Recall that the mixed-type quadratic and cubic functional equation (1.9) was introduced in [41] and the authors proved the following theorem.
Theorem 2.1
Let X and Y be vector spaces. A function \(h:X\longrightarrow Y\) satisfies functional equation (1.9) if and only if there exist a quadratic function \(A^{2}:X\longrightarrow Y\), a cubic function \(A^{3}:X\longrightarrow Y\) and a constant \(A^{0}\) such that
$$ h(x)= A^{0}+A^{2}(x)+A^{3}(x) $$
for all \(x\in X\).
Next, by using an alternative method we obtain the general solution of (1.9), which is a tool to reach some of our goals in this section.
Lemma 2.2
Let X and Y be real vector spaces. Suppose that \(h:X\longrightarrow Y\) satisfies (1.9) for all \(x,y\in X\).
-
(i)
If h is even and \(h(0)=0\), then it is quadratic;
-
(ii)
If h is odd, then it is cubic.
Proof
(i) We first note that the evenness of h converts (1.9) to
$$ h(x+2y)-3h(x+y)+3h(x)-h(x-y)=0, $$
(2.1)
for all \(x,y\in X\) (here and in the rest of the proof). Letting \(x=0\) in (2.1), we have
Interchanging x with \(x-y\) in (2.1), we find
$$ h(x+y)-3h(x)+3h(x-y)-h(x-2y)=0. $$
(2.3)
A difference computation of (2.1) and (2.3) shows that
$$ h(x+2y)+h(x-2y)-4h(x+y)-4h(x-y)+6h(x)=0. $$
(2.4)
Replacing x by 2x in (2.4) and using (2.2), we obtain
$$ h(x+y)+h(x-y)-h(2x+y)-h(2x-y)+6h(x)=0. $$
(2.5)
Switching \((x,y)\) by \((y,x)\) in (2.5) and applying again the evenness of f, we obtain
$$ h(x+y)+h(x-y)-h(x+2y)-h(x-2y)+6h(y)=0. $$
(2.6)
Inserting (2.4) into (2.6), we find
$$ h(x+y)+h(x-y)=2h(x)+2h(y). $$
(ii) Using our assumption on (1.9), we obtain
$$ h(x+2y)-3h(x+y)+3h(x)-h(x-y)-6h(y)=0. $$
(2.7)
Putting \(x=0\) in (2.7) and applying from the oddness of h, we find
On the other hand, if we replace \(x-y\) instead of x in (2.6), then,
$$ h(x+y)-3h(x)+3h(x-y)-h(x-2y)-6h(y)=0. $$
(2.9)
Now, it follows from (2.7) and (2.9) that
$$ h(x+2y)+h(x-2y)=4\bigl[h(x+y)+h(x-y)\bigr]-6h(x). $$
This completes the proof. □
Henceforth, let V and W be vector spaces over \(\mathbb{Q}\), \(n\in \mathbb{N}\) and \(x_{i}^{n}=(x_{i1},x_{i2},\ldots ,x_{in})\in V^{n}\), where \(i\in \{1,2\}\). We denote \(x_{i}^{n}\) by \(x_{i}\) if there is no risk of a mistake. Let \(x_{1},x_{2}\in V^{n}\) and \(m\in \mathbb{N}_{0}\) with \(0\leq m\leq n\). Put \(\mathcal{M}^{n}= \{ \mathfrak{N}_{n}=(N_{1},\ldots ,N_{n})\mid N_{j} \in \{x_{1j}\pm x_{2j},x_{1j}\} \} \), where \(j\in \{1,\ldots ,n\}\). Consider
$$ \mathcal{M}_{m}^{n} := \bigl\{ \mathfrak{N}_{n} \in \mathcal{M}^{n}\mid \operatorname{Card}\{N_{j}: N_{j}=x_{1j}\}=m \bigr\} . $$
For \(r\in \mathbb{Q}\), we put \(r\mathcal{M}_{m}^{n}= \{ r\mathfrak{N}_{n}: \mathfrak{N}_{n}\in \mathcal{M}_{m}^{n} \} \) in which \(r\mathfrak{N}_{n}=(rN_{1},\ldots ,rN_{n})\).
