1 Introduction

Basing on an observation that every continuous map of a convex set satisfying a restricted Lipschitz condition must be Lipschitz continuous (see [12]), we present some generalizations of Browder–Göhde–Kirk fixed point (Browder [1], Göhde [7], Kirk [9], see also [4, 6, 15, 16]), and propose their applications, including an extension of the classical Radamacher theorem.

Section 2 contains the auxiliary results characterizing the Lipschitz continuous functions with the aid of some weaker conditions.

In Sect. 3, we present two fixed point theorems. Let X be a uniformly convex Banach space, \(C\subset X\) a nonempty bounded convex closed set, and T a selfmapping of C. Theorem 1 says that T has a fixed point, if for some function \(\beta :\left( 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) satisfying the conditions

$$\begin{aligned} \limsup _{t\rightarrow 0+}\frac{\beta \left( t\right) }{t}<+\infty ,\quad \liminf _{t\rightarrow 0+}\frac{\beta \left( t\right) }{t}=1, \end{aligned}$$

we have

$$\begin{aligned} \left\| Tx-Ty\right\| \le \beta \left( \left\| x-y\right\| \right) ,\quad x,y\in C,\ x\ne y. \end{aligned}$$

The second result, Theorem 2, says that the last inequality can be significantly weakened, namely, T has a fixed point, if T is continuous and, for a certain function \(\beta :\left( 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) and a zero sequence \(\left( t_{n}\right) \) of positive numbers such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\beta \left( t_{n}\right) }{t_{n}}=1, \end{aligned}$$

we have, for all \(n\in N\) and for all \(x,y\in C,\)

$$\begin{aligned} \left\| x-y\right\| =t_{n}\Longrightarrow \left\| T\left( x\right) -T\left( y\right) \right\| \le \beta \left( \left\| x-y\right\| \right) . \end{aligned}$$

In Sect. 4, we use Theorem 1 to get a result on the existence and uniqueness of \(L^{p}\)-solutions (\(1<p<+\infty \)) of the iterative functional equation

$$\begin{aligned} \varphi \left( x\right) =h\left( x,\varphi \left[ f\left( x\right) \right] \right) . \end{aligned}$$

In Sect. 5, we give some refinements of the classical Radamacher theorem on the differentiability of the Lipschitz mappings.

2 Some auxiliary results on Lipschitz continuity

We begin with the following

Lemma 1

Let X, Y be normed spaces,  \(C\subset X\) a convex set,  \(T:C\rightarrow Y\) a mapping,  and \(\beta :\left( 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) a real function such that

$$\begin{aligned} \left\| Tx-Ty\right\| \le \beta \left( \left\| x-y\right\| \right) ,\quad x,y\in C,\ x\ne y. \end{aligned}$$
(1)

If

$$\begin{aligned} \limsup _{t\rightarrow 0+}\frac{\beta \left( t\right) }{t}<+\infty , \end{aligned}$$
(2)

then

$$\begin{aligned} \left\| Tx-Ty\right\| \le L\left\| x-y\right\| , \quad x,y\in C, \end{aligned}$$

where

$$\begin{aligned} L:=\liminf _{t\rightarrow 0+}\frac{\beta \left( t\right) }{t}. \end{aligned}$$
(3)

Proof

Note that conditions (2) and (1) imply that T is continuous. Indeed, from (2) there are some real positive M and \(\delta \) such that \(\beta \left( t\right) \le Mt\) for all \(t\in \left( 0,\delta \right) ,\) and (1) implies that \(\left\| Tx-Ty\right\| \le M\left\| x-y\right\| \) for all \(x,y\in C\) such that \(\left\| x-y\right\| <\delta .\)

Take arbitrary \(x,y\in C,\) \(x\ne y.\) By (3), for every \(\varepsilon >0\) there is a \(t_{\varepsilon }>0\) and a unique \(n=n_{\varepsilon }\in {\mathbb {N}}_{0}\) such that

$$\begin{aligned}&\frac{\beta \left( t_{\varepsilon }\right) }{t_{\varepsilon }}\le L+\varepsilon ,\end{aligned}$$
(4)
$$\begin{aligned}&0\le \left\| x-y\right\| -n\left( \varepsilon \right) t_{\varepsilon }<t_{\varepsilon }, \end{aligned}$$
(5)

and

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}t_{\varepsilon }=0. \end{aligned}$$
(6)

Put

$$\begin{aligned} z_{k}=x+kt_{\varepsilon }\frac{y-x}{\left\| y-x\right\| },\quad k=0,1,\ldots ,n\left( \varepsilon \right) . \end{aligned}$$

