval frames. Then, these identities have been generalized by some researchers for the alternate dual frames and other types of frames [2, 11, 19, 20].

Recently, $K$-frames were introduced by G$\breve{\textrm{a}}$vruţa [13] to study atomic systems with respect to a bounded operator $K\in B(\mathcal{H})$. Recall that atomic decomposition for a closed subspace $\mathcal{H}_{0}$ of a Hilbert space $\mathcal{H}$ introduced by Feichtinger et al. in [10] with frame-like properties. Although, the sequences in atomic decompositions do not necessarily belong to $\mathcal{H}_{0}$. This interesting property, which was the main motivation of introducing atomic decomposition and $K$-frame theory, comes from some problems in sampling theory [16, 17]. In this work, we extend and improve these results to obtain some equalities and inequalities for $K$-frames. First, we are going to state some preliminaries of $K$-frames and their duals, which are used in our main results. Let $\mathcal{H}$ be a separable Hilbert space and $I$ a countable index set and $K\in B(\mathcal{H})$. A sequence $F:=\{f_{i}\}_{i\in I}\subseteq\mathcal{H}$ is called a $K$-frame for $\mathcal{H}$, if there exist constants $A,B>0$ such that

[TABLE]

Obviously, $F$ is an ordinary frame if $K=I_{\mathcal{H}}$ and so $K$-frames are a generalization of the ordinary frames [7, 8, 13]. The constants $A$ and $B$ in $(\ref{001})$ are called the lower and the upper bounds of $F$, respectively. A $K$-frame $F$ is called a Parseval $K$-frame, whenever $\sum_{i\in I}|\langle f,f_{i}\rangle|^{2}=\|K^{*}f\|^{2}$. Since $F$ is a Bessel sequence, similar to ordinary frames the synthesis operator can be defined as $T_{F}:l^{2}\rightarrow\mathcal{H}$; $T_{F}(\{c_{i}\}_{i\in I})=\sum_{i\in I}c_{i}f_{i}$. The operator $T_{F}$ is bounded and its adjoint which is called the analysis operator is given by $T_{F}^{*}(f)=\{\langle f,f_{i}\rangle\}_{i\in I}$, and the frame operator is given by $S_{F}:\mathcal{H}\rightarrow\mathcal{H}$; $S_{F}f=T_{F}T_{F}^{*}f=\sum_{i\in I}\langle f,f_{i}\rangle f_{i}$. Unlike ordinary frames, the frame operator of a K-frame is not invertible in general. However, if $K$ has closed range, then $S_{F}$ from $R(K)$ onto $S_{F}(R(K))$ is an invertible operator [18].

In [1], the notion of duality for $K$-frames is introduced and several approaches for construction and characterization of $K$-frames and their dual are presented. Indeed, a Bessel sequence $\{g_{i}\}_{i\in I}\subseteq\mathcal{H}$ is called a $K$-dual of $\{f_{i}\}_{i\in I}$ if

[TABLE]

Theorem 1.1** (Douglas [9]).**

Let $L_{1}\in B(\mathcal{H}_{1},\mathcal{H})$ and $L_{2}\in B(\mathcal{H}_{2},\mathcal{H})$ be bounded linear mappings on given Hilbert spaces. Then the following assertions are equivalent:

• (i)

$R(L_{1})\subseteq R(L_{2})$;

• (ii)

$L_{1}L_{1}^{*}\leq\lambda^{2}L_{2}L_{2}^{*}$,  for some $\lambda>0$;

• (iii)

There exists a bounded linear mapping $X\in L(\mathcal{H}_{1},\mathcal{H}_{2})$, such that $L_{1}=L_{2}X$.

Moreover, if (i), (ii) and (iii) are valid, then there exists a unique operator $X$ so that $L_{1}=L_{2}X$

• (a)

$\|X\|^{2}=\inf\{\alpha>0,L_{1}L_{1}^{*}\leq\alpha L_{2}L_{2}^{*}\}$;

• (b)

$N(L_{1})=N(X)$;

• (c)

$R(X)\subset\overline{R(L_{2}^{*})}$.