Definition 2.3
A mapping \(f:V^{n}\longrightarrow W\) is n-multicubic or multicubic if f satisfies (1.3) in each variable.
For a multicubic mapping f, we use the notations
$$ f \bigl(\mathcal{M}_{m}^{n} \bigr):= \sum_{\mathfrak{N}_{n}\in \mathcal{M}_{m}^{n}}f(\mathfrak{N}_{n}), $$
(2.10)
and
$$ f \bigl(\mathcal{M}_{m}^{n},z \bigr):=\sum _{\mathfrak{N}_{n}\in \mathcal{M}_{m}^{n}}f(\mathfrak{N}_{n},z)\quad (z\in V). $$
We recall that \(\binom{n}{m}:=\frac{n!}{m!(n-m)!}\) is the binomial coefficient, where \(n, m\in \mathbb{N}_{0}\) with \(n\geq m\). In the upcoming result, we find a necessary and sufficient condition for a several-variable mapping to be multicubic.
Proposition 2.4
For a mapping \(f: V^{n}\longrightarrow W\), the following assertions are equivalent.
-
(i)
f is multicubic;
-
(ii)
f satisfies
$$ \sum_{t\in \{-2,2\}^{n}} f(x_{1}+tx_{2})= \sum_{m=0}^{n}4^{n-m}(-6)^{m}f \bigl( \mathcal{M}_{m}^{n}\bigr), $$
(2.11)
where \(f (\mathcal{M}_{m}^{n} )\) is defined in (2.10).
Proof
\(\mbox{(i)}\Rightarrow \mbox{(ii)}\) The proof of this implication is by induction on n. For \(n=1\), it is clear that f fulfills (1.3). Suppose that (2.11) holds for some positive integer \(n>1\). Then,
$$\begin{aligned} &\sum_{t\in \{-2,2\}^{n+1}} f\bigl(x_{1}^{n+1}+tx_{2}^{n+1} \bigr) \\ &\quad =4\sum_{t\in \{-2,2\}^{n}}\sum _{s\in \{-1,1\}} f\bigl(x_{1}^{n}+tx_{2}^{n},x_{1,n+1}+sx_{2,n+1} \bigr)-6 \sum_{t\in \{-2,2\}^{n}} f\bigl(x_{1}^{n}+tx_{2}^{n},x_{1,n+1} \bigr) \\ &\quad =4\sum_{m=0}^{n}\sum _{s\in \{-1,1\}}4^{n-m}(-6)^{m}f\bigl( \mathcal{M}_{m}^{n},x_{1,n+1}+sx_{2,n+1} \bigr)-6 \sum_{m=0}^{n}4^{n-m}(-6)^{m}f \bigl(\mathcal{M}_{m}^{n},x_{1,n+1}\bigr) \\ &\quad =\sum_{m=0}^{n+1}4^{n+1-m}(-6)^{m}f \bigl(\mathcal{M}_{m}^{n+1}\bigr). \end{aligned}$$
\(\mbox{(ii)}\Rightarrow \mbox{(i)}\) Let \(j\in \{1,\ldots ,n\}\) be arbitrary and fixed. It is enough to prove that f is cubic in the jth variable. Set
$$ f^{*}_{j}(z):=f (z_{1},\ldots ,z_{j-1},z,z_{j+1},\ldots ,z_{n} ). $$
Assuming \(x_{2m}=0\) for all \(m\in \{1,\ldots ,n\}\backslash \{j\}\), \(x_{2j}=w\) and \(x_{1}= (z_{1},\ldots ,z_{j-1},z,z_{j+1},\ldots ,z_{n} )\) in (2.11), we obtain
