1 Introduction

The lexicographic product of two graphs G and H is the graph \(G \circ H\) whose vertex set is \(V(G \circ H)= V(G) \times V(H )\) and \((g,h)(g',h') \in E(G \circ H)\) if and only if \(gg' \in E(G)\) or \(g=g'\) and \(hh' \in E(H)\). For simplicity, the neighbourhood of \((x,y)\in V(G)\times V(H)\) will be denoted by N(xy) instead of N((xy)). Analogously, for any function f on \(G\circ H\), the image of \((x,y)\in V(G)\times V(H)\) will be denoted by f(xy) instead of f((xy)). For basic properties of the lexicographic product of two graphs, we cite the books [18, 23]. In particular, for results on domination theory of lexicographic product graphs we suggest the following works: standard domination [25, 26], Roman domination [27], weak Roman domination [6, 24, 29], total Roman domination [8, 12], total weak Roman domination [6, 11], rainbow domination [28], super domination [14], Italian domination [5], secure domination [6, 24], secure total domination [6, 11], double domination [9] and doubly connected domination [2].

In particular, the next theorem merges two results obtained in [27] and [30]. The result states that the domination number of \(G\circ H\) equals the domination number of G whenever H has domination number equal to one, while the domination number of \(G\circ H\) equals the total domination number of G for the remaining cases.

Theorem 1

([27] and [30]) For any graph G with no isolated vertex and any non-trivial graph H,

$$\begin{aligned} \gamma (G\circ H)=\left\{ \begin{array}{ll} \gamma (G), &{} \text{ if } \gamma (H)=1\text{, } \\ \gamma _{_t}(G), &{} \text{ if } \gamma (H)\ge 2\text{. } \end{array} \right. \end{aligned}$$

Another interesting result obtained in [11] concerns the case of total domination.

Theorem 2

[11] For any graph G with no isolated vertex and any non-trivial graph H,

$$\begin{aligned} \gamma _{_t}(G\circ H)=\gamma _{_t}(G). \end{aligned}$$

These two theorems suggest to consider the following problem.

Problem 1

Let G be a graph and let \(\gamma _y\) be a domination parameter well defined on \(G\circ H\) for any non-trivial graph H. Determine if for each graph H, there exists a domination parameter \(\gamma _x\) such that

$$\begin{aligned} \gamma _y(G\circ H)=\gamma _x(G). \end{aligned}$$

We proceed to show other cases for which this problem has been solved. To this end, we need to formalize the notion of w-domination introduced in [5], where \(w=(w_0,w_1, \dots ,w_l)\) is a vector of non-negative integers such that \( w_0\ge 1\). Given a graph G, a function \(f: V(G)\longrightarrow \{0,1,\dots ,l\}\) is said to be a w-dominating function if \(\sum _{u\in N(v)}f(u)\ge w_i\) for every vertex v with \(f(v)=i\), where N(v) denotes the open neighbourhood of \(v\in V(G)\). For every \(i\in \{0,\dots , l\}\), we define \(V_i=\{v\in V(G):\; f(v)=i\}\), and we will identify the function f with the subsets \(V_0,\dots ,V_l\) associated with it. So, we will use the unified notation \(f(V_0,\dots , V_l)\) for the function and these associated subsets. The weight of f is defined to be \(\omega (f)=\sum _{v\in V(G)} f(v)\), while the w-domination number of G, denoted by \(\gamma _{_w}(G)\), is defined as the minimum weight among all w-dominating functions on G. A w-dominating function of weight \(\gamma _{_w}(G)\) will be called a \(\gamma _{_w}(G)\)-function.

It was shown in [5] that a wide range of well-known domination parameters can be defined and studied through this approach. For instance, the vector \(w=(1,0)\) corresponds to standard domination, \(w=(1,1)\) corresponds to total domination, \(w=(2,0,0)\) corresponds to Italian domination, \(w=(2,0,1)\) corresponds to quasi-total Italian domination, \(w=(2,1,1)\) corresponds to total Italian domination, while \(w=(k,k-1,\dots ,1,0)\) corresponds to \(\{k\}\)-domination.

As the next result shows, Problem 1 was solved for the case of the Italian domination number, which is a well-known parameter introduced in [13] under the name of Roman \(\{2\}\)-domination number. As mentioned above, in terms of w-domination, the Italian domination number of a graph G is defined as \(\gamma _{_I}(G)=\gamma _{_{(2,0,0)}}(G)\).

Theorem 3

[5] For any graph G with no isolated vertex and any non-trivial graph H,

$$\begin{aligned} \gamma _{_I}(G\circ H)=\left\{ \begin{array}{cl} \gamma _{_{(2,1,0)}}(G) &{} \text {if } \, \gamma (H)=1,\\ \gamma _{_{(2,2,0)}}(G) &{} \text {if } \, \gamma _2(H)=\gamma (H)=2,\\ \gamma _{_{(2,2,1)}}(G) &{} \text {if } \, \gamma _2(H)>\gamma (H)=2,\\ \gamma _{_{(2,2,2,0)}}(G) &{} \text {if } \, \gamma _{_{I}}(H)=\gamma (H)=3,\\ \gamma _{_{(2,2,2)}}(G) &{} \text {if } \, \gamma _{_{I}}(H)\ne 3 \, \text { and } \, \gamma (H)\ge 3. \end{array}\right. \end{aligned}$$

In addition, Problem 1 was solved for the case of the \(\{2\}\)-domination number, which was introduced in [15]. In terms of w-domination, the \(\{2\}\)-domination number of a graph G is defined as \(\gamma _{_{\{2\}}}(G)=\gamma _{_{(2,1,0)}}(G)\).

