Returning functions with closed graph are continuous

TL;DR

This paper proves that on path-inductive spaces, returning functions with closed graphs are continuous, extending classical results and solving a problem related to weakly Swiatkowski functions.

Contribution

It establishes a new characterization of continuity for returning functions on path-inductive spaces, linking closed graphs to continuity.

Findings

Returning functions with closed graphs are continuous on path-inductive spaces.

The result applies to weakly Swiatkowski functions on real numbers.

Answers a problem posed in the Lviv Scottish Book.

Abstract

A function $f:X\to \mathbb R$ defined on a topological space $X$ is called returning if for any point $x\in X$ there exists a positive real number $M_x$ such that for every path-connected subset $C_x\subset X$ containing the point $x$ and any $y\in C_x\setminus\{x\}$ there exists a point $z\in C_x\setminus\{x,y\}$ such that $|f(z)|\le \max\{M_x,|f(y)|\}$. A topological space $X$ is called path-inductive if a subset $U\subset X$ is open if and only if for any path $\gamma:[0,1]\to X$ the preimage $\gamma^{-1}(U)$ is open in $[0,1]$. The class of path-inductive spaces includes all first-countable locally path-connected spaces and all sequential locally contractible space. We prove that a function $f:X\to \mathbb R$ defined on a path-inductive space $X$ is continuous if and only of it is returning and has closed graph. This implies that a (weakly) \'Swi\c atkowski function $f:\mathbb R\to\mathbb R$ is continuous if and only if it has closed graph, which answers a problem of Maliszewski, inscibed to Lviv Scottish Book.

Equations

$${\Gamma }_f:=\{⟨x,f(x)⟩:x\in X\}$$

$$\underset{x\to b-0}{\mathrm{liminf}}|f(x)|=\infty .$$

$$H(x,t)=\begin{cases}h(x,1-t)&\mbox{if $t\in[0,1]$}\\ h(x,t-1)&\mbox{if $t\in[1,2]$}\end{cases}\mbox{ \ for $(x,t)\in V\times[0,2]$},$$

$$\gamma ([0,\frac{1}{2^m}])\subset \{x_{\omega }\}\cup \bigcup _{n\ge m}H(\{x_n\}\times [0,2])\subset H(U\times [0,2])\subset W,$$

$$X=\mathbb{R}$$

Full text

Let $X$ and $Y$ be a topological spaces. We say that function $f\colon X\rightarrow Y$ has closed graph if its graph

[TABLE]

is closed in the product $X\times Y$. It is well-known that each continuous function $f:X\to Y$ to a Hausdorff topological space $Y$ has closed graph. Trivial examples show that the converse statement is not true in general.

There exist many (algebraic or topological) properties of functions, which being combined with the closedness of the graph yield the continuity, see Tao’s blog [21]. For example, by the classical Closed Graph Theorem, a linear operator between Banach spaces is continuous if and only if it has closed graph.

Known topological properties implying the continuity of functions with closed graph include the Darboux property, the subcontinuity, the almost continuity, the near continuity, the $B$-quasicontinuity, the weak Gibson property, etc. (see [1], [3], [4], [5], [6], [7], [8], [13], [16], [17], [18], [19], [22]).

The inspiration for our present investigation was a problem, written by the third author to the Lviv Scotish Book111http://www.math.lviv.ua/szkocka/ on 01.05.2018. Actually, the problem was originally asked several years ago by Aleksander Maliszewski: Is any Świa̧tkowski function with closed graph continuous? Our main theorem gives an affirmative answer to this question and a posteriori to many its modifications (for peripherally continuous functions, peripherally bounded functions, Darboux functions, etc.)

One of the most general properties, responsible for the continuity of real-valued functions with closed graph is introduced in the following definition.

