### Definition 3.1

A superharmonic multifunction \(F:(X,\tau )\rightarrow(Y,\sigma)\) is said to be:

(a) upper superharmonic-continuous at a point \(x\in X \) if for each open set *V* containing \(F(x)\), there exists \(W \in\tau^{*}(x)\) such that

(b) lower superharmonic-continuous at a point \(x \in X\) if for each open set *V* containing \(F(x)\), there exists \(W \in\tau^{*}(x)\) such that

$$F(W)\cap V \neq\phi; $$

(c) upper (resp. lower) superharmonic-continuous if *F* has this property at every point of *X*.

Any single-valued superharmonic function \(f:(X, \tau)\rightarrow (Y,\sigma)\) can be considered as a multi-valued one which assigns to any \(x \in X\) the singleton \(\{f(x)\}\). We apply the above definitions of both upper and lower superharmonic-continuous multifunctions to the single-valued case. It is clear that they coincide with the notion of *S*-continuous due to Mashhour *et al.* [5]. One characterization of the above superharmonic multifunction is established throughout the following result, of which the proof is straightforward, so it is omitted.

### Remark 3.1

For a superharmonic multifunction \(F:(X,\tau )\rightarrow(Y,\sigma)\), many properties of upper (resp. lower) semicontinuity [7] (resp. upper (lower)) *F*-continuity [9], upper (resp. lower) sub-continuity [1], upper (resp. lower) precontinuity [10] and upper (resp. lower) (*G*-continuity [10]) can be deduced from the upper (resp. lower) superharmonic-continuity by considering \(\tau^{*}= \tau\) (resp. \(\tau^{*} = \tau^{\alpha}\), \(\tau^{*}= SO(X,\tau)\), \(\tau^{*}= PO(X,\tau)\) and \(\tau^{*}= \beta O(X,\tau)\)).

### Proposition 3.1

*A superharmonic multifunction*
\(F:(X,\tau )\rightarrow(Y,\sigma)\)
*is upper* (*resp*. *lower*) *superharmonic*-*continuous at a point*
\(x\in X\)
*if and only if for*
\(V \in\sigma\)
*with*
\(F(x)\subseteq V \) (*resp*. \(F(x)\cap V\neq\phi\)). *Then*
\(x\in \operatorname{superharmonic-int}(F^{+} (V))\) (*resp*. \(x \in \operatorname{superharmonic-int}(F^{-} (V))\).

### Lemma 3.1

*For any*
\(A \in(X,\tau)\), *we have*

$$\operatorname{\tau-int}(A)\subseteq \operatorname{superharmonic-int}(A)\subseteq A \subseteq\operatorname{superharmonic-cl}(A) \subseteq\operatorname{\tau-cl}(A). $$

### Theorem 3.1

*The following are equivalent for a superharmonic multifunction*
\(F:(X,\tau)\rightarrow(Y,\sigma)\):

(i) *F*
*is upper superharmonic*-*continuous*;

(ii) *for each*
\(x\in X\)
*and each*
\(V \in \sigma(F(x))\), *we have*
\(F^{+} (V)\in\tau^{*}(x)\);

(iii) *for each*
\(x \in X\)
*and each*
\(V \in \sigma(F(x))\), *there exists*
\(W\in\tau^{*}\)
*such that*

(iv) \(F^{+} (V)\in\tau^{*}\)
*for every*
\(V\in\sigma\);

(v) \(F^{-} (K)\)
*is superharmonic*-*closed for every closed set*
\(K\subseteq Y\);

(vi) \(\operatorname{superharmonic-cl}(F^{-} (B))\subseteq F^{-} (\operatorname{\tau-cl}(B))\)
*for every*
\(B\subseteq Y\);

(vii) \(F^{+}(\operatorname{\tau-int}(B))\subseteq \operatorname{superharmonic-int}(F^{+} (B))\)
*for every*
\(B\subseteq Y\);

(viii) \(\operatorname{superharmonic-fr}(F^{-}(B))\subseteq F^{-}(\operatorname{fr}(B))\)
*for every*
\(B\subseteq Y\);

(ix) \(F:(X, \tau^{*}) \rightarrow (Y,\sigma)\)
*is upper superharmonic*-*continuous*.