Theorem 4

[4] For any graph G with no isolated vertex and any non-trivial graph H,

$$\begin{aligned} \gamma _{_{\{2\}}}(G\circ H)=\left\{ \begin{array}{cl} \gamma _{_{(2,1,0)}}(G) &{} \text {if } \, \gamma (H)=1,\\ \gamma _{_{(2,2,1)}}(G) &{} \text {if } \, \gamma (H)=2,\\ \gamma _{_{(2,2,2)}}(G) &{} \text {if } \, \gamma (H)\ge 3. \end{array}\right. \end{aligned}$$

We refer the reader to [5] for general results on w-domination, as well as for specific results on the domination parameters given in Theorems 3 and 4.

In this paper, we solve Problem 1 for the particular cases in which \(\gamma _y\) corresponds to the following parameters. Although we will use the standard notation for these parameters, we will define them in terms of w-domination.

  • The k-domination number of a graph G, introduced in [16, 17], can be defined as \(\gamma _{_{k}}(G)=\gamma _{_{(k,0)}}(G)\). In this paper, we are interested in the case \(k=2\), which is probably the most studied. In this case, if \(f(V_0,V_1)\) is a \(\gamma _{(2,0)}(G)\)-function, then we will say that \(V_1\) is a \(\gamma _{_2}(G)\)-set.

  • The double domination number of a graph G with no isolated vertex is defined to be \(\gamma _{_{\times 2}}(G)=\gamma _{_{(2,1)}}(G)\). If \(f(V_0,V_1)\) is a \(\gamma _{_{(2,1)}}(G)\)-function, then we will say that \(V_1\) is a \(\gamma _{_{\times 2}}(G)\)-set. This parameter was introduced in two different papers [19, 20]. Moreover, the general version of this parameter, the k-tuple domination number, is defined to be \(\gamma _{_{\times k}}(G)=\gamma _{_{(k,k-1)}}(G)\).

  • The double total domination number of a graph G with minimum degree \(\delta (G)\ge 2\) is defined to be \(\gamma _{_{\times 2,t}}(G)=\gamma _{_{(2,2)}}(G)\). If \(f(V_0,V_1)\) is a \(\gamma _{_{(2,2)}}(G)\)-function, then we will say that \(V_1\) is a \(\gamma _{_{\times 2,t}}(G)\)-set. This domination parameter was introduced in [21], and its general version is the k-tuple total domination number, which is defined to be \(\gamma _{_{\times k, t}}(G)=\gamma _{_{(k,k)}}(G)\).

  • The quasi-total Italian domination number of a graph G, recently introduced in [7], is defined to be \(\gamma _{_{I^*}}(G)=\gamma _{_{(2,0,1)}}(G)\). A (2, 0, 1)-dominating function of weight \(\gamma _{_{I^*}}(G)\) will be called a \(\gamma _{_{I^*}}(G)\)-function.

  • The total Italian domination number of a graph G with no isolated vertex is defined to be \(\gamma _{_{tI}}(G)=\gamma _{_{(2,1,1)}}(G)\). This parameter was introduced in [3], and independently in [1], under the name of total Roman \(\{2\}\)-domination number. A (2, 1, 1)-dominating function of weight \(\gamma _{_{tI}}(G)\) will be called a \(\gamma _{_{tI}}(G)\)-function.

  • The total \(\{2\}\)-domination number of a graph G of minimum degree \(\delta (G)\ge 2\) is defined as \(\gamma _{_{\{2\},t}}(G)=\gamma _{_{(2,2,2)}}(G)\). This parameter was studied in [22].

We will show that the above-mentioned domination parameters of lexicographic product graphs \(G\circ H\) are equal to \(\gamma _{w}(G)\) for some vector \(w\in \{2\}\times \{0,1,2\}^{l}\) and \(l\in \{2,3\}\). The decision on whether the equality holds for a specific vector w will depend on the value of some domination parameters of H.

Notice that if G is a graph with no isolated vertex and H is a non-trivial graph, then the following domination chain is deduced by the definition of the parameters involved in it.

$$\begin{aligned} \gamma _{_{I}}(G\circ H)\le \gamma _{_{I^*}}(G\circ H)\le \gamma _{_{2}}(G\circ H)\le \gamma _{_{\times 2}}(G\circ H)\le \gamma _{_{\times 2,t}}(G\circ H). \end{aligned}$$
(1)

Furthermore, the equality \(\gamma _{_{tI}}(G\circ H)=\gamma _{_{\times 2}}(G\circ H)\) was deduced in [9], while the equality \(\gamma _{_{I^*}}(G\circ H)=\gamma _{_{2}}(G\circ H)\) will be proved in Sect. 2 and the equality \(\gamma _{_{\{2\},t}}(G\circ H)=\gamma _{_{\times 2,t}}(G\circ H)\) will be proved in Sect. 4. Therefore, the following domination chain holds whenever G is a graph with no isolated vertex and H is a non-trivial graph.

$$\begin{aligned} \gamma _{_{I}}(G\circ H) \begin{array}{l} \le \gamma _{_{I^*}}(G\circ H)=\gamma _{_{2}}(G\circ H) \\ \ \le \gamma _{_{tI}}(G\circ H) =\gamma _{_{\times 2}}(G\circ H) \\ \ \le \gamma _{_{\{2\},t}}(G\circ H) = \gamma _{_{\times 2,t}}(G\circ H). \end{array} \end{aligned}$$
(2)

2 Double Domination and Total Italian Domination

To get our results, we need to set up some tools and introduce some known results.