Then, by the convexity of C, 

$$\begin{aligned} z_{k}\in C,\quad k=0,1,\ldots ,n\left( \varepsilon \right) ; \end{aligned}$$

clearly

$$\begin{aligned} \left\| z_{k}-z_{k+1}\right\| =t_{\varepsilon },\quad k=0,1,\ldots ,n\left( \varepsilon \right) ; \end{aligned}$$
(7)

and, by (5),

$$\begin{aligned} \left\| z_{n\left( \varepsilon \right) }-y\right\| =\left( \left\| y-x\right\| -nt_{\varepsilon }\right) <t_{\varepsilon }. \end{aligned}$$
(8)

Hence, applying in turn: the triangle inequality, condition (1), some obvious identities, (4), (7) and (5), we get

$$\begin{aligned} \left\| Tx-Ty\right\|= & {} \left\| \sum _{k=0}^{n\left( \varepsilon \right) -1}\left( Tz_{k}-Tz_{k+1}\right) +\left( Tz_{n\left( \varepsilon \right) }-Ty\right) \right\| \\\le & {} \sum _{k=0}^{n\left( \varepsilon \right) -1}\left\| Tz_{k}-Tz_{k+1}\right\| +\left\| Tz_{n\left( \varepsilon \right) }-Ty\right\| \\\le & {} \sum _{k=0}^{n\left( \varepsilon \right) -1}\beta \left( \left\| z_{k}-z_{k+1}\right\| \right) +\left\| Tz_{n\left( \varepsilon \right) }-Ty\right\| \\= & {} \sum _{k=0}^{n\left( \varepsilon \right) -1}\beta \left( t_{\varepsilon }\right) +\left\| Tz_{n\left( \varepsilon \right) }-Ty\right\| \\= & {} n\left( \varepsilon \right) \beta \left( t_{\varepsilon }\right) +\left\| Tz_{n\left( \varepsilon \right) }-Ty\right\| \\= & {} \frac{\beta \left( t_{\varepsilon }\right) }{t_{\varepsilon }}\left( n\left( \varepsilon \right) t_{\varepsilon }\right) +\left\| Tz_{n\left( \varepsilon \right) }-Ty\right\| \\\le & {} \left( L+\varepsilon \right) n\left( \varepsilon \right) t_{\varepsilon }+\left\| Tz_{n\left( \varepsilon \right) }-Ty\right\| \\\le & {} \left( L+\varepsilon \right) \left\| x-y\right\| +\left\| Tz_{n\left( \varepsilon \right) }-Ty\right\| , \end{aligned}$$

that is

$$\begin{aligned} \left\| Tx-Ty\right\| \le \left( L+\varepsilon \right) \left\| x-y\right\| +\left\| Tz_{n\left( \varepsilon \right) }-Ty\right\| . \end{aligned}$$

Since the continuity of T and the conditions (6) and (8) imply that

$$\begin{aligned} \lim _{\varepsilon \rightarrow 0}\left\| Tz_{n\left( \varepsilon \right) }-Ty\right\| =0, \end{aligned}$$

letting \(\varepsilon \rightarrow 0\) in the above inequality, we obtain

$$\begin{aligned} \left\| Tx-Ty\right\| \le L\left\| x-y\right\| , \end{aligned}$$

which completes the proof. \(\square \)

Remark 1

If \(\beta :\left( 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) is subadditive, then

$$\begin{aligned} \limsup _{t\rightarrow 0+}\frac{\beta \left( t\right) }{t}=\liminf _{t \rightarrow 0+}\frac{\beta \left( t\right) }{t} \end{aligned}$$

(see [8, p. 250, Theorem 7.11.1], also [11]). In this case, instead of (3) it is enough to assume that \(L<+\infty .\)

Remark 2

The reasoning in the proof of Lemma 1 simplifies, if

$$\begin{aligned} \lim _{t\rightarrow 0+}\frac{\beta \left( t\right) }{t}=L. \end{aligned}$$

Indeed, if this condition holds, then for every \(\varepsilon >0\) there is a \( \delta >0\) such that

$$\begin{aligned} \beta \left( t\right) \le \left( L+\varepsilon \right) t,\quad t\in \left( 0,\delta \right) . \end{aligned}$$

Take arbitrary \(x,y\in C,\) \(x\ne y,\) choose \(n\in {\mathbb {N}}\) such that

$$\begin{aligned} \frac{\left\| x-y\right\| }{n}<\delta , \end{aligned}$$

and put

$$\begin{aligned} z_{k}:=x+\frac{k}{n}\left( y-x\right) ,\quad k=0,1,\ldots ,n. \end{aligned}$$