Remark 1.2*.*

Suppose $F=\{f_{i}\}_{i\in I}$ is a $K$-frame for $\mathcal{H}$ with the optimal bounds $A$ and $B$, respectively. Clearly $B=\|S_{F}\|$. Also, using Douglas’ theorem, there exists an operator $X\in B(\mathcal{H},l^{2})$ such that

[TABLE]

and $\{X^{*}\delta_{i}\}_{i\in I}$ is a $K$-dual of $F$, where $\{\delta_{i}\}_{i\in I}$ is the standard orthonormal basis of $l^{2}$, see [13]. Moreover, by Douglas’ theorem the equation (1.3) has a unique solution as $X_{F}$ such that

[TABLE]

Now, let $A$ be the optimal lower bound of $F$. Then, we obtain

[TABLE]

which coincides with ordinary frames. Indeed, if $K=I_{\mathcal{H}}$, we easily obtain $X_{F}=T_{F}^{*}S_{F}^{-1}$ and so $\|X_{F}\|^{-2}=\|T_{F}^{*}S_{F}\|^{-2}=\|S_{F}^{-1}\|^{-1}$.

For further information in $K$-frame theory we refer the reader to [1, 10, 13, 18]. Throughout this paper, we suppose $\mathcal{H}$ is a separable Hilbert space, $K^{{\dagger}}$ the pseudo inverse of operator $K$, $I$ a countable index set and for every $J\subset I$, we show the complement of $J$ by $J^{c}$. Also $I_{\mathcal{H}}$ denotes the identity operator on $\mathcal{H}$. For two Hilbert spaces $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$ we denote by $B(\mathcal{H}_{1},\mathcal{H}_{2})$ the set of all bounded linear operators between $\mathcal{H}_{1}$ and $\mathcal{H}_{2}$, and we abbreviate $B(\mathcal{H},\mathcal{H})$ by $B(\mathcal{H})$. Also we denote the range of $K\in B(\mathcal{H})$ by $R(K)$, and the orthogonal projection of $\mathcal{H}$ onto a closed subspace $V$ of $\mathcal{H}$ by $\pi_{V}$. Finally, we use of $T_{J}$ and $S_{J}$ to denote the synthesis operator and frame operator, whenever the index set is limited to $J\subset I$.

2. Main Results

In this section we obtain several equalities and inequalities for $K$-frames, $K$-duals and Parseval $K$-frames. These results extend and improve some important results of [2, 11, 20].

Theorem 2.1**.**

Let $F=\{f_{i}\}_{i\in I}$ be a $K$-frame for $\mathcal{H}$ and $\{g_{i}\}_{i\in I}$ be a $K$-dual of $F$. Then for every $J\subseteq I$ and $f\in\mathcal{H}$,

[TABLE]

Proof.

Suppose that $J\subseteq I$, and consider the operator

[TABLE]

One can easily show that $M_{J}$ is a well defined and bounded operator on $\mathcal{H}$. Moreover, we have $M_{J}+M_{J^{c}}=K$. Hence,

[TABLE]

∎

Using Theorem 2.1 we immediately obtain the following corollary.

Corollary 2.2**.**

Let $F=\{f_{i}\}_{i\in I}$ be a $K$-frame with a $K$-dual $\{g_{i}\}_{i\in I}$. Then for every $J\subseteq I$ and $f\in\mathcal{H}$,

[TABLE]

where $X_{F}$ is as in $(\ref{XW})$.

Theorem 2.3**.**

Let $F=\{f_{i}\}_{i\in I}$ be a $K$-frame for $\mathcal{H}$ and $\{g_{i}\}_{i\in I}$ be a $K$-dual of $F$. Then for every bounded sequence $\{\alpha_{i}\}_{i\in I}$ and $f\in\mathcal{H}$,

[TABLE]

In the sequel, we survey some inequalities on Parseval $K$-frames. These results also extend some important inequalites for frames in [2, 11].

Theorem 2.4**.**

Assume that $F=\{f_{i}\}_{i\in I}$ is a Parseval $K$-frame for $\mathcal{H}$. For every $f\in\mathcal{H}$, $J\subseteq I$ and $E\subseteq J^{c}$ we have

[TABLE]

Proof.

First note that for every $J\subseteq I$, we have $S_{J}+S_{J^{c}}=KK^{*}$. Hence $S_{J}^{2}-S_{J^{c}}^{2}=KK^{*}S_{J}-S_{J^{c}}KK^{*}$. In fact

[TABLE]

Hence, for every $f\in\mathcal{H}$ we obtain

[TABLE]

and consequently

[TABLE]

∎

Theorem 2.5**.**

Let $F=\{f_{i}\}_{i\in I}$ be a Parseval $K$-frame for $\mathcal{H}$. Then for every $J\subseteq I$ and $f\in\mathcal{H}$,

[TABLE]

Proof.