Lemma 1

Let G be a graph with no isolated vertex and H a non-trivial graph. If \(\gamma _{_{\times 2}}(H)=2\), then \(\gamma _{_{\times 2}}(G\circ H)\le \gamma _{_{(2,1,0)}}(G)\).

Proof

Let \(S=\{v_1,v_2\}\) be a \(\gamma _{_{\times 2}}(H)\)-set and \(g(W_0,W_1,W_2)\) a \(\gamma _{_{(2,1,0)}}(G)\)-function. Since \(W=(W_1\times \{v_1\})\cup (W_2\times S)\) is a double dominating set of \(G\circ H\), we conclude that \(\gamma _{_{\times 2}}(G\circ H)\le \vert W\vert =\omega (g)=\gamma _{_{(2,1,0)}}(G)\). \(\square \)

Notice that for any \(u\in V(G)\) the subgraph of \(G\circ H\) induced by \(\{u\}\times V(H)\) is isomorphic to H. For simplicity, we will denote this subgraph by \(H_u\).

Theorem 5

[9] The following statements hold for any graph G with no isolated vertex and any non-trivial graph H.

  1. (i)

    \(\gamma _{_{\times 2}}(G\circ H)=\gamma _{_{tI}}(G\circ H).\)

  2. (ii)

    If \(\gamma _{_2}(H)\ge 3\) and \(\gamma (H)=1\), then \(\gamma _{_{\times 2}}(G\circ H)=\gamma _{_{tI}}(G).\)

  3. (iii)

    There exists a \(\gamma _{_{\times 2}}(G\circ H)\)-set S such that \(\vert S \cap V(H_u)\vert \le 2\), for every \(u\in V(G)\).

By Theorem 5 (i), we will restrict the proof of the next result to obtain the values of \( \gamma _{_{\times 2}}(G\circ H)\).

Theorem 6

For any graph G with no isolated vertex and any non-trivial graph H,

$$\begin{aligned} \gamma _{_{\times 2}}(G\circ H)=\gamma _{_{tI}}(G\circ H)=\left\{ \begin{array}{cl} \gamma _{_{(2,1,0)}}(G) &{} \text {if } \, \gamma _{_{\times 2}}(H)=2,\\ \gamma _{_{(2,1,1)}}(G) &{} \text {if } \, \gamma _{_{2}}(H)\ge 3 \, \text { and } \, \gamma (H)=1,\\ \gamma _{_{(2,2,1)}}(G) &{} \text {if } \, \gamma (H)=2,\\ \gamma _{_{(2,2,2)}}(G) &{} \text {if } \, \gamma (H)\ge 3. \end{array}\right. \end{aligned}$$

Proof

First, we assume that \(\gamma (H)=1\). Since \(\gamma _{_{ I}}(G\circ H)\le \gamma _{_{\times 2}}(G\circ H)\), if \(\gamma _{_{\times 2}}(H)=2\), then Theorem 3 and Lemma 1 lead to \(\gamma _{_{(2,1,0)}}(G)= \gamma _{_{I}}(G\circ H)\le \gamma _{_{\times 2}}(G\circ H)\le \gamma _{_{(2,1,0)}}(G)\). Therefore, in this case we conclude that \(\gamma _{_{\times 2}}(G\circ H)=\gamma _{_{(2,1,0)}}(G)\). Now, if \(\gamma _{_{2}}(H)\ge 3\), then Theorem 5 (ii) leads to \( \gamma _{_{\times 2}}(G\circ H)=\gamma _{_{tI}}(G)=\gamma _{_{(2,1,1)}}(G)\).

From now on we assume that \(\gamma (H)\ge 2\). Let S be a \(\gamma _{_{\times 2}}(G\circ H)\)-set which satisfies Theorem 5 (iii). Let \(f(X_0,X_1,X_2)\) be the function defined on G by \(X_i=\{x\in V(G): \vert S\cap V(H_x)\vert =i\}\) for every \(i \in \{0,1,2\}\). Notice that \(\gamma _{_{\times 2}}(G\circ H)=\vert S\vert =\omega (f)\). We claim that f is a \(\gamma _{_{(2,2,w)}}(G)\)-function, where \(w\in \{1,2\}\). In order to prove this claim and find the exact value of w, we differentiate the following two cases.

Case 1. \(\gamma (H)=2\). Assume that \(x\in X_0\cup X_1\). Since \(\gamma (H)=2\), there exists a vertex \(z\in V(H)\) such that \((x,z)\notin S\) and \(\vert S\cap N(x,z)\cap V(H_x)\vert =0\). Hence, \(\vert S\cap (N(x,z)\setminus V(H_x))\vert \ge 2\), which implies that \(f(N(x))\ge 2\). Now, assume that \(x\in X_2\). In this case, there exists a vertex \(y\in V(H)\) such that \(\vert S\cap N(x,y)\cap V(H_x)\vert \le 1\), and so \(f(N(x))\ge 1\). Therefore, f is a (2, 2, 1)-dominating function on G and, as a consequence, \(\gamma _{_{\times 2}}(G\circ H)=\vert S\vert =\omega (f)\ge \gamma _{_{(2,2,1)}}(G)\).