$$\begin{aligned} &2^{n-1} \bigl[f^{*}_{j}(z+2w)+f^{*}_{j}(z-2w) \bigr] \\ &\quad =\sum_{m=0}^{n-1}\left [\begin{pmatrix} n-1 \\ m \end{pmatrix}2^{n-1-m}4^{n-m} (-6)^{m}\right ] \bigl[f^{*}_{j}(z+w)+f^{*}_{j}(z-w) \bigr] \\ &\qquad {}+\sum_{m=1}^{n}\left [ \begin{pmatrix} n \\ m \end{pmatrix}2^{n-m}4^{n-m} (-6)^{m} \right ]f^{*}_{j}(z) \\ &\quad =4\sum_{m=0}^{n-1}\left [\begin{pmatrix} n-1 \\ m \end{pmatrix}8^{n-1-m}(-6)^{m}\right ] \bigl[f^{*}_{j}(z+w)+f^{*}_{j}(z-w) \bigr] \\ &\qquad {}-6\sum_{m=0}^{n-1}\left [\begin{pmatrix} n \\ m \end{pmatrix}8^{n-1-m}(-6)^{m}\right ]f^{*}_{j}(z) \\ &\quad =4\times 2^{n-1} \bigl[f^{*}_{j}(z+w)+f^{*}_{j}(z-w) \bigr]-6 \times 2^{n-1}f^{*}_{j}(z). \end{aligned}$$
(2.12)
It follows from (2.12) that
$$ f^{*}_{j}(z+2w)+f^{*}_{j}(z-2w)=4 \bigl[f^{*}_{j}(z+w)+f^{*}_{j}(z-w) \bigr]-6f^{*}_{j}(z). $$
Now, the proof is completed. □
Definition 2.5
Let \(n\in \mathbb{N}\) and \(k\in \{0,\ldots ,n\}\). A mapping \(f:V^{n}\longrightarrow W\) is called k-quadratic and \(n-k\)-cubic (briefly, multiquadratic–cubic) if f is quadratic (see equation (1.2)) in each of some k variables and is cubic in each of the other variables (see equation (1.3)).
In Definition 2.5, we suppose for simplicity that f is quadratic in each of the first k variables, but one can obtain analogous results without this assumption. It is obvious that for \(k=n\) (resp., \(k=0\)), the above definition leads to the so-called multiquadratic (resp., multicubic) mappings; some basic facts on the mentioned mappings can be found, for instance, in [8, 13] and [45].
To reach our results in the rest of this section, we identify \(x=(x_{1},\ldots ,x_{n})\in V^{n}\) with \((x^{k},x^{n-k})\in V^{k}\times V^{n-k}\), where \(x^{k}:=(x_{1},\ldots ,x_{k})\) and \(x^{n-k}:=(x_{k+1},\ldots ,x_{n})\). Put \(x_{i}^{k}=(x_{i1},\ldots ,x_{ik})\in V^{k}\) and \(x_{i}^{n-k}=(x_{i,k+1},\ldots ,x_{in})\in V^{n-k}\), where \(i\in \{1,2\}\). For a multiquadratic–cubic mapping f, we also recall the notation
$$ f \bigl(x_{i}^{k},\mathcal{M}_{m}^{n-k} \bigr):=\sum_{\mathfrak{N}_{n} \in \mathcal{M}_{m}^{n-k}}f \bigl(x_{i}^{k}, \mathfrak{N}_{n-k} \bigr), $$
where
$$ \mathcal{M}_{m}^{n-k} := \bigl\{ \mathfrak{N}_{n-k} \in \mathcal{M}^{n-k}\mid \operatorname{Card}\{N_{j}: N_{j}=x_{1j}\}=m \bigr\} , $$
in which
$$ \mathcal{M}^{n-k}= \bigl\{ \mathfrak{N}_{n-k}=(N_{k+1}, \ldots ,N_{n})\mid N_{j}\in \{x_{1j}\pm x_{2j},x_{1j}\} \bigr\} . $$
In the following result, we describe a multiquadratic–cubic mapping as an equation. The proof is similar to the proof of [8, Proposition 2.1], but we include some parts for the sake of completeness.
