**2. Main Result**

We begin this section with the following useful result of total Roman dominating functions given in [16].

**Lemma 1** ([16])**.** *If G is a graph with no isolated vertex, then there exists a <sup>γ</sup>tR*(*G*)*-function f*(*<sup>V</sup>*0, *V*1, *<sup>V</sup>*2) *such that either V*2 *is a dominating set of G, or the set S of vertices not dominated by V*2 *satisfies G*[*S*] = *kK*2 *for some k* ≥ 1*, where S* ⊆ *V*1 *and ∂*(*S*) ⊆ *V*0*.*

It is known from [9] that for any graph *G*, *<sup>γ</sup>R*(*G*) ≤ <sup>2</sup>*γ*(*G*) and also, from Inequality chain (1) that *γ*(*G*) ≤ *<sup>γ</sup>t*<sup>2</sup>(*G*). Hence, and as consequence of both inequalities above, we deduce that the following result improves the lower and upper bounds given in Inequality chain (2) for the total Roman domination number of graphs.

**Theorem 1.** *For any graph G with neither isolated vertex nor components isomorphic to K*2*,*

$$
\gamma\_{t2}(G) + \gamma(G) \le \gamma\_{tR}(G) \le \gamma\_R(G) + \gamma(G).
$$

**Proof.** We first prove the lower bound. By Lemma 1, there exists a *<sup>γ</sup>tR*(*G*)-function *<sup>g</sup>*(*Vg*0 , *Vg*1 , *Vg*2 ) such that either *Vg*2 is a dominating set of *G*, or *<sup>V</sup>g*1,1 satisfies *<sup>G</sup>*[*Vg*1,1] ∼= *kK*2 for some *k* ≥ 1. Hence, *Vg*2 is a dominating set of *G* − *<sup>V</sup>g*1,1 and can be extended to a dominating set of *G* by adding to it the set *<sup>I</sup>*(*Vg*1,1). So *γ*(*G*) ≤ |*Vg*2 ∪ *<sup>I</sup>*(*Vg*1,1)| = |*Vg*2 | + |*Vg*1,1|/2. Moreover, *Vg*2 ∪ *<sup>V</sup>g*1,2 is a total dominating set of *G* − *<sup>V</sup>g*1,1 and it is easy to check that *Vg*2 ∪ *<sup>V</sup>g*1,2 ∪ *<sup>I</sup>*(*Vg*1,1) is a semitotal dominating set of *G*. Therefore *<sup>γ</sup>t*2(*G*) ≤ |*Vg*2 ∪ *<sup>V</sup>g*1,2 ∪ *<sup>I</sup>*(*Vg*1,1)| = |*Vg*2 | + |*Vg*1,2| + |*I*(*Vg*1,1)| = |*Vg*2 | + |*Vg*1,2| + |*Vg*1,1|/2 and so,

$$
\gamma\_{t2}(\mathcal{G}) + \gamma(\mathcal{G}) \le (|V\_2^{\mathcal{G}}| + |V\_{1,2}^{\mathcal{G}}| + |V\_{1,1}^{\mathcal{G}}|/2) + (|V\_2^{\mathcal{G}}| + |V\_{1,1}^{\mathcal{G}}|/2) = 2|V\_2^{\mathcal{G}}| + |V\_1^{\mathcal{G}}| = \gamma\_{tR}(\mathcal{G})\_\*
$$

which completes the proof of the lower bound.

Now, in order to prove the upper bound, let *D* be a *γ*(*G*)-set and *f*(*<sup>V</sup>*0, *V*1, *<sup>V</sup>*2) be a *<sup>γ</sup>R*(*G*)-function. Also, we consider *V*1,0 = {*v* ∈ *V*1 : *<sup>N</sup>*(*v*) ⊆ *<sup>V</sup>*0} and let *f* (*V*0, *<sup>V</sup>*1, *<sup>V</sup>*2) be a function defined as follows.


Since *f* is an RDF on *G*, by construction we have that *f* is a TRDF on *G*. Therefore,

$$\begin{aligned} \gamma\_{t\mathbb{R}}(G) &\leq \quad \omega(f')\\ &\leq \quad |V\_1| + 2|V\_2| + |(V\_{1,0} \cup V\_2) \cap D| + |V\_0 \cap D|\\ &\leq \quad (|V\_1| + 2|V\_2|) + |D|\\ &= \quad \gamma\_{\mathbb{R}}(G) + \gamma(G), \end{aligned}$$

which completes the proof.