Moreover, let \(h(Y_0,Y_1,Y_2)\) be a \(\gamma _{_{(2,2,1)}}(G)\)-function and \(S=\{v_1,v_2\}\) a \(\gamma (H)\)-set. Notice that the set \(Y=(Y_1\times \{v_1\})\cup (Y_2\times S)\) is a double dominating set of \(G\circ H\), which implies that \(\gamma _{_{\times 2}}(G\circ H)\le \vert Y\vert =\omega (h)=\gamma _{_{(2,2,1)}}(G)\).

Case 2. \(\gamma (H)\ge 3\). Let \(x\in V(G)\). Since \(\gamma (H)\ge 3\), there exists \(y\in V(H)\) such that \((x,y)\notin S\) and \(\vert S\cap N(x,y)\cap V(H_x)\vert =0\), which implies that \(\vert S\cap (N(x,y)\setminus V(H_x))\vert \ge 2\), and so \(f(N(x))\ge 2\). Therefore, f is a (2, 2, 2)-dominating function on G and, as a consequence, \(\gamma _{_{\times 2}}(G\circ H)=\vert S\vert =\omega (f)\ge \gamma _{_{(2,2,2)}}(G)\).

It remains to show that \(\gamma _{_{\times 2}}(G\circ H)\le \gamma _{_{(2,2,2)}}(G)\). To see this we only need to observe that for any \(\gamma _{_{(2,2,2)}}(G)\)-function \(g(W_0,W_1,W_2)\) and any pair of vertices \(v_1,v_2\in V(H)\), the set \(W=(W_2\times \{v_1,v_2\})\cup (W_1\times \{v_1\})\) is a double dominating set of \(G\circ H\), which implies that \(\gamma _{_{\times 2}}(G\circ H)\le \vert W\vert = \omega (g)=\gamma _{_{(2,2,2)}}(G)\), as required. \(\square \)

3 Quasi-total Italian Domination and 2-Domination

To begin this section, we will introduce some basic tools.

Lemma 2

For any graph G with no isolated vertex and any non-trivial graph H with \(\gamma (H)=1\), there exists a \(\gamma _{_{2}}(G\circ H)\)-set D satisfying that \(\vert D\cap V(H_u)\vert \le 2\) for every \(u\in V(G)\).

Proof

Given a \(\gamma _{_{2}}(G\circ H)\)-set D, we define the set \(R_D=\{x\in V(G): \, \vert D\cap V(H_x)\vert \ge 3\}\). Now, we assume that D is a \(\gamma _{_{2}}(G\circ H)\)-set such that \(\vert R_D\vert \) is minimum among all \(\gamma _{_{2}}(G\circ H)\)-sets. Suppose that \(\vert R_D\vert \ge 1\). Let v be a universal vertex of H and \(u\in R_D\). Now, we take \(u'\in N(u)\) and \(v'\in N(v)\), and consider a set \(D'\subseteq V(G)\times V(H)\) satisfying the following properties.

  • \(D'\cap V(H_u)=\{(u,v),(u,v')\}\);

  • \(\vert D'\cap V(H_{u'})\vert =\min \{2, \vert D\cap V(H_{u'})\vert +1\}\);

  • \(D'\cap V(H_x)=D\cap V(H_x)\) for every \(x\in V(G){\setminus } \{u,u'\}\).

Observe that \(D'\) is a 2-dominating set of \(G\circ H\) satisfying \(\vert D'\vert \le \vert D\vert \) and \(\vert R_{D'}\vert <\vert R_D\vert \), which is a contradiction. Therefore, \(R_D=\varnothing \), as required. \(\square \)

Lemma 3

Let G be a graph with no isolated vertex and H a non-trivial graph. If \(\gamma _{_{2}}(H)\ge 3\) and \(\gamma (H)=1\), then \(\gamma _{_{2}}(G\circ H)\ge \gamma _{_{(2,1,1)}}(G).\)

Proof

Let D be a \(\gamma _{_{2}}(G\circ H)\)-set which satisfies Lemma 2. Let \(f(X_0,X_1,X_2)\) be the function defined on G by \(X_i=\{x\in V(G): \vert D\cap V(H_x)\vert =i\}\) for every \(i\in \{0,1,2\}\). Notice that \(\gamma _{_{2}}(G\circ H)=\vert D\vert =\omega (f)\). We claim that f is a (2, 1, 1)-dominating function on G. Assume that \(x\in X_0\). Since \(D\cap V(H_x)=\varnothing \), we have that \(\vert D\cap (N(x)\times V(H)\vert \ge 2\), which implies that \(f(N(x))\ge 2\). Now, assume that \(x\in X_1\cup X_2\). Since \(\vert D\cap V(H_x)\vert \le 2\) and \(\gamma _{_{2}}(H)\ge 3\), there exists \(y\in V(H)\) such that \((x,y)\not \in D\) and \(\vert D\cap V(H_x) \cap N(x,y) \vert \le 1\), which implies that \(\vert D\cap (N(x)\times V(H))\vert \ge 1\), and so \(f(N(x))\ge 1\). Therefore, f is a (2, 1, 1)-dominating function on G and, as a consequence, \(\gamma _{_{2}}(G\circ H)=\vert D\vert =\omega (f)\ge \gamma _{_{(2,1,1)}}(G)\). \(\square \)

Theorem 7

The following statements hold for any graph G with no isolated vertex and any non-trivial graph H.

