Proof. The proof of the first part may be safely left to the reader. For the second part, the binomial series expansion of $(1+|t|^{q_1})^p$ in a small neighbourhood $\Omega$ of $0$ is given by

(2.6)\begin{align} (1+|t|^{q_1})^p=\sum_{k=0}^{\infty} \left(\begin{matrix} p\\ k \end{matrix}\right)|t|^{kq_1},\quad t\in \Omega, \text{ where }\left(\begin{matrix} p\\ k \end{matrix}\right)=\frac{p(p-1)\cdots(p-k+1)}{k!}. \end{align}

Let $\boldsymbol {m}$ be the least positive integer such that $\boldsymbol mq_1> 1$. Since the series in (2.6) is absolutely convergent in $\Omega$, we may write

\[ (1+|t|^{q_1})^p=\sum_{k=0}^{\boldsymbol{m}-1} \left(\begin{matrix} p\\ k \end{matrix}\right)|t|^{kq_1}+\mathcal{O}(|t|^{\boldsymbol mq_1}),\quad t\in \Omega. \]

Therefore,

(2.7)\begin{equation} (1+|t|^{q_1})^p-a|t|^{q_2}=\left\{\sum_{k=0}^{\boldsymbol{m}-1} \left(\begin{matrix} p\\ k \end{matrix}\right)|t|^{kq_1}-a|t|^{q_2}\right\}+\mathcal{O}(|t|^{\boldsymbol mq_1}), \quad t\in \Omega. \end{equation}

Note that $\left \{\sum \limits _{k=0}^{\boldsymbol {m}-1} (\begin {matrix} p\\ k \end {matrix})|t|^{kq_1}-a|t|^{q_2}\right \}$ is not differentiable at $0$ but $\mathcal {O}(|t|^{\boldsymbol mq_1})$ is differentiable at $0$. Hence, from (2.7), we conclude that $(1+|t|^{q_1})^p-a|t|^{q_2}$ is not differentiable. This completes the proof of the lemma.

Proof. The result follows from the same lines of proof of [Reference Chattopadhyay, Hong, Pal, Pradhan and Ray6, proposition 2.6].

Proof. Suppose $AB=0$. Then, $A$ and $B$ have disjoint supports. It follows that $\|A+tB\|_p^p=\|A\|_p^p+|t|^p\|B\|_p^p$. Note that $\|A\|_p=\|B\|_p=1$. Thus $\|A+tB\|_p^p=1+|t|^p.$ But for each $t\in \mathbb {R}$, the identity $(1+|t|^q)^{1/q}=(1+|t|^p)^{1/p}$ is not true, which leads to a contradiction.

Proof. On the contrary, suppose that there exists an isometric embedding $S_q^m$ into $S_p^n$. Then, by lemma 2.2, we have $p< q$. Therefore, by lemma 4.4, we may assume that there exists a diagonal self-adjoint matrix $A$ and a self-adjoint matrix $B$ in $S_p^n$ such that $(A, B)$ has $(\mathbf {I}_{q,p})$. In particular, for any real number $t$, we have the identity

(4.4)\begin{equation} \left(1+|t|^q\right)^{\frac{p}{q}}=\|A+tB\|_p^p. \end{equation}

In a small neighbourhood $\Omega$ of zero, by Kato–Rellich theorem 2.4, there exist $n$ real-analytic functions $\lambda _1(\cdot ), \lambda _2(\cdot ),\ldots, \lambda _n(\cdot )$ on $\Omega$ such that they are complete set of eigenvalues of $A+tB$ for $t\in \Omega$. So for $t\in \Omega$, from (4.4), we have

(4.5)\begin{equation} \left(1+|t|^q\right)^{\frac{p}{q}}=\sum_{k=1}^{n}|\lambda_k(t)|^p. \end{equation}

If $\lambda _k(0)\neq 0$ for all $1\leq k\leq n$, that is if $A$ is invertible, then the right-hand side of (4.5) is a real-analytic function in $\Omega$ but the left-hand side of (4.5) is not analytic in $\Omega$. So $A$ must be a singular matrix. Again from (4.5), it is clear that not all $\lambda _k(0)$ are zero. So there is a non-empty proper subset $O\subset \{1,2,\ldots,n\}$ such that $\lambda _k(0)=0$ for $k\in O$, and $\lambda _k(0)\neq 0$ for $k\in \{1,2,\ldots,n\}\setminus O$. Since the quasi-norm $\|\cdot \|_p$ is unitary invariant, we may assume that there exists an $l\in \{1,2,\ldots,n-1\}$ such that $\lambda _k(0)=0$ for $1\leq k\leq l$, and $\lambda _k(0)\neq 0$ for $l+1\leq k\leq n$. Note that if some $\lambda _i(t)$ are identically zero, we can neglect those as they have no contribution in $\|A+tB\|_p$. Therefore, we may also assume that all $\lambda _k(t)$ are non-zero analytic functions in $\Omega$. Let $m_k$ be the multiplicity of zero at $0$ of the function $\lambda _k(t)$ for $1\leq k\leq l$. Then,

(4.6)\begin{equation} \lambda_k(t)=t^{m_k}\,\mu_k(t),\quad \mu_k(0)\neq 0,\quad 1\leq k\leq l, \end{equation}

where $\mu _k(\cdot )$ are analytic in $\Omega$. For further progression, let us define

