In this section, let X and Y be real normed spaces and let V and W be real vector spaces. In the following theorem, we prove that if, for any given mapping f, there exists a mapping F (near f) with some properties possessed by cubic-quadratic-additive mappings, then the mapping F must be uniquely determined.
Theorem 2.1
Let
\(a > 1\)
be a real constant, let
\(\Phi: V \backslash\{ 0 \} \to[0,\infty)\)
be a function satisfying one of the following conditions:
$$\begin{aligned}& \lim_{n \to\infty} \frac{\Phi ( a^{n}x )}{a^{n}} = 0, \end{aligned}$$
(1)
$$\begin{aligned}& \lim_{n \to\infty} a^{n} \Phi \biggl( \frac{x}{a^{n}} \biggr) = \lim_{n \to\infty} \frac{\Phi ( a^{n}x )}{a^{2n}} = 0, \end{aligned}$$
(2)
$$\begin{aligned}& \lim_{n \to\infty} a^{2n} \Phi \biggl( \frac{x}{a^{n}} \biggr) = \lim_{n \to\infty} \frac{\Phi ( a^{n}x )}{a^{3n}} = 0, \end{aligned}$$
(3)
$$\begin{aligned}& \lim_{n \to\infty} a^{3n} \Phi \biggl( \frac{x}{a^{n}} \biggr) = 0 \end{aligned}$$
(4)
for all
\(x \in V \backslash\{ 0 \}\), and let
\(f : V \to Y\)
be a given mapping. If there exists a mapping
\(F : V \to Y\)
such that
$$ \bigl\Vert f(x) - F(x) \bigr\Vert \leq\Phi(x) $$
(5)
for all
\(x \in V \backslash\{ 0 \}\)
and
$$ F_{o}^{(1)}(ax) := aF_{o}^{(1)}(x), \qquad F_{e}(ax) := a^{2}F_{e}(x),\qquad F_{o}^{(2)}(ax) := a^{3}F_{o}^{(2)}(x) $$
(6)
for all
\(x \in V\), then
F
is given by
$$ F(x) = \left \{ \textstyle\begin{array}{l@{\quad}l} \lim_{n \to\infty} ( \frac{f_{o}^{(1)} ( a^{n}x )}{a^{n}} + \frac{f_{e} ( a^{n}x )}{a^{2n}} + \frac{f_{o}^{(2)} ( a^{n}x )}{a^{3n}} ) & \textit{if }\Phi \textit{ satisfies }\text{(1)}, \\ \lim_{n \to\infty} ( a^{n} f_{o}^{(1)} ( \frac{x}{a^{n}} ) + \frac{f_{e} ( a^{n}x )}{a^{2n}} + \frac{f_{o}^{(2)} ( a^{n}x )}{a^{3n}} ) & \textit{if }\Phi\textit{ satisfies }\text{(2)}, \\ \lim_{n \to\infty} ( a^{n} f_{o}^{(1)} ( \frac{x}{a^{n}} ) + a^{2n} f_{e} ( \frac{x}{a^{n}} ) + \frac{f_{o}^{(2)} ( a^{n}x )}{a^{3n}} ) & \textit{if }\Phi\textit{ satisfies }\text{(3)}, \\ \lim_{n \to\infty} ( a^{n} f_{o}^{(1)} ( \frac{x}{a^{n}} ) + a^{2n} f_{e} ( \frac{x}{a^{n}} ) + a^{3n} f_{o}^{(2)} ( \frac{x}{a^{n}} ) ) & \textit{if }\Phi\textit{ satisfies }\text{(4)} \end{array}\displaystyle \right . $$
(7)
for all
\(x \in V \backslash\{ 0 \}\). In other words, F
is the unique mapping satisfying the conditions (5) and (6).