Of course

$$\begin{aligned}&\left\| z_{k}-z_{k+1}\right\| =\frac{\left\| x-y\right\| }{n},\quad k=0,1,\ldots ,n-1; \\&z_{0} =x,\quad z_{k}=y, \end{aligned}$$

and, by the convexity of C, 

$$\begin{aligned} z_{k}\in C,\quad k=0,1,\ldots ,n. \end{aligned}$$

Applying the triangle inequality and (1), we hence get

$$\begin{aligned} \left\| Tx-Ty\right\|= & {} \left\| \sum _{k=0}^{n-1}\left( Tz_{k}-Tz_{k+1}\right) \right\| \le \sum _{k=0}^{n-1}\left\| Tz_{k}-Tz_{k+1}\right\| \le \sum _{k=0}^{n-1}\beta \left( \left\| z_{k}-z_{k+1}\right\| \right) \\= & {} \sum _{k=0}^{n-1}\beta \left( \frac{\left\| x-y\right\| }{n}\right) \le \sum _{k=0}^{n-1}\left( L+\varepsilon \right) \frac{\left\| x-y\right\| }{n}=\left( L+\varepsilon \right) \left\| x-y\right\| , \end{aligned}$$

that is \(\left\| Tx-Ty\right\| \le \left( L+\varepsilon \right) \left\| x-y\right\| .\) Since \(\varepsilon >0\) is chosen arbitrarily, letting \(\varepsilon \rightarrow 0,\) we conclude that T is L-Lipschitzian.

A much weaker necessary and sufficient condition for a continuous map to be Lipschitz continuous gives the following

Lemma 2

Let X and Y be real normed spaces and \(C\subset X\) a bounded convex set. Suppose that \(T:C\rightarrow Y\) is continuous. If there are a nonnegative real L and two positive sequences \(\left( t_{n}\right) ,\left( c_{n}\right) ,\)

$$\begin{aligned} \lim _{n\rightarrow \infty }t_{n}=0,\quad \lim _{n\rightarrow \infty }c_{n}=L, \end{aligned}$$

such that for every \(n\in {\mathbb {N}}\) and for all \(x,y\in C,\)

$$\begin{aligned} \left\| x-y\right\| =t_{n}\Longrightarrow \left\| T\left( x\right) -T\left( y\right) \right\| \le c_{n}t_{n}, \end{aligned}$$
(9)

then T is Lipschitz continuous,  and

$$\begin{aligned} \left\| Tx-Ty\right\| \le L\left\| x-y\right\| , \quad x,y\in C. \end{aligned}$$

Proof

Take arbitrary \(x,y\in C,\) \(x\ne y.\) For every \(n\in {\mathbb {N}}\), there is a unique \(m_{n}\in {\mathbb {N}}\cup \left\{ 0\right\} \) such that

$$\begin{aligned} m_{n}t_{n}\le \left\| x-y\right\| <\left( m_{n}+1\right) t_{n}. \end{aligned}$$

Put

$$\begin{aligned} z_{k}:=x+\frac{kt_{n}}{\left\| y-x\right\| }\left( y-x\right) ,\quad k=0,1,\ldots ,m_{n}. \end{aligned}$$
(10)

Since

$$\begin{aligned} 0\le \frac{m_{n}t_{n}}{\left\| y-x\right\| }\le 1, \end{aligned}$$

and, for each \(k=0,1,\ldots ,m_{n},\)

$$\begin{aligned} z_{k}=\left( 1-\frac{kt_{n}}{\left\| y-x\right\| }\right) x+\frac{ kt_{n}}{\left\| y-x\right\| }\left( y-x\right) , \end{aligned}$$

the convexity of C implies that

$$\begin{aligned} z_{k}\in C,\quad k=0,1,\ldots ,m_{n}. \end{aligned}$$

Moreover, by (10),

$$\begin{aligned} \left\| z_{k}-z_{k+1}\right\| =t_{n},\quad k=0,1,\ldots ,m_{n}-1, \end{aligned}$$
(11)

and, for \(k=m_{n},\)

$$\begin{aligned} z_{m_{n}}-y=\left( x+\frac{kt_{n}}{\left\| y-x\right\| }\left( y-x\right) \right) -y=\left( \frac{kt_{n}}{\left\| y-x\right\| } -1\right) \left( y-x\right) , \end{aligned}$$

we have

$$\begin{aligned} \left\| z_{m_{n}}-y\right\| =\left\| y-x\right\| -m_{n}t_{n}<t_{n}. \end{aligned}$$
(12)