Since $S_{J}^{2}-S_{J^{c}}^{2}=KK^{*}S_{J}-S_{J^{c}}KK^{*}$, we can write

[TABLE]

and consequently

[TABLE]

Thus, we obtain

[TABLE]

This completes the proof. ∎

Notice that, in Theorem 2.5 if we take $K=I_{\mathcal{H}}$ it reduce to Theorem 2.2 in [20]. Also, Theorem 2.5 leads to the following concept, which is a generalization of [11] for Parseval frames. Let $F=\{f_{i}\}_{i\in I}$ be a Parseval $K$-frame. Then we consider

[TABLE]

Now, we are going to present some properties of these notations. For this we need the next lemma.

Lemma 2.6**.**

Let $K$ be a closed range operator and $F=\{f_{i}\}_{i\in I}$ be a $K$-frame for $\mathcal{H}$. Then

• (i)

$\left\|\sum_{i\in I}\langle f,f_{i}\rangle f_{i}\right\|^{2}\leq\|S_{F}\|\sum_{i\in I}|\langle f,f_{i}\rangle|^{2},\quad(f\in\mathcal{H})$.

• (ii)

$\sum_{i\in I}|\langle f,f_{i}\rangle|^{2}\leq\|K^{\dagger}\|^{2}\|X_{F}\|^{2}\left\|\sum_{i\in I}\langle f,f_{i}\rangle f_{i}\right\|^{2},\quad(f\in R(K))$.

Proof.

The first part is a known result for every Bessel sequence [8] and so for $K$-frames. Hence, we only need to show $(ii)$. Let $f\in R(K)$. Then we have

[TABLE]

∎

Theorem 2.7**.**

Suppose $F=\{f_{i}\}_{i\in I}$ is a Parseval $K$-frame for $\mathcal{H}$. The following assertion hold:

• (i)

$\dfrac{3}{4}\leq v_{-}(F,K,J)\leq v_{+}(F,K,J)\leq\|K\|\|K^{\dagger}\|(1+\|K\|)$**

• (ii)

$v_{+}(F,K,J)=v_{+}(F,K,J^{c})$, and $v_{-}(F,K,J)=v_{-}(F,K,J^{c})$

• (iii)

$v_{+}(F,K,I)=v_{-}(F,K,I)=1$, and $v_{+}(F,K,\emptyset)=v_{-}(F,K,\emptyset)=1.$

Proof.

For (i), it is sufficient to show the upper inequality. Since $F$ is a Bessel sequence, so by applying Lemma 2.6 (i) we obtain

[TABLE]

Moreover,

[TABLE]

Hence,

[TABLE]

On the other hand, in the proof of Theorem 2.4 we observed that

[TABLE]

and consequently we have

[TABLE]

for every $f\in\mathcal{H}$. Thus

[TABLE]

This shows (ii). Finally (iii) is easy to check. ∎

Using the result above-mentioned, we stablish some equivalent results for Parseval $K$-frames.

Corollary 2.8**.**

Let $\{f_{i}\}_{i\in I}$ be a Parseval $K$-frame for $\mathcal{H}$. Then for every $J\subset I$ and $f\in\mathcal{H}$ the following conditions are equivalent.

• (i)

$v_{+}(F,K,J)=v_{-}(F,K,J)=1$.

• (ii)

$\left\|\sum_{i\in J}\langle f,f_{i}\rangle f_{i}\right\|^{2}=Re\sum_{i\in J}\langle f,f_{i}\rangle\overline{\langle KK^{*}f,f_{i}\rangle}.$**

• (iii)

$\left\|\sum_{i\in J^{c}}\langle f,f_{i}\rangle f_{i}\right\|^{2}=Re\sum_{i\in J^{c}}\langle f,f_{i}\rangle\overline{\langle KK^{*}f,f_{i}\rangle}$.

Proof.

$(ii)\Leftrightarrow(iii)$ is clear. Also, $(i)\Rightarrow(ii)$ holds by a direct computation. Now, let $(ii)$ holds, then

[TABLE]

i.e., $(i)$ holds. Hence $(i)\Leftrightarrow(iii)$ and similarly $(i)\Leftrightarrow(iii)$. ∎

Finally, we obtain the following more general result.

Corollary 2.9**.**

Let $\{f_{i}\}_{i\in I}$ be a Parseval $K$-frame for $\mathcal{H}$. Then for every $J\subset I$ and $f\in\mathcal{H}$ the following conditions are equivalent.

• (i)

$\left\|\sum_{i\in J}\langle f,f_{i}\rangle f_{i}\right\|^{2}=\sum_{i\in J}\langle f,f_{i}\rangle\overline{\langle KK^{*}f,f_{i}\rangle}.$**

• (ii)

$\left\|\sum_{i\in J^{c}}\langle f,f_{i}\rangle f_{i}\right\|^{2}=\sum_{i\in J^{c}}\langle f,f_{i}\rangle\overline{\langle KK^{*}f,f_{i}\rangle}$.