Now, we show a family of graphs *Gp*,*<sup>q</sup>* given by Cabrera et al. in [18], which satisfy that *<sup>γ</sup>tR*(*Gp*,*<sup>q</sup>*) = *<sup>γ</sup>t*<sup>2</sup>(*Gp*,*<sup>q</sup>*) + *<sup>γ</sup>*(*Gp*,*<sup>q</sup>*) (observe that *<sup>γ</sup>*(*Gp*,*<sup>q</sup>*) = *p*, *<sup>γ</sup>t*<sup>2</sup>(*Gp*,*<sup>q</sup>*) = *p* + 1 and *<sup>γ</sup>tR*(*Gp*,*<sup>q</sup>*) = 2*p* + 1). Let *p*, *q* be two integers such that *q* ≥ *p* ≥ 2. From the complete bipartite graph *Kp*,*<sup>q</sup>* and the empty graph *Np*, we construct the graph *Gp*,*<sup>q</sup>* as follows. We add *p* new edges which form a matching between the vertices of *Np* and the vertices of degree *q* in *Kp*,*q*. Figure 1 shows the graph *G*3,4 and a *<sup>γ</sup>tR*(*<sup>G</sup>*3,4)-function *g*(*<sup>V</sup>*0, *V*1, *<sup>V</sup>*2).

**Figure 1.** The graph *G*3,4.

Next, we provide some useful properties that satisfies a specific TRDF for the graphs *G* with *<sup>γ</sup>tR*(*G*) = *<sup>γ</sup>t*2(*G*) + *<sup>γ</sup>*(*G*).

**Theorem 2.** *For any graph G such that <sup>γ</sup>tR*(*G*) = *<sup>γ</sup>t*2(*G*) + *<sup>γ</sup>*(*G*)*, there exists a <sup>γ</sup>tR*(*G*)*-function f*(*<sup>V</sup>*0, *V*1, *<sup>V</sup>*2) *satisfying the following conditions.*


**Proof.** Let *f*(*<sup>V</sup>*0, *V*1, *<sup>V</sup>*2) be a *<sup>γ</sup>tR*(*G*)-function that satisfies Lemma 1. Hence, condition (i) holds.

Now, we proceed to prove (ii). First, we notice that *A* = *V*2 ∪ *<sup>I</sup>*(*<sup>V</sup>*1,1) and *B* = *V*2 ∪ *V*1,2 ∪ *<sup>I</sup>*(*<sup>V</sup>*1,1) are a dominating set and a semitotal dominating set, respectively. Hence, *γ*(*G*) ≤ |*A*| and *<sup>γ</sup>t*2(*G*) ≤ |*B*|. Since |*A*| + |*B*| = *<sup>γ</sup>tR*(*G*) and *<sup>γ</sup>tR*(*G*) = *<sup>γ</sup>t*2(*G*) + *<sup>γ</sup>*(*G*), we obtain that |*B*| + |*A*| = *<sup>γ</sup>t*2(*G*) + *<sup>γ</sup>*(*G*). If |*A*| > *<sup>γ</sup>*(*G*), then |*B*| < *<sup>γ</sup>t*<sup>2</sup>(*G*), which is a contradiction. Therefore, |*A*| = *γ*(*G*) and so, |*B*| = *<sup>γ</sup>t*<sup>2</sup>(*G*), which completes the proof of (ii).

Finally, we proceed to prove (iii). Let *v* ∈ *V*1,2. Clearly, *<sup>N</sup>*(*v*) ∩ *V*2 = ∅. If *<sup>N</sup>*(*v*) ∩ *V*1,2 = ∅ or |*N*(*v*) ∩ *<sup>V</sup>*2| > 1, then (*<sup>V</sup>*2 ∪ *V*1,2 ∪ *<sup>I</sup>*(*<sup>V</sup>*1,1)) \ {*v*} is a semitotal dominating set of *G*, which is a contradiction with the fact that *V*2 ∪ *V*1,2 ∪ *<sup>I</sup>*(*<sup>V</sup>*1,1) is a *<sup>γ</sup>t*2(*G*)-set by (ii). Therefore, *<sup>N</sup>*(*v*) ∩ *V*1,2 = ∅ and |*N*(*v*) ∩ *<sup>V</sup>*2| = 1, which implies that *<sup>G</sup>*[*<sup>V</sup>*1,2] is isomorphic to an empty graph, and that |*N*(*v*) ∩ *<sup>V</sup>*2| = 1, which completes the proof.