Proposition 2.6
Let \(n\in \mathbb{N}\) and \(k\in \{0,\ldots ,n\}\). If a mapping \(f: V^{n}\longrightarrow W\) is k-quadratic and \(n-k\)-cubic mapping, then f satisfies the equation
$$ \sum_{s\in \{-1,1\}^{k}}\sum _{t\in \{-2,2\}^{n-k}} f \bigl(x_{1}^{k}+sx_{2}^{k},x_{1}^{n-k}+tx_{2}^{n-k} \bigr)=2^{k}\sum_{m=0}^{n-k}4^{n-k-m}(-6)^{m} \sum_{i\in \{1,2\}}f \bigl(x_{i}^{k}, \mathcal{M}_{m}^{n-k} \bigr) $$
(2.13)
for all \(x_{i}^{k}=(x_{i1},\ldots ,x_{ik})\in V^{k}\) and \(x_{i}^{n-k}=(x_{i,k+1},\ldots ,x_{in})\in V^{n-k}\), where \(i\in \{1,2\}\).
Proof
Since for \(k\in \{0,n\}\) our assertion follows from Proposition 2.4 and [45, Theorem 3], we can assume that \(k\in \{1,\ldots ,n-1\}\). Let \(x^{n-k}\in V^{n-k}\) be arbitrary and fixed. Consider the mapping \(g_{x^{n-k}}:V^{k}\longrightarrow W\) defined via \(g_{x^{n-k}} (x^{k} ):=f (x^{k},x^{n-k} )\) for \(x^{k}\in V^{k}\). Similar to the proof of Proposition 2.1 from [8], one can show that
$$ \sum_{s\in \{-1,1\}^{k}}f \bigl(x_{1}^{k}+sx_{2}^{k},x^{n-k} \bigr)=2^{k} \sum_{j_{1},\ldots ,j_{k}\in \{1,2\}}f \bigl(x_{j_{1}1},\ldots ,x_{j_{k}k},x^{n-k} \bigr) $$
(2.14)
for all \(x_{1}^{k},x_{2}^{k}\in V^{k}\) and \(x^{n-k}\in V^{n-k}\). Similar to the above, we obtain from Proposition 2.4 that
$$ \sum_{t\in \{-2,2\}^{n-k}} f \bigl(x^{k},x_{1}^{n-k}+tx_{2}^{n-k} \bigr)=\sum_{m=0}^{n-k}4^{n-k-m}(-6)^{m}f \bigl(x^{k},\mathcal{M}_{m}^{n-k} \bigr) $$
(2.15)
for all \(x_{1}^{n-k},x_{2}^{n-k}\in V^{n-k}\) and \(x^{k}\in V^{k}\). Now, equalities (2.14) and (2.15) show that (2.13) holds for f. □
It is easily seen that the mapping \(f:\mathbb{R}^{n}\longrightarrow \mathbb{R}\) defined by \(f(r_{1},\ldots ,r_{n})=\prod_{j=1}^{k}\prod_{i=k+1}^{n}r_{j}^{2}r_{i}^{3}\) is multiquadratic–cubic and hence (2.13) is valid for f by Proposition 2.6. Therefore, this equation is called a multiquadratic–cubic functional equation. Note that in the case \(k=n\) and \(k=0\), equation (2.13) converts to (1.7) and (2.11), respectively.
Definition 2.7
Let V and W be vector spaces over \(\mathbb{Q}\), \(n\in \mathbb{N}\). A mapping \(f:V^{n}\longrightarrow W\) is said to be n-multimixed quadratic–cubic, or briefly multimixed quadratic–cubic, if f satisfies (1.9) in each variable.