### Proof

(i) ⇔ (ii) and (i) ⇒ (iv): Follow from Proposition 3.1.

(ii) ⇔ (iii): This is obvious, since the arbitrary union of superharmonic-open set is superharmonic-open.

(iv) = (v): Let *K* be closed in *Y*, the result satisfies

$$F^{+}(Y\backslash K)=X\backslash F^{-}(K). $$

(v) ⇒ (vi): By putting \(K = \operatorname{\sigma-cl}(B)\) and applying Lemma 3.1.

(vi) ⇒ (vii): Let \(B\Rightarrow Y\), then \(\operatorname{\sigma-int}(B) \in \sigma\) and so \(Y \backslash\operatorname{\sigma-int}(B)\) is super-closed in \((Y,\sigma)\). Therefore by (vi) we get

$$X\backslash\operatorname{super-int}\bigl(F^{+} (B)\bigr)=\operatorname{super-cl}\bigl(X\backslash F^{+} (B)\bigr)\subseteq \operatorname{sub-cl}(X\backslash F^{+} \bigl(\operatorname{\sigma-int}(B)\bigr) $$

and

$$\operatorname{supra-cl}(F^{-}\bigl(Y \operatorname{\sigma-int}(B)\bigr) \subseteq F-\bigl(Y\backslash\operatorname{\sigma-int}(B)\bigr)\subseteq X \backslash F^{+} \bigl(\operatorname{\sigma-int}(B)\bigr). $$

This implies that

$$F^{+} \bigl(\operatorname{\sigma-int}(B)\bigr) \subseteq \operatorname{supra-int}\bigl(F^{+}(B)\bigr). $$

(vii) ⇒ (ii): Let \(x\in X\) be arbitrary and each \(V\in \sigma(F(x))\) then

$$F^{+} (V) \subseteq \operatorname{supra-int}\bigl(F^{+} (V)\bigr). $$

Hence \(F^{+} (V) \in\tau^{*}(x)\).

(viii) ⇔ (v): Clearly, a suprafrontier and frontier of any set is superharmonic-closed and closed, respectively.

(ix) ⇔ (iv): Follows immediately. □

### Theorem 3.2

*For a superharmonic multifunction*
\(F:(X,\tau )\rightarrow(Y,\sigma)\), *the following statements are equivalent*:

(i) *F*
*is lower superharmonic*-*continuous*;

(ii) *for each*
\(X\in X\)
*and each*
\(V \in\sigma\)
*such that*

$$F(x)\cap V \neq\phi\quad\textit{and}\quad F^{-} (V) \in \tau^{*}(x); $$

(iii) *for each*
\(x\in X\)
*and each*
\(V \in\sigma\)
*with*
\(F(x)\cap V \neq \phi\), *there exists*
\(W \in\tau^{*}\)
*such that*

$$F(W)\cap V \neq\phi; $$

(iv) \(F^{-} (V)\in\tau^{*}\)
*for every*
\(V \in\sigma\);

(v) \(F^{+} (K)\)
*is superharmonic*-*closed for every closed set*
\(K\subseteq Y\);

(vi) \(\operatorname{superharmonic-cl}(F^{+} (B)) \subseteq F^{+} (\sigma \operatorname{cl-}(B))\)
*for any*
\(B\subseteq Y\);

(vii) \(F^{-} (\operatorname{\sigma-{int}}(B))\subseteq \operatorname{superharmonic-int}(F^{-} (B))\)
*for any*
\(B\subseteq Y\);

(viii) \(\operatorname{superharmonic-fr}(F^{+} (B)) \subseteq F^{+} (\operatorname{fr}(B))\)
*for every*
\(B \subseteq Y\);

(ix) \(F:(X, \tau^{*})\rightarrow(Y, \sigma)\)
*is lower superharmonic*-*continuous*.

