1 Introduction

Let H, K be unitary spaces. It is easy to check that, if \(f:H\rightarrow K\) satisfies \(\langle f(x)|f(y)\rangle =\langle x|y\rangle \), then f is an linear isometry. The above equation was generalized in normed spaces X, Y by considering a norm derivative \(\rho '_+(x,y){:}{=}\Vert x\Vert \cdot \lim \limits _{t\rightarrow 0^+}\frac{\Vert x+ty\Vert -\Vert x\Vert }{t}\) instead of inner product, i.e.

$$\begin{aligned} \rho '_+(f(x),f(y))=\rho '_+(x,y), \quad x,y\in X, \end{aligned}$$
(1)

with an unknown function \(f:X\rightarrow Y\). Note that if the norm comes from an inner product \(\langle \cdot ,\cdot \rangle \), we obtain \(\rho '_+(x,y)=\langle x|y\rangle \). Another generalization of the orthogonality equation in Hilbert spaces H, K is to look for the solutions of

$$\begin{aligned} \langle f(x)|g(y)\rangle =\langle x|y\rangle ,\quad x,y\in H, \end{aligned}$$
(2)

where \(f,g:H\!\rightarrow \!K\) are unknown functions. Solutions of (1) and (2) can be found in the authors’ previous papers [3, 4, 6]. Another generalization of (2) we can find in the paper [5] where the author studies the equation

$$\begin{aligned} \langle f(x)|g(y^* )\rangle =\langle x|y^* \rangle ,\quad x\in E,y^* \in F^*, \end{aligned}$$

where \(f:E\rightarrow F\), \(g:E^* \rightarrow F^*\), E, F are Banach spaces, \(E^*,F^*\) are spaces dual to E and F respectively, and \(\langle a|\varphi \rangle {:}{=}\varphi (a)\).

In this paper we will give a natural generalization of such functional equations in the case of abelian groups. In this case we will consider biadditive mappings instead of inner products.

2 Preliminaries

We start by recalling here some notions and results from the theory of groups and semigroups (see [2, Appendix A]).

Definition 1

A group is torsion if every element has the finite order.

A group is torsion-free if every element except the identity has the infinite order.

Definition 2

A semigroup \((H,+)\) is said to be divisible if

$$\begin{aligned} \forall _{x\in H}\, \forall _{n\in {\mathbb {N}}}\, \exists _{y\in H}\, x =ny. \end{aligned}$$

Let p be a prime number. The Prüfer p-group is the unique p-group in which every element has p different p-th roots. Alternatively we can write \({\mathbb {Z}}(p^\infty ) = {\mathbb {Z}}[1/p] /{\mathbb {Z}}\), where \({\mathbb {Z}}[1/p]= \{ \frac{m}{p^n}: m\in {\mathbb {Z}},\ n\in {\mathbb {N}}_0 \}\), \({\mathbb {N}}_0{:}{=}{\mathbb {N}} \cup \{0\}\). It is known fact that Prüfer p-groups are divisible and torsion.

Definition 3

Let \(A_i\), \(i \in I\), be groups. The direct sum \(\bigoplus \limits _{i \in I} A_i\) is the set of tuples \((a_i)_{i \in I} \in \prod \limits _{i\in I}A_i\) such that \(a_i\ne 0\) for finitely many \(i\in I\).

Remark 1

There exist an abelian divisible group G and divisible subgroups D, K of G such that \(D\cap K\) is not divisible.

Lemma 1

Let G be an abelian group, D, K be divisible subgroups of G. Then \(D+K\) is divisible.

Proof

Let \(x\in D\), \(y\in K\), \(n\in {\mathbb {N}}\). Then there exist \(u\in D\) and \(v\in K\) such that \(x=nu\) and \(y=nv\). Hence \(x+y=n(u+v)\). \(\square \)

Theorem 1

Let G be an abelian divisible group, \(D_1, D_2\) be divisible subgroups of G and \(D_1\cap D_2\) be divisible. Then there exist divisible groups \(K_0, K_1, K_2, K_3\) such that \(G=\bigoplus \limits _{i=0}^{3} K_i\), \(D_2 =K_0 \oplus K_1\), \(D_1 =K_0 \oplus K_2\).