• (iii)

$S_{J}f\perp S_{J^{c}}f$.

• (iv)

$f\perp S_{J^{c}}S_{J}f$*. *

Proof.

Since,

[TABLE]

Thus $(i)\Leftrightarrow(ii)$. Moreover $S_{J}$ and $S_{J^{c}}$ are positive, so we have

[TABLE]

This follows that, $(iii)\Leftrightarrow(iv)$ and $(i)\Leftrightarrow(iv)$. ∎

3. Some new inequalities

In this section, applying Jensen’s operator inequality we obtain some new inequalities for $K$-frames. First, recall that a continuous function $h$ defined on an interval $J$ is said to be operator convex if

[TABLE]

for all $\lambda\in[0,1]$ and all self-adjoint operators $A$,$B$ with spectra in $J$. A general formulation of Jensen’s operator inequality is given as follows (see [12]).

Theorem 3.1**.**

Let $\mathcal{H}$, $\mathcal{K}$ be Hilbert spaces and $h:J\rightarrow\mathbb{R}$ be an operator convex function. For each $n\in\mathbb{N}$, the inequality

[TABLE]

holds for every $n$-tuple $(A_{1},...,A_{n})$ of self-adjoint operators in $B(\mathcal{H})$ with spectra in $J$ and every $n$-tuple $(\Phi_{1},...,\Phi_{n})$ of positive linear mappings $\Phi_{i}:B(\mathcal{H})\rightarrow B(\mathcal{K})$ such that $\sum_{i=1}^{n}\Phi_{i}(I_{\mathcal{H}})=I_{\mathcal{K}}$.

Also, a variant of Jensen’s operator inequality (3.1) for convex functions is presented as follows (see [15]).

Theorem 3.2**.**

Let $A_{1},...,A_{n}$ be self-adjoin operators with spectra in $[m,M]$ for some scalars $m<M$ and $\Phi_{1},...,\Phi_{n}$ be positive linear mappings from $B(\mathcal{H})$ into $B(\mathcal{K})$ with $\sum_{i=1}^{n}\Phi_{i}(I_{\mathcal{H}})=I_{\mathcal{K}}$, where $I_{\mathcal{H}}$ and $I_{\mathcal{K}}$ are the identity mappings on $\mathcal{H}$ and $\mathcal{K}$, respectively. If $h:[m,M]\rightarrow\mathbb{R}$ is a continuous convex function, then

[TABLE]

Now, we apply inequalities $(\ref{jensen})$ and $(\ref{jensen2})$ to obtain some inequalities for $K$-frames.

Theorem 3.3**.**

Suppose $F=\{f_{i}\}_{i\in I}$ is a Parseval $K$-frame for $\mathcal{H}$. Then for every subset $J$ of $I$,

• (i)

if $h:[0,\|K\|^{2}]\rightarrow\mathbb{R}$ is an operator convex function, then

[TABLE]

• (ii)

if $h:[0,\|K\|^{2}]\rightarrow\mathbb{R}$ is a convex function, then

[TABLE]

Proof.

Let $A_{1}=S_{J}$ and $A_{2}=S_{J^{c}}$. Since $S_{J}$ and $S_{J^{c}}$ are positive operators with $S_{J}+S_{J^{c}}=KK^{*}$, the spectra of $A_{1}$ and $A_{2}$ are in $[0,\|K\|^{2}]$. Now, if we define $\Phi_{1}=\dfrac{1}{2}Id$ and $\Phi_{2}=\dfrac{1}{2}Id$, where $Id$ is the identity mapping on $B(\mathcal{H})$, then $(i)$ and $(ii)$ follow from inequalities $(\ref{jensen})$ and $(\ref{jensen2})$, respectively. ∎

Corollary 3.4**.**

Suppose $F=\{f_{i}\}_{i\in I}$ is a Parseval $K$-frame for $\mathcal{H}$. Then for every subset $J$ of $I$ and $f\in\mathcal{H}$, we obtain

[TABLE]

Proof.

Since the function $f(x)=x^{2}$ is convex on $[0,\infty)$, by Theorem 3.3 we have

[TABLE]

and

[TABLE]

and so

[TABLE]

Therefore, it follows from $(\ref{proofcor1})$ and $(\ref{proofcor2})$ that

[TABLE]

Hence for every $f\in\mathcal{H}$, we obtain inequality $(\ref{cor1})$. ∎