Maximal classes for families of lower and upper semicontinuous functions with a closed graph

Abstract

In this paper we characterize the following maximal classes for families of lower and upper semicontinuous functions with a closed graph: the maximal additive class, the maximal multiplicative class and the maximal classes with respect to maximum and minimum.

1 Introduction

The letters ${\mathbb R}$, $\mathbb {Q}$ and ${\mathbb N}$ denote the real line, the set of rationals and the set of positive integers, respectively. The family of all functions from a set X into Y is denoted by $Y^X$. For each set $A \subset X$ its characteristic function is denoted by ${\chi }_{A}$. In particular, ${\chi }_{\emptyset }$ stands for the zero constant function.

Let X be a topological space. The symbol $X^d$ denotes the set of all accumulation points of X. For each set $A\subset X$ the symbols ${{\mathrm{int}}}A$ and ${{\mathrm{cl}}}A$ denote the interior and the closure of A, respectively. The spaces ${\mathbb R}$ and $X\times {\mathbb R}$ are considered with their standard topologies. We say that a function $f:X \rightarrow {\mathbb R}$ has a closed graph, if the graph of f, i.e., the set $\{ (x, f(x)) : x \in X \}$ is a closed subset of the product $X\times {\mathbb R}$. We say that a function $f: X \rightarrow {\mathbb R}$ is lower (upper) semicontinuous at a point $x \in X$, if for each $\varepsilon >0$ there is an open neighborhood U of x such that $f(z)> f(x) - \varepsilon$ ( $f(z) < f(x)+ \varepsilon$, respectively) for each $z \in U$. If $f: X \rightarrow {\mathbb R}$ is lower (upper) semicontinuous at each point $x \in X$, then we say that the function f is lower (upper, respectively) semicontinuous. Let ${{\mathscr {C}}onst}(X)$, ${\mathcal C}(X)$, ${\mathcal U}(X)$, ${lsc}(X)$, ${usc}(X)$ denote the class of all real-valued functions on X that are constant, continuous, have a closed graph, are lower and upper semicontinuous, respectively. Obviously ${\mathcal C}(X) \subset {\mathcal U}(X)$ (see also e.g. [5]) and $C(X)={lsc}(X) \cap {usc}(X)$. For ${\mathcal F}(X)$ and $\mathcal G(X)$ nonempty subsets of ${\mathbb R}^X$ the symbol ${\mathcal F}\mathcal G(X)$ denotes the class ${\mathcal F}(X) \cap \mathcal G(X)$. Further denote by ${\mathcal F}^+(X)$ the family of all nonnegative functions from ${\mathcal F}(X)$. Let $f\in {\mathbb R}^X$. The symbol G(f) denotes the graph of f and the symbols C(f) and D(f) denote the sets of points of continuity and discontinuity of f, respectively. For each $y\in {\mathbb R}$ let $[f=y]=\bigl \{ x\in X :f(x)=y \bigr \}$. Similarly we define the symbols $[f>y]$, $[f<y]$.

If ${\mathcal F}\subset {\mathbb R}^X$ is a family of functions, denote by

$$\begin{aligned} {\mathcal F}+{\mathcal F}&\mathop {=}\limits ^{df }&\bigl \{ f \in {\mathbb R}^X : f = g+h \; \text {for some}\, g, h \in {\mathcal F}\bigr \},\\ {\mathcal M_a}({\mathcal F})&\mathop {=}\limits ^{df }&\bigl \{ f \in {\mathbb R}^X : \bigl (\forall _{g \in {\mathcal F}}\bigr )\; f+g \in {\mathcal F}\bigr \},\\ {\mathcal M_m}({\mathcal F})&\mathop {=}\limits ^{df }&\bigl \{ f \in {\mathbb R}^X : \bigl (\forall _{g \in {\mathcal F}}\bigr )\; f\cdot g \in {\mathcal F}\bigr \},\\ {\mathcal M_{\max }}({\mathcal F})&\mathop {=}\limits ^{df }&\bigl \{ f \in {\mathbb R}^X : \bigl (\forall _{g \in {\mathcal F}}\bigr ) \; \max (f, g) \in {\mathcal F}\bigr \},\\ {\mathcal M_{\min }}({\mathcal F})&\mathop {=}\limits ^{df }&\bigl \{ f \in {\mathbb R}^X : \bigl (\forall _{g \in {\mathcal F}}\bigr ) \; \min (f, g) \in {\mathcal F}\bigr \}. \end{aligned}$$