(4.7)\begin{equation} \Psi(t)=\sum_{k=l+1}^{n}|\lambda_k(t)|^p-1, \quad t\in\Omega. \end{equation}

It is clear that $\Psi (\cdot )$ is analytic in $\Omega$ and $\Psi (0)=0$. Then, $\Psi (t)=t\Psi _{1}(t)$, where $\Psi _{1}(\cdot )$ is a analytic function in $\Omega$. Applying (4.6) and (4.7) in (4.5), we have

(4.8)\begin{equation} \left(1+|t|^q\right)^{\frac{p}{q}}=\sum_{k=1}^{l}|t|^{m_k p}|\,\mu_k(t)|^p+t\Psi_{1}(t)+1, \quad t\in\Omega. \end{equation}

Now we will reach our goal through case-by-case analysis.

Case I: Let $m_kp>1$ for $1\leq k\leq l$. Then, the right-hand side of (4.8) is differentiable at $0$ but the left-hand side of (4.8) is not differentiable at $0$. So this case is not possible.

Case II: Suppose at least one of $m_kp=1$ for $1\leq k\leq l$. Without loss of generality, we assume that $m_1p=1.$ Note that as $\mu _k(0)\neq 0$, $|\mu _k(t)|^p$ are analytic in $\Omega$ for $1\leq k\leq l$. So we can express $|\mu _k(t)|^p$ as

(4.9)\begin{equation} |\mu_k(t)|^p=|\mu_k(0)|^p+ t\,\xi_k(t), \quad 1\leq k\leq l, \quad t\in\Omega, \end{equation}

where $\xi _k(\cdot )$ are some analytic function in the small neighbourhood $\Omega$ of $0$. Let $\boldsymbol {m}$ be the least positive integer such that $\boldsymbol mq> 1$.

Sub-case I: Suppose $l\geq 3$, $m_1p=m_2p=\cdots =m_jp=1$, $m_{j+1}p, \ldots, m_ip<1$ and $m_{i+1}p,\ldots,m_lp>1$ for some $j\in \{1,2,\ldots, l-2\}$, and $i\in \{j+1\ldots, l-1\}$. Combining (4.8) and (4.9), we have

(4.10)\begin{align} & 1+\alpha_1\,|t|^{q}+\alpha_2\,|t|^{2q}+\cdots+\alpha_{\boldsymbol{m}-1}\,|t|^{(\boldsymbol{m}-1)q}+\mathcal{O}(|t|^{\boldsymbol mq}) =1+\sum_{k=1}^{j}|t|\,|\mu_k(0)|^p\nonumber\\ & \quad+\sum_{k=j+1}^{i} |t|^{m_kp}\,|\mu_k(0)|^p +\sum_{k=1}^{i} |t|^{m_kp}t\xi_k(t)+\sum_{k=i+1}^{l} |t|^{m_kp}\,|\mu_k(t)|^p+t\Psi_{1}(t), \enspace t\in\Omega, \end{align}

where $\alpha _k$ is the coefficient of $|t|^{kq}$ in the binomial expansion of $(1+|t|^q)^{p/q}$ for $|t|<1$. That is, $\alpha _k=(\begin {matrix} p/q\\ k \end {matrix})=\frac {p/q(p/q-1)\cdots (p/q-k+1)}{k!}$, $k\in \mathbb {N}$.

From (4.10), we have

(4.11)\begin{align} & \left\{\alpha_1\,|t|^{q}+\alpha_2\,|t|^{2q}+\cdots+\alpha_{\boldsymbol{m}-1}\,|t|^{(\boldsymbol{m}-1)q}\right\}\nonumber\\ & \qquad -\left\{\sum_{k=1}^{j}|t|\,|\mu_k(0)|^p + \sum_{k=j+1}^{i} |t|^{m_kp}\,|\mu_k(0)|^p \right\}\nonumber\\ & \quad = \sum_{k=1}^{i} |t|^{m_kp}t\xi_k(t) + \sum_{k=i+1}^{l} |t|^{m_kp}\,|\mu_k(t)|^p+\mathcal{O}(|t|^{\boldsymbol mq})+t\Psi_{1}(t), \quad t\in\Omega. \end{align}

In this case, the right-hand side of the above equation (4.11) is differentiable at $0$. Since $p< q$, so $\alpha _{2}<0$. Therefore, by the similar lines of argument as given in the proof of lemma 2.5, the left-hand side of (4.11) is not differentiable at $0$. Hence, this case does not exist.

Sub-case II: Suppose $l\geq 2$, $m_1p=\cdots =m_jp=1$ and $m_{j+1}p,\ldots,m_lp<1$ for some $j\in \{1,\ldots,l-1\}$. Then, for $t\in \Omega$, from (4.10), we get

(4.12)\begin{align} & \alpha_1\,|t|^{q}+\cdots+\alpha_{\boldsymbol{m}-1}\,|t|^{(\boldsymbol{m}-1)q} -\sum_{k=1}^{j}|t|\,|\mu_k(0)|^p-\sum_{k=j+1}^{l} |t|^{m_kp}\,|\mu_k(0)|^p\nonumber\\ & = \sum_{k=1}^{l} |t|^{m_kp}t\xi_k(t)+\mathcal{O}(|t|^{\boldsymbol mq})+t\Psi_{1}(t). \end{align}

It is clear that the left side of equation (4.12) is not differentiable at $0$, but the other side is differentiable at $0$. So equation (4.12) never exists in $\Omega$.

