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 HLindex for both nonbipartite and bipartite graphs. According to [29] (see Theorem 2.2.1), the following relations hold:
$$ \frac{\lambda_{nk+1}}{d_{1}} \leq 1\gamma_{k}\leq\frac{\lambda_{nk+1}}{d_{n}}. $$
(11)
Proposition 1
For a simple, connected, and nonbipartite 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
$$\begin{aligned} &\alpha_{1}=\frac{1}{n1} \biggl(n+\sqrt{\frac{n2}{n} \bigl(b_{1}(n1)n^{2} \bigr)} \biggr),\\ & \beta_{1}= \frac{1}{n1} \biggl(n\sqrt{\frac{n2}{n} \bigl(b_{1}(n1)n^{2} \bigr)} \biggr),\\ &\alpha_{2}=\frac{1}{n1} \biggl(n+\sqrt{\frac{(n4)}{(n+2)} \bigl(b_{1}(n1)n^{2}\bigr)} \biggr),\\ & \beta_{2}= \frac{n}{n1} \biggl(1\sqrt{\frac{b_{1}(n1)n^{2}}{n(n2)}} \biggr), \end{aligned}$$
and
$$ 0 \leq R(G) \leq d_{1}\max \bigl(1\alpha_{3},1 \beta_{3} \bigr) $$
(13)
when
n
is odd with
$$ \alpha_{3}= \frac{1}{n1} \biggl(n+\sqrt{\frac{(n3)}{(n+1)} \bigl(b_{1}(n1)n^{2}\bigr)} \biggr), \qquad\beta_{3}= \frac{1}{n1} \bigl(n\sqrt{b_{1}(n1)n^{2}} \bigr). $$
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:
$$ S_{b}^{1}= \Biggl\{ \boldsymbol{\gamma} \in \mathbb{R}^{n1}:\sum_{i=1}^{n1} \gamma_{i}=n,g(\boldsymbol{\gamma})=\sum_{i=1}^{n1} \gamma_{i}^{2}=b_{1}=n+2\sum _{(i,j)\in E}\frac{1}{d_{i}d_{j}} \Biggr\} . $$
It is well known that, for every connected graph of order n (see [30]), we have
$$ \frac{1}{n1} \leq \frac{2}{n} \sum_{(i,j)\in E} \frac{1}{d_{i}d_{j}} < 1 $$
and, consequently,
$$ \frac{n^{2}}{n1} \leq b_{1} < 2n, $$
(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 \(a=n\), \(N=n1\), \(b=b_{1}\), and^{Footnote 1}
\(p=2\). By Lemma 1 we have for a noncomplete graph
$$h^{\ast}= \biggl\lfloor \frac{n^{2}}{b_{1}} \biggr\rfloor $$
with
$$ \biggl\lfloor \frac{n}{2} \biggr\rfloor < h^{\ast} < n1. $$
(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
$$ \alpha_{1}=\frac{1}{n1} \biggl(n+\sqrt{\frac{n2}{n} \bigl(b_{1}(n1)n^{2} \bigr)} \biggr). $$
In virtue of (15), we have \(\frac{n}{n1} \leq \alpha_{1} < 2\) 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
$$ \beta_{1}=\frac{1}{n1} \biggl(n\sqrt{\frac{n2}{n} \bigl(b_{1}(n1)n^{2} \bigr)} \biggr). $$
From (15), we have \(\frac{2}{n1} < \beta_{1} \leq \frac{n}{n1}\) 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
$$ \alpha_{2}=\frac{1}{n1} \biggl(n+\sqrt{\frac{(n4)}{(n+2)} \bigl(b_{1}(n1)n^{2}\bigr)} \biggr). $$
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
$$ \beta_{2}=\frac{n}{n1} \biggl(1\sqrt{\frac{b_{1}(n1)n^{2}}{n(n2)}} \biggr). $$
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:
$$ \alpha_{3}= \frac{1}{n1} \biggl(n+\sqrt{\frac{(n3)}{(n+1)} \bigl(b_{1}(n1)n^{2}\bigr)} \biggr) $$
and
$$ \beta_{3}=\frac{1}{n1} \bigl(n\sqrt{b_{1}(n1)n^{2}} \bigr), $$
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
$$ 0 \leq R(G) \leq d_{1}\bigl1\beta^{\mathrm{bip}}_{1}\bigr, $$
(17)
where
$$ \beta^{\mathrm{bip}}_{1}=1\sqrt{\frac{b_{2}}{n2}1}. $$
Proof
When G is bipartite, the HLindex 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
$$ S^{2}_{b}= \Biggl\{ \boldsymbol{\gamma} \in \mathbb{R}^{n2}:\sum_{i=2}^{n1} \gamma_{i}=n2,g(\boldsymbol{\gamma})=\sum _{i=2}^{n1}\gamma_{i}^{2}=b_{2}=n+2 \sum_{(i,j)\in E}\frac{1}{d_{i}d_{j}}4 \Biggr\} . $$
By (15) and \(b_{2}=b_{1}4\), we have
$$ \frac{(n2)^{2}}{n1} \leq b_{2} < 2(n2), $$
(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 \(a=n2\), \(N=n2\), \(b=b_{2}\), and \(p=2\). By Lemma 1 we have for a noncomplete graph
$$h^{\ast}= \biggl\lfloor \frac{(n2)^{2}}{b_{2}} \biggr\rfloor $$
with
$$ \frac{n}{2}1 < h^{\ast} \leq n1, $$
(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:
$$ \alpha^{\mathrm{bip}}_{1}=1+\sqrt{ \biggl(\frac{n4}{n} \biggr) \biggl(\frac{b_{2}}{n2}1 \biggr)} . $$

