Angle in the space of $p$-summable sequences

1.   Introduction

In an inner product space, we can calculate angles between two subspaces. Let $(X, \langle\cdot, \cdot\rangle)$ be a real inner product space. If $U = \text{span}\; \{u\}$ is a $1$-dimensional subspace and $V = \text{span}\; \{v_{1}, \cdots, v_{t}\}$ is a $t$-dimensional subspace of $X$, then the angle between subspaces $U$ and $V$ is defined by $\theta$ with $0\leq\theta\leq\frac{\pi}{2}$ and $\cos ^{2}\theta = \frac{\langle u, u_{V}\rangle ^{2}}{\left\Vert u\right\Vert ^{2}\left\Vert u_{V}\right\Vert ^{2}}$. In formula, $u_{V}$ denotes the (orthogonal) projection of $u$ on $V$ and $\left\Vert \cdot\right\Vert = \langle \cdot, \cdot\rangle ^{2}$. Gunawan et al. ^[3] show that the value of $\cos \theta$ is equal to the ratio between the length of the projection of $u$ on $V$ and the length of $u$ $\left(\cos ^{2}\theta = \frac{\left\Vert u_{V}\right\Vert ^{2}}{\left\Vert u\right\Vert ^{2}}\right)$. Likewise, if $U = \text{span}\; \{u, w_{2}, \cdots, w_{s}\}$ and $V = \text{span}\; \{v, w_{2}, \cdots, w_{s}\}$ are $s$-dimensional subspaces of $X$ that intersects on $(s-1)$-dimensional subspace $W = \text{span}\; \{w_{2}, \cdots, w_{s}\}$ with $s\geq 2$ then the angle between $U$ dan $V$ is defined by $\theta$ with $0\leq\theta\leq\frac{\pi}{2}$ and $\cos ^{2}\theta = \frac{\langle u_{W}^{\perp }, v_{W}^{\perp }\rangle ^{2}}{\left\Vert u_{W}^{\perp }\right\Vert ^{2}\left\Vert v_{W}^{\perp }\right\Vert ^{2}}$ with $u_{W}^{\perp }, v_{W}^{\perp }$ are the orthogonal complement of $u$ and $v$, respectively, on $W$. Gunawan et al. ^[3] show that the value of $\cos \theta$ is equal to the ratio between the volume of the $s$-dimensional parallelepiped spanned by the projection of $u, w_{2}, \cdots, w_{s}$ on $V$ and the volume of the $s$-dimensional parallelepiped spanned by $u, w_{2}, \cdots, w_{s}$.

Now suppose $(X, \left\Vert \cdot\right\Vert)$ is a normed space. As it is known, not all normed spaces are inner product spaces. For instance, the space $\ell^{p}$ for $1\leq p < \infty$ with norm $\left\Vert x\right\Vert_{p} = \left[ \underset{n = 1}{\overset{\infty }{\sum\limits }}\left\vert x_{n}\right\vert ^{p}\right] ^{\frac{1}{p}}$ is not an inner product space except for $p = 2$ (see ^[6]). Konca et al. ^[5] define a (weighted) inner product $\langle\cdot, \cdot\rangle_{v}$ on $\ell^{p}$ where $2 < p < \infty$, but $\left(\ell^{p}, \langle\cdot, \cdot\rangle_{v}\right)$ is not Banach space. They also find that the inner product is actually defined on a larger space namely $\ell_{2, v}$ that contains $\ell^{p}$ and $\left(\ell_{2, v}, \langle\cdot, \cdot\rangle_{v}\right)$ is Banach space. More recent works may be found in ^[4,8,12]

Now suppose $(A, \left\Vert \cdot\right\Vert)$ is a normed space where $A$ is subspace of $X$. Two questions arise: What are $A$ complete? Can we define a norm on $A$ which satisfies the parallelogram law? The reason why we are interested in the parallelogram law is because we eventually wish to define an inner product, so that we can define angle between two subspaces and many other notions on this space. The concept of the angle between two vectors and two subspaces in normed space has been studied intensively, see ^{[1,2,7,9,10,11]}.

