3.1 Bounds on median eigenvalues of the normalized Laplacian matrix
In this section we localize median eigenvalues of the normalized Laplacian matrix by applying the methodology (see [19]) recalled in Section 2 (Theorems 1 and 2). The additional information on median eigenvalues and the interlacing between eigenvalues of normalized Laplacian and adjacency matrices turned out to be a handy tool for bounding the HL-index for both non-bipartite and bipartite graphs. According to [29] (see Theorem 2.2.1), the following relations hold:
$$ \frac{|\lambda_{n-k+1}|}{d_{1}} \leq |1-\gamma_{k}|\leq\frac{|\lambda_{n-k+1}|}{d_{n}}. $$
(11)
Proposition 1
For a simple, connected, and non-bipartite graphs
$$ 0 \leq R(G) \leq d_{1}\max\bigl(|1-\alpha_{1}|,|1- \beta_{1}|,|1-\alpha_{2}|,|1-\beta_{2}| \bigr) $$
(12)
when
n
is even with
$$\begin{aligned} &\alpha_{1}=\frac{1}{n-1} \biggl(n+\sqrt{\frac{n-2}{n} \bigl(b_{1}(n-1)-n^{2} \bigr)} \biggr),\\ & \beta_{1}= \frac{1}{n-1} \biggl(n-\sqrt{\frac{n-2}{n} \bigl(b_{1}(n-1)-n^{2} \bigr)} \biggr),\\ &\alpha_{2}=\frac{1}{n-1} \biggl(n+\sqrt{\frac{(n-4)}{(n+2)} \bigl(b_{1}(n-1)-n^{2}\bigr)} \biggr),\\ & \beta_{2}= \frac{n}{n-1} \biggl(1-\sqrt{\frac{b_{1}(n-1)-n^{2}}{n(n-2)}} \biggr), \end{aligned}$$
and
$$ 0 \leq R(G) \leq d_{1}\max \bigl(|1-\alpha_{3}|,|1- \beta_{3}| \bigr) $$
(13)
when
n
is odd with
$$ \alpha_{3}= \frac{1}{n-1} \biggl(n+\sqrt{\frac{(n-3)}{(n+1)} \bigl(b_{1}(n-1)-n^{2}\bigr)} \biggr), \qquad\beta_{3}= \frac{1}{n-1} \bigl(n-\sqrt{b_{1}(n-1)-n^{2}} \bigr). $$
Proof
From (11), we can easily derive the following bounds:
$$ 0 \leq R(G) \leq \left \{ \textstyle\begin{array}{l@{\quad}l} d_{1}\max (|1-\gamma_{\frac{n+2}{2}}|,|1-\gamma_{\frac{n}{2}}| ) & \text{ if $n$ is even,} \\ d_{1} (|1-\gamma_{\frac{n+1}{2}}| ) & \text{ if $n$ is odd.} \end{array}\displaystyle \right . $$
(14)
By applying majorization, we are able to bound the median eigenvalues \(\gamma_{i}\) (with \(i=\frac{n}{2},\frac{n+1}{2},\frac{n+2}{2}\)) considered in (14).
To this aim, we face the set:
$$ S_{b}^{1}= \Biggl\{ \boldsymbol{\gamma} \in \mathbb{R}^{n-1}:\sum_{i=1}^{n-1} \gamma_{i}=n,g(\boldsymbol{\gamma})=\sum_{i=1}^{n-1} \gamma_{i}^{2}=b_{1}=n+2\sum _{(i,j)\in E}\frac{1}{d_{i}d_{j}} \Biggr\} . $$
It is well known that, for every connected graph of order n (see [30]), we have
$$ \frac{1}{n-1} \leq \frac{2}{n} \sum_{(i,j)\in E} \frac{1}{d_{i}d_{j}} < 1 $$
and, consequently,
$$ \frac{n^{2}}{n-1} \leq b_{1} < 2n, $$
(15)
where the left inequality is attained for the complete graph \(G=K_{n}\).
