Theorem 4
Let
\(\mathcal{A}\)
be a
\((p,q)\)
th order
\(n\times n\)
dimensional nonnegative rectangular tensor, S
be a nonempty proper subset of
N, S̄
be the complement of
S
in
N. Then
$$\lambda_{0}\leq\Psi^{S}(\mathcal{A})=\max\bigl\{ \Psi_{1}^{S}(\mathcal{A}),\Psi _{1}^{\bar{S}}( \mathcal{A}),\Psi_{2}^{S}(\mathcal{A}),\Psi_{2}^{\bar {S}}( \mathcal{A}),\Psi_{3}^{S}(\mathcal{A}),\Psi_{3}^{\bar{S}}( \mathcal {A}),\Psi_{4}^{S}(\mathcal{A}),\Psi_{4}^{\bar{S}}( \mathcal{A})\bigr\} , $$
where
$$\begin{aligned} \Psi_{1}^{S}(\mathcal{A}) =&\max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{S}}( \mathcal {A})+r_{j}^{\overline{\Delta^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{S}}( \mathcal{A})-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{\overline{\Delta^{S}}}( \mathcal{A})r_{j}^{\Delta^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Psi_{1}^{\bar{S}}(\mathcal{A}) =&\max_{i\in\bar{S},j\in S} \frac {1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{\bar{S}}}( \mathcal{A})+r_{j}^{\overline{\Delta^{\bar{S}}}}(\mathcal {A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{\bar {S}}}( \mathcal{A})-r_{j}^{\overline{\Delta^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{\overline{\Delta^{\bar{S}}}}( \mathcal{A})r_{j}^{\Delta^{\bar {S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Psi_{2}^{S}(\mathcal{A}) =&\max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+c_{i}^{\Omega ^{S}}( \mathcal{A})+c_{j}^{\overline{\Omega^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+c_{i}^{\Omega ^{S}}( \mathcal{A})-c_{j}^{\overline{\Omega^{S}}}(\mathcal {A})\bigr)^{2}+4c_{i}^{\overline{\Omega^{S}}}( \mathcal{A})c_{j}^{\Omega^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Psi_{2}^{\bar{S}}(\mathcal{A}) =&\max_{i\in\bar{S},j\in S} \frac {1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+c_{i}^{\Omega ^{\bar{S}}}( \mathcal{A})+c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+c_{i}^{\Omega^{\bar {S}}}( \mathcal{A})-c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4c_{i}^{\overline{\Omega^{\bar{S}}}}( \mathcal{A})c_{j}^{\Omega^{\bar {S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Psi_{3}^{S}(\mathcal{A}) =&\max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{S}}( \mathcal {A})+c_{j}^{\overline{\Omega^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{S}}( \mathcal{A})-c_{j}^{\overline{\Omega^{S}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{\overline{{\Delta^{S}}}}( \mathcal{A})c_{j}^{\Omega^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Psi_{3}^{\bar{S}}(\mathcal{A}) =&\max_{i\in\bar{S},j\in S} \frac {1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline {\Delta^{\bar{S}}}}( \mathcal{A})+c_{i}^{\Omega^{\bar{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline {\Delta^{\bar{S}}}}( \mathcal{A})+c_{i}^{\Omega^{\bar{S}}}(\mathcal {A})\bigr)^{2}+4r_{j}^{\Delta^{\bar{S}}}( \mathcal{A})c_{i}^{{\overline{\Omega ^{\bar{S}}}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} , \\ \Psi_{4}^{S}(\mathcal{A}) =&\max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline{\Delta ^{S}}}( \mathcal{A})+c_{i}^{\Omega^{S}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline {\Delta^{S}}}( \mathcal{A})+c_{i}^{\Omega^{S}}(\mathcal{A})\bigr)^{2}+4r_{j}^{\Delta ^{S}}( \mathcal{A})c_{i}^{{\overline{\Omega^{S}}}}(\mathcal{A}) \bigr]^{\frac {1}{2}} \bigr\} , \\ \Psi_{4}^{\bar{S}}(\mathcal{A}) =&\max_{i\in\bar{S},j\in S} \frac {1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{\bar{S}}}( \mathcal{A})+c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{\bar{S}}}( \mathcal{A})-c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{{\overline{\Delta^{\bar{S}}}}}( \mathcal{A})c_{j}^{\Omega ^{\bar{S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$
Proof
Because \(\lambda_{0}\) is the largest singular value of \(\mathcal{A}\), from Theorem 2 in [6], there are nonnegative nonzero vectors \(x=(x_{1},x_{2},\ldots,x_{n})^{\mathrm{T}}\) and \(y=(y_{1},y_{2},\ldots,y_{n})^{\mathrm{T}}\), such that
$$\begin{aligned}& \mathcal{A}x^{p-1}y^{q}=\lambda_{0} x^{[l-1]}, \end{aligned}$$
(1)
$$\begin{aligned}& \mathcal{A}x^{p}y^{q-1}=\lambda_{0} y^{[l-1]}. \end{aligned}$$
(2)
Let
$$\begin{aligned}& x_{t}=\max\{x_{i}:i\in S\},\qquad x_{h}=\max \{x_{i}:i\in\bar{S}\};\\& y_{f}=\max\{y_{i}:i\in S\},\qquad y_{g}=\max\{y_{i}:i\in\bar{S}\}; \\& w_{i}=\max\{x_{i},y_{i}\}, \quad i\in N,\qquad w_{S}=\max\{w_{i}:i\in S\},\qquad w_{\bar{S}}=\max\{ w_{i}:i\in\bar{S}\}. \end{aligned}$$
Then at least one of \(x_{t}\) and \(x_{h}\) is nonzero, and at least one of \(y_{f}\) and \(y_{g}\) is nonzero. We next divide into four cases to prove.
Case I: If \(w_{S}=x_{t}, w_{\bar{S}}=x_{h}\), then \(x_{t}\geq y_{t}, x_{h}\geq y_{h}\).
(i) If \(x_{h}\geq x_{t}\), then \(x_{h}=\max\{w_{i}:i\in N\}\). From (3) of Theorem 2.2 in [10], we have
$$\begin{aligned} & \bigl(\lambda_{0} -a_{h\cdots hh\cdots h}-r_{h}^{\overline{\Delta^{S}}}( \mathcal {A})\bigr)x_{h}^{l-1}\leq r_{h}^{\Delta^{S}}( \mathcal{A})x_{t}^{l-1}. \end{aligned}$$
(3)
If \(x_{t}=0\), by \(x_{h}>0\), we have \(\lambda_{0} -a_{h\cdots hh\cdots h}-r_{h}^{\overline{\Delta^{S}}}(\mathcal{A})\leq0\). Then \(\lambda_{0}\leq a_{h\cdots hh\cdots h}+r_{h}^{\overline{\Delta^{S}}}(\mathcal {A})\leq\Psi_{1}^{S}(\mathcal{A})\). Otherwise, \(x_{t}>0\). From (1), we have
$$\begin{aligned} (\lambda_{0} -a_{t\cdots tt\cdots t})x_{t}^{l-1} \leq& \lambda_{0} x_{t}^{l-1}-a_{t\cdots tt\cdots t}x_{t}^{p-1}y_{t}^{q} \\ =&\sum_{(i_{2},\ldots, i_{p},j_{1},\ldots, j_{q})\in\Delta^{S} \atop \delta_{ti_{2}\cdots i_{p}j_{1}\cdots j_{q}}=0}a_{ti_{2}\cdots i_{p}j_{1}\cdots j_{q}}x_{i_{2}} \cdots x_{i_{p}}y_{j_{1}}\cdots y_{j_{q}} \\ &{}+\sum_{(i_{2},\ldots, i_{p},j_{1},\ldots, j_{q})\in\overline{\Delta ^{S}}}a_{ti_{2}\cdots i_{p}j_{1}\cdots j_{q}}x_{i_{2}} \cdots x_{i_{p}}y_{j_{1}}\cdots y_{j_{q}} \\ \leq&\sum_{(i_{2},\ldots, i_{p},j_{1},\ldots, j_{q})\in\Delta^{S} \atop \delta_{ti_{2}\cdots i_{p}j_{1}\cdots j_{q}}=0}a_{ti_{2}\cdots i_{p}j_{1}\cdots j_{q}}x_{t}^{l-1} +\sum_{(i_{2},\ldots, i_{p},j_{1},\ldots, j_{q})\in\overline{\Delta^{S}}} a_{ti_{2}\cdots i_{p}j_{1}\cdots j_{q}}x_{h}^{l-1} \\ =& r_{t}^{\Delta^{S}}(\mathcal{A})x_{t}^{l-1}+r_{t}^{\overline{\Delta ^{S}}}( \mathcal{A})x_{h}^{l-1}, \end{aligned}$$