We consider again the family of graphs *Gp*,*q*. Let *g*(*<sup>V</sup>*0, *V*1, *<sup>V</sup>*2) be a *<sup>γ</sup>tR*(*Gp*,*<sup>q</sup>*)-function defined as *V*2 = {*v*} and *V*1 = (*S*(*Gp*,*<sup>q</sup>*) ∪ *<sup>L</sup>*(*Gp*,*<sup>q</sup>*)) \ {*v*}, for some *v* ∈ *<sup>S</sup>*(*G*). Notice that *g* satisfies the conditions given in Theorem 2. For an example, see the *<sup>γ</sup>tR*(*<sup>G</sup>*3,4)-function *g* showed in the Figure 1.

Next, we will show a family of graphs *Gr* that satisfy the upper bound in the Theorem 1. In this case we have that *γ*(*Gr*) = *r*, *<sup>γ</sup>R*(*Gr*) = 2*r* and *<sup>γ</sup>tR*(*Gr*) = 3*<sup>r</sup>*, where *r* ≥ 2 is an integer. The graph *Gr* is constructed from the path graph *P*3*<sup>r</sup>*−<sup>2</sup> = *v*1*v*2 ··· *v*3*<sup>r</sup>*−2 and the empty graph *N*2 by taking one copy of *P*3*<sup>r</sup>*−<sup>2</sup> and *r* copies of *N*2 and adding edges between the vertex *v*3*i*−2 and the *i*-th copy of *N*2, for *i* ∈ {1, . . . ,*<sup>r</sup>*}. Figure 2 shows the graph *G*3.

**Figure 2.** The graph *G*3.

### **3. Graphs** *G* **with** *<sup>γ</sup>tR***(***G***) = <sup>2</sup>***γ***(***G***)**

We begin this section with a simple characterization, which is a direct consequence of Theorem 1 and the Inequality chains (1) and (2).

**Theorem 3.** *Let G be a graph with no isolated vertex. Then <sup>γ</sup>tR*(*G*) = <sup>2</sup>*γ*(*G*) *if and only if <sup>γ</sup>tR*(*G*) = *<sup>γ</sup>t*2(*G*) + *γ*(*G*) *and <sup>γ</sup>t*2(*G*) = *<sup>γ</sup>*(*G*)*.*

We observe that the condition *<sup>γ</sup>tR*(*G*) = *<sup>γ</sup>t*2(*G*) + *γ*(*G*) is a necessary condition but is not sufficient to satisfy the equality *<sup>γ</sup>tR*(*G*) = <sup>2</sup>*γ*(*G*). For instance, see the graph *G*3,4 shown in Figure 1.

Next, we give another characterization for the graphs *G* satisfying *<sup>γ</sup>tR*(*G*) = <sup>2</sup>*γ*(*G*). It is important to emphasize that this characterization depends only of the existence of a *γ*(*G*)-set which satisfies some specific conditions.

**Theorem 4.** *Let G be a graph with no isolated vertex. Then <sup>γ</sup>tR*(*G*) = <sup>2</sup>*γ*(*G*) *if and only if there exist a γ*(*G*)*-set S and a set D* ⊆ *S such that*


**Proof.** First, we suppose that *<sup>γ</sup>tR*(*G*) = <sup>2</sup>*γ*(*G*). By Lemma 1, there exists a *<sup>γ</sup>tR*(*G*)-function *<sup>g</sup>*(*Vg*0 , *Vg*1 , *Vg*2 ) such that either *Vg*2 is a dominating set of *G*, or *<sup>V</sup>g*1,1 satisfies *<sup>G</sup>*[*Vg*1,1] ∼= *kK*2 for some *k* ≥ 1. By proceeding analogously as the proof of the lower bound of Theorem 1 and since *<sup>γ</sup>tR*(*G*) = <sup>2</sup>*γ*(*G*) = *<sup>γ</sup>t*2(*G*) + *<sup>γ</sup>*(*G*), we obtain *γ*(*G*) = |*Vg*2 | + |*Vg*1,1|/2 and *<sup>γ</sup>t*2(*G*) = |*Vg*2 | + |*Vg*1,2| + |*Vg*1,1|/2. Therefore *<sup>V</sup>g*1,2 = ∅.

Let *D* be the set formed by taking one vertex from each *K*2-componen<sup>t</sup> of *<sup>G</sup>*[*Vg*1,1]. Notice that *D* ∪ *Vg*2 is a dominating set of *G*. Hence <sup>2</sup>*γ*(*G*) = *<sup>γ</sup>tR*(*G*) = |*Vg*1 | + <sup>2</sup>|*Vg*2 | = 2|*D*| + <sup>2</sup>|*Vg*2 |, which implies that *S* = *D* ∪ *Vg*2 is a *γ*(*G*)-set. Thus, by construction of sets *S* and *D*, it is easy to see that Statements (a) and (b) hold.