From (11) and (9), we get

$$\begin{aligned} \left\| Tz_{k}-Tz_{k+1}\right\| \le c_{n}t_{n},\quad k=0,1,\ldots ,m_{n}-1, \end{aligned}$$

so, by the triangle inequality,

$$\begin{aligned} \left\| Tx-Ty\right\|= & {} \left\| \sum _{k=0}^{m_{n}-1}\left( Tz_{k}-Tz_{k+1}\right) +\left( Tz_{m_{n}}-Ty\right) \right\| \\\le & {} \sum _{k=0}^{m_{n}-1}\left\| Tz_{k}-Tz_{k+1}\right\| +\left\| Tz_{m_{n}}-Ty\right\| \\\le & {} m_{n}c_{n}t_{n}+\left\| Tz_{m_{n}}-Ty\right\| \\= & {} c_{n}\left( m_{n}t_{n}\right) +\left\| Tz_{m_{n}}-Ty\right\| \end{aligned}$$

whence, taking into account that \(m_{n}t_{n}\le \left\| x-y\right\| \), by (12), we get

$$\begin{aligned} \left\| Tx-Ty\right\| \le c_{n}\left\| x-y\right\| +\left\| Tz_{m_{n}}-Ty\right\| . \end{aligned}$$
(13)

Since, by (12), \(\left\| z_{m_{n}}-y\right\| <t_{n},\) we have

$$\begin{aligned} \lim _{n\rightarrow \infty }\left\| z_{m_{n}}-y\right\| =0, \end{aligned}$$

and, in view of the assumed continuity of T, 

$$\begin{aligned} \lim _{n\rightarrow \infty }\left\| Tz_{m_{n}}-Ty\right\| =0. \end{aligned}$$

Hence, letting \(n\rightarrow \infty \) in (13), and taking into account that \( \lim _{n\rightarrow \infty }c_{n}=L,\) we conclude that

$$\begin{aligned} \left\| Tx-Ty\right\| \le L\left\| x-y\right\| , \end{aligned}$$

which was to be shown. \(\square \)

3 Fixed-point theorems

Recall that a real normed vector space \(\left( X,\left\| \cdot \right\| \right) \) is called uniformly convex, if for every \( \varepsilon \in \left( 0,2\right] \) there is some \(\delta >0\) such that for any two vectors \(x,y\in X\) with \(\left\| x\right\| =\left\| y\right\| =1,\) the condition \(\left\| x-y\right\| \ge \varepsilon \) implies that \(\left\| \frac{x+y}{2}\right\| \le 1-\delta \) (Goebel and Reich [6]; see also [13] for a generalization).

Applying Lemma 1 with \(L=1\) we obtain the following generalization of the Browder–Göhde–Kirk theorem.

Theorem 1

Let X be a uniformly convex Banach space,  \(C\subset X\) a nonempty bounded convex closed set and T a selfmapping of C. If there exists a function \(\beta :\left( 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) such that

$$\begin{aligned} \left\| Tx-Ty\right\| \le \beta \left( \left\| x-y\right\| \right) ,\quad x,y\in C,\ x\ne y, \end{aligned}$$

and

$$\begin{aligned} \limsup _{t\rightarrow 0+}\frac{\beta \left( t\right) }{t}<+\infty ,\quad \liminf _{t\rightarrow 0+}\frac{\beta \left( t\right) }{t}=1, \end{aligned}$$

then T has a fixed point in C.

Proof

Applying Lemma 1 with \(L=1,\) we get

$$\begin{aligned} \left\| Tx-Ty\right\| \le L\left\| x-y\right\| , \quad x,y\in C, \end{aligned}$$

that is T is nonexpansive, and the result follows from the original version of the Browder–Göhde–Kirk theorem. \(\square \)

In particular, the thesis of Browder–Göhde–Kirk theorem remains true, if the nonexpansivity of the mapping T is replaced for instance, by the inequality

$$\begin{aligned} \left\| Tx-Ty\right\| \le \exp \left( \left\| x-y\right\| \right) -1,\quad x,y\in C,\ x\ne y. \end{aligned}$$

Lemma 2 and the Browder–Göhde–Kirk theorem yield the following:

Proposition 1

Let X be a uniformly convex Banach space and \(C\subset X\) a nonempty bounded closed and convex set. Suppose that \(T:C\rightarrow C\) is continuous. If there exist two positive sequences \(\left( t_{n}\right) ,\left( c_{n}\right) ,\)

$$\begin{aligned} \lim _{n\rightarrow \infty }t_{n}=0,\quad \lim _{n\rightarrow \infty }c_{n}=1, \end{aligned}$$

such that for every \(n\in {\mathbb {N}}\) and for all \(x,y\in C,\)

$$\begin{aligned} \left\| x-y\right\| =t_{n}\Longrightarrow \left\| T\left( x\right) -T\left( y\right) \right\| \le c_{n}t_{n}, \end{aligned}$$

then T has a fixed point.