Proof
Assume that F is a mapping satisfying (5) and (6) for a given mapping \(f : V \to Y\). First, we consider the mapping \(F_{o}^{(1)}\). If \(\Phi: V \backslash\{ 0 \} \to[0,\infty)\) satisfies the condition (1), then it follows from (6) that
$$\begin{aligned}& \biggl\Vert F_{o}^{(1)}(x) - \frac{f_{o}^{(1)} ( a^{n}x )}{a^{n}} \biggr\Vert \\& \quad = \frac{1}{a^{n}} \bigl\Vert F_{o}^{(1)} \bigl( a^{n}x \bigr) - f_{o}^{(1)} \bigl( a^{n}x \bigr) \bigr\Vert \\& \quad = \frac{1}{2(a^{3}-a)a^{n}} \bigl\Vert a^{3}F \bigl( a^{n}x \bigr) - a^{3}f \bigl( a^{n}x \bigr) - a^{3}F \bigl( -a^{n}x \bigr) + a^{3}f \bigl( -a^{n}x \bigr) \\& \qquad {}-F \bigl( a^{n+1}x \bigr) + f \bigl( a^{n+1}x \bigr) + F \bigl( -a^{n+1}x \bigr) - f \bigl( -a^{n+1}x \bigr) \bigr\Vert \\& \quad \leq \frac{1}{2(a^{3}-a)a^{n}} \bigl( a^{3} \bigl\Vert F \bigl( a^{n}x \bigr) - f \bigl( a^{n}x \bigr) \bigr\Vert + a^{3} \bigl\Vert F \bigl( -a^{n}x \bigr) - f \bigl( -a^{n}x \bigr) \bigr\Vert \\& \qquad {}+ \bigl\Vert F \bigl( a^{n+1}x \bigr) - f \bigl( a^{n+1}x \bigr) \bigr\Vert + \bigl\Vert F \bigl( -a^{n+1}x \bigr) - f \bigl( -a^{n+1}x \bigr) \bigr\Vert \bigr) \\& \quad \leq \frac{a^{3} \Phi( a^{n}x ) + a^{3} \Phi( -a^{n}x ) + \Phi( a^{n+1}x ) + \Phi( -a^{n+1}x )}{2(a^{3}-a) a^{n}} \\& \quad \to 0,\quad \mbox{as } n\to\infty \end{aligned}$$
for all \(x \in V \backslash\{ 0 \}\); that is, we see that \(F_{o}^{(1)}(x) = \lim_{n \to\infty} \frac{1}{a^{n}} f_{o}^{(1)}(a^{n}x)\) for all \(x \in V \backslash\{ 0 \}\).
If \(\Phi: V \backslash\{ 0 \} \to[0,\infty)\) satisfies the condition (2), (3), or (4), then it follows from (6) that
$$\begin{aligned}& \biggl\Vert F_{o}^{(1)}(x) - a^{n} f_{o}^{(1)} \biggl( \frac{x}{a^{n}} \biggr) \biggr\Vert \\& \quad = a^{n} \biggl\Vert F_{o}^{(1)} \biggl( \frac{x}{a^{n}} \biggr) - f_{o}^{(1)} \biggl( \frac{x}{a^{n}} \biggr) \biggr\Vert \\& \quad = \frac{a^{n}}{2(a^{3}-a)} \biggl\Vert a^{3} F \biggl( \frac{x}{a^{n}} \biggr) - a^{3} f \biggl( \frac{x}{a^{n}} \biggr) - a^{3} F \biggl( \frac{-x}{a^{n}} \biggr) + a^{3} f \biggl( \frac{-x}{a^{n}} \biggr) \\& \qquad {}- F \biggl( \frac{x}{a^{n-1}} \biggr) + f \biggl( \frac{x}{a^{n-1}} \biggr) + F \biggl( \frac{-x}{a^{n-1}} \biggr) - f \biggl( \frac{-x}{a^{n-1}} \biggr) \biggr\Vert \\& \quad \leq \frac{a^{n}}{2(a^{3}-a)} \biggl( a^{3} \biggl\Vert F \biggl( \frac{x}{a^{n}} \biggr) - f \biggl( \frac{x}{a^{n}} \biggr) \biggr\Vert + a^{3} \biggl\Vert F \biggl( \frac{-x}{a^{n}} \biggr) - f \biggl( \frac{-x}{a^{n}} \biggr) \biggr\Vert \\& \qquad {}+ \biggl\Vert F \biggl( \frac{x}{a^{n-1}} \biggr) - f \biggl( \frac{x}{a^{n-1}} \biggr) \biggr\Vert + \biggl\Vert F \biggl( \frac{-x}{a^{n-1}} \biggr) - f \biggl( \frac{-x}{a^{n-1}} \biggr) \biggr\Vert \biggr) \\& \quad \leq \frac{1}{2(a^{3}-a)} \biggl( a^{n+3} \Phi \biggl( \frac{x}{a^{n}} \biggr) + a^{n+3} \Phi \biggl( \frac{-x}{a^{n}} \biggr) + a^{n} \Phi \biggl( \frac{x}{a^{n-1}} \biggr) + a^{n} \Phi \biggl( \frac{-x}{a^{n-1}} \biggr)\biggr) \\& \quad \to 0,\quad \mbox{as } n \to\infty \end{aligned}$$
for all \(x \in V \backslash\{ 0 \}\); that is, we see that \(F_{o}^{(1)}(x) = \lim_{n \to\infty} a^{n} f_{o}^{(1)} ( \frac{x}{a^{n}} )\) for all \(x \in V \backslash\{ 0 \}\).