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returning at a point $x\in X$ if there exists a positive real number $M_{x}$ such that for every path-connected subset $C_{x}\subset X$ containing the point $x$ and any $y\in C_{x}\setminus\{x\}$ there exists a point $z\in C_{x}\setminus\{x,y\}$ such that $|f(z)|\leq\max\{M_{x},|f(y)|\}$;

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returning if $f$ is returning at each point.

The following theorem is the main result of this paper.

For the proof of this theorem we shall need some auxiliary notions and results.

For two points $a<b$ on the real line let $[a,b]$ and $(a,b)$ be the closed and open intervals with end-points $a,b$, respectively. To avoid confusion between notations for open intervals and ordered pairs, an ordered pair of real numbers $a,b$ will be denoted by $\langle a,b\rangle$.

A function $f:X\to\mathbb{R}$ defined on a topological space $X$ is called

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weakly discontinuous if for any non-empty closed subset $A\subset\mathbb{R}$ the set $C(f{\restriction}A)$ of continuity points of the restriction $f{\restriction}A$ has non-empty interior in $A$;

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locally bounded at a point $x\in X$ if $x$ has a neighborhood $U_{x}\subset X$ such that the set $f(U_{x})$ is bounded in $\mathbb{R}$.

The following lemma is known (see [2], [6]) and its proof is included for the convenience of the reader.

1. (1)

$f$* is weakly discontinuous;* 2. (2)

$f$* is continuous at a point $x\in X$ if and only if $f$ is locally bounded at $x$;* 3. (3)

the set $C(f)$ of continuity points is open and dense in $X$.

Assuming that $f$ is locally bounded at some point $x\in X$, we can find a bounded neighborhood $U_{x}\subset X$ such that $f(U_{x})$ is bounded and hence $U_{x}\cup f(U_{x})\subset(-n,n)$. Then $U_{x}\subset K_{n}$ and the continuity of $f{\restriction}U_{x}$ follows from the continuity of $f{\restriction}K_{n}$.

The weak discontinuity of $f$ implies the density of the set $C(f)$ in $X$. To see that $C(f)$ is open in $X$, take any continuity point $x\in X$ of $f$ and find a neighborhood $U_{x}\subset X$ whose image $f(U_{x})$ is bounded in the real line. By Lemma 1(2), $f{\restriction}U_{x}$ is continuous and hence $U_{x}\subset C(f)$. ∎

Let $\mathcal{C}$ be the family of connected components of the set $C(f)$. It follows that each set $C\in\mathcal{C}$ is an open connected subset of the real line.

To derive a contradiction, assume that $f|(-\infty,b]$ is not locally bounded at $b$. We claim that

[TABLE]

Assuming that $\liminf_{x\to b-0}|f(x)|<\infty$, we can fix any $M>\max\{|f(b)|,\liminf_{x\to b-0}|f(x)|\}$ and conclude that for every $\varepsilon>0$ the interval $(b-\varepsilon,b)$ contains a point $x$ such that $|f(x)|<M$.

Since the graph $\Gamma_{f}$ is closed and the points $\langle b,M\rangle$ and $\langle b,-M\rangle$ do not belong to $\Gamma_{f}$, there exists $\varepsilon>0$ such that the set $[b-\varepsilon,b]\times\{-M,M\}$ is disjoint with $\Gamma_{f}$. Since $f{\restriction}(-\infty,b]$ is not locally bounded at $b$, there exists a point $y\in(b-\varepsilon,b)$ such that $|f(y)|>M>\liminf_{x\to b-0}|f(x)|$. Now the definition of $\liminf_{x\to b-0}|f(x)|<M$ yields a point $x\in(y,b)$ such that $|f(x)|<M<|f(y)|$. By the Mean Value Theorem, for some point $z\in[x,y]\subset(-\infty,b)\subset C(f)$ we have $|f(z)|=M$ and hence \langle z,f(z)\rangle\in\Gamma_{f}\cap\big{(}[b-\varepsilon,b]\times\{-M,M\}\big{)}, which contradicts the choice of $\varepsilon$. This contradiction completes the proof of the equality $\liminf_{x\to b-0}|f(x)|=\infty$.