This proposition improves the relevant result in [11] where the uniform continuity of T is assumed.

The main result of this section reads as follows.

Theorem 2

Let X be a uniformly convex Banach space and \(C\subset X\) a nonempty bounded closed and convex set. Suppose that \(T:C\rightarrow C\) is continuous. If there exist a function \(\beta :\left( 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) and a sequence of positive real \( \left( t_{n}\right) ,\) \(\lim _{n\rightarrow \infty }t_{n}=0\) satisfying the condition

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\beta \left( t_{n}\right) }{t_{n}}=1, \end{aligned}$$

such that for every \(n\in {\mathbb {N}}\) and for all \(x,y\in C,\)

$$\begin{aligned} \left\| x-y\right\| =t_{n}\Longrightarrow \left\| T\left( x\right) -T\left( y\right) \right\| \le \beta \left( \left\| x-y\right\| \right) , \end{aligned}$$

then T has a fixed point.

Proof

Setting \(c_{n}:=\frac{\beta \left( t_{n}\right) }{t_{n}}\) we have \( \lim _{n\rightarrow \infty }c_{n}=1\) and for every \(n\in {\mathbb {N}}\) and for all \(x,y\in C,\) if \(\left\| x-y\right\| =t_{n},\) then

$$\begin{aligned} \left\| T\left( x\right) -T\left( y\right) \right\| \le \beta \left( t_{n}\right) =c_{n}t_{n}, \end{aligned}$$

and the result follows from Proposition 1. \(\square \)

4 An application in the theory of iterative functional equations

For a measure space \(\left( \Omega ,\Sigma ,\mu \right) \) and a real \(p>1,\) denote by \(\left( L^{p}\left( \Omega \right) ,\left\| \cdot \right\| _{p}\right) \) the Banach space of all (equivalence classes with respect to the \(\mu \)-a.e. equality) of \(\Sigma \)-measurable functions \(\varphi :\Omega \rightarrow {\mathbb {R}}\) such that \(\left| \varphi \right| ^{p}\) is \( \mu \)-integrable, and

$$\begin{aligned} \left\| \varphi \right\| _{p}=\left( \int _{\Omega }\left| \varphi \right| ^{p}d\mu \right) ^{1/p}. \end{aligned}$$

It is well known that \(\left( L^{p}\left( \Omega \right) ,\left\| \cdot \right\| _{p}\right) \) is a uniformly convex Banach space (Clarkson [3]).

In this section, we consider solutions \(\varphi \in L^{p}\left( \Omega \right) \) of the iterative-type functional equation

$$\begin{aligned} \varphi \left( x\right) =h\left( x,\varphi \left[ f\left( x\right) \right] \right) . \end{aligned}$$

We assume that \(\left( \Omega ,\Sigma ,\mu \right) \) the given functions f and h satisfy the following conditions:

  1. (i)

    \(k\in {\mathbb {N}}\); \(\Omega \subset {\mathbb {R}}^{k}\) is an open set; \(\mu \) is the Lebesgue measure, \(\mu \left( \Omega \right) =1;\) and \(f:\Omega \rightarrow \Omega ,\) \(f=\left( f_{1},\ldots ,f_{k}\right) ,\) is a locally Lipschitzian homeomorphic mapping;

  2. (ii)

    \(h:\Omega \times \mathbb {R\rightarrow R}\) is such that: for every \(y\in R\) the function \(\Omega \ni x\longmapsto h\left( x,y\right) \) is Lebesgue measurable, and \({\mathbb {R}}\ni y\longmapsto h\left( x,y\right) \) is continuous for almost all \( x\in \Omega \) (with respect to \(\mu );\)

  3. (iii)

    \(p\in {\mathbb {R}},\) \(p>1,\) and there are \(g_{1},g_{2}\in L^{p}\left( \Omega \right) ,\) \(g_{1}\le g_{2}\) a.e. in \(\Omega \) such that for all \(x\in \Omega \) and \(y\in {\mathbb {R}},\) the following implication holds true:

$$\begin{aligned} g_{1}\left( f\left( x\right) \right) \le y\le g_{2}\left( f\left( x\right) \right) \Longrightarrow g_{1}\left( x\right) \le h\left( x,y\right) \le g_{2}\left( x\right) . \end{aligned}$$