It is noteworthy to see that \(S_{b}^{1}\) is derived from the general set S with \(a=n\), \(N=n-1\), \(b=b_{1}\), andFootnote 1
\(p=2\). By Lemma 1 we have for a non-complete graph
$$h^{\ast}= \biggl\lfloor \frac{n^{2}}{b_{1}} \biggr\rfloor $$
with
$$ \biggl\lfloor \frac{n}{2} \biggr\rfloor < h^{\ast} < n-1. $$
(16)
We distinguish the following cases:
-
1.
Considering \(\gamma_{\frac{n}{2}}\) for n even, by (16) we have \(h < h^{\ast}\). Hence, by solving equation (8) of Theorem 1, we can derive the unit root \(\alpha_{1}\) such as \(\gamma_{\frac{n}{2}}\leq \alpha_{1}\). Therefore
$$ \alpha_{1}=\frac{1}{n-1} \biggl(n+\sqrt{\frac{n-2}{n} \bigl(b_{1}(n-1)-n^{2} \bigr)} \biggr). $$
In virtue of (15), we have \(\frac{n}{n-1} \leq \alpha_{1} < 2\) with the left inequality attained only for the complete graph \(G=K_{n}\).
In a similar way, we can evaluate the value \(\beta_{1} \leq \gamma_{\frac{n}{2}} \), through the solution of equation (10) of Theorem 2 (where \(h< h^{\ast}+1\)), where
$$ \beta_{1}=\frac{1}{n-1} \biggl(n-\sqrt{\frac{n-2}{n} \bigl(b_{1}(n-1)-n^{2} \bigr)} \biggr). $$
From (15), we have \(\frac{2}{n-1} < \beta_{1} \leq \frac{n}{n-1}\) with the right inequality attained only for the complete graph \(G=K_{n}\).
Having \(\beta_{1} \leq \gamma_{\frac{n}{2}} \leq \alpha_{1}\), then \(\vert 1-\gamma_{\frac{n}{2}}\vert \leq \max(|1-\alpha_{1}|, |1-\beta_{1}|)\).
-
2.
Picking now \(\gamma_{\frac{n+2}{2}}\), where \(h \leq h^{\ast}\) by equation (8) of Theorem 1 we deduce
$$ \alpha_{2}=\frac{1}{n-1} \biggl(n+\sqrt{\frac{(n-4)}{(n+2)} \bigl(b_{1}(n-1)-n^{2}\bigr)} \biggr). $$
Taking into account the lower bound of \(\beta_{2} \leq \gamma_{\frac{n+2}{2}}\), by (16) and for n even, we have \(h < h^{\ast}+1\) and then
$$ \beta_{2}=\frac{n}{n-1} \biggl(1-\sqrt{\frac{b_{1}(n-1)-n^{2}}{n(n-2)}} \biggr). $$
In this case, we derive \(\vert 1-\gamma_{\frac{n+2}{2}}\vert \leq \max(|1-\alpha_{2}|, |1-\beta_{2}|)\).
-
3.
For a graph with n odd number of vertices, we need to study only \(\gamma_{\frac{n+1}{2}}\) with \(\beta_{3} \leq \gamma_{\frac{n+1}{2}} \leq \alpha_{3}\), and we have by Theorem 1 and Theorem 2, respectively:
$$ \alpha_{3}= \frac{1}{n-1} \biggl(n+\sqrt{\frac{(n-3)}{(n+1)} \bigl(b_{1}(n-1)-n^{2}\bigr)} \biggr) $$
and
$$ \beta_{3}=\frac{1}{n-1} \bigl(n-\sqrt{b_{1}(n-1)-n^{2}} \bigr), $$
where \(h < h^{\ast}+1\).
Hence, \(\vert 1-\gamma_{\frac{n+1}{2}}\vert \leq \max(|1-\alpha_{3}|, |1-\beta_{3}|)\).