The above classes ${\mathcal M_a}({\mathcal F})$, ${\mathcal M_m}({\mathcal F})$, ${\mathcal M_{\max }}({\mathcal F})$ and ${\mathcal M_{\min }}({\mathcal F})$ are called the maximal additive class for ${\mathcal F}$, the maximal multiplicative class for ${\mathcal F}$, the maximal class with respect to maximum and minimum for ${\mathcal F}$, respectively.

In 1987 Menkyna [7] characterized the maximal additive and multiplicative classes for the family of functions with a closed graph. He proved that ${\mathcal M_a}({\mathcal U}(X))={\mathcal C}(X)$ for a topological space X [7, Theorem 1] and ${\mathcal M_m}({\mathcal U}(X)) = \{ f \in {\mathcal C}(X): [f=0]\ is\ an~open~set\}$ for a locally compact normal topological space X [7, Theorem 2]. Let ${\mathcal Q}(X)$ denote the family of all quasi-continuous functions from a topological space X to ${\mathbb R}$. Recall that $f \in {\mathcal Q}(X)$ if and only if for each $x \in X$, $\varepsilon > 0$ and for each neighbourhood U of x there is a nonempty open set $V \subset U$ such that $|f(x) - f(y)| < \varepsilon$ for each $y\in V$. In 2008 Sieg [8] considered real functions defined on ${\mathbb R}$ and showed that ${\mathcal M_a}({\mathcal Q}{\mathcal U}({\mathbb R})) = {\mathcal C}({\mathbb R})$, ${\mathcal M_m}({\mathcal Q}{\mathcal U}({\mathbb R})) = \{f \in {\mathcal C}({\mathbb R}): {f ={\chi }_{\emptyset } \,\,\mathrm{or }\, f(x) \ne 0 \,\,\mathrm{for~all }\, x \in {\mathbb R}} \}$ and ${\mathcal M_{\max }}({\mathcal Q}{\mathcal U}({\mathbb R}))= {\mathcal M_{\min }}({\mathcal Q}{\mathcal U}({\mathbb R}))=\emptyset$. In 2014 Szczuka (see  [9, 10]) characterized the following maximal classes for lower and upper semicontinuous strong Świa̧tkowski functions and lower and upper semicontinuous extra strong Świa̧tkowski functions: the maximal additive class, the maximal multiplicative class and the maximal classes with respect to maximum. She proved, among others, that if ${\mathcal F}$ denotes the family of lower semicontinuous strong Świa̧tkowski real functions defined on ${\mathbb R}$, then ${\mathcal M_a}({\mathcal F})={{\mathscr {C}}onst}$ [9, Theorems 3.1], ${\mathcal M_m}({\mathcal F})={{\mathscr {C}}onst}^+$ [9, Theorem 3.2] and ${\mathcal M_{\max }}({\mathcal F}) ={{\mathscr {C}}onst}$ [9, Theorem 3.3].