  1. (i)

    \(\gamma _{_{I^*}}(G\circ H)=\gamma _{_{2}}(G\circ H).\)

  2. (ii)

    If \(\gamma (H)\ge 2\), then \(\gamma _{_{I^*}}(G\circ H)=\gamma _{_{I}}(G\circ H).\)

Proof

By definition, \(\gamma _{_{I^*}}(G\circ H)\le \gamma _{_{2}}(G\circ H)\). Hence, it remains to show that \(\gamma _{_{I^*}}(G\circ H)\ge \gamma _{_{2}}(G\circ H)\). Let \(f(V_0,V_1,V_2)\) be a \(\gamma _{_{I^*}}(G\circ H)\)-function such that \(\vert V_2\vert \) is minimum among all \(\gamma _{_{I^*}}(G\circ H)\)-functions. If \(V_2=\varnothing \), then \(V_1\) is a 2-dominating set of \(G\circ H\), and so \(\gamma _{_{2}}(G\circ H)\le \vert V_1\vert = \gamma _{_{I^*}}(G\circ H)\). We assume that \(V_2\ne \varnothing \) and, in that case, we differentiate the next two cases for a fixed vertex \((u,v)\in V_2\). Obviously, \(N(u,v)\cap (V_1\cup V_2)\ne \varnothing \).

Case 1. \(N(u,v)\cap (V_1\cup V_2)\subseteq V(H_u)\). In this case, for any \((u',v')\in N(u)\times V(H)\) we define the function \(f'(V_0',V_1',V_2')\) where \(V_0'=V_0\setminus \{(u',v')\}\), \(V_1'=V_1\cup \{(u,v),(u',v')\}\) and \(V_2'=V_2\setminus \{(u,v)\}\). Observe that \(\omega (f')=\omega (f)\), every vertex in \(V_2'\) has a neighbour in \(V_1'\cup V_2'\) and every vertex \(w\in V_0'\subseteq V_0\) satisfies that \(f'(N(w))\ge 2\). Hence, \(f'\) is a \(\gamma _{_{I^*}}(G\circ H)\)-function and \(\vert V_2'\vert <\vert V_2\vert \), which is a contradiction.

Case 2. \((N(u)\times V(H))\cap (V_1\cup V_2)\ne \varnothing \). If \(V(H_u)\subseteq V_1\cup V_2\), then the function h, defined by \(h(u,v)=1\) and \(h(x,y)=f(x,y)\) whenever \((x,y)\in V(G\circ H)\setminus \{(u,v)\}\), is a quasi-total Italian dominating function on \(G\circ H\) with \(\omega (h)<\omega (f)\), which is a contradiction. Hence, there exists \(v'\in V(H)\) such that \((u,v')\in V_0\). In that case, let \(f'(V_0',V_1',V_2')\) be a function defined by \(V_0'=V_0\setminus \{(u,v')\}\), \(V_1'=V_1\cup \{(u,v),(u,v')\}\) and \(V_2'=V_2{\setminus } \{(u,v)\}\). As in the previous case, \(\omega (f')=\omega (f)\), every vertex in \(V_2'\) has a neighbour in \(V_1'\cup V_2'\) and every vertex \(w\in V_0'\subseteq V_0\) satisfies that \(f'(N(w))\ge 2\). Thus, \(f'\) is a \(\gamma _{_{I^*}}(G\circ H)\)-function with \(\vert V_2'\vert <\vert V_2\vert \), which is a contradiction again.

According to the two cases above, we deduce that \(V_2=\varnothing \), which implies that \(\gamma _{_{2}}(G\circ H)\le \gamma _{_{I^*}}(G\circ H).\) Therefore, the proof of (i) is complete.

Finally, we proceed to prove (ii). By definition, \(\gamma _{_{I}}(G\circ H)\le \gamma _{_{I^*}}(G\circ H)\). Thus, it remains to show that \(\gamma _{_{I}}(G\circ H)\ge \gamma _{_{I^*}}(G\circ H)\) whenever \(\gamma (H)\ge 2\). Let \(g(W_0,W_1,W_2)\) be a \(\gamma _{_{I}}(G\circ H)\)-function such that \(\vert W_2\vert \) is the minimum among all \(\gamma _{_{I}}(G\circ H)\)-functions. Obviously, if \(W_2=\varnothing \) or \(N(u,v)\not \subseteq W_0\) for every \((u,v)\in W_2\), then g is a \(\gamma _{_{I^*}}(G\circ H)\)-function and we are done. Suppose to the contrary that there exists a vertex \((u,v)\in W_2\) such that \(N(u,v)\subseteq W_0\). Notice that \(g(V(H_u))\ge 3\), as \(\gamma (H)\ge 2\). Thus, we differentiate the next two cases.

