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. □