Sub-case III: Suppose $l\geq 2$, $m_1p=\cdots =m_jp=1$, $m_{j+1}p,\ldots,m_lp>1$ for some $j\in \{1,\ldots,l-1\}$. Again from (4.10), we have

(4.13)\begin{align} & \alpha_1\,|t|^{q}+\cdots+\alpha_{\boldsymbol{m}-1}\,|t|^{(\boldsymbol{m}-1)q}-\sum_{k=1}^{j}|t|\,|\mu_k(0)|^p\nonumber\\ & \quad = \sum_{k=1}^{j}|t|\, t\, \xi_k(t)+\sum_{k=j+1}^{l} |t|^{m_kp}\,|\mu_k(t)|^p+t\Psi_{1}(t)+\mathcal{O}(|t|^{\boldsymbol mq}),\quad t\in\Omega. \end{align}

By a similar argument as given for the non-existence of equation (4.12), we conclude the non-existence of equation (4.13).

Sub-case IV: Let $m_1p=\cdots =m_lp=1$. By similar argument as given in the above sub-case, we conclude that this situation never exists.

Case III: Let $m_kp<1$ for $1\leq k\leq l$. From (4.10), we have

(4.14)\begin{align} & \alpha_1\,|t|^{q}+\cdots+\alpha_{\boldsymbol{m}-1}\,|t|^{(\boldsymbol{m}-1)q} -\sum_{k=1}^{l} |t|^{m_kp}\,|\mu_k(0)|^p\nonumber\\& \quad =\sum_{k=1}^{l} |t|^{m_kp}t\xi_k(t)+t\Psi_{1}(t)+\mathcal{O}(|t|^{\boldsymbol mq}), \quad t\in\Omega. \end{align}

Note that the right-hand side of (4.14) is differentiable at $0$. Therefore, the existence of equation (4.14) implies that the left-hand side of (4.14) is also differentiable at $0$.

Sub-case I: Let $2q\leq 1$. The necessary condition for differentiability of the left-hand side of (4.14) is that there exists a non-empty proper subset $I\subset \{1,2,\ldots,l\}$ such that

1. (a) $m_kp=2q,\ k\in I,$ and $\alpha _2=\sum \limits _{k\in I}|\mu _k(0)|^p.$

By lemma 2.2, we obtain $\frac {p}{q}<1$. Hence, $\alpha _2=\frac {1}{2!}\frac {p}{q}(\frac {p}{q}-1)<0$, but $\sum \limits _{k\in I}|\mu _k(0)|^p>0$. So the above identity (A) is not true. Hence, equation (4.14) never exists in the small neighbourhood $\Omega$, which is a contradiction.

Sub-case II: Let $2q> 1$. In this case, the necessary condition for differentiability of the left-hand side of (4.14) is that

1. (a) $m_kp=q,\ k\in \{1,2,\ldots,l\}, \text {and}\ \alpha _1=\sum \limits _{k=1}^{l}|\mu _k(0)|^p$.

Suppose the above necessary condition (a) holds, then from (4.14), we have,

(4.15)\begin{align} & \alpha_{2}\,|t|^{2q-1}\operatorname{sgn}(t)+\alpha_{3}\,|t|^{3q-1}\operatorname{sgn}(t)+\mathcal{O}(|t|^{4q-1})\nonumber\\& \quad =\sum_{k=1}^{l} |t|^{q}\xi_k(t)+\Psi_{1}(t), \quad t\in\Omega\setminus\{0\}.\end{align}

Taking limit $t\to 0$ on both sides of (4.15), we have $\Psi _{1}(0)=0$ which implies $\Psi _{1}(t)=t\Psi _{2}(t)$ for some analytic function $\Psi _{2}(t)$ in $\Omega$ of $0$. Let $\xi _k(t)=\xi _k(0)+t\,\eta _k(t), 1\leq k\leq l$ and $\eta _k$ are some analytic functions in $\Omega$. Then, from (4.15), we have

(4.16)\begin{align} & \alpha_{2}\,|t|^{2q-2}+\alpha_{3}\,|t|^{3q-2}-|t|^{q-1}\operatorname{sgn}(t)\sum_{k=1}^{l} \xi_k(0)\nonumber\\& \quad =\sum_{k=1}^{l}|t|^q \eta_k(t)+\Psi_{2}(t)+\mathcal{O}(|t|^{4q-2}), \quad t\in\Omega_2\setminus\{0\}. \end{align}

Observe that the right-hand side of (4.16) is bounded around $0$ but the left-hand side is unbounded around $0$, which leads to a contradiction.

Hence, case III does not exist.