i.e.,
$$\begin{aligned} & \bigl(\lambda_{0} -a_{t\cdots tt\cdots t}-r_{t}^{\Delta^{S}}( \mathcal {A})\bigr)x_{t}^{l-1}\leq r_{t}^{\overline{\Delta^{S}}}( \mathcal{A})x_{h}^{l-1}. \end{aligned}$$
(4)
If \(\lambda_{0} -a_{t\cdots tt\cdots t}-r_{t}^{\Delta^{S}}(\mathcal{A})\leq 0\), then \(\lambda_{0}\leq a_{t\cdots tt\cdots t}+r_{t}^{\Delta^{S}}(\mathcal{A})\leq \Psi_{1}^{S}(\mathcal{A})\). If \(\lambda_{0} -a_{t\cdots tt\cdots t}-r_{t}^{\Delta^{S}}(\mathcal{A})> 0\), multiplying (3) with (4) and noting that \(x_{t}^{l-1}x_{h}^{l-1}>0\), we have
$$\begin{aligned} & \bigl(\lambda_{0}-a_{t\cdots tt\cdots t}-r_{t}^{\Delta^{S}}( \mathcal{A})\bigr) \bigl(\lambda _{0} -a_{h\cdots hh\cdots h}-r_{h}^{\overline{\Delta^{S}}}( \mathcal{A})\bigr) \leq r_{t}^{\overline{\Delta^{S}}}(\mathcal{A})r_{h}^{\Delta^{S}}( \mathcal{A}). \end{aligned}$$
(5)
Solving \(\lambda_{0}\) in (5) gives
$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{t\cdots tt\cdots t}+a_{h\cdots hh\cdots h}+r_{t}^{\Delta^{S}}( \mathcal{A})+r_{h}^{\overline{\Delta ^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{t\cdots tt\cdots t}+r_{t}^{\Delta^{S}}( \mathcal{A})-a_{h\cdots hh\cdots h}-r_{h}^{\overline{\Delta^{S}}}(\mathcal{A}) \bigr)^{2}+4r_{t}^{\overline {\Delta^{S}}}(\mathcal{A})r_{h}^{\Delta^{S}}( \mathcal{A}) \bigr]^{\frac {1}{2}} \bigr\} \\ \leq&\max_{i\in S,j\in\bar{S}}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{S}}( \mathcal {A})+r_{j}^{\overline{\Delta^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{S}}( \mathcal{A})-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{\overline{\Delta^{S}}}( \mathcal{A})r_{j}^{\Delta^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{1}^{S}(\mathcal{A}). \end{aligned}$$
(ii) If \(x_{t}\geq x_{h}\), similar to the proof of (i), we have
$$\begin{aligned} & \bigl(\lambda_{0} -a_{h\cdots hh\cdots h}-r_{h}^{\Delta^{\bar{S}}}( \mathcal {A})\bigr) \bigl(\lambda_{0} -a_{t\cdots tt\cdots t}-r_{t}^{\overline{\Delta^{\bar {S}}}}( \mathcal{A})\bigr)\leq r_{h}^{\overline{\Delta^{\bar{S}}}}(\mathcal {A})r_{t}^{\Delta^{\bar{S}}}(\mathcal{A}), \end{aligned}$$
and
$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{h\cdots hh\cdots h}+a_{t\cdots tt\cdots t}+r_{h}^{\Delta^{\bar{S}}}( \mathcal{A})+r_{t}^{\overline{\Delta ^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{h\cdots hh\cdots h}-a_{t\cdots tt\cdots t}+r_{h}^{\Delta^{\bar {S}}}( \mathcal{A})-r_{t}^{\overline{\Delta^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4r_{h}^{\overline{\Delta^{\bar{S}}}}( \mathcal{A})r_{t}^{\Delta^{\bar {S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \leq&\max_{i\in\bar{S},j\in S}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{\bar{S}}}( \mathcal {A})+r_{j}^{\overline{\Delta^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{\bar {S}}}( \mathcal{A})-r_{j}^{\overline{\Delta^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{\overline{\Delta^{\bar{S}}}}( \mathcal{A})r_{j}^{\Delta^{\bar {S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{1}^{\bar{S}}(\mathcal{A}). \end{aligned}$$