Let \(x_{1},x_{2}\in V^{n}\) and \(p_{l}\in \mathbb{N}_{0}\) with \(0\leq p_{l}\leq n\), where \(l\in \{1,2,3,4\}\). Set
$$ \mathbb{M}^{n}= \bigl\{ \mathfrak{M}_{n}=(M_{1}, \ldots ,M_{n})\mid M_{j} \in \{x_{1j}\pm x_{2j},x_{1j},x_{2j},-x_{2j}\} \bigr\} , $$
for all \(j\in \{1,\ldots ,n\}\). Consider the subset \(\mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n}\) of \(\mathbb{M}^{n}\) as follows:
$$\begin{aligned} \mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n} : =&\bigl\{ \mathfrak{M}_{n} \in \mathbb{M}^{n}\mid \operatorname{Card}\{M_{j}: M_{j}=x_{1j}\}=p_{1}, \\ & \operatorname{Card}\{M_{j}: M_{j}=x_{2j} \}=p_{2}, \operatorname{Card}\{M_{j}: M_{j}=-x_{2j} \}=p_{3}, \\ & \operatorname{Card}\{M_{j}: M_{j}=x_{1j}+x_{2j} \}=p_{4}\bigr\} . \end{aligned}$$
Hereafter, for a multimixed quadratic–cubic mapping f, we use the notations
$$ f \bigl(\mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n} \bigr):= \sum_{ \mathfrak{M}_{n}\in \mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n}}f ( \mathfrak{M}_{n} ), $$
(2.16)
and
$$ f \bigl(\mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},z \bigr):=\sum _{ \mathfrak{M}_{n}\in \mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n}}f ( \mathfrak{M}_{n},z )\quad (z\in V). $$
For each \(x_{1},x_{2}\in V^{n}\), we consider the equation
$$ f(x_{1}+2x_{2})=\sum _{p_{1}=0}^{n}\sum_{p_{2}=0}^{n-p_{1}} \sum_{p_{3}=0}^{n-p_{1}-p_{2}} \sum _{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f \bigl(\mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n} \bigr), $$
(2.17)
where \(f (\mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n} )\) is defined in (2.16).
Definition 2.8
Given a mapping \(f: V^{n}\longrightarrow W\). We say f
-
(i)
has zero condition if \(f(x)=0\) for any \(x\in V^{n}\) with at least one component that is equal to zero;
-
(ii)
is even in the jth variable if
$$ f(z_{1},\ldots ,z_{j-1},-z_{j},z_{j+1}, \ldots , z_{n})=f(z_{1}, \ldots ,z_{j-1},z_{j},z_{j+1}, \ldots , z_{n}); $$
-
(iii)
is odd in the jth variable if
$$ f(z_{1},\ldots ,z_{j-1},-z_{j},z_{j+1}, \ldots , z_{n})=-f(z_{1}, \ldots ,z_{j-1},z_{j},z_{j+1}, \ldots , z_{n}). $$
In what follows, it is assumed that every mapping \(f: V^{n}\longrightarrow W\) satisfying (2.17) has zero condition. With this assumption, we unify the general system of mixed-type quadratic and cubic functional equations defining a multimixed quadratic–cubic mapping to an equation and indeed this functional equation describe a multimixed quadratic–cubic mapping as well.
Proposition 2.9
A mapping \(f: V^{n}\longrightarrow W\) is multimixed quadratic–cubic if and only if it satisfies equation (2.17).