Since the function $f$ is returning at $b$, there exists a positive real constant $M$ such that for every $\lambda<b$ there exists a point $x\in(\lambda,b)$ such that $|f(x)|\leq\max\{M,|f(\lambda)|\}$. Since $\liminf_{x\to b-0}|f(x)|=\infty$, there exists a point $c\in(-\infty,b)$ such that $|f(c)|>M$. Taking into account that the function $f{\restriction}(-\infty,b)$ is continuous and tends to infinity at $b$, we conclude that the set $L:=\{x\in[c,b]:|f(x)|\leq|f(c)|\}$ is compact and hence has the largest element $\lambda\in L$. By the choice of the constant $M$, there exists a point $x\in(\lambda,b)$ such that $|f(x)|\leq\max\{M,|f(\lambda)|\}\leq|f(c)|$. Then $x\in L$ and hence $x\leq\lambda$, which contradicts the inclusion $x\in(\lambda,b)$. This contradiction completes the proof of the continuity of $f{\restriction}(-\infty,b]$.

By analogy we can consider the cases $C=(a,+\infty)$ and $C=(a,b)$ for some $a<b$. ∎

Claim 2 implies that the complement $F=\mathbb{R}\setminus C(f)$ has no isolated points. By Lemma 1, the function $f$ is weakly discontinuous, so there exist points $a<b$ such that $\emptyset\neq[a,b]\cap F\subset C(f{\restriction}F)$. By the continuity of $f$ on the compact set $K:=[a,b]\cap F$, there exists a number $M\in\mathbb{N}$ such that $|f(x)|<M$ for all $x\in K$. The restriction $f{\restriction}[a,b]$ is discontinuous and hence unbounded. Then we can choose a sequence $(x_{n})_{n\in\omega}$ of points of $[a,b]$ such that $|f(x_{n})|>M+n$ for all $n\in\omega$. For every $n\in\omega$ let $C_{n}\in\mathcal{C}$ be the unique connected component of $C(f)$, containing the point $x_{n}$. Replacing $x_{n}$ by a suitable subsequence, we can assume that $|f(x_{n})|>\max\{|f(x)|:x\in[a,b]\cap\bigcup_{k<n}\bar{C}_{k}\}$ for all $n\in\omega$. This ensures that the components $C_{n}$ are pairwise disjoint and hence $\mathrm{diam}(C_{n})\to 0$.

Replacing $(x_{n})_{n\in\omega}$ by a suitable subsequence, we can assume that $C_{n}\subset[a,b]$ for all $n\in\omega$ and $(x_{n})_{n\in\omega}$ converges to some point $c\in[a,b]$, which is a discontinuity point of $f$ as $\lim_{n\to\infty}|f(x_{n})|=\infty\neq|f(c)|$. Then $|f(c)|<M$ by the choice of $M$. The graph $\Gamma_{f}$ of $f$ is closed and hence is disjoint with the set $[c-\varepsilon,c+\varepsilon]\times\{-M,M\}$ for some $\varepsilon>0$. For every $n\in\omega$ write the component $C_{n}$ as $C_{n}=(a_{n},b_{n})$ for some points $a_{n}<b_{n}$ in the set $[a,b]\cap F$. Since $x_{n}\to c$ and $|b_{n}-a_{n}|\to 0$, there exists $n\in\omega$ such that $(a_{n},b_{n})\subset[c-\varepsilon,c+\varepsilon]$.

Since $|f(a_{n})|<M<|f(x_{n})|$, the Mean Value Theorem applied to the continuous function $f{\restriction}[a_{n},x_{n}]$ yields a point $z_{n}\in[a_{n},z_{n}]$ with $|f(z_{n})|=M$. Then \langle z_{n},f(z_{n})\rangle\in\Gamma_{f}\cap\big{(}[c-\varepsilon,c+\varepsilon]\times\{-M,M\}\big{)}=\emptyset, which is a desired contradiction completing the proof of Theorem 1. ∎

Theorem 1 admits a generalization to real-valued functions defined on path-inductive topological spaces.