### Proof

The proof is a quite similar to that of Theorem 3.1. Recall that the net \((\chi_{i})_{(i\in l)}\) is superharmonic-convergent to \(x_{0}\), if for each \(W \in\tau^{*} (x_{O})\) there exists a \(i_{o} \in I\) such that for each \(i\ge i_{o}\) it implies \(x_{i} \in W\). □

### Theorem 3.3

*A superharmonic multifunction*
\(F : (X, \tau )\rightarrow(Y,\sigma)\)
*is upper superharmonic*-*continuous if and only if for each net*
\((\chi_{i})_{(i\in l)}\)
*superharmonic*-*convergent to*
\(x_{o}\)
*and for each*
\(V\in\sigma\)
*with*
\(F(x_{o})\subseteq V\)
*there is*
\(i_{o} \in I\)
*such that*
\(F(X_{i}) \subseteq V\)
*for all*
\(i \ge i_{o}\).

### Proof

Necessity, let \(V\in\sigma\) with \(F(x_{o})\subseteq V\). By upper superharmonic-continuity of *F*, there is \(W\in\tau^{*}(X_{O})\) such that \(F(W)\subseteq V\). Since from the hypothesis a net \((\chi _{i})_{(i\in l)}\) is superharmonic-convergent to \(x_{o}\) and \(W \in\tau ^{*}(x_{o})\) there is one \(i_{o} \in I\) such that \(x_{i} \in W\) for all \(i > i_{o}\) and then \(F(X_{i}) \subseteq V\) for all \(i > i_{o}\). As regards sufficiency, assume the converse, *i.e.* there is an open set *V* in *Y* with \(F(x_{o} )\subseteq V\) such that for each \(W\in\tau^{*}\) under inclusion we have the relation \(F(W)\nsubseteq V\), *i.e.* there is \(x_{w} \in W \) such that \(F(x_{w}) \nsubseteq V\). Then all of \(x_{w}\) will form a net in *X* with directed set *W* of \(\tau^{*}(x_{o})\), clearly this net is superharmonic-convergent to \(x_{o}\). But \(F(x_{w})\nsubseteq V\) for all \(W \in\tau^{*}(x_{o})\). This leads to a contradiction which completes the proof. □

### Theorem 3.4

*A superharmonic multifunction*
\(F : (X,\tau )\rightarrow(Y, \sigma)\)
*is lower superharmonic*-*continuous if and only if for each*
\(y_{o} \in F(x_{o})\)
*and for every net*
\((\chi_{i})_{(i\in l)}\)
*superharmonic*-*convergent to*
\(x_{o}\), *there exists a subnet*
\((Z_{j})_{(j\in J)}\)
*of the net*
\((\chi_{i})_{(i\in l)}\)
*and a net*
\((y_{i})_{(j,v)\in J}\)
*in*
*Y*
*so that*
\((y_{i})_{(j,v)\in J}\)
*superharmonic*-*convergent to*
*y*
*and*
\(y_{j} \in F(z_{j})\).

### Proof

For necessity, suppose *F* is lower superharmonic-continuous, \((\chi_{i})_{(i\in l)}\) is a net superharmonic-convergent to \(x_{o}\), \(y \in F(x_{o})\) and \(V \in \sigma(y)\). So we have \(F(x_{o}) \cap V \ne\phi\), by lower superharmonic-continuity of *F* at \(x_{o}\), there is a superharmonic-open set \(W \subseteq X\) containing \(x_{o}\) such that \(W \subseteq F^{-}(V)\). We have superharmonic-convergence of a net \((\chi _{i})_{(i\in l)}\) to \(x_{0}\) and for this *W*, there is a \(i_{o} \in I\) such that, for each \(i > i_{o}\), we have \(x_{i} \in W\) and therefore \(x_{i} \in F^{-}(V)\). Hence, for each \(V\in\sigma(y)\), define the sets