Proof
Suppose that f is a multimixed quadratic–cubic mapping. We proceed with the proof by induction on n. For \(n=1\), it is obvious that f satisfies equation (1.9). Let (2.17) be true for some fixed and positive integer \(n>1\). Then,
$$\begin{aligned} &f \bigl(x_{1}^{n+1}+2x_{2}^{n+1},z \bigr) \\ &\quad =\sum_{p_{1}=0}^{n} \sum _{p_{2}=0}^{n-p_{1}} \sum_{p_{3}=0}^{n-p_{1}-p_{2}} \sum_{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f \bigl(\mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},z \bigr), \end{aligned}$$
(2.18)
for all \(x_{1},x_{2}\in V^{n}\) and \(z\in V\). Using (2.18) and the fact that (2.17) holds for the case \(n=1\), we obtain
$$\begin{aligned} &f \bigl(x_{1}^{n+1}+2x_{2}^{n+1} \bigr) \\ &\quad =f \bigl(x_{1}^{n}+2x_{2}^{n},x_{1,n+1}+2x_{2,n+1} \bigr) \\ &\quad =3f \bigl(x_{1}^{n}+2x_{2}^{n},x_{1,n+1}+x_{2,n+1} \bigr)+f \bigl(x_{1}^{n}+2x_{2}^{n},x_{1,n+1}-x_{2,n+1} \bigr) \\ &\qquad {} -3f \bigl(x_{1}^{n}+2x_{2}^{n},x_{1,n+1} \bigr)+3f \bigl(x_{1}^{n}+2x_{2}^{n},x_{2,n+1} \bigr)-3f \bigl(x_{1}^{n}+2x_{2}^{n},-x_{2,n+1} \bigr) \\ &\quad =3\sum_{p_{1}=0}^{n} \sum _{p_{2}=0}^{n-p_{1}} \sum _{p_{3}=0}^{n-p_{1}-p_{2}} \sum_{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f \bigl(\mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},x_{1,n+1}+x_{2,n+1} \bigr) \\ &\qquad {} +\sum_{p_{1}=0}^{n} \sum _{p_{2}=0}^{n-p_{1}} \sum _{p_{3}=0}^{n-p_{1}-p_{2}} \sum_{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f \bigl(\mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},x_{1,n+1}-x_{2,n+1} \bigr) \\ &\qquad {} -3\sum_{p_{1}=0}^{n} \sum _{p_{2}=0}^{n-p_{1}} \sum _{p_{3}=0}^{n-p_{1}-p_{2}} \sum_{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f \bigl(\mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},x_{1,n+1} \bigr) \\ &\qquad {} +3\sum_{p_{1}=0}^{n} \sum _{p_{2}=0}^{n-p_{1}} \sum _{p_{3}=0}^{n-p_{1}-p_{2}} \sum_{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f \bigl(\mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},x_{2,n+1} \bigr) \\ &\qquad {} -3\sum_{p_{1}=0}^{n} \sum _{p_{2}=0}^{n-p_{1}} \sum _{p_{3}=0}^{n-p_{1}-p_{2}} \sum_{p_{4}=0}^{n-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f \bigl(\mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},-x_{2,n+1} \bigr) \\ &\quad =\sum_{p_{1}=0}^{n+1} \sum _{p_{2}=0}^{n+1-p_{1}} \sum _{p_{3}=0}^{n+1-p_{1}-p_{2}} \sum_{p_{4}=0}^{n+1-p_{1}-p_{2}-p_{3}}(-3)^{p_{1}+p_{3}}3^{p_{2}+p_{4}}f \bigl(\mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n+1} \bigr). \end{aligned}$$
This means that (2.17) holds for \(n+1\).
Conversely, let \(j\in \{1,\ldots ,n\}\) be arbitrary and fixed. Set
$$ f^{*}_{j}(z):=f (z_{1},\ldots ,z_{j-1},z,z_{j+1},\ldots ,z_{n} ). $$
Putting \(x_{2k}=0\) for all \(k\in \{1,\ldots ,n\}\backslash \{j\}\), \(x_{2j}=w\) and \(x_{1}= (z_{1},\ldots ,z_{j-1},z,z_{j+1},\ldots ,z_{n} )\) in (2.17), we obtain