Second, we consider the mapping \(F_{e}\). If \(\Phi: V \backslash\{ 0 \} \to[0,\infty)\) satisfies the condition (1) or (2), then it follows from (6) that
$$\begin{aligned}& \biggl\Vert F_{e}(x) - \frac{f_{e} ( a^{n}x )}{a^{2n}} \biggr\Vert \\& \quad = \frac{1}{a^{2n}} \bigl\Vert F_{e} \bigl( a^{n}x \bigr) - f_{e} \bigl( a^{n}x \bigr) \bigr\Vert = \frac{1}{2a^{2n}} \bigl\Vert F \bigl( a^{n}x \bigr) - f \bigl( a^{n}x \bigr) + F \bigl( -a^{n}x \bigr) - f \bigl( -a^{n}x \bigr) \bigr\Vert \\& \quad \leq \frac{1}{2a^{2n}} \bigl\Vert F \bigl( a^{n}x \bigr) - f \bigl( a^{n}x \bigr) \bigr\Vert + \frac{1}{2a^{2n}} \bigl\Vert F \bigl( -a^{n}x \bigr) - f \bigl( -a^{n}x \bigr) \bigr\Vert \\& \quad \leq \frac{\Phi( a^{n}x ) + \Phi( -a^{n}x )}{ 2a^{2n}} \\& \quad \to 0, \quad \mbox{as } n \to\infty \end{aligned}$$
for all \(x \in V \backslash\{0\}\); that is, we see that \(F_{e}(x) = \lim_{n \to\infty} \frac{1}{a^{2n}} f_{e}(a^{n}x)\) for all \(x \in V \backslash\{ 0 \}\).
If \(\Phi: V \to[0,\infty)\) satisfies the condition (3) or (4), we get
$$\begin{aligned}& \biggl\Vert F_{e}(x)- a^{2n} f_{e} \biggl( \frac{x}{a^{n}} \biggr) \biggr\Vert \\& \quad = a^{2n} \biggl\Vert F_{e} \biggl( \frac{x}{a^{n}} \biggr) - f_{e} \biggl( \frac{x}{a^{n}} \biggr) \biggr\Vert \\& \quad = \frac{a^{2n}}{2} \biggl\Vert F \biggl( \frac{x}{a^{n}} \biggr) - f \biggl( \frac{x}{a^{n}} \biggr) + F \biggl( \frac{-x}{a^{n}} \biggr) - f \biggl( \frac{-x}{a^{n}} \biggr) \biggr\Vert \\& \quad \leq \frac{a^{2n}}{2} \biggl\Vert F \biggl( \frac{x}{a^{n}} \biggr) - f \biggl( \frac{x}{a^{n}} \biggr) \biggr\Vert + \frac{a^{2n}}{2} \biggl\Vert F \biggl( \frac{-x}{a^{n}} \biggr) - f \biggl( \frac{-x}{a^{n}} \biggr) \biggr\Vert \\& \quad \leq \frac{a^{2n}}{2} \biggl( \Phi \biggl( \frac{x}{a^{n}} \biggr) + \Phi \biggl( \frac{-x}{a^{n}} \biggr) \biggr) \\& \quad \to 0, \quad \mbox{as } n \to\infty \end{aligned}$$
for all \(x \in V \backslash\{ 0 \}\). Then \(F_{e}(x) = \lim_{n \to\infty} a^{2n} f_{e} ( \frac{x}{a^{n}} )\) for all \(x \in V \backslash\{ 0 \}\) holds.
Finally, we consider the mapping \(f_{o}^{(2)}\). If \(\Phi: V\backslash\{ 0 \} \to[0,\infty)\) satisfies the condition (1), (2), or (3), then it follows from (6) that
$$\begin{aligned}& \biggl\Vert F_{o}^{(2)}(x) - \frac{f_{o}^{(2)} ( a^{n}x )}{a^{3n}} \biggr\Vert \\& \quad = \frac{1}{a^{3n}} \bigl\Vert F_{o}^{(2)} \bigl( a^{n}x \bigr) - f_{o}^{(2)} \bigl( a^{n}x \bigr) \bigr\Vert \\& \quad = \frac{1}{2(a^{3}-a)a^{3n}} \bigl\Vert -aF \bigl( a^{n}x \bigr) + af \bigl( a^{n}x \bigr) + aF \bigl( -a^{n}x \bigr) - af \bigl( -a^{n}x \bigr) \\& \qquad {}+F \bigl( a^{n+1}x \bigr) - f \bigl( a^{n+1}x \bigr) -F \bigl( -a^{n+1}x \bigr) + f \bigl( -a^{n+1}x \bigr) \bigr\Vert \\& \quad \leq \frac{1}{2(a^{3}-a)a^{3n}} \bigl( a \bigl\Vert F \bigl( a^{n}x \bigr) - f \bigl( a^{n}x \bigr) \bigr\Vert + a \bigl\Vert F \bigl( -a^{n}x \bigr) - f \bigl( -a^{n}x \bigr) \bigr\Vert \\& \qquad {}+ \bigl\Vert F \bigl( a^{n+1}x \bigr) - f \bigl( a^{n+1}x \bigr) \bigr\Vert + \bigl\Vert F \bigl( -a^{n+1}x \bigr) - f \bigl( -a^{n+1}x \bigr) \bigr\Vert \bigr) \\& \quad \leq \frac{a\Phi ( a^{n}x ) + a\Phi ( -a^{n}x ) + \Phi ( a^{n+1}x ) + \Phi ( -a^{n+1}x )}{2(a^{3}-a)a^{3n}} \\& \quad \to 0,\quad \mbox{as } n \to\infty \end{aligned}$$
for all \(x \in V \backslash\{ 0 \}\); that is, we see that \(F_{o}^{(2)}(x) = \lim_{n \to\infty} \frac{1}{a^{3n}} f_{o}^{(2)}(a^{n}x)\) for all \(x \in V \backslash\{ 0 \}\).