$$I_{v} =\bigl\{ i_{o} \in I:i >i_{o} \Rightarrow x_{i} \in F^{-}(V)\bigr\} $$

and

$$J =\bigl\{ (i,V):V\in D(y),i \in I_{v}\bigr\} $$

and an order ≥ on *J* given as \((i^{\prime},V^{\prime}) \ge(i,V)\) if and only if \(i^{\prime}> i\) and \(V^{\prime}\subseteq V\). Also, define \(\zeta: J \rightarrow I\) by \(\zeta((j,V))= j\). Then *ζ* is increasing and cofinal in *I*, so *ζ* defines a subset of \((\chi_{i})_{(i\in l)}\), denoted by \((z_{i})_{(j,v)\in J}\). On the other hand for any \((j,V) \in J\), since \(j > j_{o} \) implies \(x_{j} \in F^{-}(V)\) we have \(F(Z_{j})\cap V = F(X_{j}) \cap V\ne\phi\). Pick \(y_{j} \in F(Z_{j}) \cap V \ne\phi\). Then the net \((y_{i})_{(j,v)\in J}\) is supraconvergent to *y*. To see this, let \(V_{0} \in\sigma(y)\); then there is \(j_{0} \in I\) with \(j_{o} = \zeta( j_{o}, V_{o} )\); \((j_{o}, V_{o}) \in J\) and \(y_{jo} \in V\). If \((j,V) > (j_{o},V_{o})\) this means that \(j > j_{o}\) and \(V \subseteq V_{o}\). Therefore

$$y_{j} \in F(z_{j}) \cap V \subseteq F(x_{j}) \cap V \subseteq F(x_{j}) \cap V_{o}. $$

So \(y_{j}\in V_{o} \). Thus \((y_{i})_{(j,v)\in J}\) is superharmonic-convergent to *y*, which shows the result.

To show the sufficiency, assume the converse, *i.e.*
*F* is not lower superharmonic-continuous at \(x_{o}\). Then there exists \(V \in\sigma\) such that \(F(x_{o}) \cap V\ne\phi\) and for any superharmonic-neighborhood \(W \subseteq X\) of \(x_{o}\), there exists \(x_{w} \in W\) for which \(F(x_{w}) \cap V = \phi\). Let us consider the net \((\chi_{w})_{W\in \tau^{*}(\chi_{0})}\), which is obviously superharmonic-convergent to \(x_{o}\). Suppose \(y_{o} \in F(x_{o}) \cap V\), by hypothesis there is a superset \((z_{k})_{k\in K}\) of \((\chi_{w})_{W\in\tau^{*}(\chi_{0})}\) and \(y_{k} \in F(z_{k})\) like \((y_{k})_{k\in K}\) superharmonic-convergent to \(y_{o}\). As \(y_{o} \in V \in \sigma\) there is \(k_{0}^{\prime}\in K\) so that \(k>k_{0}^{\prime}\) implies \(y_{k} \in V\). On the other hand \((z_{k})_{kEK} \) is a superset of the net \((\chi^{w})_{W\in\tau ^{*}(\chi_{0})}\) and so there exists a superharmonic function \(\Omega:K \rightarrow\tau^{*}(x_{o}) \) such that \(z_{k}=\chi_{\Omega(k)}\) and for each \(W \in \tau^{*}(x_{o})\) there exists \(k_{0}^{\prime\prime}\in K\) such that \(\Omega(k_{0}^{\prime\prime}) \ge W\). If \(k\ge k_{0}^{\prime\prime}\) then \(\Omega(k) \ge\Omega(k_{0}^{\prime\prime}) \ge W \). Considering \(k_{0} \in K\) so that \(k_{o} \ge k_{0}^{\prime}\) and \(k_{o} \ge k_{0}^{\prime\prime}\). Therefore \(y_{k} \in V\) and by the meaning of the net \((\chi_{W})_{W\in\tau^{*}(\chi_{0})}\), we have

$$F(z_{k}) \cap V = F(\chi_{\Omega(K)}) \cap V = \phi. $$

This gives \(y_{k} \notin V\), which contradicts the hypothesis and so the requirement holds. □

### Definition 3.2

A subset *W* of a space \((X, \tau)\) is called superharmonic-regular, if for any \(x \in W\) and any \(H \in\tau^{*}(x)\) there exists \(U \in\tau\) such that

$$x \in U \subseteq \operatorname{\tau-cl}(U) \subseteq H . $$

Recall that \(F: (X, \tau) \rightarrow(Y,\sigma)\) is punctually superharmonic-regular, if for each \(X\in X\), \(F(x)\) is superharmonic-regular.