By using the bounds on the eigenvalues of the normalized Laplacian matrix computed above, bounds (12) and (13) follow. □
Proposition 2
For a simple, connected and bipartite graphs with
n
even we have
$$ 0 \leq R(G) \leq d_{1}\bigl|1-\beta^{\mathrm{bip}}_{1}\bigr|, $$
(17)
where
$$ \beta^{\mathrm{bip}}_{1}=1-\sqrt{\frac{b_{2}}{n-2}-1}. $$
Proof
When G is bipartite, the HL-index is defined as
$$ R(G)=\left \{\textstyle\begin{array}{l@{\quad}l} |\lambda_{\frac{n}{2}}|=|\lambda_{\frac{n+2}{2}}| & \text{ if $n$ is even,} \\ 0 & \text{ if $n$ is odd} . \end{array}\displaystyle \right . $$
(18)
In virtue of (11), we can derive the following bound when n is even:
$$ R(G) \leq d_{1}\vert 1-\gamma_{\frac{n}{2}}\vert . $$
(19)
By applying majorization techniques, we are able to bound \(\gamma_{\frac{n}{2}}\) considered in (19).
To this aim, we now face the set
$$ S^{2}_{b}= \Biggl\{ \boldsymbol{\gamma} \in \mathbb{R}^{n-2}:\sum_{i=2}^{n-1} \gamma_{i}=n-2,g(\boldsymbol{\gamma})=\sum _{i=2}^{n-1}\gamma_{i}^{2}=b_{2}=n+2 \sum_{(i,j)\in E}\frac{1}{d_{i}d_{j}}-4 \Biggr\} . $$
By (15) and \(b_{2}=b_{1}-4\), we have
$$ \frac{(n-2)^{2}}{n-1} \leq b_{2} < 2(n-2), $$
(20)
where the left inequality is attained for the complete graph \(G=K_{n}\).
The set \(S_{b}^{2}\) is derived from the general set S with \(a=n-2\), \(N=n-2\), \(b=b_{2}\), and \(p=2\). By Lemma 1 we have for a non-complete graph
$$h^{\ast}= \biggl\lfloor \frac{(n-2)^{2}}{b_{2}} \biggr\rfloor $$
with
$$ \frac{n}{2}-1 < h^{\ast} \leq n-1, $$
(21)
for n even.
We now consider \(\gamma_{\frac{n}{2}}\):
-
1.
By (21), we have \(h \leq h^{\ast}\). In virtue of equation (8) of Theorem 1, we deduce the following upper bound:
$$ \alpha^{\mathrm{bip}}_{1}=1+\sqrt{ \biggl(\frac{n-4}{n} \biggr) \biggl(\frac{b_{2}}{n-2}-1 \biggr)} . $$
-
2.
In a similar way, we can evaluate the value \(\beta^{\mathrm{bip}}_{1} \leq \gamma_{\frac{n}{2}} \), where \(h < h^{\ast}+1\). Applying Theorem 2 entails
$$ \beta^{\mathrm{bip}}_{1}=1-\sqrt{\frac{b_{2}}{n-2}-1}. $$
It is easy to show that \(|1-\alpha^{\mathrm{bip}}_{1}| \leq |1-\beta^{\mathrm{bip}}_{1}|\). Hence bound (17) follows. □
3.2 Bounds on \(R(G)\) through the energy index
In the following we obtain bounds on the HL-index starting from (2). Our aim is to bound the energy index making use of additional information on the first eigenvalue of \(A(G)\). In [24] the authors show that, if a tighter bound k on \(\lambda_{1}\) such as \(\lambda_{1} \geq k \geq \frac{2m}{n}\) is available, then the energy index for a non-bipartite graph is bounded as
$$ E(G)\leq k + \sqrt{(n-1) \bigl(2m-k^{2}\bigr)}, $$
while for a bipartite graph it is
$$ E(G)\leq 2k + \sqrt{(n-2) \bigl(2m-2k^{2}\bigr)}. $$
In order to find the value of k, we can introduce new variables \(x_{i}=\lambda_{i}^{2}\), facing the set:
$$ S^{2}_{b}= \Biggl\{ \boldsymbol{x} \in \mathbb{R}_{+}^{n}: \sum_{i=1}^{n}x_{i}=2m \Biggr\} . $$
In virtue of (2), we are now able to derive the following bounds for non-bipartite and bipartite graphs, respectively.