By a path in a topological space $X$ we understand any continuous function $\gamma:[0,1]\to X$.

We define a topological space $X$ to be path-inductive if a subset $U\subset X$ is open if and only if for any path $\gamma:[0,1]\to X$ the preimage $f^{-1}(U)$ is open in $[0,1]$.

By the sequentiality or the first-countability of $X$, there exists a sequence $\{x_{n}\}_{n\in\omega}\subset X\setminus A$ that converges to a point $x_{\omega}\in A$.

If $X$ is locally contractible, then there exists a neighborhood $V\subset X$ of the point $x_{\omega}$ and a continuous map $h:V\times[0,1]\to X$ such that $h(x,0)=x$ and $h(x,1)=x_{\omega}$ for all $x\in V$. Replacing $(x_{n})_{n\in\omega}$ by a suitable subsequence, we can assume that $\{x_{n}\}_{n\in\omega}\subset V$. Consider the homotopy $H:V\times[0,2]\to X$ defined by

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and observe that $H(x,0)=H(x,2)=x_{\omega}$ and $H(x,1)=x$ for every $x\in V$.

Define a path $\gamma:[0,1]\to X$ by $\gamma(0)=x_{\omega}$ and $\gamma(t)=H(x_{n},2^{n+2}t-2)$ where $n\in\omega$ is the unique number such that $\frac{1}{2^{n+1}}<t\leq\frac{1}{2^{n}}$. It is clear that the function $\gamma$ is continuous at each point $t\in(0,2]$. To see that $\gamma$ is continuous at [math], fix any neighborhood $W\subset X$ of $\gamma(0)=x_{\omega}$ and using the continuity of the function $H$ at points of the compact set $\{x_{\omega}\}\times[0,2]\subset H^{-1}(x_{\omega})\subset H^{-1}(W)$, find a neighborhood $U\subset V$ of the point $x_{\omega}$ such that $H(U\times[0,2])\subset W$. Since the sequence $(x_{n})_{n\in\omega}$ converges to $x_{\omega}$, there exists a number $m\in\omega$ such that $\{x_{n}\}_{n\geq m}\subset U$. Then

[TABLE]

witnessing that $\gamma$ is continuous at [math].

Next, assume that the space $X$ is first-countable and locally path-connected. In this case we can find a countable neighborhood base $\{U_{n}\}_{n\in\omega}$ at $x_{\omega}$ such that for every $n\in\omega$ and point $x\in U_{n+1}$ there exists a path $\gamma_{x}:[0,1]\to U_{n}$ such that $\gamma_{x}(0)=x$ and $\gamma_{x}(1)=x_{\omega}$. Since the sequence $(x_{k})_{k\in\omega}$ converges to $x_{\omega}$, for every $n\in\omega$ there exists a number $k_{n}\in\omega$ such that $x_{k_{n}}\in U_{n+1}$. By our assumption, there exists a path $\gamma_{n}:[\tfrac{1}{2^{n+1}},\frac{1}{2^{n}}]\to U_{n}$ such that $\gamma_{n}(\frac{1}{2^{n+1}})=\gamma_{n}(\frac{1}{2^{n}})=x_{\omega}$ and $\gamma_{n}(\frac{3}{2^{n+2}})=x_{k_{n}}$. The paths $\gamma_{n}$, $n\in\omega$, compose a continuous path $\gamma:[0,1]\to X$ such that $\gamma(0)=x_{\omega}$ and $\gamma{\restriction}[\frac{1}{2^{n+1}},\frac{1}{2^{n}}]=\gamma_{n}{\restriction}[\frac{1}{2^{n+1}},\frac{1}{2^{n}}]$ for all $n\in\omega$.