Case IV: Suppose $l\geq 2$ and there are some $m_kp$ which are strictly less than $1$ and rest are strictly greater than $1$. We may assume $m_1p,\ldots,m_ip<1$ and $m_{i+1}p,\ldots,m_lp>1$ for some $i\in \,\{1,2,\ldots,l-1\}$. From (4.10), we obtain

(4.17)\begin{align} & \left\{\alpha_1\,|t|^{q}+\alpha_2\,|t|^{2q}+\cdots+\alpha_{\boldsymbol{m}-1}\,|t|^{(\boldsymbol{m}-1)q}\right\}-\sum_{k=1}^{i} |t|^{m_kp}\,|\mu_k(0)|^p\nonumber\\ & \quad = \sum_{k=1}^{i} |t|^{m_kp}t\xi_k(t) + \sum_{k=i+1}^{l} |t|^{m_kp}\,|\mu_k(t)|^p+\mathcal{O}(|t|^{\boldsymbol mq})+t\Psi_{1}(t), \quad t\in\Omega. \end{align}

Then, we again consider the following two sub-cases:

Sub-case I: Let $2q\leq 1$. By the similar lines of argument as given in sub-case I of sase III of this proof, we conclude that this situation never exists.

Sub-case II: Let $2q> 1$. In this case, the necessary condition for differentiability of the left-hand side of (4.14) is that

1. (1) $m_kp=q,~~ k\in \{1,2,\ldots,i\}, \text { and } \alpha _1=\sum \limits _{k=1}^{i}|\mu _k(0)|^p$.

Suppose the above necessary condition (1) holds, then from (4.17), we have,

(4.18)\begin{align} & \alpha_{2}\,|t|^{2q-1}\operatorname{sgn}(t)+\alpha_{3}\,|t|^{3q-1}\operatorname{sgn}(t)+\mathcal{O}(|t|^{4q-1})\nonumber\\ & \quad = \sum_{k=1}^{i} |t|^{q}\xi_k(t) + \sum_{k=i+1}^{l} |t|^{m_kp-1}\,\operatorname{sgn}(t)|\mu_k(t)|^p+\Psi_{1}(t), \quad t\in\Omega. \end{align}

Since $2q>1$ and $m_kp>1$ for all $i+1\leq k\leq l$, then taking limit $t\to 0$ on both sides of (4.18), we have $\Psi _1(0)=0$. Let $\Psi _{1}(t)=t\Psi _{2}(t)$. Let $\xi _k(t)=\xi _k(0)+t\,\eta _k(t), 1\leq k\leq l$ and $\eta _k$ are some analytic functions in $\Omega$. Then, from (4.18), we get

(4.19)\begin{align} & \alpha_{2}\,|t|^{2q-2}+\alpha_{3}\,|t|^{3q-2}-\operatorname{sgn}{(t)}\,|t|^{q-1}\,\sum_{k=1}^{i} \xi_k(0)-\sum_{k=i+1}^{l} |t|^{m_kp-2}\,|\mu_k(0)|^p\nonumber\\ & \quad = \sum_{k=1}^{i} |t|^{q}\eta_k(t) + \sum_{k=i+1}^{l} |t|^{m_kp-2}\,t\,\xi_k(t)+\Psi_{2}(t) +\mathcal{O}(|t|^{4q-2}), \quad t\in\Omega\setminus\{0\}. \end{align}

Note that as $\alpha _{2}<0$, the left-hand side of (4.19) is unbounded near $0$ but the right-hand side of (4.19) is bounded near $0$ and tends to $\Psi _2(0)$ as $t\to 0$, which implies that this case never exists.

Therefore, from all the above cases, we conclude that there does not exist any isometric embedding of $S_q^m$ into $S_p^n$ for $0< q\neq p<1$. This completes the proof of the theorem.

Proof. Suppose that there is an isometric embedding of $S_q^m$ into $S_p^n$. Then, there exist $A,B\in S_p^n$ such that $(A,B)$ has $(\mathbf {I}_{q,p})$. Here, we use the same notations used in the proof of theorem 4.8. So the existence of isometric embedding implies that identity (4.5) is true in a small neighbourhood $\Omega$ of $0$. Note that if $A$ is invertible, then the right-hand side of (4.5) is real-analytic but the left-hand side of (4.5) is not real-analytic in $\Omega$. So the matrix $A$ is not invertible. Therefore, by the similar argument as given in the proof of theorem (4.8), we can deduce (4.8) from identity (4.5). In other words, we have the following.

(4.20)\begin{equation} \left(1+|t|^q\right)^{\frac{p}{q}}=\sum_{k=1}^{l}|t|^{m_k p}|\,\mu_k(t)|^p+t\Psi_{1}(t)+1,\quad t\in\Omega. \end{equation}

Note that the left-hand side of (4.20) is one time differentiable at $0$, and one time differentiability of (4.20) implies that some of $m_kp$ are in $(1,2)$ and rest belong to $[2,\infty )$. Since $\|\cdot \|_p$ is unitary invariant, we may assume that there exists $i\in \{1,2,\ldots,l-1\}$ such that

\[ 1< m_kp<2 \text{ for } 1\leq k\leq i, \text{ and } 2\leq m_kp<\infty\text{ for }i+1\leq k\leq l. \]

A simple calculation shows that $\Psi _{1}(0)=0$. Let $|\mu _k(t)|^p=|\mu _k(0)|^p+ t\xi _k(t)$, $1\leq k\leq l$, and $\Psi _{1}(t)=t\Psi _{2}(t)$, where $\xi _k(\cdot )$ and $\Psi _{2}(\cdot )$ are some analytic functions in $\Omega$. Then, from (4.20), we have