### Lemma 3.2

*In a superharmonic space*
\((X,\tau)\), *if*
\(W \subseteq X\)
*is superharmonic*-*regular and contained in a superharmonic*-*open set*
*H*, *then there exists*
\(U \in\tau\)
*such that*

$$W \subseteq U \subseteq \operatorname{\tau-cl}(U) \subseteq H . $$

*For a superharmonic multifunction*
\(F:(X, \tau)\rightarrow(Y,\sigma)\), *a superharmonic multifunction*
\(\operatorname{superharmonic-cl}(F):(X, \tau)\rightarrow (Y,\sigma)\)
*is defined as follows*:

$$(\operatorname{superharmonic-cl} F) (x) =\operatorname{superharmonic-cl}\bigl(F(x)\bigr) $$

*for each*
\(x \in X\).

### Proposition 3.2

*For a punctually*
*α*-*paracompact and punctually superharmonic*-*regular superharmonic multifunction*
\(F: (X, \tau) \rightarrow(Y, \sigma)\), *we have*

$$\bigl(\operatorname{superharmonic-cl}(F)^{+} (W)\bigr) = F^{+} (W) $$

*for each*
\(W\in \sigma^{*}\).

### Proof

Let \(x \in(\operatorname{superharmonic-cl}(F))^{+}(W)\) for any \(W\in \sigma^{*}\), this means

$$F(x) \subseteq \operatorname{superharmonic-cl}\bigl(F(x)\bigr) \subseteq W, $$

which leads to \(x \in F^{+} (W)\). Hence one inclusion holds. To show the other, let \(X\in F^{+} (W)\) where \(W \in \sigma^{*} (x)\). Then \(F(x) \subseteq W\), by the hypothesis of *F* and the fact that \(\sigma\subseteq\sigma^{*}\), applying Lemma 3.2, there exists \(G \in\sigma\) such that

$$F(x)\subseteq G\in\operatorname{\sigma-cl}(G)\subseteq W. $$

Therefore

$$\operatorname{superharmonic-cl}\bigl(F(x)\bigr) \subseteq W. $$

This means that \(x \in(\operatorname{superharmonic-cl} F)^{+} (W)\). Hence the equality holds. □

### Theorem 3.5

*Let*
\(F (X, \tau)\rightarrow(Y, \sigma)\)
*be a punctually a*-*paracompact and punctually superharmonic*-*regular superharmonic multifunction*. *Then*
*F*
*is upper superharmonic*-*continuous if and only if*

$$(\operatorname{superharmonic-cl} F): (X, \tau)\rightarrow(Y, \sigma) $$

*is upper superharmonic*-*continuous*.

### Proof

As regards necessity, suppose \(V \in\sigma\) and \(x \in(\operatorname{superharmonic-cl} F)^{+} (V) = F^{+} (V)\) (see Proposition 3.2). By upper superharmonic-continuity of *F*, there exists \(H \in\tau^{*}(x)\) such that \(F(H) \subseteq V\). Since \(\sigma\in\sigma^{*}\), by Lemma 3.2 and the assumption of *F*, there exists \(G \in\sigma\) such that

$$F(h) \subseteq G \subseteq\operatorname{\sigma-cl}(G) \subseteq W $$

for each \(h \in H\).