Let $(\ell^{p}, \left\Vert \cdot\right\Vert_{p})$ be a normed space and $A$ is subspace in $\ell^{p}$ for $1\le p < \infty$. In this paper, we discuss completness of $A$ that equipped with usual norm on $\ell^{p}$. We also introduce a new inner product $\langle \cdot, \cdot\rangle _{b, 2}$ on $A$ such that $\left(A, \left\Vert \cdot\right\Vert_{b, 2}\right)$ is complete where $\left\Vert x\right\Vert_{b, 2} = \left(\langle x, x\rangle_{b, 2}\right)^{\frac{1}{2}}$ denotes the induced norm in $A$. Motivated by this fact, we shall discuss the angle between two subspaces in $A$.

2.   Main results

2.1.   The completness of subspaces on $\ell^{p}$

Let $\{x_{1}, \ldots, x_{n}\}$ be linearly independent set on $\ell^{p}$ for $1\le p < \infty$. Define $A = \text{span}\; \{x_{1}, \ldots, x_{n}\}$. Then we observe that $\left(A, \left\Vert \cdot\right\Vert_{p}\right)$ is a subspace of $\left(\ell^{p}, \left\Vert \cdot\right\Vert_{p}\right)$. In ^[6], we know that $\ell^{p}$ is Banach space. Here, $A$ will be proved as Banach space. Before this result, we have the following lemma.

Lemma 2.1. Let $C = \sum\limits_{i = 1}^{n}a_{i}$ where $a_{1}, \ldots, a_{n}\in R$.

(a) If $sign(a_{j}) = sign(C)$ for $j = 1, \ldots, n$ then the equality

holds for some $t\in \left(0, 1\right)$.

(b) If $sign(a_{j}) = -sign(C)$ for $j = 1, \ldots, n$ then the equality

holds for some $t\in \left(0, 1\right)$.

Proof. (a) Without loss of generality, writing $j = 1.$ Suppose that $\text{sign}(a_{1}) = \text{sign}(C)$. Choose $t\in \left(0, 1\right)$ such that

and $\text{sign}(D) = \text{sign}(C)$. We can write $C = ta_{1}+D$. Hence, $\left\vert C\right\vert = \left\vert ta_{1}\right\vert +\left\vert D\right\vert.$

(b) Without loss of generality, writing $j = 1$ and $\text{sign}(C) = 1.$ Suppose that $\text{sign}(a_{1}) = -\text{sign}(C).$ Choose $t\in \left(0, 1\right)$ such that

and $\text{sign}(E) = \text{sign}(C).$ We observe that $C,$ $-ta_{1}$, $E$ are positive real. Hence, we can write $\left\vert C\right\vert = \left\vert ta_{1}\right\vert +\left\vert E\right\vert$.

Using Lemma 2.1, We shall now show that $\left(A, \left\Vert \cdot\right\Vert_{p}\right)$ is Banach space.

Theorem 2.2. Let $\{x_{1}, \ldots, x_{n}\}$ be linearly independent set on $\ell^{p}$ where $1\leq p < \infty$. If $A = span\{x_{1}, \ldots, x_{n}\}$ then $\left(A, \left\Vert \cdot\right\Vert_{p}\right)$ is Banach space.

Proof. We consider any Cauchy sequence $(y_{k})$ in $A,$ writing $y_{k} = \alpha _{k1}x_{1}+\alpha _{k2}x_{2}+\ldots +\alpha _{kn}x_{n}$. Since $(y_{k})$ is Cauchy sequence, we have for every $\epsilon > 0$ there exist $N_{\epsilon}$ such that

for all $k, l > N_{\epsilon}$. It follows that for every $j = 1, 2, \ldots$ we have

Case 1. For some $i = 1, \ldots, n$, $\text{sign}\left(\sum\limits_{i = 1}^{n}\left(\alpha _{ki}-\alpha _{li}\right) \zeta _{ij}\right) = \text{sign}\left(\alpha _{ki}-\alpha _{li}\right) \zeta _{ij}$ holds. By Lemma 1, we can find $t\in \left(0, 1\right)$ such that $\left\vert t\left(\alpha _{ki}-\alpha _{li}\right) \zeta _{ij}\right\vert < \epsilon _{1}.$