If \(\Phi: V \backslash\{ 0 \} \to[0,\infty)\) satisfies the condition (4), then it follows from (4) and (6) that
$$\begin{aligned}& \biggl\Vert F_{o}^{(2)}(x) - a^{3n} f_{o}^{(2)} \biggl( \frac{x}{a^{n}} \biggr) \biggr\Vert \\& \quad = a^{3n} \biggl\Vert F_{o}^{(2)} \biggl( \frac{x}{a^{n}} \biggr) - f_{o}^{(2)} \biggl( \frac{x}{a^{n}} \biggr) \biggr\Vert \\& \quad = \frac{a^{3n}}{2(a^{3}-a)} \biggl\Vert -aF \biggl( \frac{x}{a^{n}} \biggr) + af \biggl( \frac{x}{a^{n}} \biggr) + aF \biggl( \frac{-x}{a^{n}} \biggr) - af \biggl( \frac{-x}{a^{n}} \biggr) \\& \qquad {}+F \biggl( \frac{x}{a^{n-1}} \biggr) - f \biggl( \frac{x}{a^{n-1}} \biggr) - F \biggl( \frac{-x}{a^{n-1}} \biggr) + f \biggl( \frac{-x}{a^{n-1}} \biggr) \biggr\Vert \\& \quad \leq \frac{a^{3n}}{2(a^{3}-a)} \biggl( a \biggl\Vert F \biggl( \frac{x}{a^{n}} \biggr) - f \biggl( \frac{x}{a^{n}} \biggr) \biggr\Vert + a \biggl\Vert F \biggl( \frac{-x}{a^{n}} \biggr) - f \biggl( \frac{-x}{a^{n}} \biggr) \biggr\Vert \\& \qquad {}+ \biggl\Vert F \biggl( \frac{x}{a^{n-1}} \biggr) - f \biggl( \frac{x}{a^{n-1}} \biggr) \biggr\Vert + \biggl\Vert F \biggl( \frac{-x}{a^{n-1}} \biggr) - f \biggl( \frac{-x}{a^{n-1}} \biggr) \biggr\Vert \biggr) \\& \quad \leq \frac{a^{3n}}{2(a^{3}-a)} \biggl( a \Phi \biggl( \frac{x}{a^{n}} \biggr) + a \Phi \biggl( \frac{-x}{a^{n}} \biggr) + \Phi \biggl( \frac{x}{a^{n-1}} \biggr) + \Phi \biggl( \frac{-x}{a^{n-1}} \biggr)\biggr) \\& \quad \to 0,\quad \mbox{as } n \to\infty \end{aligned}$$
for all \(x \in V \backslash\{ 0 \}\); that is, we see that \(F_{o}^{(2)}(x) = \lim_{n \to\infty} a^{3n} f_{o}^{(2)} ( \frac{x}{a^{n}} )\) for all \(x \in V \backslash\{ 0 \}\). Since \(F(x) = F_{o}^{(1)}(x) + F_{e}(x) + F_{o}^{(2)}(x)\), F is given by the equalities in (7) and F is uniquely determined for any case. □
In general, it is not easy to apply Theorem 2.1 for practical applications. Hence, we introduce a couple of corollaries which are useful for investigating the uniqueness problems in the stability of the cubic-quadratic-additive functional equations.