In this paper we deal with the families of lower and upper semicontinuous functions with a closed graph. We obtain the following results:

• ${\mathcal M_a}({\mathcal U}{lsc}(X)) ={\mathcal U}{lsc}(X)$, where X is a topological space (Theorem 2.5),

• ${\mathcal M_a}({\mathcal U}{usc}(X))={\mathcal U}{usc}(X)$, where X is a topological space (Theorem 3.3),

• ${\mathcal M_m}({\mathcal U}{lsc}(X)) = \{f \in {\mathcal C}(X): [f=0] \,\text {is an open set and}\, f(x)\ge 0 \,\text {for all}\, x \in X \} = {\mathcal M_m}({\mathcal U}{usc}(X))$, where X is a perfectly normal topological space such that $X=X^d$ (Theorems 2.7, 3.4),

• ${\mathcal M_{\max }}({{\mathcal U}lsc}(X))={{\mathcal U}lsc}(X)$, where X is a topological space (Theorem 2.10),

• ${\mathcal M_{\min }}({{\mathcal U}usc}(X))={{\mathcal U}usc}(X)$, where X is a topological space (Theorem 3.5),

• ${\mathcal M_{\min }}({\mathcal U}{lsc}(X)) ={\mathcal M_{\max }}({\mathcal U}{usc}(X))= \emptyset$, where X is a perfectly normal topological space such that $X^d \ne \emptyset$ (Corollary 2.15, Theorems 3.6).

2 Lower semicontinuous functions with a closed graph

We start with a following proposition.

Proposition 2.1

Let X be a topological space. A function $f:X \rightarrow {\mathbb R}$ has the closed graph if and only if for each $x\in X$ and for each $m \in {\mathbb N}$ there is a neighbourhood V of x such that $f(z)\in (-\infty , -m) \cup (f(x)-1/m, f(x)+ 1/m) \cup (m, \infty )$ for each $z \in V$.

Proof

The implication ($\Leftarrow$) we can find in [2] (see p. 118, lines 11–14). The implication ($\Rightarrow$) immediately follows from [6] or [1, Proposition 1]: if $f\in {\mathcal U}(X)$, then for each $x \in X$ and each neighborhood U of f(x) such that $Y {\setminus } U$ is compact there is an neighborhood V of x such that $f(V)\subset U$. Now, it is sufficient to take $U = (-\infty , -m) \cup (f(x)-1/m, f(x)+ 1/m) \cup (m, \infty )$. Observe that, the equivalence of this proposition also immediately follows from [1, Proposition 2]. $\square$

From above and the definitions of the class ${lsc}$ we obtain:

Lemma 2.2

Let X be a topological space. A function $f:X \rightarrow {\mathbb R}$ is lower semicontinuous function with a closed graph if and only if for each $x\in X$ and for each $m \in {\mathbb N}$ there is a neighbourhood V of x such that $f(z)\in (f(x)-1/m, f(x)+ 1/m) \cup (m, \infty )$ for each $z \in V$.

Proof

First, assume that for each $x\in X$ and for each $m \in {\mathbb N}$ there is a neighbourhood V of x such that $f(z)\in (f(x)-1/m, f(x)+ 1/m) \cup (m, \infty )$ for each $z \in V$. Then, by Proposition 2.1, $f \in {\mathcal U}(X)$. Now, we will show that $f \in {lsc}(X)$. Let $x\in X$ and $\varepsilon >0$. We choose $m \in {\mathbb N}$ such that $m \ge \max \{\frac{1}{\varepsilon }, f(x)-\varepsilon \}$. There is a neighbourhood V of x such that $f(z)\in (f(x)-1/m, f(x)+ 1/m) \cup (m, \infty ) \subset (f(x) -\varepsilon , \infty )$ for each $z \in V$ and consequently $f \in {lsc}(X)$.

Now, let $f \in {{\mathcal U}lsc}(X)$. Fix $x \in X$ and $m \in {\mathbb N}$. Since $f \in {lsc}(X)$, there is a neighbourhood $V_1$ of x such that $f(z)\in (f(x)-1/m, \infty )$ for each $z \in V_1$. We consider two cases.

First, assume that $f(x) \ge 0$. Since $f \in {\mathcal U}(X)$, there is a neighbourhood $V_2$ of x such that $f(z)\in (-\infty , -m) \cup (f(x)-1/m, f(x)+ 1/m) \cup (m, \infty )$ for each $z \in V_2$ (see Proposition 2.1). Let $V =V_1 \cap V_2$ and let $z \in V$. Then $f(z)\in (f(x)-1/m, f(x)+ 1/m) \cup (m, \infty )$.