Case 2. For some $i = 1, \cdots, n,$ $\text{sign}\left(\sum\limits_{i = 1}^{n}\left(\alpha _{ki}-\alpha _{li}\right) \zeta _{ij}\right) = \text{sign}\left(\alpha _{ki}-\alpha _{li}\right) \zeta _{ij}$ holds. By Lemma 1, we can find $t\in \left(0, 1\right)$ such that $\left\vert t\left(\alpha _{ki}-\alpha _{li}\right) \zeta _{ij}\right\vert < \epsilon _{1}.$

So, we obtain $\left\vert t\zeta _{ij}\right\vert \left\vert \left(\alpha _{ki}-\alpha _{li}\right) \right\vert < \epsilon _{1}$ for all $k, l\geq N_{\epsilon }$. Thus, for each fixed $i\in \left\{ 1, \ldots, n\right\}$, $\left(\alpha _{ki}\right)$ is a Cauchy sequence of real numbers. Hence, it is convergent, say $\alpha _{ki}\rightarrow \alpha _{i}$ as $k\rightarrow \infty$. Now, we can view that $\alpha = \left(\alpha _{1}, \ldots, \alpha _{n}\right)$ and we define $y: = \alpha _{1}x_{1}+\ldots +\alpha _{n}x_{n}$. Its obvious that $y\in A.$ Because $\alpha _{k}$ converge to $\alpha$ then $y_{k}$ converge $y$. Hence, $\left(A, \left\Vert \cdot\right\Vert_{p}\right)$ is Banach Space.

Next, we discuss a new inner product in $\ell^{p}$ and prove that $A$ with the new inner product is Hilbert space. Write $b = \underset{i = 1}{\overset{n}{\sum\limits }}\left\vert x_{i}\right\vert$. We define a following mapping

for every $y, z\in \ell^{p}$. Using Hölder's inequality, we have

Hence, the mapping $\langle \cdot, \cdot\rangle _{b, 2}$ is defined on $\ell^{p}$. Next, we have the following proposition.

Proposition 2.3. The mapping $\langle \cdot, \cdot\rangle _{b, 2}$ in $(2.1)$ defines an inner product on $\ell^{p}$.

Proof. For every $y, z, w \in \ell^p,$ we will verify that $\langle \cdot, \cdot \rangle _{b, 2}$ satisfies the four properties of an inner product.

1) Since $\underset{k = 1}{\overset{\infty }{\sum\limits }}b_{k}^{p-2}y_{k}^{2} \geq 0$, we have $\langle y, y\rangle _{b, 2}\geq 0$.

Next, we show that $\langle y, y\rangle _{b, 2} = 0$ if only if $y = 0$. Suppose that $y = 0,$ then

Conversely, if $\langle y, y\rangle _{b, 2} = 0$, then $\underset{k = 1 }{\overset{\infty }{\sum\limits }}b_{k}^{p-2}y_{k}^{2} = 0$. Since $b_{k}^{p-2}\neq 0$, we obtain $y = 0.$

2) Observe that

3) Observe that

4) Observe that

Therefore $\langle \cdot, \cdot \rangle _{b, 2}$ defines an inner product on $\ell^{p}$.

Corollary 2.4. The following function

defines a norm that corresponds to the inner product $\langle \cdot, \cdot\rangle _{b, 2 }$ on $\ell^{p}$.

Using a norm that corresponds to the inner product $\langle \cdot, \cdot\rangle _{b, 2 }$, we have the following result.

Theorem 2.5. The space $\left(A, \left\Vert \cdot \right\Vert _{b, 2}\right)$ is complete. Accordingly, $\left(A, \langle \cdot, \cdot\rangle _{b, 2}\right)$ is a Hilbert space.