(4.21)\begin{align} & 1+\alpha_1\,|t|^{q}+\alpha_2\,|t|^{2q}+\mathcal{O}(|t|^{3q})\nonumber\\ & \quad =\sum_{k=1}^{i} |t|^{m_kp}\,|\mu_k(0)|^p+\sum_{k=1}^{i} |t|^{m_kp}t\xi_k(t)\nonumber\\ & \qquad+\sum_{k=i+1}^{l} |t|^{m_kp}\,|\mu_k(t)|^p+t^2\Psi_{2}(t)+1,\quad t\in\Omega_1,\nonumber\\ & \quad \implies\alpha_1\,|t|^{q-2}-\sum_{k=1}^{i} |t|^{m_kp-2}\,|\mu_k(0)|^p \nonumber\\ & \quad = \sum_{k=1}^{i} |t|^{m_kp-2}t\xi_k(t)\nonumber\\ & \qquad+\sum_{k=i+1}^{l} |t|^{m_kp-2}\,|\mu_k(t)|^p+\Psi_{2}(t) -\alpha_2\,|t|^{2q-2}+\mathcal{O}(|t|^{3q-2}),\quad t\in\Omega\setminus\{0\}. \end{align}

Observe that if $m_kp\neq q$ for some $k\in \{1,\ldots,i\}$, then the left-hand side of above equation (4.21) becomes unbounded near $0$ but the right-hand side of (4.21) is bounded and going to $\Psi _2(0)$ as $t\to \infty$. So $m_kp=q$ for all $k\in \{1,\ldots,i\}$. Again by the same reason, we have $\alpha _1=\sum \limits _{k=1}^{i}|\mu _k(0)|^p$. Therefore, from (4.21), we have

(4.22)\begin{align} \alpha_2\,|t|^{2q-2}=\sum_{k=1}^{i} |t|^{q-2}t\xi_k(t)+\sum_{k=i+1}^{l} |t|^{m_kp-2}\,|\mu_k(t)|^p+\Psi_{2}(t) +\mathcal{O}(|t|^{3q-2}),\quad t\in\Omega. \end{align}

Now if there is a non-empty set $I\subseteq \{i+1,\ldots,l\}$ such that $m_kp=2$ for $k\in I$, then we can rewrite (4.22) as

\begin{align*} \alpha_2\,|t|^{2q-2}& =\sum_{k=1}^{i} |t|^{q-2}t\xi_k(t)\\& \quad +\sum_{k\in \{i+1,\ldots,l\}\setminus I} |t|^{m_kp-2}\,|\mu_k(t)|^p+\tilde{\Psi}_{2}(t) +\mathcal{O}(|t|^{3q-2}),\quad t\in\Omega, \end{align*}

where $\tilde {\Psi }_{2}(t)=\Psi _{2}(t)+\sum _{k\in I} |\mu _k(t)|^p,\, t\in \Omega$. Clearly, $\tilde {\Psi }_2(\cdot )$ is analytic in $\Omega$. Therefore, in (4.22), without loss of generality, we may assume that $m_kp>2$ for all $i+1\leq k\leq l$ as $\Psi _2$ can be replaced by $\tilde {\Psi }_2$ and other term can be modified accordingly. Taking limit $t\to 0$ on both sides of (4.22), we conclude that $\Psi _{2}(0)=0$. Let $\xi _k(t)=\xi _k(0)+t\eta _k(t), 1\leq k\leq l$, and $\Psi _{2}(t)=t\Psi _{3}(t)$, where $\eta _k(\cdot )$, and $\Psi _{3}(\cdot )$ are some analytic functions in $\Omega$. Then, from (4.22), we have

(4.23)\begin{align} & \alpha_2\,|t|^{2q-2}-\sum_{k=1}^{i} |t|^{q-2}\, t \, \xi_k(0)\nonumber\\ & \quad = \sum_{k=i+1}^{l} |t|^{m_kp-2}\,|\mu_k(0)|^p+\sum_{k=1}^{i} |t|^{q-2}\, t^2\, \eta_k(t)\nonumber\\ & \qquad+\sum_{k=i+1}^{l} |t|^{m_kp-2}t \,\xi_k(t)+t\Psi_{3}(t) +\mathcal{O}(|t|^{3q-2}),\quad t\in\Omega. \end{align}

Case I: Let $1< q\leq \frac {3}{2}$, then $2q-2\leq 1$. Therefore, $|t|^{2q-1}$ is not differentiable at $0$. Now if $m_kp>3$ for all $k\in \{i+1,\ldots,l\}$, then the right-hand side of (4.23) is differentiable at $0$ but the left-hand side of (4.23) is not differentiable at $0$. So without loss of any generality, we may assume that there exists $j\in \{i+1,\ldots,l-1\}$ such that $m_kp\leq 3$ for $i+1\leq k\leq j$ and $m_kp>3$ for $j+1\leq k\leq l$. Then, from (4.23), we have

(4.24)\begin{align} & \alpha_2\,|t|^{2q-2}-|t|^{q-2}\, t\sum_{k=1}^{i} \xi_k(0)-\sum_{k=i+1}^{j} |t|^{m_kp-2}\,|\mu_k(0)|^p\nonumber\\ & \quad = \sum_{k=j+1}^{l} |t|^{m_kp-2}\,|\mu_k(0)|^p+\sum_{k=1}^{i} |t|^{q-2}\, t^2\, \eta_k(t)\nonumber\\ & \qquad+\sum_{k=i+1}^{l} |t|^{m_kp-2}t \,\xi_k(t)+t\Psi_{3}(t) +\mathcal{O}(|t|^{3q-2}),\quad t\in\Omega. \end{align}

Note that the right-hand side of (4.24) is differentiable at $0,$ which forces us to conclude that there exists a non-empty subset $J\subseteq \{i+1,\ldots,j\}$ such that $m_kp=2q$ for $k\in J$ and $\alpha _{2}=\sum \limits _{k\in J} |\mu _k(0)|^p$. But $\alpha _{2}<0$ and $\sum \limits _{k\in J} |\mu _k(0)|^p>0$, which leads to a contradiction.

