Proof. The proof is by induction on . The statement is trivial for . Suppose that and assume that the statement is true for any tree of order less than n. If , then statement is trivial. Let . Clearly T is a double star. Thus, and hence with equality if and only if . So, in the following we may assume that . Let be a diametral path in T satisfying that is as large as possible, where . Root T at . First suppose that . Let f be a -function, where . Since and . Define the function by for each and otherwise. Obviously h is a -RDF on T and . By the induction hypothesis, we obtain as desired. Suppose that . By the choice of diametral path, we have that the degree of each child of with depth 1, is 2.
If , then clearly any -function can be extended to a -RDF on T by assigning 0 to and 5 to , and so by the induction hypothesis, Assume that . If is a strong support vertex or is a support vertex and , then by the induction hypothesis, as desired.
Case 1. is a support vertex of degree 3.
If
, then by assigning 5 to
, 4 to all leaves at distance 2 from
and 0 to remaining vertices we obtain a
-RDF on
T with weight 13, implying that
. Thus, let
. If
, then by assigning 5 to
, 4 to
and 0 to other vertices of
, a
-function is extended to a
-function and so by induction hypothesis,
. Now assume that
, it is obvious that
and as above
. By induction hypothesis on
, we have
as desired.
Let the equality
hold. Then we must have equality throughout in Equation (
1). In particular, we must have
. By the induction hypothesis,
. Hence
can be built from
copies
and
copies
by adding
edges between
to connect the graph. Without loss of generality, we may assume that
. First let
. If
is a leaf of
, then there exists a
-function that assigns 4 to
and 5 to the other support vertex of
, and the function
h defined by
,
and
otherwise, is a [4]-RDF on
T and so
which is impossible. Hence
is a support vertex. If
, then
. Suppose that
. If
, then the function
h defined by
for
,
for
,
,
for
and
otherwise, is a [4]-RDF of
T and so
, a contradiction. Thus,
and so
.
Now let . If is a leaf of , say , then the function g defined on T by , for , for , , and otherwise, is a quadruple Roman dominating function of T of weight less than which is a contradiction. If is a support vertex of , say , then the function g defined on T by , for , for , , and otherwise, is a quadruple Roman dominating function of T of weight less than which is a contradiction. Thus, and so
Case 2. is not a support vertex and .
It is clear that is a healthy spider with at least three feet. We first assume that , then and applying the fact and the induction hypothesis, we have . Now, we distinguish two situations.
Subcase 2.1..
First let
and let
. Then
and every
-function can be extended to a
-function by assigning 5 to
, 4 to leaves of
except
and 0 to remaining vertices. Thus, by induction hypothesis,
Let the equality
hold. Then we must have equality throughout in Equation (
1). In particular, we must have
and that
is a leaf in
. If
, then clearly
, which is impossible. By the induction hypothesis, we have
. Clearly
has a
-function
h assigning 4 to
. Now
h can be extended to a
-RDF on
T by assigning 4 to
and any leaf of
at distance 2 from
and 0 to remaining vertices, implying that
which is a contradiction.
Assume now that . Then and each -function can be extended to a -function by assigning 5 to support vertices in and 0 to remaining vertices. Thus, by induction hypothesis, , as desired.
Subcase 2.2..
Assume first that
and let
. Then
and each
-function can be extended to a
-function by assigning 5 to
, 4 to leaves of
except
and 0 to remaining vertices. Thus,
. Therefore, by induction hypothesis,
Assume now that
. Then
and each
-function can be extended to a
-function by assigning 5 to support vertices in
and 0 to the remaining vertices. Thus, by induction hypothesis,
Let the equality
hold. Let
. Then we must have equality throughout in Equation (
2). Moreover, we must have
and that
is not a leaf of
. By the induction hypothesis,
. As a result,
can be built from
copies
and
copies
by adding
edges between
to connect the graph. Without loss of generality, we may assume that
. First let
. Then
is a support vertex. If
, then
. Let
. If
, then the function
h defined by
,
for each
,
for
,
for
,
for each
, and
otherwise, is a [4]-RDF on
T and so
, a contradiction. Thus,
and so
.
Now let . If is a support vertex of , say , then the function h defined on T by , for , for , , for and otherwise, is a quadruple Roman dominating function of T of weight less than which is a contradiction. Thus, and so .
Case 3. is not a support vertex and .
If , then assigning 5 to support vertices of T and 0 to other vertices, introduces a [4]-RDF on T, implying that , as desired. Let , then , and each -function can be extended to a -function by assigning 5 to support vertices in and 0 to the remaining vertices. Thus, by induction hypothesis, , as desired. □
Proof. The proof is by induction on n. If , then . If , then .
Suppose that and assume that the statement is true for any tree of order less than n. If , then . If , then . Let and be a diametral path of T such that is as large as possible. We root T at . Now let f be a -function. If , then and so f restricted to is a [4]-RDF. As a result, by induction hypothesis, . Now, let . By the choice of diametral path, the degree of every child of with depth 1, is two.
Case 1. is a strong support vertex or has a child w with depth one different from .
If is a strong support vertex, then as above we have . Assume that w is a child of with depth one different from . Suppose that . Let . Clearly and the function f, restricted to is a [4]-RDF of , implying that . By the above inequality and the induction hypothesis, , as desired.
Case 2. and is a support vertex.
Considering Case 1, we can assume that is a support vertex. Assume that w is a leaf adjacent to . Let . Obviously, . Without loss of generality, we may assume that and . Now the result follows as in Case 1.
Case 3..
Let . Without loss of generality, we may assume that and . Hence the function f restricted to is a [4]-RDF of , implying that . As a result, by the induction hypothesis, , and this completes the proof. □