Case II: Assume that \(w_{S}=y_{f}, w_{\bar{S}}=y_{g}\). If \(y_{g}\geq y_{f}\), similar to the proof of (i), we have
$$\begin{aligned} & \bigl(\lambda_{0} -a_{f\cdots ff\cdots f}-c_{f}^{\Omega^{S}}( \mathcal {A})\bigr) \bigl(\lambda_{0} -a_{g\cdots gg\cdots g}-c_{g}^{\overline{\Omega ^{S}}}( \mathcal{A})\bigr)\leq c_{f}^{\overline{\Omega^{S}}}(\mathcal {A})c_{g}^{\Omega^{S}}(\mathcal{A}), \end{aligned}$$
and
$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{f\cdots ff\cdots f}+a_{g\cdots gg\cdots g}+c_{f}^{\Omega^{S}}( \mathcal{A})+c_{g}^{\overline{\Omega ^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{f\cdots ff\cdots f}-a_{g\cdots gg\cdots g}+c_{f}^{\Omega ^{S}}( \mathcal{A})-c_{g}^{\overline{\Omega^{S}}}(\mathcal {A})\bigr)^{2}+4c_{f}^{\overline{\Omega^{S}}}( \mathcal{A})c_{g}^{\Omega^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \leq&\max_{i\in S,j\in\bar{S}}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+c_{i}^{\Omega^{S}}( \mathcal {A})+c_{j}^{\overline{\Omega^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+c_{i}^{\Omega ^{S}}( \mathcal{A})-c_{j}^{\overline{\Omega^{S}}}(\mathcal {A})\bigr)^{2}+4c_{i}^{\overline{\Omega^{S}}}( \mathcal{A})c_{j}^{\Omega^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{2}^{S}(\mathcal{A}). \end{aligned}$$
If \(y_{f}\geq y_{g}\), similarly, we have
$$\begin{aligned} & \bigl(\lambda_{0} -a_{g\cdots gg\cdots g}-c_{g}^{\Omega^{\bar{S}}}( \mathcal {A})\bigr) \bigl(\lambda_{0} -a_{f\cdots ff\cdots f}-c_{f}^{\overline{\Omega^{\bar {S}}}}( \mathcal{A})\bigr)\leq c_{g}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})c_{f}^{\Omega^{\bar{S}}}(\mathcal{A}) \end{aligned}$$
and
$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{g\cdots gg\cdots g}+a_{f\cdots ff\cdots f}+c_{g}^{\Omega^{\bar{S}}}( \mathcal{A})+c_{f}^{\overline{\Omega ^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{g\cdots g\cdots g}-a_{f\cdots ff\cdots f}+c_{g}^{\Omega^{\bar {S}}}( \mathcal{A})-c_{f}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4c_{g}^{\overline{\Omega^{\bar{S}}}}( \mathcal{A})c_{f}^{\Omega^{\bar {S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \leq&\max_{i\in\bar{S},j\in S}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+c_{i}^{\Omega^{\bar{S}}}( \mathcal {A})+c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+c_{i}^{\Omega^{\bar {S}}}( \mathcal{A})-c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4c_{i}^{\overline{\Omega^{\bar{S}}}}( \mathcal{A})c_{j}^{\Omega^{\bar {S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{2}^{\bar{S}}(\mathcal{A}). \end{aligned}$$
Case III: Assume that \(w_{S}=x_{t}, w_{\bar{S}}=y_{g}\). If \(y_{g}\geq x_{t}\), similar to the proof of (i), we have