Proof. We consider any Cauchy sequence $(y_{k})$ in $A,$ writing $y_{k} = \alpha _{k1}x_{1}+\ldots +\alpha _{kn}x_{n}$ and $\alpha _{k} = \left(\alpha _{k1}, \ldots, \alpha _{kn}\right)$. Since $y_{k}$ is Cauchy, for every $\epsilon > 0$ there is an $N_{\epsilon }$ such that for all $k, l > N_{\epsilon }$,

It follows that for every $j = 1, 2, \ldots$ we have

So, we obtain $\left\vert \zeta _{ij}\right\vert b_{j}^{\frac{p-2}{2}}\left\vert \alpha _{ki}-\alpha _{li}\right\vert < \epsilon$ for all $k, l\geq N_{\epsilon }$. Thus, for each fixed $i\in \left\{ 1, \ldots, n\right\}$, $\left(\alpha _{ki}\right)$ is a Cauchy sequence of real numbers. Hence, it is convergent, say $\alpha _{ki}\rightarrow \alpha _{i}$ as $k\rightarrow \infty$. Now, we can view that $\alpha = \left(\alpha _{1}, \ldots, \alpha _{n}\right)$ and we define $y: = \alpha _{1}x_{1}+\ldots +\alpha _{n}x_{n}$. Its obvious that $y\in A.$ Because $\alpha _{k}$ converge to $\alpha$ then $y_{k}$ converge $y.$ Hence, $\left(A, \left\Vert \cdot \right\Vert _{b, 2}\right)$ is complete.

2.2.   Angle betweeen two subspaces on $A$

We know that $\left(A, \langle \cdot, \cdot\rangle _{b, 2}\right)$ is a Hilbert space with $A = \text{span}\; \{x_{1}, \cdots, x_{n}\}$ and, as before, $\{x_{1}, \cdots, x_{n}\}$ is a linearly independent set on $\ell^{p}$. Using the Gram-Schmidt process, we have an orthonormal set $\{y_{1}, \cdots, y_{n}\}$. As a consequence, $span\{x_{1}, \cdots, x_{n}\} = \text{span}\; \{y_{1}, \cdots, y_{n}\}$. For every $u, v\in A$, we can write $u = \sum\limits_{i = 1}^{n}c_{i}y_{i}$ and $v = \sum\limits_{i = 1}^{n}d_{i}y_{i}$ where $c_{i}, d_{i}\in\mathbb{R}$ for every $i = 1, \ldots, n$. Moreover,

Because $\langle y_{i}, y_{j}\rangle _{b, 2} = 0$ for $i\neq j$ and $\langle y_{i}, y_{i}\rangle _{b, 2} = \left\Vert y_{i}\right\Vert _{b, 2}^{2}$ then

with $c = (c_{1}, \cdots, c_{n})$ and $d = (d_{1}, \cdots, d_{n})$. If $u = v$, then $\left\Vert u\right\Vert _{b, 2} = \left(\sum\limits_{i = 1}^{n}c_{i}^{2}\right) ^{\frac{1}{2}}$.

According to the above form, we conclude that an inner product on $A$ can be viewed as the inner product of the Euclidean space $\mathbb{R}^{n}$. Hence, an explisit formula of the angle between $u$ and $v$ of $A$, denoted $\theta$, is given by

Note that the angle between two vectors of $A$ is also the angle between two lines of $A$. Moreover, the angle between two vectors of $A$ coincides with the angle between two vectors of $\mathbb{R}^{n}$.

Example 2.6. Let $A = \text{span}\; \{x_{1}, \cdots, x_{5}\}$ with $\{x_{1}, \cdots, x_{5}\}$ be the orthonormal set on $\ell^{p}$. If $u = x_1+2x_2+3x_4+4x_5$ and $v = x_2+3x_3+5x_4+2x_5$ then the angle between $u$ and $v$ of $A$ is

Next, we can discuss angle between $1$-dimensional subspace and $(n-1)$-dimensional subspace where $n\ge 2$ of $A$. The result is shown as follows.