Applying Theorem 1, we prove the following:

Theorem 3

Let conditions (i)–(iii) be satisfied. Assume that there are a Lebesgue measurable function \(\alpha :\Omega \rightarrow \left[ 0,\infty \right) ,\) and a function \(\beta :\left[ 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) such that

$$\begin{aligned}&\left| h\left( x,y_{1}\right) \right| \le \alpha \left( x\right) \beta \left( \left| y_{1}-y_{2}\right| \right) ,\quad x\in \Omega , \ y_{1},y_{2}\in {\mathbb {R}};\end{aligned}$$
(14)
$$\begin{aligned}&\left[ \alpha \left( x\right) \right] ^{p}\le \left| J_{f}\left( x\right) \right| \quad \text {a.e. in }\Omega , \end{aligned}$$
(15)

where,  for \(f=\left( f_{1},\ldots ,f_{k}\right) \) and \(x=\left( x_{1},\ldots ,x_{k}\right) ,\) the symbol \(J_{f}\left( x\right) :=\frac{\partial \left( f_{1},\ldots ,f_{k}\right) }{\partial \left( x_{1},\ldots ,x_{k}\right) }\) stands for the Jacobian of f; 

the function

$$\begin{aligned} \left[ 0,\infty \right) \ni t\longmapsto \left[ \beta \left( t^{\frac{1}{p} }\right) \right] ^{p}\ \text {is concave}; \end{aligned}$$
(16)

and

$$\begin{aligned} \lim _{t\rightarrow 0}\frac{\beta \left( t\right) }{t}\le 1. \end{aligned}$$
(17)

Then there exists \(\varphi \in L^{p}\left( \Omega \right) ,\) \(g_{1}\le \varphi \le g_{2}\) a.e. in \(\Omega ,\) such that

$$\begin{aligned} \varphi \left( x\right) =h\left( x,\varphi \left[ f\left( x\right) \right] \right) \quad \text {a.e. for }x\in \Omega ; \end{aligned}$$
(18)

moreover,  if \(\beta \left( t_{n}\right) \ne t_{n}\) for a sequence of \( t_{n}>0,\) \(\lim _{n\rightarrow \infty }t_{n}=0,\) then such \(\varphi \) is unique,  and for any \(\varphi _{0}\in L^{p}\left( \Omega \right) ,\) \( g_{1}\le \varphi _{0}\le g_{2}\) a.e. in \(\Omega ,\) the sequence \(\left( \varphi _{n}\right) \) defined recursively by

$$\begin{aligned} \varphi _{n}\left( x\right) =h\left( x,\varphi _{n-1}\left[ f\left( x\right) \right] \right) \quad \text {a.e. for }x\in \Omega ; \ n\in {\mathbb {R}}, \end{aligned}$$

converges to \(\varphi \) in the norm \(\left\| \cdot \right\| _{p}.\)

Proof

Put

$$\begin{aligned} C:=\left\{ \varphi \in L^{p}\left( \Omega \right) :g_{1}\le \varphi \le g_{2}\text { a.e. in }\Omega \right\} . \end{aligned}$$

It is easy to see that C is a nonempty, convex and closed subset of \( L^{p}\left( \Omega \right) .\) Define the mapping T on C by

$$\begin{aligned} T\left( \varphi \right) \left( x\right) :=h\left( x,\varphi \left[ f\left( x\right) \right] \right) ,\quad x\in \Omega . \end{aligned}$$

Take an arbitrary \(\varphi \in C.\) Then, in view of Carathéodory theorem [2], conditions (ii) imply that the function \( T\left( \varphi \right) \) is Lebesgue measurable. Since \(g_{1}\le \varphi \le g_{2}\) a.e. in \(\Omega \) we have, for a.e. \(x\in \Omega \)

$$\begin{aligned} g_{1}\left[ f\left( x\right) \right] \le \varphi \left[ f\left( x\right) \right] \le g_{2}\left[ f\left( x\right) \right] , \end{aligned}$$

whence, in view of condition (iii),

$$\begin{aligned} g_{1}\left( x\right) \le h\left( x,\varphi \left[ f\left( x\right) \right] \right) \le g_{2}\left( x\right) \end{aligned}$$

for a.e. \(x\in \Omega ,\) that is \(T\left( \varphi \right) \in C,\) which proves that T maps C into itself.