Proof

Let \(K_0 =D_1\cap D_2\). Then there exist divisible groups \(K_1, K_2\) such that \(D_2 =K_0 \oplus K_1\), \(D_1 =K_0 \oplus K_2\). We show that \(K_1 \cap K_2 =\{ 0 \}\). Let \(x\in K_1 \cap K_2\), then \(x\in D_1 \cap D_2 =K_0\). Hence \(x=0\). Finally, there exists a divisible group \(K_3\) such that \(G=\left( \bigoplus \limits _{i=0}^{2} K_i \right) \oplus K_3\). \(\square \)

After these preparations we may now pass to multi-additive functions. By \(\mathrm {Perm}(n)\) we denote the set of all bijections of the set \(\{1, \dots , n\}\).

Definition 4

Let S be a semigroup, H be a group, \(n\in {\mathbb {N}}\). The function \(A :S^n\rightarrow H\) is called n-additive if

$$\begin{aligned}&A(x_1,\ldots ,x_{i-1}, x_i + y, x_{i+1}, \ldots , x_n)\\&\quad = A(x_1, \ldots , x_n) + A(x_1,\ldots ,x_{i-1},y,x_{i+1}, \ldots , x_n),\\ \end{aligned}$$

for all \(y, x_1, \ldots , x_n \in S\) and \(i\in \{1, \ldots , n\}\).

Moreover, A is called symmetric if

$$\begin{aligned}&A(x_1, \ldots , x_n) = A_n(x_{\sigma (1)} , \ldots , x_{\sigma (n)}) \end{aligned}$$

for all \(x_1, \ldots , x_n \in S\) and \(\sigma \in \mathrm {Perm}(n)\).

Lemma 2

Let H, X be groups, H be divisible. Let further \(B:H^2 \rightarrow X\) be a biadditive function. Then for every element \(x\in H\) of the finite order we have

$$\begin{aligned} B(x,y)=B(y,x)=0,\ y\in H. \end{aligned}$$

Remark 2

The previous lemma can be easily extended to the n-additive functions for \(n\ge 2\).

We use following two lemmas to show the existence of some biadditive map from \({\mathbb {Q}}^2\) to \({\mathbb {Z}}(2^\infty )\).

Lemma 3

Let \(k\in {\mathbb {N}}\), \(l\in 2{\mathbb {N}}-1\). Then there exists exactly one number \(\varphi (2^k,l)\in \{1,3,\ldots 2^k -1\}\) such that \(l\varphi (2^k,l)\equiv 1 (\bmod 2^k)\).

Proof

Let \(l(2i-1)\equiv r_i (\bmod 2^k)\), \(1\le r_i< 2^k\) for \(i\in \{1,2,\ldots 2^{k-1}\}\) . We observe that \(r_i\in 2{\mathbb {N}} -1\) and \(r_i \ne r_j\) for \(i\ne j\). Indeed, if \(r_i=r_j\), then \(l(2i-2j) \equiv 0 (\bmod 2^k)\) which means that \(i=j\). Hence there exists exactly one j such that \(l(2j-1)\equiv 1 (\bmod 2^k)\). \(\square \)

Lemma 4

Let \(k,m\in {\mathbb {N}}\), \(l,n\in 2{\mathbb {N}}-1\). Then

$$\begin{aligned}&n\varphi (2^k,ln) \equiv \varphi (2^k,l) (\bmod 2^k),\\&\varphi (2^{k+m},l)\equiv \varphi (2^k,l) (\bmod 2^k). \end{aligned}$$

Proof

We have

$$\begin{aligned}&l\Big ( n\varphi (2^k,ln) - \varphi (2^k,l)\Big )= ln\varphi (2^k,ln) - l\varphi (2^k,l)\equiv 0 (\bmod 2^k),\\&l\varphi (2^{k+m},l)=1+c2^{k+m}=1+(c2^m)2^k \equiv 1 (\bmod 2^k)\equiv l\varphi (2^k,l) (\bmod 2^k), \end{aligned}$$

for some \(c\in {\mathbb {N}}_0\) so

$$\begin{aligned}&l\Big ( \varphi (2^{k+m},l)- \varphi (2^k,l)\Big ) \equiv 0 (\bmod 2^k), \end{aligned}$$

which means that

$$\begin{aligned}&\varphi (2^{k+m},l)- \varphi (2^k,l) \equiv 0 (\bmod 2^k). \end{aligned}$$

Theorem 2

There exists a biadditive and symmetric function \(C:{\mathbb {Q}}^2 \rightarrow {\mathbb {Z}} (2^\infty )\) such that \(C(1,1)=\frac{1}{2} +{\mathbb {Z}}\).