Proposition 3
-
1.
For a simple, connected and non-bipartite graph
G
$$ R(G)\leq \frac{k}{n} + \frac{1}{n}\sqrt{(n-1) \bigl(2m-k^{2} \bigr)}. $$
(22)
-
2.
For a simple, connected and bipartite graph
G
$$ R(G)\leq \frac{2k}{n} + \frac{1}{n}\sqrt{(n-2) \bigl(2m-2k^{2} \bigr)}, $$
(23)
where, by means of Theorem
2,
$$k=\frac{1}{1+h^{\ast}} \biggl(n+\sqrt{\frac{2m(1+h^{\ast})-n^{2}}{h^{\ast}}} \biggr),\quad h^{\ast}= \biggl\lfloor \frac{n^{2}}{2m} \biggr\rfloor .$$
Remark 1
Bounds (22) and (23) are tighter than or equal to (3) and (4), respectively.
Proof of Proposition 3
-
1.
Non-bipartite graphs
We start by proving that the condition \((2m-k^{2})\geq0\) required in bound (22) is always satisfied for simple and connected graphs.
We have \(k \in ( \frac{2m}{n},\frac{1}{2} (n+\sqrt{4m-n^{2}} ) )\). Indeed, by the basic concepts of calculus it is easy to see that k increases when m increases and then \(h^{\ast}\) tends to 1. Hence, k is limited from above by \(\frac{1}{2} (n+\sqrt{4m-n^{2}} )\) and where \(\frac{1}{2} (n+\sqrt{4m-n^{2}} )\leq \sqrt{2m}\) the required condition is satisfied.
We now show how bound (22) improves bound (3) presented in [14].
We need to prove that the following inequality holds:
$$ \biggl( k-\frac{2m}{n} \biggr)\leq \biggl( \sqrt{ ( n-1 ) \biggl( 2m- \frac{4m^{2}}{n^{2}} \biggr) }-\sqrt{ ( n-1 ) \bigl( 2m-k^{2} \bigr) } \biggr). $$
By simple algebraic rules we obtain
$$\begin{aligned} &k^{2}n-4k\frac{m}{n}-4mn+4m+4 \frac{m^{2}}{n} \\ &\quad\leq-2\sqrt{ \bigl( 2mn-2m+k^{2}-k^{2}n \bigr) \biggl( 2mn-2m-4 \frac{m^{2}}{n}+4\frac{m^{2}}{n^{2} } \biggr) }. \end{aligned}$$
(24)
The left-hand side term of (24) can be represented by
$$ f(k)=k^{2}n-4k\frac{m}{n}-4mn+4m+4\frac{m^{2}}{n}. $$
The function \(f(k)\) is a convex parabola that assumes negative values in the range of k we are interested in. Indeed we have \(f(\frac{2m}{n}) \leq 0\) and \(f(\sqrt{2m}) \leq 0\).
Both sides of (24) being negative we can apply some basic concepts of algebra, getting
$$ k^{2}n^{2}+k ( 4mn-8m ) +8m-8mn+4m^{2}\geq 0 . $$
(25)
The function \(t(k)=k^{2}n^{2}+k ( 4mn-8m ) +8m-8mn+4m^{2}\) is again a convex parabola with vertex \(( \frac{2m ( 2-n ) }{n^{2}},\frac{8m ( 2m-n^{2} ) ( n-1 )}{n^{2}} ) \). Both coordinates are less than zero (then \(\frac{2m (2-n ) }{n^{2}} < \frac{2m}{n}\)). Having \(t(\frac{2m}{n})=8m(n-2m)(1-n)\geq 0\), inequality (25) is satisfied.