Proposition 2.7. If $U = span\{x_{k}\}$ where $k = 1, \ldots, n$ and $V = span\{x_{i_{2}(k)}, \ldots, x_{i_{n}(k)}\}$ where $\{i_{2}(k), \ldots, i_{n}(k)\} = \{1, 2, \ldots, n\}-\{k\}$ of $A$, then the angle between subspaces $U$ and $V$ is $\theta$ $(0\leq \theta \leq \frac{ \pi }{2})$ with

Proof. Without loss of generality, write $k = 1$. Using the Gram-Schmidt process, we have an orthonormal set $\{y_{2}, \cdots, y_{n}\}$. As a consequence, $\text{span}\; \{x_{2}, \cdots, x_{n}\} = \text{span}\; \{y_{2}, \cdots, y_{n}\}.$ The projection of $x_{1}$ on $V$ is given by

Then, we have

Hence, we obtain

Example 2.8. Let $A = \text{span}\; \{x_{1}, x_{2}, x_{3}\}$ with $\{x_{1}, x_{2}, x_{3}\}$ be the orthonormal set in $\ell^{1}$. Take $x_1 = (1, 0, 0, 0, \ldots)$, $x_2 = (0, 1, 0, 0, \ldots)$ and $x_{3} = (0, 0, 1, 0, \ldots)$, so that we have $b = (1, 1, 1, 0, \ldots)$. Clearly $\left\Vert x_{1}\right\Vert _{b, 2}^{2} = 1$, $\langle x_{1}, x_{2}\rangle _{b, 2}^{2} = 0$ and $\langle x_{1}, x_{3}\rangle _{b, 2}^{2} = 0$. If $U = \text{span}\{x_{1}\}$ and $V = \text{span}\{x_{2}, x_{3}\}$ of $A$ then angle between $U$ and $V$ is $\theta$ with

Hence $\theta = \frac{\pi}{2}.$

Before we can discuss an explicit formula for the cosine of the angle between two subspaces of $A = \text{span}\; \{x_{1}, \cdots, x_{n}\}$, we recall definition of angle two subspace in inner product space as follows.

Definition 2.9. ^[3] Let $(X, \langle\cdot, \cdot\rangle)$ be a real inner product space. If $U = \text{span}\; \{u_{1}, \ldots, u_{t}\}$ is a $t$-dimensional subspace and $V = \text{span}\; \{v_{1}, \ldots, v_{s}\}$ is a $s$-dimensional subspace of $X$ with $t\leq s$, then the angle between subspaces $U$ and $V$ is defined by $\theta$ $(0\leq \theta \leq \frac{ \pi }{2})$ and

where $u_{i}^{V}$ denote the projection of $u_{i}$ on $V$ for each $i = 1, \ldots, t$.

Using Definition 2.9, we have the following theorem.

Theorem 2.10. If $U = span\{x_{1}, \ldots, x_{n_{1}}\}$ is a $n_{1}$-dimensional subspace and $V = span\{x_{n_{1}+1}, \ldots, x_{n}\}$ is a $(n-n_{1})$-dimensional subspace of $A$ with $n_{1}\leq \frac{n}{2}$, then the angle between subspaces $U$ and $V$ is defined by $\theta$ and

where $[\langle x_{i}, y_{l}\rangle _{b, 2}]$ is a $(n_{1}\times \left(n-n_{1}\right))$ matrix and $[\langle x_{i}, y_{l}\rangle _{b, 2}]^{T}$ is its transpose for $i, j = 1, \ldots, n_{1}$.

Proof. Suppose that $\{ x_{n_{1}+1}, \ldots, x_{n}\}$ is linearly independent. Using the Gram-Schmidt process, we obtain the orthonormal set $\{y_{n_{1}+1}, \ldots, y_{n}\}$. Here $\text{span}\; \{ x_{n_{1}+1}, \ldots, x_{n}\} = \text{span}\; \{y_{n_{1}+1}, \ldots, y_{n}\}$. For each $i = 1, \ldots n_{1}$, the projection of $x_{i}$ on V is given by

So, for $i, j = 1, \ldots, n_{1}$, we have

Next, using formula angle in ^[3], we obtain

where $[\langle x_{i}, y_{l}\rangle _{b, 2}]$ is a $(n_{1}\times \left(n-n_{1}\right))$ matrix and $[\langle x_{i}, y_{l}\rangle _{b, 2}]^{T}$ is its transpose. Therefore, cosine of the angle between U and V is

where $[\langle x_{i}, x_{j}\rangle _{b, 2}]$ is a $(n_{1}\times n_{1})$ matrix.