Corollary 2.2
Let
\(a > 1\)
be a real constant and let
\(\phi: V \backslash\{ 0 \} \to[0,\infty)\)
be a function satisfying either
$$ \Phi(x) := \sum_{i=0}^{\infty}\frac{\phi ( a^{i}x )}{a^{i}} < \infty $$
(8)
or
$$ \Phi(x) := \sum_{i=0}^{\infty}a^{3i} \phi \biggl( \frac{x}{a^{i}} \biggr) < \infty $$
(9)
for all
\(x \in X \backslash\{ 0 \}\). For any given mapping
\(f : V \to Y\), if there exists a mapping
\(F : V \to Y\)
satisfying the inequality
$$ \bigl\Vert f(x) - F(x) \bigr\Vert \leq\Phi(x) $$
(10)
for all
\(x \in V \backslash\{ 0 \}\)
and the condition (6) for all
\(x \in V\), then
F
is a unique mapping satisfying the conditions (6) and (10).
Proof
If ϕ satisfies (8), then we have
$$ \lim_{n \to\infty} \frac{\Phi ( a^{n}x )}{a^{n}} = \lim_{n \to\infty} \sum_{i=0}^{\infty}\frac{\phi ( a^{n+i}x )}{a^{n+i}} = \lim _{n \to\infty} \sum_{i=n}^{\infty}\frac{\phi ( a^{i}x )}{a^{i}} = 0, $$
i.e., Φ satisfies the condition (1) for all \(x \in V \backslash\{ 0 \}\).
For the case when ϕ satisfies (9), it holds that
$$ \lim_{n \to\infty} a^{3n} \Phi \biggl( \frac{x}{a^{n}} \biggr) = \lim_{n \to\infty} \sum_{i=0}^{\infty}a^{3n+3i} \phi \biggl( \frac{x}{a^{n+i}} \biggr) = \lim _{n \to\infty} \sum_{i=n}^{\infty}a^{3i} \phi \biggl( \frac{x}{a^{i}} \biggr) = 0, $$
i.e., Φ satisfies the condition (4) for all \(x \in V \backslash\{ 0 \}\). Hence, our assertion is true in view of Theorem 2.1. □
Corollary 2.3
Let
\(a > 1\)
be a real constant, let
\(\phi, \psi: V \backslash\{ 0 \} \to[0,\infty)\)
be functions satisfying each of the following conditions:
$$ \begin{aligned} &\sum_{i=0}^{\infty}a^{i} \psi \biggl( \frac{x}{a^{i}} \biggr) < \infty,\qquad \sum _{i=0}^{\infty}\frac{\phi ( a^{i}x )}{a^{2i}} < \infty, \\ &\tilde{\Phi}(x) := \sum_{i=0}^{\infty}a^{i} \phi \biggl( \frac{x}{a^{i}} \biggr) < \infty, \qquad \tilde{ \Psi}(x) := \sum_{i=0}^{\infty}\frac{\psi ( a^{i}x )}{a^{2i}} < \infty \end{aligned} $$
(11)
for all
\(x \in V \backslash\{ 0 \}\), and let
\(f : V \to Y\)
be an arbitrarily given mapping. If there exists a mapping
\(F : V \to Y\)
satisfying the inequality
$$ \bigl\Vert f(x) - F(x) \bigr\Vert \leq\tilde{\Phi}(x) + \tilde{\Psi}(x) $$
(12)
for all
\(x \in V \backslash\{ 0 \}\)
and the condition (6) for all
\(x \in V\), then
F
is a unique mapping satisfying the conditions (6) for all
\(x \in V\)
and the inequality (12) for all
\(x \in V \backslash\{ 0 \}\).