Therefore, bound (22) performs better than or equal to bound (3).
Furthermore, we see that both bounds perform equally when \(h^{\ast}= \frac{n^{2}}{2m}\) (i.e.
\(\frac{n^{2}}{2m}\) is an integer). It is noteworthy that:
-
(a)
when n is odd, \(\frac{n^{2}}{2m}\) is never an integer (\(\lfloor \frac{n^{2}}{2m} \rfloor \neq \frac{n^{2}}{2m}\));
-
(b)
when n is even, \(\frac{n^{2}}{2m}\) is an integer when \(m=\frac{n^{2}}{2x}\) with \(x|\frac{n^{2}}{2}\), \(2 \leq x \leq \frac{n}{2}\) (where \(x|\frac{n^{2}}{2}\) is shorthand for ‘x divides \(\frac{n^{2}}{2}\)’). In this case \(k=\frac{2m}{n}\) and we derive bound (3).
-
2.
Bipartite graphs
As for non-bipartite graphs, we start by proving that the condition \((m-k^{2})\geq0\) required in bound (23) is always satisfied for simple and connected graphs. In this case \(h^{\ast}\) tends to 2 for complete graphs (where \(m \leq \frac{n^{2}}{4}\)).
By some basic algebraic concepts, we see that \((m-k^{2})\geq0\) entails:
$$m^{2}\bigl(h^{4}+2h^{3}-3h^{2}-4h+4\bigr)+m\bigl(2n^{2}h-4n^{2}-2n^{2}h^{2}\bigr)+n^{4} \geq 0. $$
The function \(f(m)=m^{2}(h^{4}+2h^{3}-3h^{2}-4h+4)+m(2n^{2}h-4n^{2}-2n^{2}h^{2})+n^{4}\) is concave, decreasing, and non-negative on the interval \(m \in (\frac{n}{2},\frac{n^{2}}{4} )\). Therefore, the required condition is satisfied.
We now show how bound (23) improves bound (4) presented in [14].
We need to prove that the following inequality holds:
$$ 2 \biggl( k-\frac{2m}{n} \biggr)\leq \biggl( \sqrt{ ( n-2 ) \biggl( 2m- \frac{8m^{2}}{n^{2}} \biggr) }-\sqrt{ ( n-2 ) \bigl( 2m-2k^{2} \bigr) } \biggr). $$
By simple algebra, we have
$$\begin{aligned} &4mn-8km+4m^{2}-2mn^{2}+k^{2}n^{2} \\ &\quad\leq-n\sqrt{ \bigl(2mn-4m+4k^{2}-2k^{2}n \bigr) \biggl(2mn-4m-8 \frac{m^{2}}{n}+16\frac{m^{2}}{n^{2}} \biggr)}. \end{aligned}$$
(26)
The left-hand side term of (26) can be represented by
$$ f(k)=k^{2}n^{2}-8km+4mn+4m^{2}-2mn^{2}. $$
The function \(f(k)\) is a convex parabola that assumes negative values in the range of k we are interested in. Indeed we have \(f(\frac{2m}{n}) \leq 0\) and \(f(\sqrt{m}) \leq 0\).
Both sides of (26) being negative, by some manipulations we obtain
$$ k^{2}n^{2}+4km(n-4)+4m^{2}+16m-8mn\geq 0 . $$
(27)
The function \(t(k)=k^{2}n^{2}+4km(n-4)+4m^{2}+16m-8mn\) is again a convex parabola with vertex \(( \frac{2m (4-n ) }{n^{2}},\frac{8m(4m-n^{2})(n-2)}{n^{2}} )\). Both coordinates are less than zero (then \(\frac{4m ( 4-n ) }{2n^{2}} < \frac{2m}{n}\)). Having \(t(\frac{2m}{n})=\frac{8m}{n}(n+2m(n-2))\geq 0\), inequality (27) is satisfied.
Hence bound (23) performs better than or equal to bound (4).
□