Take arbitrary \(\varphi _{1},\varphi _{2}\in C.\) Making use in turn of: the definition of T;  (14) (we use here the measurability of \(\alpha \) and \(\beta \)); (15); the theorem on change of the variables under integral (see Łojasiewicz [10]), the inclusion \(f\left( \Omega \right) \subset \Omega ;\) an obvious equality; the assumption that the Lebesgue measure of \(\Omega \) is 1 and the Jensen integral inequality for the concave function (16); and the definition of the norm \(\left\| \cdot \right\| _{p},\) we obtain

$$\begin{aligned} \left\| T\left( \varphi _{1}\right) -T\left( \varphi _{2}\right) \right\| _{p}= & {} \left( \int _{\Omega }\left| h\left( x,\varphi _{1} \left[ f\left( x\right) \right] \right) -h\left( x,\varphi _{2}\left[ f\left( x\right) \right] \right) \right| ^{p}{\mathrm{d}}x\right) ^{1/p} \\\le & {} \left( \int _{\Omega }\left| \alpha \left( x\right) \beta \left( \left| \varphi _{1}\left[ f\left( x\right) \right] -\varphi _{2}\left[ f\left( x\right) \right] \right| \right) \right| ^{p}{\mathrm{d}}x\right) ^{1/p}\\\le & {} \left( \int _{\Omega }\left| J_{f}\left( x\right) \right| \left| \beta \left( \left| \varphi _{1}\left[ f\left( x\right) \right] -\varphi _{2}\left[ f\left( x\right) \right] \right| \right) \right| ^{p}{\mathrm{d}}x\right) ^{1/p} \\= & {} \left( \int _{f\left( \Omega \right) }\left[ \beta \left( \left( \left| \varphi _{1}\left( x\right) -\varphi _{2}\left( x\right) \right| \right) \right) \right] ^{p}{\mathrm{d}}x\right) ^{1/p} \\\le & {} \left( \int _{\Omega }\left[ \beta \left( \left| \varphi _{1}\left( x\right) -\varphi _{2}\left( x\right) \right| \right) \right] ^{p}{\mathrm{d}}x\right) ^{1/p} \\= & {} \left( \int _{\Omega }\left[ \beta \left( \left( \left| \varphi _{1}\left( x\right) -\varphi _{2}\left( x\right) \right| ^{p}\right) ^{ \frac{1}{p}}\right) \right] ^{p}{\mathrm{d}}x\right) ^{1/p} \\\le & {} \left( \left[ \beta \left( \left( \int _{\Omega }\left| \varphi _{1}\left( x\right) -\varphi _{2}\left( x\right) {\mathrm{d}}x\right| ^{p}\right) ^{ \frac{1}{p}}\right) \right] ^{p}\right) ^{1/p} \\= & {} \beta \left( \left\| \varphi _{1}-\varphi _{2}\right\| _{p}\right) , \end{aligned}$$

which, taking into account (17), proves that T satisfies the conditions of Theorem 1. Since \(\left( L^{p}\left( \Omega \right) ,\left\| \cdot \right\| _{p}\right) \) is a uniformly convex Banach space, all the assumptions of Theorem 1 are satisfied. Consequently, there is a function \( \varphi \in C\) such that \(\varphi =T\left( \varphi \right) .\)

If \(\beta \left( t_{n}\right) \ne t_{n}\) for a sequence of \(t_{n}>0\) such that \(\lim _{n\rightarrow \infty }t_{n}=0,\) then, by the concavity of \(\beta ,\) it is increasing and

$$\begin{aligned} \beta \left( t\right) <t,\quad t>0. \end{aligned}$$

Since \(\lim _{n\rightarrow \infty }\beta ^{n}\left( t\right) =0\) for every \( t>0,\) the “moreover” result follows from Theorem 1.2 in [14]. \(\square \)

In one-dimensional case, if \(\beta =\hbox {id}|_{\left[ 0,\infty \right) }\) and \(p\ge 1,\) the theory of \(L^{p}\)-solutions of the considered functional equation simplifies. Namely, from [14], Corollary 3.1 and Theorem 3.2, we have the following

Remark 3

[14] Let \(\Omega =\left( 0,a\right) \) where \(0<a\le \infty ,\) and \(p\ge 1.\) Assume that:

\(f:\Omega \rightarrow \Omega \) is absolutely continuous and

$$\begin{aligned} 0<f\left( x\right) <x,\quad x\in \Omega ; \end{aligned}$$

\(h:\Omega \times {\mathbb {R}}\rightarrow {\mathbb {R}}\) is such that for every \(y\in {\mathbb {R}}\) the function \(\Omega \ni x\longmapsto h\left( x,y\right) \) is Lebesgue measurable, the function \({\mathbb {R}}\ni y\longmapsto h\left( x,y\right) \) is continuous for almost all \(x\in \Omega ;\)

moreover, for some \(x_{0}\in \Omega \) and a function \(\alpha :\left( 0,x_{0}\right) \rightarrow \left[ 0,\infty \right) \); we have

$$\begin{aligned} \left| h\left( x,y_{1}\right) \right| \le \alpha \left( x\right) \left| y_{1}-y_{2}\right| ,\quad x\in \left( 0,x_{0}\right) , \ y_{1},y_{2}\in {\mathbb {R}}. \end{aligned}$$

Then

  1. (a)

    if for some \(x_{0}\in \Omega \) we have

    $$\begin{aligned} \alpha ^{p}\left( x\right) \le f^{\prime }\left( x\right) \quad \text {a.e. in }\left( 0,x_{0}\right) , \end{aligned}$$

    then there exists at most one solution \(\varphi \in L^{p}\left( \Omega \right) \) of (18);

  2. (b)

    if for some \(x_{0}\in \Omega \) and \(c\in \left[ 0,1\right) \) we have

    $$\begin{aligned} \alpha ^{p}\left( x\right) \le cf^{\prime }\left( x\right) \quad \text {a.e. in }\left( 0,x_{0}\right) , \end{aligned}$$

    then there exists exactly one solution \(\varphi \in L^{p}\left( \Omega \right) \) of Eq. (18).

5 A refinement of Radamacher’s theorem

Applying Lemma 1 we obtain the following refinement of the classical Radamacher’s theorem (see, for instance [5, Theorem 3.1.6], or [10, p.161]).

Theorem 4

Let \(\Omega \subset {\mathbb {R}}^{k}\) be an open convex set and \(f:\Omega \rightarrow {\mathbb {R}}^{m}\) for some \(k,m\in {\mathbb {N}}.\) If there is a function \(\beta :\left( 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) such that

$$\begin{aligned} \limsup _{t\rightarrow 0+}\frac{\beta \left( t\right) }{t}<+\infty , \end{aligned}$$

and

$$\begin{aligned} \left\| f\left( x\right) -f\left( y\right) \right\| \le \beta \left( \left\| x-y\right\| \right) ,\quad x,y\in \Omega ,\ x\ne y, \end{aligned}$$

where \(\left\| \cdot \right\| \) denotes the respective Euclidean norm,  then f is differentiable almost everywhere in \(\Omega ;\) that is,  the points in \(\Omega \) at which f is not differentiable form a set of Lebesgue measure zero.

Proof

Assume first that \(\Omega \) is convex. By Lemma 1, we have

$$\begin{aligned} \left\| f\left( x\right) -f\left( y\right) \right\| \le L\left\| x-y\right\| ,\quad x,y\in \Omega , \end{aligned}$$

where

$$\begin{aligned} L:=\liminf _{t\rightarrow 0+}\frac{\beta \left( t\right) }{t}<+\infty . \end{aligned}$$

In view of Radamacher’s theorem, the function f is differentiable almost everywhere in \(\Omega .\)

To end the proof, it is enough to note that every open set \(\Omega \subset {\mathbb {R}}^{k}\) is a countable sum of convex sets. \(\square \)

Similarly, making use of Lemma 2, we obtain the following improvement of Radamacher’s theorem.

Theorem 5

Let \(k,m\in {\mathbb {N}},\) \(\Omega \subset {\mathbb {R}}^{k}\) be an open convex set and \(f:\Omega \rightarrow {\mathbb {R}}^{m}\) be continuous. If for some function \(\beta :\left( 0,\infty \right) \rightarrow \left[ 0,\infty \right) \) there is a positive sequence \(\left( t_{n}\right) \) with \( \lim _{n\rightarrow \infty }t_{n}=0,\) such that

$$\begin{aligned} \lim _{n\rightarrow \infty }\frac{\beta \left( t_{n}\right) }{t_{n}}<\infty , \end{aligned}$$

the function f is such that for every \(n\in {\mathbb {N}}\) and for all \( x,y\in \Omega ,\)

$$\begin{aligned} \left\| x-y\right\| =t_{n}\Longrightarrow \left\| f\left( x\right) -f\left( y\right) \right\| \le \beta \left( \left\| x-y\right\| \right) , \end{aligned}$$

where \(\left\| \cdot \right\| \) denotes the Euclidean norm,  then f is differentiable almost everywhere in \(\Omega .\)

The results of this section show that condition (i) in Theorem 3 can be replaced by a significantly weaker one.