Proof
If we put \(\Phi(x) = \tilde{\Phi}(x) + \tilde{\Psi}(x)\), then it follows from (12) that
$$ \frac{1}{a^{4n}} \Phi \bigl( a^{2n}x \bigr) = \sum _{i=0}^{\infty}\frac{1}{a^{4n-i}} \phi \bigl( a^{2n-i}x \bigr) + \sum_{i=0}^{\infty}\frac{1}{a^{4n+2i}} \psi \bigl( a^{2n+i}x \bigr) $$
for all \(x \in V\backslash\{ 0 \}\). We make a change of the summation indices in the preceding equality with \(j = i-2n\) and \(k = 2n+i\) to get
$$\begin{aligned}& \frac{1}{a^{4n}} \Phi \bigl( a^{2n}x \bigr) \\& \quad = \frac{1}{a^{2n}} \sum_{j=-2n}^{\infty}a^{j} \phi \biggl( \frac{x}{a^{j}} \biggr) + \sum _{k=2n}^{\infty}\frac{1}{a^{2k}} \psi \bigl( a^{k}x \bigr) \\& \quad = \frac{1}{a^{2n}} \sum_{i=1}^{2n} \frac{1}{a^{i}} \phi \bigl( a^{i}x \bigr) + \frac{1}{a^{2n}} \sum _{i=0}^{\infty}a^{i} \phi \biggl( \frac{x}{a^{i}} \biggr) + \sum_{i=2n}^{\infty}\frac{1}{a^{2i}} \psi \bigl( a^{i}x \bigr) \\& \quad = \frac{1}{a^{n}} \sum_{i=1}^{n-1} \frac{a^{i}}{a^{n}} \frac{1}{a^{2i}} \phi \bigl( a^{i}x \bigr) + \sum _{i=n}^{2n} \frac{a^{i}}{a^{2n}} \frac{1}{a^{2i}} \phi \bigl( a^{i}x \bigr) + \frac{1}{a^{2n}} \tilde{\Phi}(x) + \sum_{i=2n}^{\infty}\frac{1}{a^{2i}} \psi \bigl( a^{i}x \bigr) \\& \quad \leq \frac{1}{a^{n}} \sum_{i=1}^{\infty}\frac{1}{a^{2i}} \phi \bigl( a^{i}x \bigr) + \sum _{i=n}^{\infty}\frac{1}{a^{2i}} \phi \bigl( a^{i}x \bigr) + \frac{1}{a^{2n}} \tilde{\Phi}(x) + \sum _{i=2n}^{\infty}\frac{1}{a^{2i}} \psi \bigl( a^{i}x \bigr) \end{aligned}$$
for any \(x \in V \backslash\{ 0 \}\). Hence, it follows from (11) that
$$ \lim_{n \to\infty} \frac{1}{a^{4n}} \Phi \bigl( a^{2n}x \bigr) = 0 $$
for all \(x \in V \backslash\{ 0 \}\). On the other hand, we use the above equality to get
$$ \lim_{n \to\infty} \frac{1}{a^{4n+2}} \Phi \bigl( a^{2n+1}x \bigr) = \frac{1}{a^{2}} \lim_{n \to\infty} \frac{1}{a^{4n}} \Phi \bigl( a^{2n}ax \bigr) = 0 $$
for all \(x \in V \backslash\{ 0 \}\). From the above two equalities, we conclude that
$$ \lim_{n \to\infty} \frac{1}{a^{2n}} \Phi \bigl( a^{n}x \bigr) = 0 $$
for all \(x \in V \backslash\{ 0 \}\).
Similarly, we have
$$ a^{2n} \Phi \biggl( \frac{x}{a^{2n}} \biggr) = \sum _{i=0}^{\infty}a^{2n+i} \phi \biggl( \frac{x}{a^{2n+i}} \biggr) + \sum_{i=0}^{\infty}\frac{1}{a^{2i-2n}} \psi \bigl( a^{i-2n}x \bigr) $$
for all \(x \in V \backslash\{ 0 \}\). If we make a change of the summation indices in the last equality with \(j = i+2n\) and \(k = i-2n\), then we get
$$\begin{aligned}& a^{2n} \Phi \biggl( \frac{x}{a^{2n}} \biggr) \\& \quad = \sum_{j=2n}^{\infty}a^{j} \phi \biggl( \frac{x}{a^{j}} \biggr) + \frac{1}{a^{2n}} \sum _{k=-2n}^{\infty}\frac{1}{a^{2k}} \psi \bigl( a^{k}x \bigr) \\& \quad = \sum_{i=2n}^{\infty}a^{i} \phi \biggl( \frac{x}{a^{i}} \biggr) + \frac{1}{a^{2n}} \sum _{i=1}^{2n} a^{2i} \psi \biggl( \frac{x}{a^{i}} \biggr) + \frac{1}{a^{2n}} \sum_{i=0}^{\infty}\frac{1}{a^{2i}} \psi \bigl( a^{i}x \bigr) \\& \quad = \sum_{i=2n}^{\infty}a^{i} \phi \biggl( \frac{x}{a^{i}} \biggr) + \frac{1}{a^{n}} \sum _{i=1}^{n-1} \frac{a^{i}}{a^{n}} a^{i} \psi \biggl( \frac{x}{a^{i}} \biggr) + \sum_{i=n}^{2n} \frac{a^{i}}{a^{2n}} a^{i} \psi \biggl( \frac{x}{a^{i}} \biggr) + \frac{1}{a^{2n}} \tilde{\Psi}(x) \\& \quad \leq \sum_{i=2n}^{\infty}a^{i} \phi \biggl( \frac{x}{a^{i}} \biggr) + \frac{1}{a^{n}} \sum _{i=1}^{\infty}a^{i} \psi \biggl( \frac{x}{a^{i}} \biggr) + \sum_{i=n}^{\infty}a^{i} \psi \biggl( \frac{x}{a^{i}} \biggr) + \frac{1}{a^{2n}} \tilde{\Psi}(x) \end{aligned}$$
for any \(x \in V \backslash\{ 0 \}\). Thus, it follows from (11) that
$$\begin{aligned}& \lim_{n \to\infty} a^{2n} \Phi \biggl( \frac{x}{a^{2n}} \biggr) = 0, \\& \lim_{n \to\infty} a^{2n+1} \Phi \biggl( \frac{x}{a^{2n+1}} \biggr) = a \lim_{n \to\infty} a^{2n} \Phi \biggl( \frac{1}{a^{2n}} \frac{x}{a} \biggr) = 0 \end{aligned}$$
for each \(x \in V \backslash\{ 0 \}\). Thus, we see that
$$ \lim_{n \to\infty} a^{n} \Phi \biggl( \frac{x}{a^{n}} \biggr) = 0 $$
for each \(x \in V \backslash\{ 0 \}\).