Case 1. \(g(V(H_u))\ge 4\). Let \(u'\in N(u)\) and \(v'\in V(H)\setminus \{v\}\). We define a function \(g'(W_0',W_1',W_2')\) on \(G\circ H\) as \(g'(u,v)=g'(u,v')=g'(u',v)=g'(u',v')=1\), \(g'(V(H_{u}){\setminus } \{(u,v), (u,v')\})=g'(V(H_{u'}){\setminus } \{(u',v),(u',v')\})=0\) and \(g'(x,y)=g(x,y)\) for every \(x\in V(G){\setminus } \{u,u'\}\) and \(y\in V(H)\). Notice that \(g'\) is an Italian dominating function on \(G\circ H\) with \(\omega (g')\le \omega (g)\) and \(\vert W_2'\vert <\vert W_2\vert \), which is a contradiction.

Case 2. \(g(V(H_u))=3\). In this case, since \(\gamma (H)\ge 2\), we deduce that \(\gamma _{_{I}}(H)=3\) and \(\gamma (H)=2\) by the minimality of \(W_2\). Let \(\{v_1,v_2\}\) be a \(\gamma (H)\)-set and \(u'\in N(u)\). Consider the function \(g'(W_0',W_1',W_2')\) defined as \(g'(u,v_1)=g'(u,v_2)=1\), \(g'(u,v)=0\) for every \(v\in V(H){\setminus } \{v_1,v_2\}\), \(g'(V(H_{u'}))=1\) and \(g'(x,y)=g(x,y)\) for every \(x\in V(G){\setminus } \{u,u'\}\) and \(y\in V(H)\). Notice that \(g'\) is an Italian dominating function on \(G\circ H\) with \(\omega (g')\le \omega (g)\) and \(\vert W_2'\vert <\vert W_2\vert \), which is a contradiction.

Therefore, either \(W_2=\varnothing \) or every vertex in \(W_2\) has a neighbour in \(W_1\cup W_2\), and so \(\gamma _{_{I^*}}(G\circ H)=\gamma _{_{I}}(G\circ H)\). \(\square \)

According to Theorem 7, we can restrict the proof of the next result to obtain the values of \( \gamma _{_{ 2}}(G\circ H)\).

Theorem 8

For any graph G with no isolated vertex and any non-trivial graph H,

$$\begin{aligned} \gamma _{_{2}}(G\circ H)=\gamma _{_{I^*}}(G\circ H)=\left\{ \begin{array}{cl} \gamma _{_{(2,1,0)}}(G) &{} \text {if } \, \gamma _{_{\times 2}}(H)=2,\\ \gamma _{_{(2,1,1)}}(G) &{} \text {if } \, \gamma _{_{2}}(H)\ge 3 \, \text { and } \, \gamma (H)=1,\\ \gamma _{(2,2,0)}(G) &{} \text {if } \, \gamma _2(H)=\gamma (H)=2,\\ \gamma _{_{(2,2,1)}}(G) &{} \text {if } \, \gamma _2(H)>\gamma (H)=2,\\ \gamma _{_{(2,2,2,0)}}(G) &{} \text {if } \, \gamma _{_{I}}(H)=\gamma (H)=3,\\ \gamma _{_{(2,2,2)}}(G) &{} \text {if } \, \gamma _{_{I}}(H)\ne 3 \, \text { and } \, \gamma (H)\ge 3. \end{array}\right. \end{aligned}$$

Proof

Since \(\gamma _{_{I}}(G\circ H)\le \gamma _{_{2}}(G\circ H)\), if \(\gamma _{_{\times 2}}(H)=2\), then by Lemma 1 and Theorem 3 we have that \(\gamma _{_{(2,1,0)}}(G)=\gamma _{_{I}}(G\circ H)\le \gamma _{_{2}}(G\circ H)\le \gamma _{_{(2,1,0)}}(G)\). Therefore, in this case we obtain \( \gamma _{_{2}}(G\circ H)= \gamma _{_{(2,1,0)}}(G)\).

Now, since \(\gamma _{_{2}}(G\circ H) \le \gamma _{_{\times 2}}(G\circ H)\), if \(\gamma _{_{2}}(H)\ge 3\) and \(\gamma (H)=1\), then Lemma 3 and Theorem 5 (ii) lead to \(\gamma _{_{(2,1,1)}}(G)\le \gamma _{_{2}}(G\circ H) \le \gamma _{_{\times 2}}(G\circ H)=\gamma _{_{(2,1,1)}}(G).\) Therefore, \(\gamma _{_{2}}(G\circ H)=\gamma _{_{(2,1,1)}}(G)\).

Finally, if \(\gamma (H)\ge 2\), then Theorem 7 leads to \(\gamma _{_{2}}(G\circ H)=\gamma _{_{I^*}}(G\circ H)=\gamma _{_{I}}(G\circ H)\) and so we complete the proof by Theorem 3. \(\square \)

4 Double Total Domination and Total \(\{2\}\)-Domination

Although in general, \(\gamma _{_{\{2\},t}}(G)\le \gamma _{_{\times 2,t}}(G)\), we show below that for the case of lexicographic product graphs these parameters always coincide.

