Non-Archimedean Riesz-Frechet Representation

1. Introduction

The century-old Riesz-Frechet representation (theorem) states that every bounded linear functional on a real/complex Hilbert space is completely determined by a single element in that Hilbert space.

Theorem 1.

[1,2,4,8,9,10,11,12,13,14,18,19,20,21,23,26,28,31,32,33,34,35,36,39,43,44,46,46] (Riesz-Frechet Representation) Let $\mathcal{H}$ be a complex Hilbert space. Let $f:\mathcal{H}\to \mathbb{C}$ be a bounded linear functional. Then there is a unique element $\tau \in \mathcal{H}$ satisfying the following conditions.

(i) 

$f\left(h\right)=\langle h,\tau \rangle$ for all $h\in \mathcal{H}$.

(ii) 

$\parallel f\parallel =\parallel \tau \parallel$.

Over time, Riesz-Frechet representation has been extended to Lebesgue spaces [30,37,42] and spaces of continuous functionals on compact Hausdorff spaces (celebrated Riesz-Markov-Kakutani representation) [2,3,7,8,15,24,29,35,38,45]. It is known that Riesz-Frechet representation also holds for quaternionic Hilbert spaces [16]. Riesz-Frechet representation has also been derived for certain classes of Banach spaces by Giles [17] and James [22]. The traditional proof of the Riesz-Frechet representation uses the orthogonal decomposition theorem, which rests on the parallelogram law [19,32]. If the Hilbert space is separable, then one can also give a proof of the Riesz-Frechet representation for Hilbert spaces using orthonormal bases, which relies on the Gram-Schmidt orthonormalization process [20]. We are fundamentally motivated by the following question.

Question 2.

What is the non-Archimedean Riesz-Frechet representation?

We begin by recalling the definition of non-Archimedean Banach and p-adic Hilbert spaces.

Definition 1.

[40] Let $\mathbb{K}$ be a field. A map $|·|:\mathbb{K}\to [0,\infty )$ is said to be a  non-Archimedean valuation  if the following conditions hold:

(i) 

If $\lambda \in \mathbb{K}$ satisfies $\left|\lambda \right|=0$, then $\lambda =0$.

(ii) 

$\left|\lambda \mu \right|=\left|\lambda \right|\left|\mu \right|$ for all $\lambda ,\mu \in \mathbb{K}$.

(iii) 

(Ultratriangle inequality) $|\lambda +\mu |\le max\left\{\lambda \right|,\left|\mu \right|\}$ for all $\lambda ,\mu \in \mathbb{K}$.

We say $\mathbb{K}$ isnon-Archimedean complete valued fieldif it is complete w.r.t. the ultrametric

$$\begin{array}{c}\hfill d(\lambda ,\mu )|\lambda -\mu |,\forall \lambda ,\mu \in \mathbb{K}.\end{array}$$

Definition 2.

[41] Let $\mathbb{K}$ be a non-Archimedean complete valued field. A vector space $\mathcal{X}$ over $\mathbb{K}$ is said to benon-Archimedean Banach spaceif there is a map (called as ultra-norm) $\parallel ·\parallel :\mathcal{X}\to [0,\infty )$ satisfying the following conditions:

(i) 

If $x\in \mathcal{X}$ is such that $\parallel x\parallel =0$, then $x=0$.

(ii) 

$\parallel \lambda x\parallel =\left|\lambda \right|\parallel x\parallel$ for all $\lambda \in \mathbb{K}$, for all $x\in \mathcal{X}$.

(iii) 

(Ultranorm inequality) $\parallel x+y\parallel \le max\{\parallel x\parallel ,\parallel y\parallel \}$ for all $x,y\in \mathcal{X}$.

(iv) 

$\mathcal{X}$ is complete w.r.t. the ultrametric

$$\begin{array}{c}\hfill d(x,y):=\parallel x-y\parallel ,\forall x,y\in \mathcal{X}.\end{array}$$

Definition 3.

[25] Let $\mathbb{K}$ be a non-Archimedean valued field (with valuation $|·|$) and $\mathcal{X}$ be a non-Archimedean Banach space (with norm $\parallel ·\parallel$) over $\mathbb{K}$. We say that $\mathcal{X}$ is ap-adic Hilbert spaceif there is a map (called asp-adic inner product) $\langle ·,·\rangle :\mathcal{X}\times \mathcal{X}\to \mathbb{K}$ satisfying the following conditions:

(i) 

If $x\in \mathcal{X}$ is such that $\langle x,y\rangle =0$ for all $y\in \mathcal{X}$, then $x=0$.

