Equalities and inequalities for several variables mappings – Journal of Inequalities and Applications

ByAbasalt Bodaghi

Jan 4, 2022

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:Xlongrightarrow Y) satisfies functional equation (1.9) if and only if there exist a quadratic function (A^{2}:Xlongrightarrow Y), a cubic function (A^{3}:Xlongrightarrow Y) and a constant (A^{0}) such that

$$h(x)= A^{0}+A^{2}(x)+A^{3}(x)$$

for all (xin 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:Xlongrightarrow Y) satisfies (1.9) for all (x,yin X).

1. (i)

If h is even and (h(0)=0), then it is quadratic;

2. (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,yin 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

$$h(x+2y)+h(x-2y)=4bigl[h(x+y)+h(x-y)bigr]-6h(x).$$

This completes the proof. □

Henceforth, let V and W be vector spaces over (mathbb{Q}), (nin mathbb{N}) and (x_{i}^{n}=(x_{i1},x_{i2},ldots ,x_{in})in V^{n}), where (iin {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 (min mathbb{N}_{0}) with (0leq mleq 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 (jin {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 (rin mathbb{Q}), we put (rmathcal{M}_{m}^{n}= { rmathfrak{N}_{n}: mathfrak{N}_{n}in mathcal{M}_{m}^{n} } ) in which (rmathfrak{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

$$f bigl(mathcal{M}_{m}^{n},z bigr):=sum _{mathfrak{N}_{n}in mathcal{M}_{m}^{n}}f(mathfrak{N}_{n},z)quad (zin V).$$

We recall that (binom{n}{m}:=frac{n!}{m!(n-m)!}) is the binomial coefficient, where (n, min mathbb{N}_{0}) with (ngeq 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.

1. (i)

f is multicubic;

2. (ii)

f satisfies

$$sum_{tin {-2,2}^{n}} f(x_{1}+tx_{2})= sum_{m=0}^{n}4^{n-m}(-6)^{m}f bigl( mathcal{M}_{m}^{n}bigr),$$

(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_{tin {-2,2}^{n+1}} fbigl(x_{1}^{n+1}+tx_{2}^{n+1} bigr) \ &quad =4sum_{tin {-2,2}^{n}}sum _{sin {-1,1}} fbigl(x_{1}^{n}+tx_{2}^{n},x_{1,n+1}+sx_{2,n+1} bigr)-6 sum_{tin {-2,2}^{n}} fbigl(x_{1}^{n}+tx_{2}^{n},x_{1,n+1} bigr) \ &quad =4sum_{m=0}^{n}sum _{sin {-1,1}}4^{n-m}(-6)^{m}fbigl( 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 (jin {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 (min {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 =4sum_{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 {}-6sum_{m=0}^{n-1}left [begin{pmatrix} n \ m end{pmatrix}8^{n-1-m}(-6)^{m}right ]f^{*}_{j}(z) \ &quad =4times 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 (nin mathbb{N}) and (kin {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 (iin {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 (nin mathbb{N}) and (kin {0,ldots ,n}). If a mapping (f: V^{n}longrightarrow W) is kquadratic and (n-k)cubic mapping, then f satisfies the equation

$$sum_{sin {-1,1}^{k}}sum _{tin {-2,2}^{n-k}} f bigl(x_{1}^{k}+sx_{2}^{k},x_{1}^{n-k}+tx_{2}^{n-k} bigr)=2^{k}sum_{m=0}^{n-k}4^{n-k-m}(-6)^{m} sum_{iin {1,2}}f bigl(x_{i}^{k}, mathcal{M}_{m}^{n-k} bigr)$$

(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 (iin {1,2}).

Proof

Since for (kin {0,n}) our assertion follows from Proposition 2.4 and [45, Theorem 3], we can assume that (kin {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

$$sum_{sin {-1,1}^{k}}f bigl(x_{1}^{k}+sx_{2}^{k},x^{n-k} bigr)=2^{k} sum_{j_{1},ldots ,j_{k}in {1,2}}f bigl(x_{j_{1}1},ldots ,x_{j_{k}k},x^{n-k} bigr)$$

(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

$$sum_{tin {-2,2}^{n-k}} f bigl(x^{k},x_{1}^{n-k}+tx_{2}^{n-k} bigr)=sum_{m=0}^{n-k}4^{n-k-m}(-6)^{m}f bigl(x^{k},mathcal{M}_{m}^{n-k} bigr)$$

(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}), (nin 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 (0leq p_{l}leq n), where (lin {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 (jin {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

$$f bigl(mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n},z bigr):=sum _{ mathfrak{M}_{n}in mathbb{M}_{(p_{1},p_{2},p_{3},p_{4})}^{n}}f ( mathfrak{M}_{n},z )quad (zin V).$$

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

1. (i)

has zero condition if (f(x)=0) for any (xin V^{n}) with at least one component that is equal to zero;

2. (ii)

is even in the jth variable if

$$f(z_{1},ldots ,z_{j-1},-z_{j},z_{j+1}, ldots , z_{n})=f(z_{1}, ldots ,z_{j-1},z_{j},z_{j+1}, ldots , z_{n});$$

3. (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 (zin 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 =3sum_{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 {} -3sum_{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 {} +3sum_{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 {} -3sum_{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 (jin {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 (kin {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 {}+3left [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 {}-3left [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 =3sum_{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).

1. (i)

If f is even in each variable, then it is multiquadratic. Moreover, f satisfies equation (1.7);

2. (ii)

If f is odd in each variable, then it is multicubic. In addition, equation (2.11) is true for f;

3. (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. □