Case II: Let $\frac {3}{2}< q<2$. Then, from (4.23), we have

(4.25)\begin{align} & \alpha_2\,\frac{|t|^{2q}}{t^3}-\left(\sum_{k=1}^{i} \xi_k(0)\right)|t|^{q-2} = \sum_{k=i+1}^{l} \frac{|t|^{m_kp}}{t^3}\,|\mu_k(0)|^p+\left(\sum_{k=1}^{i} \eta_k(t)\right) \frac{|t|^{q}}{t}\nonumber\\ & \quad+\sum_{k=i+1}^{l} |t|^{m_kp-2} \,\xi_k(t)+\Psi_{3}(t) +\mathcal{O}(|t|^{3q-3}), \quad t\in\Omega\setminus\{0\}. \end{align}

Let $A=\sum \limits _{k=1}^{i} \xi _k(0)$, and $\eta _k(t)=\eta _k(0)+t\,\gamma _k(t),\, 1\leq k\leq l$, for some analytic function $\gamma _k(\cdot )$ in $\Omega$. Now we rewrite the above equation (4.25) as

(4.26)\begin{align} & \alpha_2\,\frac{|t|^{2q}}{t^3}-A|t|^{q-2}-\left(\sum_{k=1}^{i} \eta_k(0)\right) \frac{|t|^{q}}{t} = \sum_{k=i+1}^{l} \frac{|t|^{m_kp}}{t^3}\,|\mu_k(0)|^p + \left(\sum_{k=1}^{i} \gamma_k(t)\right) |t|^{q}\nonumber\\ & \quad+\sum_{k=i+1}^{l} |t|^{m_kp-2} \,\xi_k(t)+\Psi_{3}(t) +\mathcal{O}(|t|^{3q-3}), \quad t\in\Omega\setminus\{0\}. \end{align}

If $m_kp\geq 4$ for $i+1\leq k\leq l$, then the right-hand side of (4.26) is differentiable but the left-hand side of (4.26) is not differentiable at $0$. So we may assume that there exist some $j\in \{i+1,\ldots,l-1\}$ such that $2\leq m_kp<4$ for $i+1\leq k\leq j,$ and $4\leq m_kp<\infty$ for $j+1\leq k\leq l.$ Then, from (4.26), we have

(4.27)\begin{align} & \alpha_2\,\frac{|t|^{2q}}{t^3}-A|t|^{q-2}-\left(\sum_{k=1}^{i} \eta_k(0)\right) \frac{|t|^{q}}{t}\nonumber\\ & \quad - \sum_{k=i+1}^{j} \frac{|t|^{m_kp}}{t^3}\,|\mu_k(0)|^p = \sum_{k=j+1}^{l} \frac{|t|^{m_kp}}{t^3}\,|\mu_k(0)|^p\nonumber\\ & \quad+ \left(\sum_{k=1}^{i} \gamma_k(t)\right) |t|^{q} + \sum_{k=i+1}^{l} |t|^{m_kp-2} \,\xi_k(t)+\Psi_{3}(t) +\mathcal{O}(|t|^{3q-3}),\quad t\in\Omega\setminus\{0\}. \end{align}

Here again observe that the right-hand side of (4.27) is differentiable at $0$, which implies the differentiability of the left-hand side of (4.27), and consequently there exists a non-empty set $J\subseteq \{i+1,\ldots,j\}$ such that $m_kp=2q$ for all $k\in J$ and $\alpha _{2}=\sum \limits _{k\in J}|\mu _k(0)|^p$. But $\alpha _{2}<0,$ and $\sum \limits _{k\in J}|\mu _k(0)|^p\geq 0$, which is a contradiction.

Hence, after analysing all the above cases, we conclude that there is no isometric embedding of $S_q^m$ into $S_p^n$. This completes the proof.

Proof. We will prove this result by contradiction. Suppose that there is an isometric embedding of $S_1^m$ into $S_p^n$. Here, in the proof, we use the same notation that is used in the proof of theorem 4.8. Therefore, we have identity (4.4) and consequently identity (4.5), that is we have

(4.28)\begin{equation} \left(1+|t|\right)^{p}=\sum_{k=1}^{n}|\lambda_k(t)|^p. \end{equation}

It is easy to observe that not all $\lambda _k(0)$ vanish. The non-differentiability of the left-hand side of (4.28) at $0$ forces that not all $\lambda _k(0)$ are non-zero. Therefore, we may assume that there is some $l\in \{1,\ldots,n-1\}$ such that $\lambda _k(0)=0$ for $1\leq k\leq l$ and $\lambda _k(0)\neq 0$ for $l+1\leq k\leq n$. Hence, we have identity (4.8). That is