Now, assume that $f(x)<0$. We choose $k \in {\mathbb N}$ such that $k \ge \max \{m, -f(x)+\frac{1}{m}\}$. Since $f \in {\mathcal U}(X)$, there is a neighbourhood $V_2$ of x such that $f(z)\in (-\infty , -k) \cup (f(x)-1/k, f(x)+ 1/k) \cup (k, \infty )$ for each $z \in V_2$. Let $V =V_1 \cap V_2$ and let $z \in V$. Since $k\ge m$, $-k \le f(x)-\frac{1}{m}$ and $\frac{1}{k}\le \frac{1}{m}$, we have $f(z)\in \bigl ((f(x)-1/k, f(x)+ 1/k) \cup (k, \infty )\bigr ) \subset \bigl ((f(x)-1/m, f(x)+ 1/m) \cup (m, \infty )\bigr )$. This completes the proof. $\square$

The next lemma follows from Proposition 2.1 and Lemma 2.2.

Lemma 2.3

Let X be a topological space. Then ${\mathcal U}^+(X) \subset {{\mathcal U}lsc}(X)$.

Now, we will characterize the class of the sums of lower semicontinuous functions with a closed graph.

Lemma 2.4

Let X be a topological space. Then ${{\mathcal U}lsc}(X) +{{\mathcal U}lsc}(X) = {{\mathcal U}lsc}(X).$

Proof

Let $f,g \in {{\mathcal U}lsc}(X)$. Fix $x \in X$ and $m \in {\mathbb N}$. Let $k \in {\mathbb N}$ be such that $1/k < 1/(2m+ |f(x)| + |g(x)|)$. By Lemma 2.2, there exists a neighbourhood V of x such that $f(z) \in (f(x) -1/k, f(x) +1/k) \cup (k, \infty )$ and $g(z) \in (g(x) -1/k, g(x) +1/k) \cup (k, \infty )$ for each $z \in V$. Let $z \in V$. We consider four cases.

If $f(z)>k$ and $g(z) >k$, then evidently $(f+g)(z) >m$.

If $f(z)> k$ and $g(z) \in (g(x) -1/k, g(x) +1/k)$, then

$$\begin{aligned} (f+g)(z)> k+g(x)- 1/k> 2m+|f(x)|+|g(x)| +g(x) -1/k >m. \end{aligned}$$

Similarly, $g(z) >k$ and $f(z) \in (f(x) -1/k, f(x) +1/k)$, implies $(f+g)(z)>m$.

Now, let $f(z) \in (f(x) -1/k, f(x) +1/k)$ and $g(z) \in (g(x) -1/k, g(x) +1/k)$. Then, we have

$$\begin{aligned} |(f+g)(z)-(f+g)(x)|\le |f(z) -f(x)| +|g(z) - g(x)|< 2/k<1/m. \end{aligned}$$

It follows that $f+g \in {{\mathcal U}lsc}(X)$. $\square$

Theorem 2.5

Let X be a topological space. Then ${\mathcal M_a}({{\mathcal U}lsc}(X))={{\mathcal U}lsc}(X).$

Proof

Since ${\chi }_{\emptyset }\in {{\mathcal U}lsc}(X)$, we conclude that ${\mathcal M_a}({{\mathcal U}lsc}(X))\subset {{\mathcal U}lsc}(X)$. The inclusion ${\mathcal U}{lsc}(X) \subset {\mathcal M_a}({{\mathcal U}lsc}(X))$ follows from Lemma 2.4. $\square$

Now, recall the following lemma [7, Lemma 2], which will be applied in this paper.