$$\begin{aligned} & \bigl(\lambda_{0}-a_{t\cdots tt\cdots t}-r_{t}^{\Delta^{S}}( \mathcal{A})\bigr) \bigl(\lambda _{0} -a_{g\cdots gg\cdots g}-c_{g}^{\overline{\Omega^{S}}}( \mathcal{A})\bigr) \leq r_{t}^{\overline{{\Delta^{S}}}}(\mathcal{A})c_{g}^{\Omega^{S}}( \mathcal{A}) \end{aligned}$$
and
$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{t\cdots tt\cdots t}+a_{g\cdots gg\cdots g}+r_{t}^{\Delta^{S}}( \mathcal{A})+c_{g}^{\overline{\Omega ^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{t\cdots tt\cdots t}-a_{g\cdots gg\cdots g}+r_{t}^{\Delta ^{S}}( \mathcal{A})-c_{g}^{\overline{\Omega^{S}}}(\mathcal {A})\bigr)^{2}+4r_{t}^{\overline{{\Delta^{S}}}}( \mathcal{A})c_{g}^{\Omega^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \leq&\max_{i\in S,j\in\bar{S}}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{S}}( \mathcal {A})+c_{j}^{\overline{\Omega^{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{S}}( \mathcal{A})-c_{j}^{\overline{\Omega^{S}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{\overline{{\Delta^{S}}}}( \mathcal{A})c_{j}^{\Omega^{S}}(\mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{3}^{S}(\mathcal{A}). \end{aligned}$$
If \(x_{t}\geq y_{g}\), similarly, we have
$$\begin{aligned} & \bigl(\lambda_{0} -a_{g\cdots gg\cdots g}-c_{g}^{\Omega^{\bar{S}}}( \mathcal {A})\bigr) \bigl(\lambda_{0} -a_{t\cdots tt\cdots t}-r_{t}^{\overline{\Delta^{\bar {S}}}}( \mathcal{A})\bigr)\leq c_{g}^{{\overline{\Omega^{\bar{S}}}}}(\mathcal {A})r_{t}^{\Delta^{\bar{S}}}(\mathcal{A}) \end{aligned}$$
and
$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{t\cdots tt\cdots t}+a_{g\cdots gg\cdots g}+r_{t}^{\overline{\Delta^{\bar{S}}}}( \mathcal{A})+c_{g}^{\Omega ^{\bar{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{t\cdots tt\cdots t}-a_{g\cdots gg\cdots g}+r_{t}^{\overline {\Delta^{\bar{S}}}}( \mathcal{A})-c_{g}^{\Omega^{\bar{S}}}(\mathcal {A})\bigr)^{2}+4r_{t}^{\Delta^{\bar{S}}}( \mathcal{A})c_{g}^{{\overline{\Omega ^{\bar{S}}}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \leq&\max_{i\in\bar{S},j\in S}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline{\Delta^{\bar {S}}}}( \mathcal{A})+c_{i}^{\Omega^{\bar{S}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline {\Delta^{\bar{S}}}}( \mathcal{A})+c_{i}^{\Omega^{\bar{S}}}(\mathcal {A})\bigr)^{2}+4r_{j}^{\Delta^{\bar{S}}}( \mathcal{A})c_{i}^{{\overline{\Omega ^{\bar{S}}}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{3}^{\bar{S}}(\mathcal{A}). \end{aligned}$$
Case IV: Assume that \(w_{S}=y_{f}, w_{\bar{S}}=x_{h}\). If \(x_{h}\geq y_{f}\), similar to the proof of (i), we have
$$\begin{aligned} & \bigl(\lambda_{0}-a_{f\cdots ff\cdots f}-c_{f}^{\Omega^{S}}( \mathcal{A})\bigr) \bigl(\lambda _{0} -a_{h\cdots hh\cdots h}-r_{h}^{\overline{\Delta^{S}}}( \mathcal{A})\bigr) \leq c_{f}^{{\overline{\Omega^{S}}}}(\mathcal{A})r_{h}^{\Delta^{S}}( \mathcal{A}) \end{aligned}$$
and