Taking into account that $0\in\gamma^{-1}(x_{\omega})\subset\gamma^{-1}(A)$ and $\frac{3}{2^{n+2}}\in\gamma^{-1}(x_{k_{n}})\subset\gamma^{-1}(X\setminus A)=X\setminus\gamma^{-1}(A)$ for all $n\in\omega$, we see that the set $\gamma^{-1}(A)$ is not open in $[0,1]$. ∎

Theorem 1 implies the following its self-generalization.

Theorems 1 and 2 motivate studying returning functions in more details. We shall show that the class of returning functions on the real line is quite wide and contains many known classes of real-valued functions possessing some generalized continuity properties.

Let us recall that a function $f:X\to\mathbb{R}$ on a topological space $X$ is called

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Świa̧tkowski if for any connected subset $C\subset X$ and points $a,b\in C$ with $f(a)<f(b)$ there exists a continuity point $x\in C$ of $f$ such that $f(a)<f(x)<f(b)$;

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weakly Świa̧tkowski if for any connected subset $C\subset X$ and points $a,b\in C$ with $f(a)<f(b)$ there exists a point $x\in C$ such that $f(a)<f(x)<f(b)$;

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Darboux if for any connected subset $C\subset X$ the image $f(C)$ is connected;

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almost continuous (or else nearly continuous) if for any open set $V\subset X$ the preimage $f^{-1}(V)$ is contained in the interior of $\overline{f^{-1}(V)}$;

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quasicontinuous if for each point $x\in X$, neighborhood $V\subset Y$ of $f(x)$ and neighborhood $O_{x}\subset X$ of $x$, there exists a non-empty open set $G\subset O_{x}$ with $f(G)\subset V$;

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$B$-quasicontinuous if for each point $x\in X$, neighborhood $V\subset Y$ of $f(x)$ and an open connected set $O\subset X$ with $x\in\overline{O}$, there exists a non-empty open set $G\subset O$ with $f(G)\subset V$;

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(weakly) Gibson if $f(\overline{O})\subset\overline{f(O)}$ for any open (connected) subset $O\subset X$;

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peripherally continuous if for any point $x\in X$ and neighborhoods $O_{x}\subset X$ and $O_{f(x)}\subset\mathbb{R}$ of $x$ and $f(x)$ there exists a neighborhood $V_{x}\subset O_{x}$ of $x$ such that $f(\partial V_{x})\subset O_{f(x)}$.

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peripherally bounded if for any point $x\in X$ there exists a bounded set $B\subset\mathbb{R}$ such that for any neighborhood $O_{x}\subset X$ of $x$ there exists a neighborhood $V_{x}\subset O_{x}$ of $x$ such that $f(\partial V_{x})\subset B$.

Here by $\partial V_{x}$ we denote the boundary of $V_{x}$ in the topological space $X$. For more information on these classes of functions, see the survey of Gibson and Natkaniec [11].

For any function $f:X\to\mathbb{R}$ these notions relate as follows (by a simple arrow we denote the implications holding under the additional assumption $X=\mathbb{R}$):

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Since each (weakly) Świa̧tkowski function is returning, Theorem 1 implies the following corollary that answers the original problem of Maliszewski.

It is interesting to compare Theorems 1 and 2 to the following known results on the continuity of functions with closed graphs.

1. (1)

$f$* is almost continuous (Long and McGehee [16]);* 2. (2)

$f$* is nearly continuous and $X$ is Baire (Moors [18]);* 3. (3)

$f$* is Darboux and $X$ is locally connected (Wójcik [23, Corollary 18]);* 4. (4)

$f$* is peripherally continuous and $X=\mathbb{R}$ (Hagan [12]);* 5. (5)

$X=\mathbb{R}$* and $f$ is bilaterally quasicontinuous (Doboš [9]);* 6. (6)

$f$* is $B$-quasicontinuous and $X$ is locally connected (Borsik [5]);* 7. (7)

$f$* is weakly Gibson (Das and Nesterenko [8]).*