Proposition 2.6

Let X be a topological space and let $f \in {\mathcal C}(X)$. Then the function $g: X \rightarrow {\mathbb R}$ defined by the formula

$$\begin{aligned} g(x)={\left\{ \begin{array}{ll} \frac{1}{f(x)},&{}\quad \mathrm {if }\, x \in [f \ne 0],\\ 0,&{}\quad \mathrm {if }\, x \in [f=0]. \end{array}\right. } \end{aligned}$$

has the closed graph.

Theorem 2.7

Let X be a normal topological space such that each singleton is $G_\delta$-set. Then

$$\begin{aligned} {\mathcal M_m}({\mathcal U}{lsc}(X)) = \{f \in {\mathcal C}(X): [f=0] \,\text {is an open set ~and }~ [f<0]^d = \emptyset \}. \end{aligned}$$

Proof

We will prove this theorem in four parts. First, we will show that ${\mathcal M_m}({\mathcal U}{lsc}(X)) \subset {\mathcal C}(X)$. Let $f \in {\mathcal M_m}({\mathcal U}{lsc}(X))$. Since ${\chi }_{{\mathbb R}}, -{\chi }_{{\mathbb R}} \in {\mathcal U}{lsc}(X)$, we have $f \in {lsc}(X)$ and $-f \in {lsc}(X)$. Consequently $f \in {lsc}(X) \cap {usc}(X) ={\mathcal C}(X)$.

Now, we assume that the function $f \in {\mathcal C}(X)$ and the set $[f=0]$ is not open. We will show that $f \notin {\mathcal M_m}({\mathcal U}{lsc}(X))$ (The proof of this part is similar to the second part of the proof of [7, Theorem 2]). Define the function $g: X \rightarrow {\mathbb R}$ by the formula

$$\begin{aligned} g(x)={\left\{ \begin{array}{ll} \frac{1}{|f(x)|},&{}\quad \mathrm {if }\, x \in [f \ne 0],\\ 0,&{}\quad \mathrm {if }\, x \in [f=0]. \end{array}\right. } \end{aligned}$$

By Proposition 2.6 the function g has the closed graph. Moreover g is non-negative function and consequently, by Lemma 2.3, $g \in {\mathcal U}{lsc}(X)$. Now, we will show that $f\cdot g \notin {\mathcal U}{lsc}(X)$.

Since the set $[f=0]$ is not open, there is $x_0 \in [f=0]$ such that for each open neighbourhood V of $x_0$ there is $x_V \in V \cap [f \ne 0]$. Notice that $(f\cdot g)(x_V) \in \{-1,1\}$ for each neighbourhood V of $x_0$ and $(f \cdot g)(x_0) =0$. By Proposition 2.1, $f \cdot g \notin {\mathcal U}(X)$.

In the third part of the proof, suppose that $f \in {\mathcal C}(X)$, the set $[f=0]$ is open and $[f<0]^d \ne \emptyset$. We will prove that $f \notin {\mathcal M_m}({\mathcal U}{lsc}(X))$. Let $x_0 \in [f<0]^d$. Then there is a net $(x_\gamma )_{\gamma \in \Gamma }$ of elements of X such that $x_\gamma \rightarrow x_0$, $x_\gamma \ne x_0$ and $f(x_\gamma ) <0$ for every $\gamma \in \Gamma$. Notice that, since $f \in {\mathcal C}(X)$ and the set $[f=0]$ is open, we have $f(x_0) <0$. By Urysohn Lemma there is a continuous function $h:X \rightarrow [0,1]$ such that $[h=0]=\{x_0\}$.

Define the function $g: X \rightarrow {\mathbb R}$ by the formula

$$\begin{aligned} g(x)={\left\{ \begin{array}{ll} \frac{1}{h(x)},&{}\quad \mathrm {if }\, x \ne x_0,\\ 0,&{}\quad \mathrm {if }\, x =x_0. \end{array}\right. } \end{aligned}$$

Observe that, by Proposition 2.6 and Lemma 2.3, $g \in {\mathcal U}{lsc}(X)$. Moreover $f\cdot g \notin {lsc}(X)$, because the net $((f\cdot g)(x_\gamma ))_{\gamma \in \Gamma }$ diverges to $-\infty$ (recall that $f(x_\gamma ) \rightarrow f(x_0) <0$).