Altogether, Φ satisfies (2) for all \(x \in V \backslash\{ 0 \}\). Hence, Theorem 2.1 implies that our conclusion of this corollary is true. □
Corollary 2.4
Let
\(a > 1\)
be a real constant, let
\(\phi, \psi: V \backslash\{ 0 \} \to[0,\infty)\)
be functions satisfying each of the following conditions:
$$ \begin{aligned} &\sum_{i=0}^{\infty}a^{2i} \psi \biggl( \frac{x}{a^{i}} \biggr) < \infty,\qquad \sum _{i=0}^{\infty}\frac{\phi ( a^{i}x )}{a^{3i}} < \infty, \\ &\tilde{\Phi}(x) := \sum_{i=0}^{\infty}a^{2i} \phi \biggl( \frac{x}{a^{i}} \biggr) < \infty,\qquad \tilde{ \Psi}(x) := \sum_{i=0}^{\infty}\frac{\psi ( a^{i}x )}{a^{3i}} < \infty \end{aligned} $$
(13)
for all
\(x \in V \backslash\{ 0 \}\), and let
\(f : V \to Y\)
be an arbitrarily given mapping. If there exists a mapping
\(F : V \to Y\)
satisfying the inequality
$$ \bigl\Vert f(x) - F(x) \bigr\Vert \leq\tilde{\Phi}(x) + \tilde{\Psi}(x) $$
(14)
for all
\(x \in V \backslash\{ 0 \}\)
and the condition (6) for all
\(x \in V\), then
F
is a unique mapping satisfying the conditions (6) for all
\(x \in V\)
and (14) for all
\(x \in V \backslash\{ 0 \}\).
Proof
If we put \(\Phi(x) = \tilde{\Phi}(x) + \tilde{\Psi}(x)\), then it follows from (13) that
$$ \frac{1}{a^{6n}} \Phi\bigl(a^{2n}x\bigr) = \sum _{i=0}^{\infty}\frac{1}{a^{6n-2i}} \phi \bigl( a^{2n-i}x \bigr) + \sum_{i=0}^{\infty}\frac{1}{a^{6n+3i}} \psi \bigl( a^{2n+i}x \bigr) $$
for all \(x \in V \backslash\{ 0 \}\). We make a change of the summation indices in the preceding equality with \(j = i-2n\) and \(k = 2n+i\) to get
$$\begin{aligned}& \frac{1}{a^{6n}} \Phi\bigl(a^{2n}x\bigr) \\& \quad = \frac{1}{a^{2n}} \sum_{j=-2n}^{\infty}a^{2j} \phi \biggl( \frac{x}{a^{j}} \biggr) + \sum _{k=2n}^{\infty}\frac{1}{a^{3k}} \psi \bigl( a^{k}x \bigr) \\& \quad = \frac{1}{a^{2n}} \sum_{i=1}^{2n} \frac{1}{a^{2i}} \phi \bigl( a^{i}x \bigr) + \frac{1}{a^{2n}} \sum _{i=0}^{\infty}a^{2i} \phi \biggl( \frac{x}{a^{i}} \biggr) + \sum_{i=2n}^{\infty}\frac{1}{a^{3i}} \psi \bigl( a^{i}x \bigr) \\& \quad = \frac{1}{a^{n}} \sum_{i=1}^{n-1} \frac{1}{a^{2i+n}} \phi \bigl( a^{i}x \bigr) + \sum _{i=n}^{2n} \frac{1}{a^{2n+2i}} \phi \bigl( a^{i}x \bigr) + \frac{1}{a^{2n}} \tilde{\Phi}(x) + \sum _{i=2n}^{\infty}\frac{1}{a^{3i}} \psi \bigl( a^{i}x \bigr) \\& \quad \leq \frac{1}{a^{n}} \sum_{i=1}^{\infty}\frac{1}{a^{3i}} \phi \bigl( a^{i}x \bigr) + \sum _{i=n}^{\infty}\frac{1}{a^{3i}} \phi \bigl( a^{i}x \bigr) + \frac{1}{a^{2n}} \tilde{\Phi}(x) + \sum _{i=2n}^{\infty}\frac{1}{a^{3i}} \psi \bigl( a^{i}x \bigr) \end{aligned}$$
for any \(x \in V \backslash\{ 0 \}\). Hence, by (13), we get
$$ \lim_{n \to\infty} \frac{1}{a^{6n}} \Phi \bigl( a^{2n}x \bigr) = 0 $$
for all \(x \in V \backslash\{ 0 \}\). On the other hand, we use the above equality to get
$$ \lim_{n \to\infty} \frac{1}{a^{6n+3}} \Phi \bigl( a^{2n+1}x \bigr) = \frac{1}{a^{3}} \lim_{n \to\infty} \frac{1}{a^{6n}} \Phi \bigl( a^{2n}ax \bigr) = 0 $$
for all \(x \in V \backslash\{ 0 \}\).