(ii) 

$\langle x,y\rangle =\langle y,x\rangle$ for all $x,y\in \mathcal{X}$.

(iii) 

$\langle \alpha x,y\rangle =\alpha \langle x,y\rangle$ for all $\alpha \in \mathbb{K}$, for all $x,y\in \mathcal{X}$.

(iv) 

$\langle x+y,z\rangle =\langle x,z\rangle +\langle y,z\rangle$ for all $x,y,z\in \mathcal{X}$.

(v)

$|\langle x,y\rangle |\le \parallel x\parallel \parallel y\parallel$ for all $x,y\in \mathcal{X}$.

Definition 3 is more general than the definition of p-adic Hilbert spaces used by Kalisch, which imposes further conditions [25]. We note that the Definition 3 is different than the p-adic Hilbert spaces defined recently by Claussnitzer and Thom [6] where an exact Riesz-Frechet representation is known [5]. We derive two different versions of non-Archimedean Riesz-Frechet representation with different assumptions and different conclusions.

2. Non-Archimedean Riesz-Frechet Representation

In the paper, we use the following notion of Schauder basis for non-Archimedean Banach spaces and orthonormal basis for p-adic Hilbert spaces.

Definition 4.

Let $\mathcal{X}$ be a non-Archimedean Banach space over $\mathbb{K}$. A collection ${\left\{{\tau }_n\right\}}_n$ in $\mathcal{X}$ is said to be a  Schauder basis  for $\mathcal{X}$ if for every $x\in \mathcal{X}$, there is a unique sequence ${\left\{{\alpha }_n\left(x\right)\right\}}_n$ in $\mathbb{K}$ such that

$$\begin{array}{c}\hfill x={\displaystyle \sum _{n=1}^{\infty }}{\alpha }_n\left(x\right){\tau }_n,\end{array}$$

series converges in the ultra-norm on $\mathcal{X}$.

Definition 5.

Let $\mathcal{X}$ be a p-adic Hilbert space over $\mathbb{K}$. A Schauder basis ${\left\{{\tau }_n\right\}}_n$ for $\mathcal{X}$ is said to be ap-adic orthonormal basisfor $\mathcal{X}$ if the following conditions hold:

(i) 

$\parallel {\tau }_n\parallel \le 1$ for all $n\in \mathbb{N}$.

(ii) 

$\langle {\tau }_n,{\tau }_m\rangle ={\delta }_{n,m}$ for all $n,m\in \mathbb{N}$.

Note that given a p-adic orthonormal basis ${\left\{{\tau }_n\right\}}_n$ for $\mathcal{X}$, condition (ii) in Definition 5 says that

$$\begin{array}{c}\hfill 1=|\langle {\tau }_n,{\tau }_n\rangle |\le \parallel {\tau }_n{\parallel }^2,\forall n\in \mathbb{N}.\end{array}$$

By using condition (i) in Definition 5 we then get $\parallel {\tau }_n\parallel =1$ for all $n\in \mathbb{N}$. Thus all elements of p-adic orthonormal basis lie in the unit sphere. Following is the standard example of p-adic Hilbert space we keep in mind.

Example 1.

Let $\mathbb{K}$ be a non-Archimedean valued field. Define

$$\begin{array}{c}\hfill c_0(\mathbb{N},\mathbb{K})\left\{{\left\{a_n\right\}}_n:a_n\in \mathbb{K},\forall n\in \mathbb{N},\underset{n\to \infty }{lim}{a}_n=0\right\}.\end{array}$$

We define

$$\begin{array}{c}\hfill \parallel {\left\{a_n\right\}}_n\parallel \underset{n\in \mathbb{N}}{max}\left|a_n\right|,\forall {\left\{a_n\right\}}_n\in c_0(\mathbb{N},\mathbb{K})\end{array}$$

and

$$\begin{array}{c}\hfill \langle {\left\{a_n\right\}}_n,{\left\{b_n\right\}}_n\rangle {\displaystyle \sum _{n=1}^{\infty }}{a}_n{b}_n,\forall {\left\{a_n\right\}}_n,{\left\{b_n\right\}}_n\in c_0(\mathbb{N},\mathbb{K}).\end{array}$$

Then $c_0(\mathbb{N},\mathbb{K})$ is a p-adic Hilbert space. The canonical Schauder basis ${\left\{e_n\right\}}_n$ is a p-adic orthonormal basis for $c_0(\mathbb{N},\mathbb{K})$.