Next, we prove Statement (c). Let *D*∗ = *v*∈*<sup>D</sup> epn*(*<sup>v</sup>*, *S*) ∪ *D*. It is readily seen that from *D* and any *γ*(*<sup>G</sup>* − *<sup>D</sup>*<sup>∗</sup>)-set we can construct a dominating set of *G*, and as *<sup>V</sup>g*1,2 = ∅, we obtain 12*<sup>γ</sup>tR*(*G*) = *γ*(*G*) ≤ *γ*(*<sup>G</sup>* − *D*∗) + |*D*|≤|*Vg*2 | + 12 |*Vg*1,1| = 12*<sup>γ</sup>tR*(*G*). Thus, we have equalities in the inequality chain above. In particular, *γ*(*<sup>G</sup>* − *D*∗) = |*Vg*2 |. Also, notice that *Vg*2 is a total dominating set of *G* − *D*∗ since *<sup>V</sup>g*1,2 = ∅. Hence, we deduce *<sup>γ</sup>t*(*<sup>G</sup>* − *D*∗) ≤ |*Vg*2 | = *γ*(*<sup>G</sup>* − *<sup>D</sup>*<sup>∗</sup>), which implies *<sup>γ</sup>t*(*<sup>G</sup>* − *D*∗) = *γ*(*<sup>G</sup>* − *<sup>D</sup>*<sup>∗</sup>), and Statement (c) holds, as desired.

Conversely, we suppose there exist a *γ*(*G*)-set *S* and a set *D* ⊆ *S* such that Statements (a), (b) and (c) hold. Let *AD* be a *<sup>γ</sup>t*(*<sup>G</sup>* − *<sup>D</sup>*<sup>∗</sup>)-set. By Statements (a) and (b) we have that *S* \ *D* is a dominating set of *G* − *D*∗ and so, by using Statement (c), we deduce that |*AD*| = *γ*(*<sup>G</sup>* − *D*∗) ≤ |*S* \ *<sup>D</sup>*|. Moreover, we observe that the function *f*(*<sup>V</sup>*0, *V*1, *<sup>V</sup>*2), defined by *V*1 = *D*∗ and *V*2 = *AD*, is a TRDF on *G*. Therefore, by Inequality chain (2) and statements above, we obtain <sup>2</sup>*γ*(*G*) ≤ *<sup>γ</sup>tR*(*G*) ≤ *ω*(*f*) = |*D*∗| + 2|*AD*| ≤ 2|*D*| + 2|*S* \ *D*| = 2|*S*| = <sup>2</sup>*γ*(*G*). Thus, we have equalities in the previous inequality chain. In particular, *<sup>γ</sup>tR*(*G*) = <sup>2</sup>*γ*(*G*), which completes the proof.

### **4. Some Necessary Conditions for the Graphs** *G* **satisfying** *<sup>γ</sup>tR***(***G***) = <sup>3</sup>***γ***(***G***)**

Analogously to the section above, we continue now with a simple characterization, which is a direct consequence of Theorem 1 and the well-know inequality *<sup>γ</sup>R*(*G*) ≤ <sup>2</sup>*γ*(*G*).

**Theorem 5.** *Let G be a graph without isolated vertices. Then <sup>γ</sup>tR*(*G*) = <sup>3</sup>*γ*(*G*) *if and only if <sup>γ</sup>tR*(*G*) = *<sup>γ</sup>R*(*G*) + *γ*(*G*) *and <sup>γ</sup>R*(*G*) = <sup>2</sup>*γ*(*G*)*.*

We want to accentuate that in all the examples in which we have observed that the upper bound of Theorem 1 is achieved, we also have that *<sup>γ</sup>R*(*G*) = <sup>2</sup>*γ*(*G*). In such a sense, we propose the following conjecture, which we could not prove.

**Conjecture 1.** Let *G* be a graph with no isolated vertex. Then *<sup>γ</sup>tR*(*G*) = <sup>3</sup>*γ*(*G*) if and only if *<sup>γ</sup>tR*(*G*) = *<sup>γ</sup>R*(*G*) + *<sup>γ</sup>*(*G*).

In order to give some necessary conditions for the graphs *G* satisfying *<sup>γ</sup>tR*(*G*) = <sup>3</sup>*γ*(*G*), we shall need the following definition and useful results.