$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{f\cdots ff\cdots f}+a_{h\cdots hh\cdots h}+r_{h}^{\overline{\Delta^{S}}}( \mathcal{A})+c_{f}^{\Omega ^{S}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{f\cdots ff\cdots f}-a_{h\cdots hh\cdots h}-r_{h}^{\overline {\Delta^{S}}}( \mathcal{A})+c_{f}^{\Omega^{S}}(\mathcal{A})\bigr)^{2}+4c_{f}^{{\overline {\Omega^{S}}}}( \mathcal{A})r_{h}^{\Delta^{S}}(\mathcal{A}) \bigr]^{\frac {1}{2}} \bigr\} \\ \leq&\max_{i\in S,j\in\bar{S}}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline{\Delta^{S}}}( \mathcal {A})+c_{i}^{\Omega^{S}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline {\Delta^{S}}}( \mathcal{A})+c_{i}^{\Omega^{S}}(\mathcal{A})\bigr)^{2}+4c_{i}^{{\overline {\Omega^{S}}}}( \mathcal{A})r_{j}^{\Delta^{S}}(\mathcal{A}) \bigr]^{\frac {1}{2}} \bigr\} \\ =&\Psi_{4}^{S}(\mathcal{A}). \end{aligned}$$
If \(y_{f}\geq x_{h}\), similarly, we have
$$\begin{aligned} & \bigl(\lambda_{0} -a_{h\cdots hh\cdots h}-r_{h}^{\Delta^{\bar{S}}}( \mathcal {A})\bigr) \bigl(\lambda_{0} -a_{f\cdots ff\cdots f}-c_{f}^{\overline{\Omega^{\bar {S}}}}( \mathcal{A})\bigr)\leq r_{h}^{{\overline{\Delta^{\bar{S}}}}}(\mathcal {A})c_{f}^{\Omega^{\bar{S}}}(\mathcal{A}) \end{aligned}$$
and
$$\begin{aligned} \lambda_{0} \leq& \frac{1}{2} \bigl\{ a_{h\cdots hh\cdots h}+a_{f\cdots ff\cdots f}+r_{h}^{\Delta^{\bar{S}}}( \mathcal{A})+c_{f}^{\overline{\Omega ^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{h\cdots hh\cdots h}-a_{f\cdots ff\cdots f}+r_{h}^{\Delta ^{\bar{S}}}( \mathcal{A})-c_{f}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4r_{h}^{{\overline{\Delta^{\bar{S}}}}}( \mathcal{A})c_{f}^{\Omega ^{\bar{S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \leq&\max_{i\in\bar{S},j\in S}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{i}^{\Delta^{\bar{S}}}( \mathcal {A})+c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}+r_{i}^{\Delta ^{\bar{S}}}( \mathcal{A})-c_{j}^{\overline{\Omega^{\bar{S}}}}(\mathcal {A})\bigr)^{2}+4r_{i}^{{\overline{\Delta^{\bar{S}}}}}( \mathcal{A})c_{j}^{\Omega ^{\bar{S}}}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ =&\Psi_{4}^{\bar{S}}(\mathcal{A}). \end{aligned}$$
The conclusion follows from Cases I, II, III and IV. □
We next give the following comparison theorem for these upper bounds in Theorems 1-4.
Theorem 5
Let
\(\mathcal{A}\)
be a
\((p,q)\)
th order
\(n\times n\)
dimensional nonnegative rectangular tensor, S be a nonempty proper subset of
N, S̄
be the complement of
S
in
N. Then
$$\Psi^{S}(\mathcal{A})\leq U^{S}(\mathcal{A})\leq\Phi( \mathcal{A})\leq\max_{i,j\in N}\bigl\{ R_{i}( \mathcal{A}),C_{j}(\mathcal{A})\bigr\} . $$
Proof
I. By Remark 2.2 in [9], \(\Phi(\mathcal{A})\leq\max_{i,j\in N}\{R_{i}(\mathcal{A}),C_{j}(\mathcal{A})\}\) holds.
II. Next, we prove \(U^{S}(\mathcal{A})\leq\Phi(\mathcal{A})\). Here, we only prove \(U_{1}^{S}(\mathcal{A})\leq\Phi(\mathcal{A})\). Similarly, we can prove \(U_{1}^{\bar{S}}(\mathcal{A}),U_{2}^{S}(\mathcal{A}), U_{2}^{\bar{S}}(\mathcal{A})\leq\Phi(\mathcal{A})\), respectively.