In the last part suppose that $f \in {\mathcal C}(X)$, the set $[f=0]$ is open, $[f<0]^d = \emptyset$ and $g \in {\mathcal U}{lsc}(X)$. Then, by [7, Theorem 2], $(f\cdot g) \in {\mathcal U}(X)$ (see also the third part of the proof of [7, Theorem 2]). It is enough to show that $(f\cdot g) \in {lsc}(X)$. Let $x_0 \in X$. If $f(x_0 )\le 0$, then the function $f\cdot g$ is continuous at $x_0$ and consequently $f \cdot g$ is a lower semicontinuous at this point. Indeed, if $f(x_0 )=0$, then by the assumption $[f=0]= {{\mathrm{int}}}[f=0]$, we have $x_0 \in {{\mathrm{int}}}[f=0] \subset {{\mathrm{int}}}[f\cdot g =0]$ and if $f(x_0)<0$, then $x_0$ is a isolated point of X. Finally, assume that $f(x_0 )> 0$. Since $f \in {\mathcal C}(X)$, there is an open neighborhood U of $x_0$ such that $U \subset [f>0]$. Since $g \in {lsc}(X)$, f is continuous and positive function on U, the function $f \cdot g$ is a lower semicontinuous at $x_0$. The proof is complete. $\square$

It is easy to see that from above for $X={\mathbb R}$ we have the following corollary.

Corollary 2.8

${\mathcal M_m}({\mathcal U}{lsc}({\mathbb R})) = \{f \in {\mathcal C}({\mathbb R}): f ={\chi }_{\emptyset } \,\,\text {or }\, f(x) > 0 \,\,\text {for all }\, x \in {\mathbb R}\}$.

Lemma 2.9

Let X be a topological space and let $f, g \in {\mathcal U}{lsc}(X)$. Then the real function $h=\max \{f,g\}$ defined on X is a lower semicontinuous function with a closed graph.

Proof

Let $f,g \in {{\mathcal U}lsc}(X)$. We will use Lemma 2.2. Fix $x \in X$ and $m \in {\mathbb N}$. Then there exists a neighbourhood V of x such that $f(z) \in (f(x) -1/m, f(x) +1/m) \cup (m, \infty )$ and $g(z) \in (g(x) -1/m, g(x) +1/m) \cup (m, \infty )$ for each $z \in V$. We assume that $f(x) \ge g(x)$ (The case $f(x) < g(x)$ is analogous). Then $h(x)=f(x)$ and it is easy to see that $h(z) \in (h(x) -1/m, h(x) +1/m) \cup (m, \infty )$ for each $z \in V$. So, $h \in {\mathcal U}{lsc}(X)$. $\square$

Theorem 2.10

Let X be a topological space. Then ${\mathcal M_{\max }}({{\mathcal U}lsc}(X))={{\mathcal U}lsc}(X)$.

Proof

The inclusion ${\mathcal U}{lsc}(X) \subset {\mathcal M_{\max }}({{\mathcal U}lsc}(X))$ follows from Lemma 2.9. So, we will only prove that ${\mathcal M_{\max }}({{\mathcal U}lsc}(X))\subset {{\mathcal U}lsc}(X)$. Let $f:X \rightarrow {\mathbb R}$ be a function such that $f \notin {{\mathcal U}lsc}(X)$. We choose $x_0 \in X$ and $m \in {\mathbb N}$, such that $m \ge f(x_0) +\frac{1}{m}$ and for each open neighborhood V of $x_0$ there is $x \in V$ such that $f(x) \le m$ and $f(x) \notin (f(x_0) -\frac{1}{m},f(x_0) +\frac{1}{m})$. We will show that $f \notin {\mathcal M_{\max }}({{\mathcal U}lsc}(X))$. Let $c = f(x_0) -\frac{1}{m}$. Define the function $g : X \rightarrow {\mathbb R}$ by $g\mathop {=}\limits ^{df }c$. Clearly $g \in {{\mathcal U}lsc}(X)$. Denote $h =\max \{f,g\}$. We will prove that $h \notin {\mathcal U}{lsc}(X)$. Notice that $h(x_0) =f(x_0)$. Observe that, for each open neighborhood V of $x_0$ there is $x_V \in V$ such that $f(x_V)\in ( -\infty , c] \cup [f(x_0) +\frac{1}{m}, m]$ and consequently $h(x_V) \in \{h(x_0) -\frac{1}{m}\} \cup [h(x_0) +\frac{1}{m}, m]$. By Proposition 2.1, $h \notin {\mathcal U}(X)$. This completes the proof. $\square$