(4.29)\begin{align} & 1+p\,|t|+\frac{p(p-1)}{2!}\,|t|^2+\mathcal{O}(|t|^3)=\sum_{k=1}^{l}|t|^{m_k p}|\,\mu_k(t)|^p+t\Psi_{1}(t)+1,\quad t\in\Omega,\nonumber\\ & \quad \implies p\,|t|=\sum_{k=1}^{l}|t|^{m_k p}|\,\mu_k(t)|^p+t\Psi_{1}(t)-p(p-1)\,|t|^2+\mathcal{O}(|t|^3)\quad t\in\Omega. \end{align}

Note that $m_kp\neq 1,\forall k\in \{1,\ldots,l\}$ as $p\in (0,1)\setminus \{\frac {1}{k}:k\in \mathbb {N}\}$. It is clear from the right-hand side of (4.29) that not all $m_kp>1$, otherwise the right-hand side of (4.29) is differentiable at $0$ but the left-hand side is not differentiable at $0$. So there exists a non-empty subset $J\subseteq \{1,\ldots,l\}$ such that $m_kp<1$ for $k\in J$ and $m_kp>1$ for $k\in \{1,\ldots,l\}\setminus J$. Therefore, from (4.29), we have

(4.30)\begin{align} & p\,|t|-\sum_{k\in J}|t|^{m_k p}|\,\mu_k(0)|^p\nonumber\\ & \quad = \sum_{k\in J}t|t|^{m_k p} \xi_k(t)+\sum_{k\in \{1,\ldots,l\}\setminus J}|t|^{m_k p} |\mu_k(t)|^p+t\Psi_{1}(t)-p(p-1)\,|t|^2+\mathcal{O}(|t|^3). \end{align}

Note that the left-hand side of (4.30) is not differentiable at $0$ but the right-hand side of (4.30) is differentiable at $0$, which contradicts our assumption. This completes the proof.

Proof. Suppose that there exists an isometric embedding $S_\infty ^m$ into $S_p^n$. Then, there exist $A$ and $B$ in $S_p^n$ such that $A, B$ have $(\mathbf {I}_{\infty,p})$. Therefore, we have the following identity

(4.31)\begin{equation} \max\{1,|t|^p\}=\|A+tB\|_p^p, \quad t\in\mathbb{R}. \end{equation}

By the Kato–Rellich theorem 2.4, there exist $n$ real-analytic functions $\lambda _1(\cdot ), \lambda _2(\cdot ),\ldots, \lambda _n(\cdot )$ in a small neighbourhood $U$ of $1$ such that they are complete set of eigenvalues of $A+tB$ for $t\in U$. So for $t\in U$, from (4.31), we have

(4.32)\begin{equation} \max\{1,|t|^p\}=\sum_{k=1}^{n}|\lambda_k(t)|^p. \end{equation}

If $\lambda _k(1)\neq 0$ for all $1\leq k\leq n$, that is if $A+B$ is invertible, then the right-hand side of (4.32) is a real-analytic function in $U$ but the left hand side of (4.32) is not analytic in $U$. So $A+B$ is a singular matrix. In (4.32), note that not all $\lambda _k(1)$ are zero as $1=\sum _{k=1}^{n}|\lambda _k(1)|^p$. So without loss of generality, we may assume that there is a natural number $l\in \{1,2,\ldots,n\}$ such that $\lambda _k(1)=0$ for $1\leq k\leq l$ and $\lambda _k(1)\neq 0$ for $l+1\leq k\leq n$. Note that if some $\lambda _i(t)$ is identically zero in $U$, we can neglect that one as it has no contribution in $\|A+tB\|_p$. Therefore, we may also assume that all $\lambda _i(t)$ are non-zero analytic functions in $U$. Let $m_k$ be the multiplicity of zero at $1$ of the function $\lambda _k(t)$ for $1\leq k\leq l$. Let $\lambda _k(t)=(t-1)^{m_k}\mu _k(t), \mu _k(1)\neq 0,1\leq k\leq l$, for some analytic function $\mu _k(\cdot )$ in $U$. Then, from (4.32), we have,

(4.33)\begin{equation} \max\{1,|t|^p\}=\sum_{k=1}^{l}|(t-1)|^{m_kp} |\mu_k(t)|^p+ \sum_{k=l+1}^{n} |\lambda_k(t)|^p, \quad t\in U. \end{equation}

Since the left-hand side of (4.33) is not differentiable at $1$, not all $m_kp$ in (4.33) are greater than $1$. So we may assume that there is a non-empty set $J\subseteq \{1,\ldots, l\}$ such that $m_kp<1$ for $k\in J$ and $m_kp>1$ for $k\in \{1,\ldots, l\}\setminus J$. Let $|\mu _k(t)|^p=|\mu _k(1)|^p+(t-1)\xi _k(t), k\in J$, where $\xi _k(\cdot )$ are some analytic function in $U$. Then, from (4.33), we have

(4.34)\begin{align} & \max\{1,|t|^p\}-\sum_{k\in J}|(t-1)|^{m_kp} |\mu_k(1)|^p\nonumber\\ & \quad = \sum_{k\in J}|(t-1)|^{m_kp} (t-1)\,\xi_k(t)\nonumber\\ & \qquad+ \sum_{k\in \{1,\ldots, l\}\setminus J}|(t-1)|^{m_kp} |\mu_k(t)|^p+\sum_{k=l+1}^{n} |\lambda_k(t)|^p, \quad t\in U. \end{align}

It is clear that the right-hand side of (4.34) is differentiable at $1$ but the left-hand side is not differentiable at $1$, which leads to a contradiction. Therefore, there is no isometric embedding of $S_\infty ^m$ into $S_p^n$. This completes the proof of the theorem.