Gram-Schmidt orthonormalization says that every separable Hilbert space has an orthonormal basis [27]. Eventhough we know from Example 1 that we have examples of p-adic orthonormal bases, we don’t know existence result of p-adic orthonormal basis in every p-adic Hilbert space.

Following is our first version of Riesz-Frechet representation.

Theorem 3.

(Non-Archimedean Riesz-Frechet Representation) Let $\mathcal{X}$ be a p-adic Hilbert space over a complete non-Archimedean valued field $\mathbb{K}$. Let $f:\mathcal{X}\to \mathbb{K}$ be a bounded linear functional. Assume that there is a p-adic orthonormal basis ${\left\{{\omega }_n\right\}}_n$ for $\mathcal{X}$ such that

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}f\left({\omega }_n\right){\omega }_n=0.\end{array}$$

Then there is a unique element $\tau \in \mathcal{X}$ satisfying the following conditions:

(i) 

$f\left(x\right)=\langle x,\tau \rangle$ for all $x\in \mathcal{X}$.

(ii) 

$\parallel f\parallel =\parallel \tau \parallel$.

Proof. 

Since ${\left\{{\omega }_n\right\}}_n$ is orthonormal,

$$\begin{array}{c}\hfill ∥{\displaystyle \sum _{k=1}^n}f\left({\omega }_k\right){\omega }_k∥=\underset{1\le k\le n}{max}\left|f\left({\omega }_k\right)\right|,\forall n\in \mathbb{N}.\end{array}$$

Since $\mathcal{X}$ is complete and ${lim}_{n\to \infty }f\left({\omega }_n\right){\omega }_n=0$, the series ${\sum }_{n=1}^{\infty }f\left({\omega }_n\right){\omega }_n$ converges in $\mathcal{X}$. Now define

$$\begin{array}{c}\hfill \tau {\displaystyle \sum _{n=1}^{\infty }}f\left({\omega }_n\right){\omega }_n.\end{array}$$

We then have

$$\begin{array}{cc}\hfill f\left(x\right)& =f\left({\displaystyle \sum _{n=1}^{\infty }}\langle x,{\omega }_n\rangle {\omega }_n\right)={\displaystyle \sum _{n=1}^{\infty }}\langle x,{\omega }_n\rangle f\left({\omega }_n\right)\hfill \\ \hfill & =〈x,{\displaystyle \sum _{n=1}^{\infty }}f\left({\omega }_n\right){\omega }_n〉=\langle x,\tau \rangle ,\forall x\in \mathcal{X}.\hfill \end{array}$$

Now

$$\begin{array}{c}\hfill \parallel f\parallel =sup\left\{{\displaystyle \frac{\left|f\right(x\left)\right|}{\parallel x\parallel }}:x\in \mathcal{X},x\ne 0\right\}=sup\left\{{\displaystyle \frac{|\langle x,\tau \rangle |}{\parallel x\parallel }}:x\in \mathcal{X},x\ne 0\right\}\le \parallel \tau \parallel .\end{array}$$

We now observe that

$$\begin{array}{c}\hfill \parallel \tau \parallel =∥{\displaystyle \sum _{n=1}^{\infty }}f\left({\omega }_n\right){\omega }_n∥=\underset{n\in \mathbb{N}}{max}|f\left({\omega }_n\right)|\le \parallel f\parallel \underset{n\in \mathbb{N}}{max}\parallel {\omega }_n\parallel =\parallel f\parallel .\end{array}$$

Uniqueness easily follows using the definition of p-adic inner product. □

Example 2.

Let p be a prime number. Define

$$\begin{array}{c}\hfill f:c_0(\mathbb{N},{\mathbb{Q}}_p)\ni {\left\{a_n\right\}}_n\mapsto f\left({\left\{a_n\right\}}_n\right){\displaystyle \sum _{n=1}^{\infty }}{p}^n{a}_n\in {\mathbb{Q}}_p.\end{array}$$

Let ${\left\{a_n\right\}}_n\in c_0(\mathbb{N},{\mathbb{Q}}_p)$. Then

$$\begin{array}{c}\hfill |p^n{a}_n{|}_p={\displaystyle \frac{1}{p^n}}{\left|a_n\right|}_p\to 0\mathrm{as}n\to \infty .\end{array}$$