(i) Suppose that
$$\begin{aligned} U_{1}^{S}(\mathcal{A}) =& \max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline{\Delta ^{S}}}( \mathcal{A})\\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}( \mathcal{A})\bigr)^{2}+4r_{i}(\mathcal {A})r_{j}^{\Delta^{S}}( \mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$
From the proof of Theorem 2.2 in [10], we can see that the bound \(U_{1}^{S}(\mathcal{A})\) is obtained by solving \(\lambda_{0}\) from
$$\begin{aligned} & (\lambda_{0}-a_{i\cdots ii\cdots i}) \bigl( \lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}( \mathcal{A})\bigr) \leq r_{i}(\mathcal{A})r_{j}^{\Delta^{S}}( \mathcal{A}). \end{aligned}$$
(6)
From the proof of Theorem 1.3 in [9], we can see that the bound
$$\begin{aligned} \Phi_{1}(\mathcal{A}) =& \max_{i,j\in N,i\neq j}\frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{i}( \mathcal{A})\\ &{} + \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{i}( \mathcal {A})\bigr)^{2}+4a_{ji\cdots ii\cdots i}r_{i}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \end{aligned}$$
is obtained by solving \(\lambda_{0}\) from
$$\begin{aligned} & (\lambda_{0}-a_{i\cdots ii\cdots i}) \bigl( \lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{i}( \mathcal{A})\bigr)\leq a_{ji\cdots ii\cdots i}r_{i}(\mathcal{A}). \end{aligned}$$
(7)
Taking \(i\in S\), \(j\in\bar{S}\) in (7), by the proof of Theorem 6 in [11], we know that if \(\lambda_{0}\) satisfies (6), then \(\lambda _{0}\) satisfies (7), which implies that
$$\begin{aligned} \Phi_{1}(\mathcal{A}) \geq&\max_{i\in S,j\in\bar{S}} \frac {1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{i}( \mathcal{A})\\ &{} + \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{i}( \mathcal {A})\bigr)^{2}+4a_{ji\cdots ii\cdots i}r_{i}(\mathcal{A}) \bigr]^{\frac{1}{2}} \bigr\} \\ \geq&U_{1}^{S}(\mathcal{A}). \end{aligned}$$
Obviously, \(U_{1}^{S}(\mathcal{A})\leq\Phi(\mathcal{A})\).
(ii) Suppose that
$$\begin{aligned} U_{1}^{S}(\mathcal{A}) =& \max_{i\in S,j\in\bar{S}} \frac{1}{2} \bigl\{ a_{i\cdots ii\cdots i}+a_{j\cdots jj\cdots j}+r_{j}^{\overline{\Delta ^{S}}}( \mathcal{A}) \\ &{}+ \bigl[\bigl(a_{i\cdots ii\cdots i}-a_{j\cdots jj\cdots j}-r_{j}^{\overline {\Delta^{S}}}( \mathcal{A})\bigr)^{2}+4c_{i}(\mathcal{A})r_{j}^{\Delta^{S}}( \mathcal {A}) \bigr]^{\frac{1}{2}} \bigr\} . \end{aligned}$$
Similar to the proof of (i), we can obtain \(U_{1}^{S}(\mathcal{A})\leq\Phi _{3}(\mathcal{A})\leq\Phi(\mathcal{A})\).
III. Finally, we prove that \(\Psi^{S}(\mathcal{A})\leq U^{S}(\mathcal{A})\). Here, we only prove \(\Psi_{1}^{S}(\mathcal{A})\leq U^{S}(\mathcal{A})\). Similarly, we can prove \(\Psi_{1}^{\bar{S}}(\mathcal{A}),\Psi_{2}^{S}(\mathcal {A}),\Psi_{2}^{\bar{S}}(\mathcal{A}),\Psi_{3}^{S}(\mathcal{A}),\Psi_{3}^{\bar {S}}(\mathcal{A}),\Psi_{4}^{S}(\mathcal{A}),\Psi_{4}^{\bar{S}}(\mathcal {A})\leq U^{S}(\mathcal{A})\), respectively.
Let \(i\in S\) and \(j\in\bar{S}\). From the proof of Theorem 4, we can see that the bound \(\Psi_{1}^{S}(\mathcal{A})\) is obtained by solving \(\lambda_{0}\) from
$$\begin{aligned} & \bigl(\lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\Delta^{S}}( \mathcal{A})\bigr) \bigl(\lambda _{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}( \mathcal{A})\bigr) \leq r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})r_{j}^{\Delta^{S}}( \mathcal{A}). \end{aligned}$$
(8)
(i) Suppose that \(r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})r_{j}^{\Delta ^{S}}(\mathcal{A})=0\). If \(\lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\Delta^{S}}(\mathcal{A})>0\), i.e., \(\lambda_{0}>a_{i\cdots ii\cdots i}+r_{i}^{\Delta^{S}}(\mathcal{A})\), then \(\lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A})\leq0\), and for any \(i\in S\),
$$(\lambda_{0}-a_{i\cdots ii\cdots i}) \bigl(\lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal{A})\bigr)\leq0\leq r_{i}(\mathcal {A})r_{j}^{\Delta^{S}}(\mathcal{A}). $$
That is to say, if \(\lambda_{0}\) satisfies (8), then \(\lambda _{0}\) satisfies (6), which implies that \(\Psi_{1}^{S}(\mathcal{A})\leq U_{1}^{S}(\mathcal{A})\leq U^{S}(\mathcal{A})\).