Theorem 9

For any graph G with no isolated vertex and any non-trivial graph H,

$$\begin{aligned} \gamma _{_{\{2\},t}}(G\circ H)=\gamma _{_{\times 2,t}}(G\circ H). \end{aligned}$$

Proof

By definition, \(\gamma _{_{\{2\},t}}(G\circ H)\le \gamma _{_{\times 2,t}}(G\circ H).\) Hence, it remains to show that \(\gamma _{_{\{2\},t}}(G\circ H)\ge \gamma _{_{\times 2,t}}(G\circ H).\) Let \(f(V_0,V_1,V_2)\) be a \(\gamma _{_{\{2\},t}}(G\circ H)\)-function such that \(\vert V_2\vert \) is minimum among all \(\gamma _{_{\{2\},t}}(G\circ H)\)-functions. If \(V_2=\varnothing \), then \(V_1\) is a double total dominating set of \(G\circ H\), and so \(\gamma _{_{\times 2,t}}(G\circ H)\le \vert V_1\vert = \gamma _{_{\{2\},t}}(G\circ H)\), as required. We assume that \(V_2\ne \varnothing \) and, in that case, we differentiate the next two cases for a fixed vertex \((u,v)\in V_2\). Obviously, \(N(u,v)\cap (V_1\cup V_2)\ne \varnothing \).

Case 1. \(N(u,v)\cap (V_1\cup V_2)\subseteq V(H_u)\). In this case, for any \((u',v),(u',v')\in N(u)\times V(H)\) we define the function \(f'(V_0',V_1',V_2')\) where \(V_0'=V_0\setminus \{(u',v),(u',v')\}\), \(V_1'=V_1\cup \{(u',v),(u',v')\}\) and \(V_2'=V_2{\setminus } \{(u,v)\}\). Observe that \(\omega (f')=\omega (f)\) and every vertex \((x,y)\in V(G\circ H)\) satisfies that \(f'(N(x,y))\ge 2\). Hence, \(f'\) is a \(\gamma _{_{\{2\},t}}(G\circ H)\)-function and \(\vert V_2'\vert <\vert V_2\vert \), which is a contradiction.

Case 2. \((N(u)\times V(H))\cap (V_1\cup V_2)\ne \varnothing \). If \(V(H_u)\subseteq V_1\cup V_2\), then the function h, defined by \(h(u,v)=1\) and \(h(x,y)=f(x,y)\) whenever \((x,y)\in V(G\circ H)\setminus \{(u,v)\}\), is a double total dominating function on \(G\circ H\) with \(\omega (h)<\omega (f)\), which is a contradiction. Hence, there exists \(v'\in V(H)\) such that \((u,v')\in V_0\). In that case, let \(f'(V_0',V_1',V_2')\) be a function defined by \(V_0'=V_0\setminus \{(u,v')\}\), \(V_1'=V_1\cup \{(u,v),(u,v')\}\) and \(V_2'=V_2{\setminus } \{(u,v)\}\). Notice that \(\omega (f')=\omega (f)\) and every vertex \((x,y)\in V(G\circ H)\) satisfies that \(f'(N(x,y))\ge 2\). Thus, \(f'\) is a \(\gamma _{_{\{2\},t}}(G\circ H)\)-function with \(\vert V_2'\vert <\vert V_2\vert \), which is a contradiction again.

According to the two cases above, we deduce that \(V_2=\varnothing \), which implies that \(V_1\) is a double total dominating set of \(G\circ H\), and so \(\gamma _{_{\times 2,t}}(G\circ H)\le \vert V_1\vert = \gamma _{_{\{2\},t}}(G\circ H)\), as required. Therefore, the proof is complete. \(\square \)

We are now in a position to formalize the tools which will allow us to calculate \(\gamma _{_{\times 2,t}}(G\circ H)\).

Lemma 4

For any graph G with no isolated vertex and any non-trivial graph H, there exists a \(\gamma _{_{\times 2,t}}(G\circ H)\)-set S satisfying that \(\vert S\cap V(H_x)\vert \le 2\) for every \(x\in V(G)\).

Proof

Given a \(\gamma _{_{\times 2,t}}(G\circ H)\)-set S, we define the set \(R_S=\{x\in V(G): \, \vert S\cap V(H_x)\vert \ge 3\}\). Assume that S is a \(\gamma _{_{\times 2,t}}(G\circ H)\)-set such that \(R_S\) has minimum cardinality among all \(\gamma _{_{\times 2,t}}(G\circ H)\)-sets. Suppose that \(R_S\ne \varnothing \) and let \(x,y\in V(G)\) be two adjacent vertices with \(x\in R_S\). Let \(S_x=S\cap V(H_x)\) and take \((x,v_1),(x,v_2)\in S_x\). Hence, there exists a set \(S'\subseteq V(G\circ H)\) satisfying the following properties.

  • \(S'\cap V(H_x)=\{(x,v_1),(x,v_2)\}\).

  • \(\vert S'\cap V(H_{y})\vert =\min \{2, \vert S\cap V(H_{y})\vert +\vert S_x\vert -2\}\).

  • \(S'\cap V(H_{z}) = S\cap V(H_{z}) \) for every \(z\in V(G){\setminus } \{x,y\}\).

Observe that \(S'\) is a double total dominating set of \(G\circ H\) with \(\vert S'\vert \le \vert S\vert \) and \(\vert R_{S'}\vert <\vert R_{S}\vert \), which is a contradiction. Therefore, the result follows. \(\square \)

Proposition 1

For any graph G with no isolated vertex and any non-trivial graph H,

$$\begin{aligned} \gamma _{_{\times 2,t}}(G\circ H)\le \gamma _{_{(2,2,2)}}(G). \end{aligned}$$

Furthermore, if H has isolated vertex or \(\gamma _{_{t}}(H)\ge 3\), then the equality holds.