Hence

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}{p}^n{a}_n=0,\forall {\left\{a_n\right\}}_n\in c_0(\mathbb{N},{\mathbb{Q}}_p).\end{array}$$

So the map f is well-defined. Define

$$\begin{array}{c}\hfill \tau {\left\{p^n\right\}}_n\in c_0(\mathbb{N},{\mathbb{Q}}_p).\end{array}$$

Then

$$\begin{array}{c}\hfill f\left(x\right)=\langle x,\tau \rangle ,\forall x\in c_0(\mathbb{N},{\mathbb{Q}}_p)\mathrm{and}\parallel f\parallel =\parallel \tau \parallel =\underset{n\in \mathbb{N}}{max}{\left|p^n\right|}_p=\underset{n\in \mathbb{N}}{max}\left\{{\displaystyle \frac{1}{p^n}}\right\}={\displaystyle \frac{1}{p}}.\end{array}$$

In the next example we show that every bounded linear functional on a p-adic Hilbert space cannot be represented by an element in the space.

Example 3.

Let p be a prime number. Define

$$\begin{array}{c}\hfill f:c_0(\mathbb{N},{\mathbb{Q}}_p)\ni {\left\{a_n\right\}}_n\mapsto f\left({\left\{a_n\right\}}_n\right){\displaystyle \sum _{n=1}^{\infty }}n{a}_n\in {\mathbb{Q}}_p.\end{array}$$

Let ${\left\{a_n\right\}}_n\in c_0(\mathbb{N},{\mathbb{Q}}_p)$. Then

$$\begin{array}{c}\hfill |n{a}_n{|}_p=\left|n\right||a_n{|}_p\le {\left|a_n\right|}_p\to 0\mathrm{as}n\to \infty .\end{array}$$

Hence

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}n{a}_n=0,\forall {\left\{a_n\right\}}_n\in c_0(\mathbb{N},{\mathbb{Q}}_p).\end{array}$$

So the map f is well-defined. We claim that there is no $\tau \in c_0(\mathbb{N},{\mathbb{Q}}_p)$ such that $f\left(x\right)=\langle x,\tau \rangle$ for all $x\in c_0(\mathbb{N},{\mathbb{Q}}_p)$. Let us suppose that this claim fails. Then thee exists a $\tau \in c_0(\mathbb{N},{\mathbb{Q}}_p)$ such that $f\left(x\right)=\langle x,\tau \rangle$ for all $x\in c_0(\mathbb{N},{\mathbb{Q}}_p)$. Set $\tau {\left\{a_n\right\}}_n\in c_0(\mathbb{N},{\mathbb{Q}}_p)$. Let $\left\{e_n\right\}$ be the standard p-adic orthonormal basis for $c_0(\mathbb{N},{\mathbb{Q}}_p)$. Now we see that

$$\begin{array}{c}\hfill n=f\left(e_n\right)=\langle e_n,\tau \rangle =a_n,\forall n\in \mathbb{N}.\end{array}$$

But then $\tau ={\left\{n\right\}}_n\notin c_0(\mathbb{N},{\mathbb{Q}}_p)$ because ${\left|n\right|}_p\nrightarrow 0$ as $n\to \infty$. Hence Riesz-Frechet representation fails.

Let $\mathcal{X}$ be a p-adic Hilbert space. A closed subspace $\mathcal{Y}$ of $\mathcal{X}$ is said to be orthogonally complementable if there exists a unique closed subspace $\mathcal{Z}$ of $\mathcal{X}$ satisfying the following: for every $x\in \mathcal{X}$, there exist unique $y\in \mathcal{Y}$ and $z\in \mathcal{Z}$ such that

(i)

$x=y+z$.

(ii)

$\langle y,z\rangle =0$.

(iii)

$\parallel x\parallel =max\{\parallel y\parallel ,\parallel z\parallel \}$.

In this case, we write $x=y\oplus z$, $\mathcal{Z}={\mathcal{Y}}^{\perp }$ and $\mathcal{X}=\mathcal{Y}\oplus {\mathcal{Y}}^{\perp }$.

Following is our second version of Riesz-Frechet representation.

Theorem 4.