Theorem 2.11

Let X be a topological space such that ${\mathcal U}(X)\ne {\mathcal C}(X)$. Then ${\mathcal M_{\min }}({\mathcal U}{lsc}(X)) = \emptyset$.

Proof

Let $f \in {\mathbb R}^X$. We will show that there is a function $g \in {\mathcal U}{lsc}(X)$ such that the function $h=\min \{f,g\}\notin {\mathcal U}{lsc}(X)$.

Let $g_1:X \rightarrow {\mathbb R}$ be a function with a closed graph and let $x_0 \in D(g_1)$. Put $g_2=|g_1|$. Then $g_2 \in {\mathcal U}{lsc}$ and there is a net $(x_\gamma )_{\gamma \in \Gamma }$ of elements of X which converges to the point $x_0$ and a net $(g_2(x_\gamma ))_{\gamma \in \Gamma }$ diverges to $\infty$. We consider two cases.

If $x_0 \in C(f)$, we define the function $g: X \rightarrow {\mathbb R}$ by $g(x) \mathop {=}\limits ^{df }g_2(x) - g_2(x_0) +f(x_0) -1$. It is easy to see that $g \in {\mathcal U}{lsc}(X)$. Let $h=\min \{f,g\}$. Then $h(x_0)=g(x_0)=f(x_0)-1$ and there is $\gamma _0 \in \Gamma$ such that $h(x_\gamma )=f(x_\gamma )$ for each $\gamma >\gamma _0$. Consequently $(x_0, f(x_0)) \in {{\mathrm{cl}}}G(h) {\setminus } G(h)$ and $h \notin {\mathcal U}(X)$.

Now, let $x_0 \in D(f)$. There is $\varepsilon >0$ such that for each neighbourhood V of $x_0$ there is $z \in V$ such that $f(z) \notin (f(x_0)- \varepsilon , f(x_0) + \varepsilon )$. Define the function $g: X \rightarrow {\mathbb R}$ by $g(x) \mathop {=}\limits ^{df }f(x_0) + \varepsilon$. Let $h=\min \{f,g\}$. Then $h(x_0)=f(x_0)$ and for each neighbourhood V of $x_0$ there is $z \in V$ such that $h(z) \in (- \infty , h(x_0) - \varepsilon ]\cup \{h(x_0)+\varepsilon \}$. By Proposition 2.2, $h \notin {\mathcal U}{lsc}(X)$ $\square$

It is easy to see that

Remark 1

Let X be a topological space such that ${\mathcal U}(X)= {\mathcal C}(X)$. Then ${\mathcal M_{\min }}({\mathcal U}{lsc}(X)) = {\mathcal C}$.

Now, we recall the definition of a P-space [4, pp. 62–63] and two propositions given by Wójtowicz and Sieg  [11, Theorem 1 and Corollary 1].

Definition 1

We say that a completely regular (Tychonoff) space X is a P-space if every $G_\delta$-subset ($F_\sigma$-subset) of X is open (closed); equivalently, every co-zero subset of X is closed.

Proposition 2.12

Let X be a completely regular space. Then ${\mathcal U}(X)= {\mathcal C}(X)$ if and only if X is a P-space.