Proof

The proof of the inequality is straightforward, as we only need to observe that for any \(\gamma _{_{(2,2,2)}}(G)\)-function \(g(W_0,W_1,W_2)\) and any pair of vertices \(v_1,v_2\in V(H)\), the set \(W=(W_2\times \{v_1,v_2\})\cup (W_1\times \{v_1\})\) is a double total dominating set of \(G\circ H\), which implies that \(\gamma _{_{\times 2,t}}(G\circ H)\le \vert W\vert = \omega (g)=\gamma _{_{(2,2,2)}}(G)\).

From now on, assume that either H has isolated vertex or \(\gamma _{_{t}}(H)\ge 3\). Notice that these assumptions imply that for any set \(S\subseteq V(H)\) of cardinality at most two, there exists a vertex \(v\in V(H)\) such that \(N(v)\cap S=\varnothing \).

Now, let D be a \(\gamma _{_{\times 2,t}}(G\circ H)\)-set satisfying Lemma 4. Since \(\vert D\cap V(H_x)\vert \le 2\) for every \(x\in V(G)\), from the assumptions above we have that there exists a vertex \(v\in V(H)\) such that \(N(x,v)\cap D \cap V(H_x)=\varnothing \). Thus, \(\vert (N(x)\times V(H))\cap D\vert \ge 2\) for every \(x\in V(G)\), which implies that any function \(f: V(G)\longrightarrow \{0,1,2\}\) such that \(f(V(H_x))=\vert D\cap V(H_x)\vert \), is a (2, 2, 2)-dominating function on G. Therefore, \(\gamma _{_{(2,2,2)}}(G)\le \omega (f)=\vert D\vert = \gamma _{_{\times 2,t}}(G\circ H)\), as required. \(\square \)

According to Theorem 9, in the proof of the following result we can restrict ourselves to determining the value of \(\gamma _{_{\times 2,t}}(G\circ H)\).

Theorem 10

For any graph G with no isolated vertex and any non-trivial graph H,

$$\begin{aligned} \gamma _{_{\times 2,t}}(G\circ H)=\gamma _{_{\{2\},t}}(G\circ H)=\left\{ \begin{array}{cl} \gamma _{_{(2,2,1)}}(G) &{} \text {if } \, \gamma _{_{t}}(H)=2,\\ \gamma _{_{(2,2,2)}}(G) &{} \text {otherwise}. \end{array}\right. \end{aligned}$$

Proof

First we assume that \(\gamma _{_{t}}(H)=2\). Let \(h(Y_0,Y_1,Y_2)\) be a \(\gamma _{_{(2,2,1)}}(G)\)-function and let \(S=\{v_1,v_2\}\) be a \(\gamma _{_{t}}(H)\)-set. Notice that the set \(Y=(Y_1\times \{v_1\})\cup (Y_2\times S)\) is a double total dominating set of \(G\circ H\), which implies that \(\gamma _{_{\times 2,t}}(G\circ H)\le \vert Y\vert =\omega (h)=\gamma _{_{(2,2,1)}}(G)\). Now, let S be a \(\gamma _{_{\times 2,t}}(G\circ H)\)-set which satisfies Lemma 4 and let \(f(X_0,X_1,X_2)\) be the function defined on G by \(X_i=\{x\in V(G): \vert S\cap V(H_x)\vert =i\}\) for every \(i\in \{0,1,2\}\). Notice that \(\gamma _{_{\times 2,t}}(G\circ H)=\vert S\vert =\omega (f)\). We claim that f is a (2, 2, 1)-dominating function on G.

Let \(x\in X_0\cup X_1\). Since \(\gamma _{_{t}}(H)=2\), there exists a vertex \(z\in V(H)\) such that \((x,z)\notin S\) and \(\vert S\cap N(x,z)\cap V(H_x)\vert =0\). Hence, as S is a \(\gamma _{_{\times 2,t}}(G\circ H)\)-set, \(\vert S\cap (N(x,z){\setminus } V(H_x))\vert \ge 2\), and so \(f(N(x))\ge 2\).

Now, let \(x\in X_2\). Since \(\gamma _{_{t}}(H)=2\) implies \(\gamma _{_{\times 2,t}}(H)\ge 3\), we have that there exists a vertex \(y\in V(H)\) such that \(\vert S\cap V(H_x)\cap N(x,y)\vert \le 1\), which leads to \(\vert S\cap (N(x,z)\setminus V(H_x))\vert \ge 1\), as S is a \(\gamma _{_{\times 2,t}}(G\circ H)\)-set, and so \(f(N(x))\ge 1\).

Therefore, f is a (2, 2, 1)-dominating function on G and, as a consequence, \(\gamma _{_{\times 2,t}}(G\circ H)=\vert S\vert =\omega (f)\ge \gamma _{_{(2,2,1)}}(G)\), concluding that \(\gamma _{_{\times 2,t}}(G\circ H)=\gamma _{_{(2,2,1)}}(G)\).

Finally, if \(\gamma _{_{t}}(H)\ge 3\) or H has isolated vertex, then by Proposition 1 we have \(\gamma _{_{\times 2,t}}(G\circ H)= \gamma _{_{(2,2,2)}}(G).\) \(\square \)