Proof. Here, we use the same notations that are used in the proof of theorem 4.8. On the contrary, suppose there exists such an isometric embedding. Then, we have identity (4.4), and consequently (4.5). Altogether, we have

(4.36)\begin{equation} \left(1+|t|^q\right)^{\frac{p}{q}}=\|A+tB\|_p^p=\sum_{k=1}^{n}|\lambda_k(t)|^p. \end{equation}

If $A$ is invertible, then the right-hand side of (4.36) is analytic in $\Omega$. Therefore, $\|A+tB\|_p^p$ has a power series representation in $\Omega$. Consequently, from (4.36), we have

(4.37)\begin{align} & 1+\frac{p}{q}|t|^q+\mathcal{O}(|t|^{2q})\nonumber\\& \quad =1+t\frac{{\rm d}}{{\rm d}t}\bigg|_{t=0}{\lVert}{A+tB}{\rVert}_p^p+\frac{t^2}{2!}\frac{{\rm d}^2}{{\rm d}t^2}\bigg|_{t=0}{\lVert}{A+tB}{\rVert}_p^p+\mathcal{O}(|t|^{3q}),\quad t\in \Omega. \end{align}

It is easy to observe that $\frac {{\rm d}}{{\rm d}t}\bigg |_{t=0}{\lVert }{A+tB}{\rVert }_p^p=0$ as $1\leq (1+|t|^q)^{\frac {p}{q}}={\lVert }{A+tB}{\rVert }_p^p$ for all $t\in \Omega$. Therefore, from (4.37), we have

(4.38)\begin{equation} \frac{p}{q}|t|^{q-2}+\mathcal{O}(|t|^{2q-2})=\frac{1}{2!}\frac{{\rm d}^2}{{\rm d}t^2}\bigg|_{t=0}{\lVert}{A+tB}{\rVert}_p^p+\mathcal{O}(|t|^{3q-2}),\quad t\in \Omega\setminus\{0\}. \end{equation}

Taking limit $t\to 0$ on both sides of (4.38), we have

\[ \frac{p}{q}~\lim_{t\to 0}|t|^{q-2}=\frac{1}{2!}\frac{{\rm d}^2}{{\rm d}t^2}\bigg|_{t=0}{\lVert}{A+tB}{\rVert}_p^p< 0, \]

which is impossible as $\frac {p}{q}>0$. We conclude that $A$ must be a singular matrix. Then, by the similar argument as given in the proof of theorem 4.9, we can deduce identity (4.8) from identity (4.5). Thus, we have

(4.39)\begin{equation} \frac{p}{q}|t|^{q-1}+\mathcal{O}(|t|^{2q-1})=\sum_{k=1}^{l}|t|^{m_k p-1}|\,\mu_k(t)|^p+\Psi_{1}(t). \end{equation}

Observe that $\Psi _{1}(0)=0$. Let $\Psi _{1}(t)=t\Psi _2(t)$, where $\Psi _{2}(\cdot )$ is some analytic function in $\Omega$. Then, from (4.39), we have,

(4.40)\begin{align} \frac{p}{q}|t|^{q-2}+\mathcal{O}(|t|^{2q-2})& =\sum_{k=1}^{l}|t|^{m_k p-2}|\,\mu_k(t)|^p+\Psi_{2}(t), \quad t\in \Omega; \end{align}

(4.41)\begin{align} \implies \frac{p}{q}~\lim_{t\to 0}|t|^{q-2}& = \lim_{t\to 0}\,\sum_{k=1}^{l}|t|^{m_k p-2}|\,\mu_k(t)|^p +\Psi_{2}(0). \end{align}

As $q\geq 2$, from the above equation (4.39), it follows that $m_kp\geq 2$ for all $k\in \{1,\ldots,l\}$. Let $J\subseteq \{1,\ldots,l\}$ be such that $m_kp=2$ for $k\in J$. Note that $J$ could be empty. Then, by repeating the similar kind of arguments as given in sub-case-II of the proof of [Reference Chattopadhyay, Hong, Pal, Pradhan and Ray6, theorem 4.2], we have

\begin{align*} & \left.\frac{{\rm d}^2}{{\rm d}t^2}\right|_{t=0}{\lVert}{A+tB}{\rVert}_p^p\\& \quad =\sum_{k\in J}|\,\mu_k(0)|^p+\Psi_2(0)< 0.\quad (\text{if } J=\emptyset,\text{ we assume }\sum_{k\in J}|\,\mu_k(0)|^p=0 ) \end{align*}

Therefore, from (4.40), we conclude that

\[ \frac{p}{q}~\lim_{t\to 0}|t|^{q-2}=\sum_{k\in J}|\,\mu_k(0)|^p+ \Psi_{2}(0)<0, \text{ but } \frac{p}{q}>0, \]

which leads to a contradiction. This completes the proof.