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
$$ \beta^{\mathrm{bip}}_{1}=1\sqrt{\frac{b_{2}}{n2}1}. $$
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 HLindex 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 nonbipartite graph is bounded as
$$ E(G)\leq k + \sqrt{(n1) \bigl(2mk^{2}\bigr)}, $$
while for a bipartite graph it is
$$ E(G)\leq 2k + \sqrt{(n2) \bigl(2m2k^{2}\bigr)}. $$
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 nonbipartite and bipartite graphs, respectively.
Proposition 3

1.
For a simple, connected and nonbipartite graph
G
$$ R(G)\leq \frac{k}{n} + \frac{1}{n}\sqrt{(n1) \bigl(2mk^{2} \bigr)}. $$
(22)

2.
For a simple, connected and bipartite graph
G
$$ R(G)\leq \frac{2k}{n} + \frac{1}{n}\sqrt{(n2) \bigl(2m2k^{2} \bigr)}, $$
(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.
Nonbipartite graphs
We start by proving that the condition \((2mk^{2})\geq0\) required in bound (22) is always satisfied for simple and connected graphs.
We have \(k \in ( \frac{2m}{n},\frac{1}{2} (n+\sqrt{4mn^{2}} ) )\). 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 \(\frac{1}{2} (n+\sqrt{4mn^{2}} )\) and where \(\frac{1}{2} (n+\sqrt{4mn^{2}} )\leq \sqrt{2m}\) 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:
$$ \biggl( k\frac{2m}{n} \biggr)\leq \biggl( \sqrt{ ( n1 ) \biggl( 2m \frac{4m^{2}}{n^{2}} \biggr) }\sqrt{ ( n1 ) \bigl( 2mk^{2} \bigr) } \biggr). $$
By simple algebraic rules we obtain
$$\begin{aligned} &k^{2}n4k\frac{m}{n}4mn+4m+4 \frac{m^{2}}{n} \\ &\quad\leq2\sqrt{ \bigl( 2mn2m+k^{2}k^{2}n \bigr) \biggl( 2mn2m4 \frac{m^{2}}{n}+4\frac{m^{2}}{n^{2} } \biggr) }. \end{aligned}$$
(24)
The lefthand side term of (24) can be represented by
$$ f(k)=k^{2}n4k\frac{m}{n}4mn+4m+4\frac{m^{2}}{n}. $$
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
$$ k^{2}n^{2}+k ( 4mn8m ) +8m8mn+4m^{2}\geq 0 . $$
(25)
The function \(t(k)=k^{2}n^{2}+k ( 4mn8m ) +8m8mn+4m^{2}\) is again a convex parabola with vertex \(( \frac{2m ( 2n ) }{n^{2}},\frac{8m ( 2mn^{2} ) ( n1 )}{n^{2}} ) \). Both coordinates are less than zero (then \(\frac{2m (2n ) }{n^{2}} < \frac{2m}{n}\)). Having \(t(\frac{2m}{n})=8m(n2m)(1n)\geq 0\), 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 nonbipartite graphs, we start by proving that the condition \((mk^{2})\geq0\) 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 \((mk^{2})\geq0\) entails:
$$m^{2}\bigl(h^{4}+2h^{3}3h^{2}4h+4\bigr)+m\bigl(2n^{2}h4n^{2}2n^{2}h^{2}\bigr)+n^{4} \geq 0. $$
The function \(f(m)=m^{2}(h^{4}+2h^{3}3h^{2}4h+4)+m(2n^{2}h4n^{2}2n^{2}h^{2})+n^{4}\) is concave, decreasing, and nonnegative 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:
$$ 2 \biggl( k\frac{2m}{n} \biggr)\leq \biggl( \sqrt{ ( n2 ) \biggl( 2m \frac{8m^{2}}{n^{2}} \biggr) }\sqrt{ ( n2 ) \bigl( 2m2k^{2} \bigr) } \biggr). $$
By simple algebra, we have
$$\begin{aligned} &4mn8km+4m^{2}2mn^{2}+k^{2}n^{2} \\ &\quad\leqn\sqrt{ \bigl(2mn4m+4k^{2}2k^{2}n \bigr) \biggl(2mn4m8 \frac{m^{2}}{n}+16\frac{m^{2}}{n^{2}} \biggr)}. \end{aligned}$$
(26)
The lefthand 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
$$ k^{2}n^{2}+4km(n4)+4m^{2}+16m8mn\geq 0 . $$
(27)
The function \(t(k)=k^{2}n^{2}+4km(n4)+4m^{2}+16m8mn\) is again a convex parabola with vertex \(( \frac{2m (4n ) }{n^{2}},\frac{8m(4mn^{2})(n2)}{n^{2}} )\). Both coordinates are less than zero (then \(\frac{4m ( 4n ) }{2n^{2}} < \frac{2m}{n}\)). Having \(t(\frac{2m}{n})=\frac{8m}{n}(n+2m(n2))\geq 0\), inequality (27) is satisfied.
Hence bound (23) performs better than or equal to bound (4).
□