An improvement of a result of Zverovich–Zverovich
Abstract.
We give an improvement of a result of Zverovich and Zverovich which gives a condition on the first and last elements in a decreasing sequence of positive integers for the sequence to be graphic, that is, the degree sequence of a finite graph.
Key words and phrases:
graph, graphic sequence,1. Statement of Results
A finite sequence of positive integers is graphic if it occurs as the sequence of vertex degrees of a graph. Here, graphs are understood to be simple, in that they have no loops or repeated edges. A result of Zverovich and Zverovich states:
Theorem 1 ([8, Theorem 6]).
Let be reals. If is a sequence of positive integers in decreasing order with and
then d is graphic.
Notice that here the term is monotonic increasing in , for and fixed , and it is also monotonic decreasing in , for and fixed . Thus any sequence that satisfies the inequality , for any pair , will also satisfy the inequality . So Theorem 1 has the following equivalent expression.
Theorem 2.
Suppose that is a decreasing sequence of positive integers with even sum. If
(1) 
then d is graphic.
The simplified form of Theorem 2 also affords a somewhat simpler proof, which we give in Section 2 below. Admittedly, the proof in [8] is already quite elementary, though it does use the strong index results of [4, 3].
The following corollary of Zverovich–Zverovich’s is obtained by taking and in Theorem 1.
Corollary 1 ([8, Corollary 2]).
Suppose that is a decreasing sequence of positive integers with even sum. If , then d is graphic.
Zverovich–Zverovich state that the bound of Corollary 1 “cannot be improved”, and they give examples to this effect. In fact, there is an improvement, as we will now describe. The subtlety here is that in the Zverovich–Zverovich examples, for a given (integer) value of , the Corollary 1 bound can’t be improved for integer . Nevertheless the bound on , for given integer , can be improved. We prove the following result in Section 2.
Theorem 3.
Suppose that is a decreasing sequence of positive integers with even sum. If , then d is graphic.
Example 1.
There are many examples of sequences that verify the hypotheses of Theorem 3 but not those of Corollary 1. For example, for every positive odd integer , consider the sequence , and for even, consider the sequence . Here, and in sequences throughout this paper, the superscripts indicate the number of repetitions of the entry.
Example 2.
Remark 1.
The fact that Theorem 2 is not sharp has also been remarked in [1], in the abstract of which the authors state that Theorem 2 is “sharp within 1”. They give the bound
(2) 
where if is odd, and otherwise. Consider any decreasing sequence with and . Note that the bound given by Theorem 2 is , the bound given by (2) is , while Theorem 3 gives the stronger bound . The paper [1] gives more precise bounds, as a function of , and the maximal gap in the sequence.
2. Proofs of Theorems 2 and 3
We will require the Erdős–Gallai Theorem, which we recall for convenience.
Erdős–Gallai Theorem.
A sequence of nonnegative integers in decreasing order is graphic if and only if its sum is even and, for each integer with ,
(EG) 
Proof of Theorem 2.
Suppose that is a decreasing sequence with even sum, satisfying (1), and which is not graphic. By the Erdős–Gallai Theorem, there exists with , such that
(3) 
For each with , replace by ; the left hand side of (3) is not decreased, while the right hand side of (3) is unchanged, so (3) still holds. Now for each with , replace by ; the left hand side of (3) is unchanged, while the right hand side of (3) has not increased, so (3) again holds. Notice that if , then (3) gives , and so . Then (1) would give , that is, . But this inequality clearly has no solutions. Hence . Thus (3) now reads , or equivalently
But this contradicts the hypothesis. ∎
The following proof uses the same general strategy as the preceding proof, but requires a somewhat more careful argument.
Proof of Theorem 3.
Suppose that satisfies the hypotheses of the theorem. First suppose that is even, say . If , then since is a strictly monotonic decreasing function of for , we have
so and hence is graphic by Theorem 2. So, assuming that is not graphic, we may suppose that . Furthermore, by Corollary 1, we may assume that , so .
Now, as in the proof of Theorem 2, by the Erdős–Gallai Theorem, there exists with , such that
(4) 
For each with , replace by ; the left hand side of (4) is not decreased, while the right hand side of (4) is unchanged, so (4) still holds. For each with , replace by ; the left hand side of (4) is unchanged, while the right hand side of (4) has not increased, so (4) again holds. Then (4) reads , and consequently, rearranging terms, . Thus . Notice that for , if any of the original terms had been less than , we would have obtained , which is impossible. Similarly, for , all the original terms must have been all equal to one. Thus . So has sum , which is odd, regardless of whether is even or odd. This contradicts the hypothesis.
Now consider the case where is odd, say . The theorem is trivial for , so we may assume that . We use essentially the same approach as we used in the even case, but the odd case is somewhat more complicated. By Corollary 1, assuming is not graphic, we have , and hence, as is odd, . Thus, since , we have or . Thus there are two cases:

,

.
By the Erdős–Gallai Theorem, there exists with , such that
(5) 
As before, for each with , replace by and for each with , replace by , and note that (5) again holds. Arguing as in the proof of Theorem 2, notice that if , then (5) gives , and so . In both cases (i) and (ii) we would have and hence , contrary to our assumption. Thus and (5) reads , and consequently, rearranging terms, we obtain in the respective cases:

.

,
In both cases we have . Consider as a quadratic in . For this to be negative, its discriminant, , must be positive. If we obtain . For we have and so , contrary to our assumption. Similarly, for we have and so again . For , the function is monotonic increasing in . So, as ,
which again gives . We conclude that .
So the two cases are:

.

,
In case (ii) we must have , but this is impossible for integer and .
In case (i), either or . Notice that for , if any of the original terms had been less than , we would have obtained , which is impossible. Similarly, for , all the original terms must have been all equal to one. Thus . Consequently, if , we have as . In this case, has sum , which is odd, regardless of whether is even or odd, contradicting the hypothesis. On the other hand, if , we have . Here, has sum , which is again odd, regardless of whether is even or odd, contrary to the hypothesis. ∎
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