Proposition 2.13

Let X be a perfectly normal or first countable space, or a locally compact space. Then ${\mathcal U}(X)\ne {\mathcal C}(X)$ if and only if X is non-discrete.

From Proposition 2.12, Theorem 2.11 and Remark 1 we obtain the following Corollary.

Corollary 2.14

Let X be a nonempty completely regular space. Then ${\mathcal M_{\min }}({\mathcal U}{lsc}(X)) = \emptyset$ if and only if X is not a P-space.

Moreover, using Proposition 2.13 and Theorem 2.11 we conclude that

Corollary 2.15

Let X be a non-discrete perfectly normal or first countable space, or a locally compact space. Then ${\mathcal M_{\min }}({\mathcal U}{lsc}(X)) = \emptyset$.

Finally, observe that we can extend the lists (see e.g. [11, Theorem 1]) of equivalent conditions for X to be a P-space as follows:

Corollary 2.16

Let X be a nonempty completely regular space. Then X is a P-space if and only if ${\mathcal M_{\min }}({\mathcal U}{lsc}(X)) \ne \emptyset$.

3 Upper semicontinuous functions with a closed graph

First, we recall some basic property of the functions with a closed graph [3, Proposition 2]

Proposition 3.1

Let X be a topological space. Let $\alpha$ be a real number. If $f \in {\mathcal U}(X)$, then $\alpha \cdot f \in {\mathcal U}(X)$.

From above and the definitions of the classes ${lsc}(X)$ and ${usc}(X)$ we obtain:

Proposition 3.2

Let X be a topological space. For each function $f \in {\mathbb R}^X$ we have $f \in {\mathcal U}{usc}(X)$ if and only if $(-f) \in {\mathcal U}{lsc}(X)$.

Now, we will characterize the following maximal classes for the family of upper semicontinuous functions with a closed graph: the maximal additive class, the maximal multiplicative class and the maximal classes with respect to maximum and minimum.

Theorem 3.3

Let X be a topological space. Then ${\mathcal M_a}({{\mathcal U}usc}(X))={{\mathcal U}usc}(X)$.

Proof

Observe that, by Proposition 3.2, $f \in {\mathcal M_a}({{\mathcal U}usc}(X))$ if and only if $-f \in {\mathcal M_a}({{\mathcal U}lsc}(X))$. Using Theorem 2.5 and again Proposition 3.2, we conclude that ${\mathcal M_a}({{\mathcal U}usc}(X))={{\mathcal U}usc}(X)$. $\square$

The next theorem follows from Proposition 3.2.

Theorem 3.4

Let X be a topological space. Then ${\mathcal M_m}({\mathcal U}{usc}(X)) = {\mathcal M_m}({\mathcal U}{lsc}(X))$.

Theorem 3.5

Let X be a topological space. Then ${\mathcal M_{\min }}({{\mathcal U}usc}(X))={{\mathcal U}usc}(X)$.

Proof

Since $-\min \{f,g\} = \max \{-f, -g\}$ for each functions $f, g \in {\mathbb R}^X$, by Proposition 3.2, we conclude that $f \in {\mathcal M_{\min }}({{\mathcal U}usc}(X))$ if and only if $-f \in {\mathcal M_{\max }}({{\mathcal U}lsc}(X))$. Now, using Theorem 2.10 and again Proposition 3.2, we obtain that ${\mathcal M_{\min }}({{\mathcal U}usc}(X))={{\mathcal U}usc}(X)$. $\square$

It is easy to see that using Theorem 2.11, Remark 1 and the equivalence $f \in {\mathcal M_{\max }}({{\mathcal U}usc}(X))$ if and only if $-f \in {\mathcal M_{\min }}({{\mathcal U}lsc}(X))$, we conclude that:

Theorem 3.6

Let X be a topological space. Then ${\mathcal M_{\max }}({\mathcal U}{usc}(X)) = {\mathcal M_{\min }}({\mathcal U}{lsc}(X))$.