(Non-Archimedean Riesz-Frechet Representation) Let $\mathcal{X}$ be a p-adic Hilbert space over a complete non-Archimedean valued field $\mathbb{K}$. Let $f:\mathcal{X}\to \mathbb{K}$ be a non-zero bounded linear functional. Assume that $Ker\left(f\right)$ is orthogonally complementable in $\mathcal{X}$ and there exists a (nonzero) vector $\omega \in Ker{\left(f\right)}^{\perp }$ such that $\langle \omega ,\omega \rangle \ne 0$. Then there is a unique element $\tau \in \mathcal{X}$ satisfying the following conditions:

(i) 

$f\left(x\right)=\langle x,\tau \rangle$ for all $x\in \mathcal{X}$.

(ii) 

$$\begin{array}{c}\hfill {\displaystyle \frac{|\langle \tau ,\tau \rangle |}{\parallel \tau \parallel }}\le \parallel f\parallel \le \parallel \tau \parallel .\end{array}$$

Moreover,

$$\begin{array}{c}\hfill \parallel \tau \parallel ={\displaystyle \frac{\left|f\right(\omega \left)\right|}{|\langle \omega ,\omega \rangle |}}\parallel \omega \parallel \le {\displaystyle \frac{{\parallel f\parallel \parallel \omega \parallel }^2}{|\langle \omega ,\omega \rangle |}}.\end{array}$$

Proof. 

Define

$$\begin{array}{c}\hfill \tau {\displaystyle \frac{f\left(\omega \right)}{\langle \omega ,\omega \rangle }}\omega .\end{array}$$

Let $x\in \mathcal{X}$. Then

$$\begin{array}{c}\hfill f\left(x-{\displaystyle \frac{f\left(x\right)}{f\left(\omega \right)}}\omega \right)=f\left(x\right)-f\left(x\right)=0.\end{array}$$

Hence

$$\begin{array}{c}\hfill x-{\displaystyle \frac{f\left(x\right)}{f\left(\omega \right)}}\omega \in Ker{\left(f\right)}^{\perp }.\end{array}$$

We then have

$$\begin{array}{cc}\hfill 0& =〈x-{\displaystyle \frac{f\left(x\right)}{f\left(\omega \right)}}\omega ,\tau 〉=〈x,\tau 〉-〈{\displaystyle \frac{f\left(x\right)}{f\left(\omega \right)}}\omega ,\tau 〉\hfill \\ \hfill & =〈x,\tau 〉-{\displaystyle \frac{f\left(x\right)}{f\left(\omega \right)}}\langle \omega ,\tau \rangle =〈x,\tau 〉-{\displaystyle \frac{f\left(x\right)}{f\left(\omega \right)}}〈\omega ,{\displaystyle \frac{f\left(\omega \right)}{\langle \omega ,\omega \rangle }}\omega 〉\hfill \\ \hfill & =\langle x,\tau \rangle -f\left(x\right).\hfill \end{array}$$

Note that

$$\begin{array}{c}\hfill |\langle \tau ,\tau \rangle |=\left|f\right(\tau \left)\right|\le \parallel f\parallel \parallel \tau \parallel .\end{array}$$

□

We next extend the non-Archimedean Riesz-Frechet representation to non-Archimedean fields admitting conjugation (involution). We use the following definition of conjugation.

Definition 6.

Let $\mathbb{K}$ be a non-Archimedean complete valued field (with valuation $|·|$). A map $\sigma :\mathbb{K}\to \mathbb{K}$ is said to be aconjugationorinvolutionif the following conditions hold:

(i) 

$\sigma (\alpha +\beta )=\sigma \left(\alpha \right)+\sigma \left(\beta \right)$ for all $\alpha ,\beta \in \mathbb{K}$.

(ii) 

$\sigma \left(\alpha \beta \right)=\sigma \left(\alpha \right)\sigma \left(\beta \right)$ for all $\alpha ,\beta \in \mathbb{K}$.

(iii) 

$\sigma \left(\sigma \right(\alpha \left)\right)=\alpha$ for all $\alpha \in \mathbb{K}$.

(iv) 

$\left|\sigma \right(\alpha \left)\right|=\left|\alpha \right|$ for all $\alpha \in \mathbb{K}$.

In this case, we say that $\mathbb{K}$ is σ-conjugated.

Definition 7.

Let $\mathbb{K}$ be a σ-conjugated non-Archimedean complete valued field (with valuation $|·|$) and $\mathcal{X}$ be a non-Archimedean Banach space (with norm $\parallel ·\parallel$) over $\mathbb{K}$. We say that $\mathcal{X}$ is ap-adic $\sigma$-Hilbert spaceif there is a map (called asp-adic $\sigma$-inner product) $\langle ·,·\rangle :\mathcal{X}\times \mathcal{X}\to \mathbb{K}$ satisfying the following:

1.

