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(x, y) instead of N((x, y)). 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(x, y) instead of f((x, y)). 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 \(\gamma _{_{\times 2,t}}(G\circ H)= \gamma _{_{(2,2,2)}}(G).\) \(\square \)