**Definition 1.** *A graph G satisfies Property* P *if for every γ*(*G*)*-set S, there exist no three vertices x*, *y*, *z* ∈ *S such that*


Notice that the families of graphs *Gp*,*<sup>q</sup>* and *Gr* given in Section 2 satisfy the Property P. Moreover, the Figure 3 shows a graph *G* that does not satisfy Property P. Observe that the set *S* = {*<sup>x</sup>*, *y*, *z*} is a *γ*(*G*)-set and also, it is easy to see that |*epn*(*<sup>x</sup>*, *S*)| = |*epn*(*<sup>z</sup>*, *S*)| = 2 and that the vertex *y* ∈ *epn*(*y*, *S*) satisfies the condition *d*(*<sup>x</sup>*, *y*) = *<sup>d</sup>*(*y*, *z*) = 2.

**Figure 3.** A graph *G* that does not satisfy the Property P.

**Lemma 2.** *Let G be a graph and let S be a γ*(*G*)*-set. If S is a packing, then for all v* ∈ *S there exists v* ∈ *S such that d*(*<sup>v</sup>*, *v*) = 3*.*

**Proof.** Suppose there exists a vertex *v* ∈ *S* such that for all vertex *v* ∈ *S* \ {*v*}, it is satisfied that *d*(*<sup>v</sup>*, *v*) > 3 (notice that *d*(*<sup>v</sup>*, *v*) ≥ 3 because *S* is a packing). Hence, every vertex at distance two of *v* is not dominated by *S*, which is a contradiction. This completes the proof.

**Proposition 1.** *If G is a graph such that every γ*(*G*)*-set is a packing, then for every γ*(*G*)*-set S and for every v* ∈ *S it is satisfied that* |*epn*(*<sup>v</sup>*, *S*)| ≥ 2*.*

**Proof.** Let *S* be a *γ*(*G*)-set and let *v* ∈ *S*. Since *S* is a packing, we have that *epn*(*<sup>v</sup>*, *S*) = ∅. If *epn*(*<sup>v</sup>*, *S*) = {*u*}, then *v* is a vertex of degree one. By using Lemma 2, we have that *S* = (*S* \ {*v*}) ∪ {*u*} is a *γ*(*G*)-set, but is not a packing, contradicting the hypothesis. So |*epn*(*<sup>v</sup>*, *S*)| ≥ 2, as desired.

**Theorem 6.** *Let G be a graph. If <sup>γ</sup>tR*(*G*) = <sup>3</sup>*γ*(*G*)*, then the following statements hold.*


**Proof.** By Theorem 5, Statement (i) follows. Moreover, Abdollahzadeh Ahangar et al. showed in [16] that every *γ*(*G*)-set is a packing, which implies that Statement (ii) holds.

Next we prove Statement (iii). In that sense, we suppose that *G* does not satisfy Property P. Hence there exist a *γ*(*G*)-set *S* and three vertices *x*, *y*, *z* ∈ *S* satisfying the conditions given in Definition 1. By Statement (ii) and Proposition 1 we have |*epn*(*<sup>v</sup>*, *S*)| ≥ 2, for every *v* ∈ *S*. Let *x* ∈ *epn*(*<sup>x</sup>*, *S*) \ *<sup>N</sup>*(*y* ) and *z* ∈ *epn*(*<sup>z</sup>*, *S*) \ *<sup>N</sup>*(*y* ). Now, we consider the function *f* defined as follows.


Notice that, by construction, *f* is a TRDF on *G*. Therefore,

$$\begin{aligned} \gamma\_{tR}(G) &\leq \quad \omega(f) \\ &\leq 2|(S \backslash \{x, z\}) \cup \{y' \}| + |S \backslash \{x, y, z\}| + |\{x, z, x', z'\}| \\ &= 2(|S| - 1) + (|S| - 3) + 4 \\ &= 3|S| - 1 \\ &< 3\gamma(G), \end{aligned}$$

which is a contradiction. Hence *G* satisfies Property P and the proof is complete.

### **5. Conclusions and Open Problems**

New results concerning the study of total Roman domination in graphs have been presented in this article. Among the main contributions, the following should be highlighted.


On the other hand, and as a consequence of this study, some open problems have arisen. Next, we expose some of the most interesting.


**Author Contributions:** All authors contributed equally to this work. Investigation, A.C.M., S.C.G. and A.C.G.; Writing—review & editing, A.C.M., S.C.G. and A.C.G. All authors have read and agreed to the published version of the manuscript.

**Funding:** This research received no external funding.

**Conflicts of Interest:** The authors declare no conflict of interest.