$$\begin{aligned} f^{*}_{j}(z+2w)&=\left [\sum_{p_{1}=0}^{n-1}\sum _{p_{4}=1}^{n-p_{1}} \begin{pmatrix} n-1 \\ p_{1} \end{pmatrix} \begin{pmatrix} n-1-p_{1} \\ p_{4}-1 \end{pmatrix}(-3)^{p_{1}}3^{p_{4}}\right ]f^{*}_{j}(z+w) \\ &\quad {} +\left [\sum_{p_{1}=0}^{n-1} \sum_{p_{4}=0}^{n-1-p_{1}}\begin{pmatrix} n-1 \\ p_{1} \end{pmatrix} \begin{pmatrix} n-1-p_{1} \\ p_{4} \end{pmatrix}(-3)^{p_{1}}3^{p_{4}}\right ]f^{*}_{j}(z-w) \\ &\quad {}+\left [\sum_{p_{1}=1}^{n} \sum_{p_{4}=0}^{n-p_{1}}\begin{pmatrix} n-1 \\ p_{1}-1 \end{pmatrix} \begin{pmatrix} n-p_{1} \\ p_{4} \end{pmatrix}(-3)^{p_{1}}3^{p_{4}}\right ]f^{*}_{j}(z) \\ &\quad {}+3\left [\sum_{p_{1}=0}^{n-1} \sum_{p_{4}=0}^{n-1-p_{1}}\begin{pmatrix} n-1 \\ p_{1} \end{pmatrix} \begin{pmatrix} n-1-p_{1} \\ p_{4} \end{pmatrix}(-3)^{p_{1}}3^{p_{4}}\right ]f^{*}_{j}(w) \\ &\quad {}-3\left [\sum_{p_{1}=0}^{n-1}\sum _{p_{4}=0}^{n-1-p_{1}}\begin{pmatrix} n-1 \\ p_{1} \end{pmatrix} \begin{pmatrix} n-1-p_{1} \\ p_{4} \end{pmatrix}(-3)^{p_{1}}3^{p_{4}}\right ]f^{*}_{j}(-w). \end{aligned}$$
(2.19)
On the other hand,
$$\begin{aligned} &\sum_{p_{1}=0}^{n-1} \begin{pmatrix} n-1 \\ p_{1} \end{pmatrix}(-3)^{p_{1}}\sum_{p_{4}=0}^{n-1-p_{1}} \begin{pmatrix} n-1-p_{1} \\ p_{4} \end{pmatrix}3^{p_{4} +1}\times 1^{n-1-p_{4}} \\ &\quad =3\sum_{p_{1}=0}^{n-1} \begin{pmatrix} n-1 \\ p_{1} \end{pmatrix}(-3)^{p_{1}}4^{n-1-p_{1}}=3(4-3)^{n-1}=3. \end{aligned}$$
(2.20)
Similarly,
$$ \sum_{p_{1}=0}^{n-1} \begin{pmatrix} n-1 \\ p_{1} \end{pmatrix}(-3)^{p_{1}}\sum_{p_{4}=0}^{n-1-p_{1}} \begin{pmatrix} n-1-p_{1} \\ p_{4} \end{pmatrix}3^{p_{4}}\times 1^{n-1-p_{4}}=1. $$
(2.21)
Moreover,
$$ \sum_{p_{1}=0}^{n-1} \begin{pmatrix} n-1 \\ p_{1} \end{pmatrix}(-3)^{p_{1}+1}\sum_{p_{4}=0}^{n-1-p_{1}} \begin{pmatrix} n-1-p_{1} \\ p_{4} \end{pmatrix}3^{p_{4}}\times 1^{n-1-p_{4}}=-3. $$
(2.22)
It follows from (2.19), (2.20), (2.21) and (2.22) that
$$ f^{*}_{j}(z+2w)=3f^{*}_{j}(z+w)+f^{*}_{j}(z-w)-3f^{*}_{j}(z)+3f^{*}_{j}(w)-3f^{*}_{j}(-w). $$
This completes the proof. □
Corollary 2.10
Suppose that a mapping \(f: V^{n}\longrightarrow W\) satisfies equation (2.17).
-
(i)
If f is even in each variable, then it is multiquadratic. Moreover, f satisfies equation (1.7);
-
(ii)
If f is odd in each variable, then it is multicubic. In addition, equation (2.11) is true for f;
-
(iii)
If f is even in each of some k variables and is odd in each of the other variables, then it is multiquadratic–cubic. In particular, f fulfilling equation (2.13).
Proof
(i) It is shown in Proposition 2.9 that for each j, \(f^{*}_{j}\) satisfies (1.9). Since it is assumed that \(f^{*}_{j}(0)=0\), the result follows from part (i) of Lemma 2.2.
(ii) This is a direct consequence of part (ii) of Lemma 2.2.
(iii) The result follows from the previous parts. □