Hence

$$\operatorname{superharmonic-cl}\bigl(F(h)\bigr) \subseteq \operatorname{superharmonic-cl} (G) \subseteq \operatorname{\sigma-cl}(G) \subseteq V $$

for each \(h \in H\), which shows that [13]

$$(\operatorname{superharmonic-cl} F) (H) \subseteq V. $$

Thus \((\operatorname{superharmonic-cl} F)\) is upper superharmonic-continuous. As regards sufficiency, assume \(V\in\sigma\) and \(X \in F^{+} (V) = (\operatorname{superharmonic-cl} F)^{+} (V)\). By the hypothesis of *F* in this case, there is \(H\in\tau^{*}(x)\) such that \((\operatorname{superharmonic-cl} F)(H) \subseteq V\), which obviously gives \(F(H) \subseteq V\). This completes the proof. □

### Lemma 3.3

*In a space*
\((X,\tau)\), *any*
\(x \in X\)
*and*
\(A\subseteq X, X \in \operatorname{superharmonic-cl}(A)\)
*if and only if*

*for each*
\(W\in \tau^{*}(x)\).

### Proposition 3.3

*For a superharmonic multifunction*
\(F: (X, \tau) \rightarrow(Y, \sigma)\),

$$(\operatorname{superharmonic-cl} F)^{-} (W) = F^{-} (W) $$

*for each*
\(W \in \sigma^{*}\).

### Proof

Let \(x \in(\operatorname{superharmonic-cl} F)^{-} (W)\). Then

$$W \cap \operatorname{superharmonic-cl}\bigl(F(x)\bigr) \neq\phi. $$

Since \(W\in\sigma^{*}\), Lemma 3.3 gives \(W\cap F(x) \neq\phi\) and hence \(x \in F^{-}(W)\). Conversely, let \(x \in F^{-}(W)\), then

$$\phi\neq F(x)\cap W\subseteq(\operatorname{supracl} F)^{-}(x) \cap W $$

and so

$$x \in(\operatorname{superharmonic-cl} F)^{-}(W). $$

Hence

$$x \in(\operatorname{superharmonic-cl} F)^{+} (W) $$

and this completes the equality. □

### Theorem 3.6

*A superharmonic multifunction*
\(F: (X, \tau )\rightarrow(Y, \sigma)\)
*is lower superharmonic*-*continuous if and only if*
\((\operatorname{superharmonic-cl} F): (X, \tau) \rightarrow(Y, \sigma)\)
*is lower superharmonic*-*continuous*.

### Proof

This is an immediate consequence of Proposition 3.2 taking in consideration that \(\tau\subseteq\tau^{*}\) and (iv) of Theorem 3.2. □

### Theorem 3.7

*If*
\(F:(X,\tau)\rightarrow(Y, \sigma)\)
*is an upper superharmonic*-*continuous surjection and for each*
\(x\in X,F(x)\)
*is compact relative to*
*Y*. *If*
\((X,\tau)\)
*is superharmonic*-*compact*, *then*
\((Y,\sigma)\)
*is compact*.

### Proof

Let

$$\{V_{i} : i \in I, V_{i} \in\sigma\} $$

be a cover of *Y*; \(F(x)\) is compact relative to *Y*, for each \(x \in X\). Then there exists a finite \(I_{o}(x)\) of *I* such that [14]

$$F(x)\subseteq U\bigl(V_{i}: i \in I_{o} (x)\bigr). $$

Upper superharmonic-continuity of *F* shows that there exists \(W(x) \in\tau^{*}(X,x)\) such that

$$F\bigl(W(x)\bigr) \subseteq\bigcup{V_{i}:i \in I_{o} (x)}. $$

Since \((X, \tau)\) is superharmonic-compact, there exists \({x_{1},x_{2}, \ldots,x_{n}}\) such that

$$X=\bigcup\bigl(W(x_{j}):1 \le j \le n\bigr). $$

Therefore

$$Y = F(X) = \bigcup\bigl(F\bigl(W(x_{j})\bigr): 1 \le j \le n \bigr)\subseteq\bigcup V_{i} : i \in I_{0} (X_{j}) \quad 1 \le j \le n. $$

Hence \((Y,\sigma)\) is compact. □