Next, we discuss angle between two subspaces that intersects on a subspace of $A$. Write $A = \text{span}\left\{ x_{1}, \ldots, x_{n_{1}}, x_{n_{1}+1}, \ldots, x_{n_{1}+n_{2}}, x_{n_{1}+n_{2}+1}, \ldots, x_{n_{1}+n_{2}+n_{3}}\right\}$ with $n = n_{1}+n_{2}+n_{3}$. Suppose now that $U = \text{span}\left\{ x_{1}, \ldots, x_{n_{1}}, x_{n_{1}+1}, \ldots, x_{n_{1}+n_{2}}\right\}$ and $V = \text{span}\left\{ x_{1}, \ldots, x_{n_{1}}, x_{n_{1}+n_{2}+1}, \ldots, x_{n_{1}+n_{2}+n_{3}}\right\}.$ We observe that $U$ and $V$ are subspace on $A\subseteq \ell^{p}$. Moreover, using Definition 2.9, we have angle between $U$ and $V$ as follows.

Theorem 2.11. If $U = span\left\{ x_{1}, \cdots, x_{n_{1}}, x_{n_{1}+1}, \cdots, x_{n_{1}+n_{2}}\right\}$ is a $(n_{1}+n_{2})$-dimensional subspace and $V = span\left\{ x_{1}, \cdots, x_{n_{1}}, x_{n_{1}+n_{2}+1}, \cdots, x_{n_{1}+n_{2}+n_{3}}\right\}$ is a $(n_{1}+n_{3})$-dimensional subspace of $A$ with $n_{2}\leq n_{3}$ that intersects on $n_{1}$-dimensional subspace $W = span\{x_{1}, \cdots, x_{n_{1}}\}$ then the angle between subspaces $U$ and $V$ is defined by $\theta$ with

where $(x_{i}^{V})_{W}^{\bot}$, $(x_{j}^{V})_{W}^{\bot}$, $(x_{i})_{W}^{\bot}$ and $(x_{j})_{W}^{\bot}$ are the orthogonal complement of $u_{i}^{V}$, $u_{j}^{V}$, $u_{i}$ and $u_{j}$, respectively, on $W$ for $i, j = n_{1}+1, \ldots, n_{1}+n_{2}$.

Proof. The projection of $x_{i}$ on $V$ is $x_{i}^{V}$. Next, we may write $x_{i}^{V} = (x_{i}^{V})_{W}+(x_{i}^{V})_{W}^{\bot}$ where $(x_{i}^{V})_{W}$ is the projection of $x_{i}^{V}$ on $W$ and $(x_{i}^{V})_{W}^{\bot}$ is the orthogonal complement of $x_{i}^{V}$ on $W$. In line with this, we may write $x_{i} = (x_{i})_{W}+(x_{i})_{W}^{\bot}$ where $(x_{i})_{W}$ is the projection of $x_{i}$ on $W$ and $(x_{i})_{W}^{\bot}$ is the orthogonal complement of $x_{i}$ on $W$ for $i = n_{1}+1, \ldots, n_{1}+n_{2}$. Using the standard $(n_{1}+n_{2})$-norm and properties of determinans, we obtain

with $[ \langle (x_{i}^{V})_{W}^{\bot }, (x_{j}^{V})_{W}^{\bot }\rangle_{b, 2} ]$ and $[ \langle (x_{i})_{W}^{\bot }, (x_{j})_{W}^{\bot }\rangle_{b, 2} ]$ are $(n_{2}\times n_{2})$ matrix.

3.   Conclusions

Based result has been given on the Sections $2$, we have known that $A$ that equipped with usual norm on $\ell^{p}$ is complete. We have introduced $\langle \cdot, \cdot\rangle _{b, 2}$ on $A$ and have shown that $\left(A, \left\Vert \cdot\right\Vert_{b, 2}\right)$ is complete. Next, we have got angle between two vectors and between $1$-dimensional subspace and $(s-1)$-dimensional subspace where $s\ge 2$ of $A$. Moreover, we have got angle between two subspaces that intersects on a subspace of $A$.

Acknowledgments

The first author is supported by PDUPT Program 2021 No. 752/UN4.22/PT.01.03/2021.

Conflict of interest

The authors declare that there is no conflict of interest.