If $x\in \mathcal{X}$ is such that $\langle x,y\rangle =0$ for all $y\in \mathcal{X}$, then $x=0$.

(i) 

$\langle x,y\rangle =\sigma (\langle y,x\rangle )$ for all $x,y\in \mathcal{X}$.

(ii) 

$\langle \alpha x,y\rangle =\alpha \langle x,y\rangle$ for all $\alpha \in \mathbb{K}$, for all $x,y\in \mathcal{X}$.

(iii) 

$\langle x+y,z\rangle =\langle x,z\rangle +\langle y,z\rangle$ for all $x,y,z\in \mathcal{X}$.

(iv) 

$|\langle x,y\rangle |\le \parallel x\parallel \parallel y\parallel$ for all $x,y\in \mathcal{X}$.

We now have p-adic versions of Theorem 3 and Theorem 4.

Theorem 5.

(Non-Archimedean Riesz-Frechet Representation) Let $\mathcal{X}$ be a p-adic σ-Hilbert space over a σ-conjugated complete non-Archimedean valued field $\mathbb{K}$. Let $f:\mathcal{X}\to \mathbb{K}$ be a bounded linear functional. Assume that there is a p-adic orthonormal basis ${\left\{{\omega }_n\right\}}_n$ for $\mathcal{X}$ such that

$$\begin{array}{c}\hfill \underset{n\to \infty }{lim}f\left({\omega }_n\right){\omega }_n=0.\end{array}$$

Then there is a unique element $\tau \in \mathcal{X}$ satisfying the following conditions:

(i) 

$f\left(x\right)=\langle x,\tau \rangle$ for all $x\in \mathcal{X}$.

(ii) 

$\parallel f\parallel =\parallel \tau \parallel$.

Proof. 

Define

$$\begin{array}{c}\hfill \tau {\displaystyle \sum _{n=1}^{\infty }}\sigma \left(f\left({\omega }_n\right)\right){\omega }_n.\end{array}$$

The other arguments are similar to those used in the proof of Theorem 3. □

Theorem 6.

(Non-Archimedean Riesz-Frechet Representation) Let $\mathcal{X}$ be a p-adic σ-Hilbert space over a σ-conjugated complete non-Archimedean valued field $\mathbb{K}$. Let $f:\mathcal{X}\to \mathbb{K}$ be a non-zero bounded linear functional. Assume that $Ker\left(f\right)$ is orthogonally complementable in $\mathcal{X}$ and there exists a (nonzero) vector $\omega \in Ker{\left(f\right)}^{\perp }$ such that $\langle \omega ,\omega \rangle \ne 0$. Then there is a unique element $\tau \in \mathcal{X}$ satisfying the following conditions:

(i) 

$f\left(x\right)=\langle x,\tau \rangle$ for all $x\in \mathcal{X}$.

(ii) 

$$\begin{array}{c}\hfill {\displaystyle \frac{|\langle \tau ,\tau \rangle |}{\parallel \tau \parallel }}\le \parallel f\parallel \le \parallel \tau \parallel .\end{array}$$

Moreover,

$$\begin{array}{c}\hfill \parallel \tau \parallel ={\displaystyle \frac{\left|f\right(\omega \left)\right|}{|\langle \omega ,\omega \rangle |}}\parallel \omega \parallel \le {\displaystyle \frac{{\parallel f\parallel \parallel \omega \parallel }^2}{|\langle \omega ,\omega \rangle |}}.\end{array}$$

Proof. 

Define

$$\begin{array}{c}\hfill \tau {\displaystyle \frac{\sigma \left(f\right(\omega \left)\right)}{\langle \omega ,\omega \rangle }}\omega .\end{array}$$

The other arguments are similar to those used in the proof of Theorem 4. □

3. Conclusions

(1)

In 1907, Riesz [34] and Frechet [13] independently showed that every continuous linear functional on a Hilbert space is completely determined by a single element in that Hilbert space [4].

(2)

In 1938, Markov [29] and in 1941, Kakutani [24] extended the Riesz-Frechet representation to function spaces.

(3)

In 1947, James [22] and 1967, Giles [17] extended the Riesz-Frechet representation to certain classes of Banach spaces.

(4)

In this article, we derived two non-Archimedean Riesz representation theorems under certain conditions for p-adic Hilbert spaces.