If \(\lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\Delta^{S}}(\mathcal{A})\leq0\), then \(\lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta ^{S}}}(\mathcal{A})\geq0\), i.e., \(\lambda_{0} \geq a_{j\cdots jj\cdots j}+r_{j}^{\overline {\Delta^{S}}}(\mathcal{A})\). From (3), we can obtain \(\lambda_{0} -a_{j\cdots jj\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A})\leq r_{j}^{\Delta^{S}}(\mathcal{A})\), i.e.,
$$\begin{aligned} & \lambda_{0} -a_{j\cdots jj\cdots j}\leq r_{j}( \mathcal{A}). \end{aligned}$$
(9)
By \(\lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\Delta^{S}}(\mathcal{A})\leq 0\leq r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})\), i.e., \(\lambda_{0}-a_{i\cdots ii\cdots i}\leq r_{i}(\mathcal{A})\), we have
$$\begin{aligned} & \lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\overline{\Delta^{\bar {S}}}}( \mathcal{A})\leq r_{i}^{\Delta^{\bar{S}}}(\mathcal{A}). \end{aligned}$$
(10)
Multiplying (9) with (10), we can obtain
$$\begin{aligned} & (\lambda_{0} -a_{j\cdots jj\cdots j}) \bigl( \lambda_{0}-a_{i\cdots ii\cdots i}-r_{i}^{\overline{\Delta^{\bar{S}}}}( \mathcal{A})\bigr)\leq r_{i}^{\Delta^{\bar {S}}}(\mathcal{A})r_{j}( \mathcal{A}), \end{aligned}$$
(11)
which implies that if \(\lambda_{0}\) satisfies (8), then \(\lambda_{0}\) satisfies (6), consequently, \(\Psi_{1}^{S}(\mathcal{A})\leq U_{1}^{\bar{S}}(\mathcal{A})\leq U^{S}(\mathcal{A})\).
(ii) Suppose that \(r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})r_{j}^{\Delta ^{S}}(\mathcal{A})>0\). Then dividing (8) by \(r_{i}^{\overline{\Delta^{S}}}(\mathcal {A})r_{j}^{\Delta^{S}}(\mathcal{A})\), we have
$$\begin{aligned} & \frac{(\lambda_{0}-a_{i\cdots i}-r_{i}^{\Delta^{S}}(\mathcal {A}))}{r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})} \frac{(\lambda_{0}-a_{j\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A}))}{r_{j}^{\Delta^{S}}(\mathcal{A})}\leq1. \end{aligned}$$
(12)
Furthermore, if \(\frac{\lambda_{0}-a_{i\cdots i}-r_{i}^{\Delta^{S}}(\mathcal {A})}{r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})}\geq1\), then by Lemma 2.3 in [12] and (12), we have
$$\begin{aligned} & \frac{(\lambda_{0}-a_{i\cdots i})}{r_{i}(\mathcal{A})} \frac{(\lambda_{0}-a_{j\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A}))}{r_{j}^{\Delta^{S}}(\mathcal{A})} \leq \frac{(\lambda_{0}-a_{i\cdots i}-r_{i}^{\Delta^{S}}(\mathcal {A}))}{r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})} \frac{(\lambda_{0}-a_{j\cdots j}-r_{j}^{\overline{\Delta^{S}}}(\mathcal {A}))}{r_{j}^{\Delta^{S}}(\mathcal{A})} \leq1. \end{aligned}$$
Thus, (6) holds, which implies that if \(\lambda_{0}\) satisfies (8), then \(\lambda_{0}\) satisfies (6), consequently, \(\Psi_{1}^{S}(\mathcal{A})\leq U_{1}^{S}(\mathcal{A})\). And if \(\frac{\lambda_{0}-a_{i\cdots i}-r_{i}^{\Delta^{S}}(\mathcal {A})}{r_{i}^{\overline{\Delta^{S}}}(\mathcal{A})}\leq1\), then (10) holds, which leads to (11) from (9). This implies that if \(\lambda_{0}\) satisfies (8), then \(\lambda_{0}\) satisfies (6), consequently, \(\Psi_{1}^{S}(\mathcal{A})\leq U_{1}^{\bar{S}}(\mathcal{A})\leq U^{S}(\mathcal{A})\). The conclusion follows immediately from what we have proved. □