From the above two equalities, we conclude that
$$ \lim_{n \to\infty} \frac{1}{a^{3n}} \Phi \bigl( a^{n}x \bigr) = 0 $$
for all \(x \in V \backslash\{ 0 \}\).
Similarly, we have
$$ a^{4n} \Phi \biggl( \frac{x}{a^{2n}} \biggr) = \sum _{i=0}^{\infty}a^{4n+2i} \phi \biggl( \frac{x}{a^{2n+i}} \biggr) + \sum_{i=0}^{\infty}\frac{1}{a^{3i-4n}} \psi \bigl( a^{i-2n}x \bigr) $$
for all \(x \in V \backslash\{ 0 \}\). If we make a change of the summation indices in the last equality with \(j = i+2n\) and \(k = i-2n\), then we get
$$\begin{aligned}& a^{4n} \Phi \biggl( \frac{x}{a^{2n}} \biggr) \\& \quad = \sum_{j=2n}^{\infty}a^{2j} \phi \biggl( \frac{x}{a^{j}} \biggr) + \frac{1}{a^{2n}} \sum _{k=-2n}^{\infty}\frac{1}{a^{3k}} \psi \bigl( a^{k}x \bigr) \\& \quad = \sum_{i=2n}^{\infty}a^{2i} \phi \biggl( \frac{x}{a^{i}} \biggr) + \frac{1}{a^{2n}} \sum _{i=1}^{2n} a^{3i} \psi \biggl( \frac{x}{a^{i}} \biggr) + \frac{1}{a^{2n}} \sum_{i=0}^{\infty}\frac{1}{a^{3i}} \psi \bigl( a^{i}x \bigr) \\& \quad = \sum_{i=2n}^{\infty}a^{2i} \phi \biggl( \frac{x}{a^{i}} \biggr) + \frac{1}{a^{n}} \sum _{i=1}^{n-1} \frac{a^{i}}{a^{n}} a^{2i} \psi \biggl( \frac{x}{a^{i}} \biggr) + \sum_{i=n}^{2n} \frac{a^{i}}{a^{2n}} a^{2i} \psi \biggl( \frac{x}{a^{i}} \biggr) + \frac{1}{a^{2n}} \tilde{\Psi}(x) \\& \quad \leq \sum_{i=2n}^{\infty}a^{2i} \phi \biggl( \frac{x}{a^{i}} \biggr) + \frac{1}{a^{n}} \sum _{i=1}^{\infty}a^{2i} \psi \biggl( \frac{x}{a^{i}} \biggr) + \sum_{i=n}^{\infty}a^{2i} \psi \biggl( \frac{x}{a^{i}} \biggr) + \frac{1}{a^{2n}} \tilde{\Psi}(x) \end{aligned}$$
for any \(x \in V \backslash\{ 0 \}\). Thus, we obtain
$$\begin{aligned}& \lim_{n \to\infty} a^{4n} \Phi \biggl( \frac{x}{a^{2n}} \biggr) = 0, \\& \lim_{n \to\infty} a^{4n+2} \Phi \biggl( \frac{x}{a^{2n+1}} \biggr) = a^{2} \lim_{n \to\infty} a^{4n} \Phi \biggl( \frac{1}{a^{2n}} \frac{x}{a} \biggr) = 0 \end{aligned}$$
for each \(x \in V \backslash\{ 0 \}\). Thus, we see that
$$ \lim_{n \to\infty} a^{2n} \Phi \biggl( \frac{x}{a^{n}} \biggr) = 0 $$
for each \(x \in V \backslash\{ 0 \}\).
Altogether, Φ satisfies (3) for all \(x \in V \backslash\{ 0 \}\). Hence, Theorem 2.1 implies that our conclusion of this corollary is true. □