Proof

A greatest common divisor in this proof will be denoted by \(\text {GCD}\). Let \(m,k\in {\mathbb {Z}}\), \(n,l\in {\mathbb {N}}\), \(\text {GCD}(m,n)=\text {GCD}(k,l)=1\). Let further \(s_n,s_l\in {\mathbb {N}}_0\) be such that \(2^{s_n} |n\), \(2^{s_n+1} \not | l\), \(2^{s_l} |l\), \(2^{s_l+1} \not | l\). We define C by the formula

$$\begin{aligned} C\left( \frac{m}{n},\frac{k}{l}\right) {:}{=}mk \frac{\varphi \left( 2^{s_n+ s_l +1},\frac{nl}{2^{s_n+ s_l}}\right) }{2^{s_n+ s_l +1}} +{\mathbb {Z}}. \end{aligned}$$

It is easy to see that C is symmetric, so we only show that C is additive in the first variable. Let \(p\in {\mathbb {Z}}\), \(q\in {\mathbb {N}}\), \(\text {GCD}(p,q)=1\), \(d=\text {GCD}(mq+np,nq)\). Let further \(s_q ,s_d\in {\mathbb {N}}_0\) be such that \(2^{s_q} |q\), \(2^{s_q+1} \not | q\) and \(2^{s_d} |d\), \(2^{s_d+1} \not | d\). Using Lemma 4 we get

$$\begin{aligned}&C\left( \frac{m}{n}+\frac{p}{q},\frac{k}{l}\right) = C\left( \frac{mq+np}{nq},\frac{k}{l}\right) =C\left( \frac{\frac{mq+np}{d}}{\frac{nq}{d}},\frac{k}{l}\right) \\&\quad = \left( \frac{mq+np}{d}\cdot k\right) \frac{\varphi \left( 2^{s_n +s_q -s_d + s_l +1},\frac{nql}{d2^{s_n +s_q -s_d +s_l}}\right) }{2^{s_n +s_q -s_d +s_l +1}} +{\mathbb {Z}}\\&\quad = \left( \frac{mq+np}{d}\cdot k \frac{d}{2^{s_d}}\right) \frac{\varphi \left( 2^{s_n +s_q -s_d + s_l +1},\frac{nql}{d2^{s_n +s_q -s_d +s_l} } \cdot \frac{d}{2^{s_d}}\right) }{2^{s_n +s_q -s_d +s_l +1}} +{\mathbb {Z}}\\&\quad = \left( \frac{mq+np}{d}\cdot k \frac{d}{2^{s_d}}\right) \frac{\varphi \left( 2^{s_n +s_q -s_d + s_l +1},\frac{nql}{2^{s_n +s_q +s_l} }\right) }{2^{s_n +s_q -s_d +s_l +1}} +{\mathbb {Z}}\\&\quad = \left( \frac{mq+np}{d}\cdot k \frac{d}{2^{s_d}}\right) \frac{\varphi \left( 2^{s_n +s_q +s_l +1},\frac{nql}{2^{s_n +s_q +s_l} }\right) }{2^{s_n +s_q -s_d +s_l +1}} +{\mathbb {Z}}\\&\quad = (mq+np)k \frac{\varphi \left( 2^{s_n +s_q+ s_l +1},\frac{nql}{2^{s_n +s_q +s_l}}\right) }{2^{s_n +s_q +s_l +1}} +{\mathbb {Z}}\\&\quad = (mqk) \frac{\varphi \left( 2^{s_n +s_q+ s_l +1},\frac{nql}{2^{s_n +s_q +s_l}}\right) }{2^{s_n +s_q +s_l +1}} +(npk)\frac{\varphi \left( 2^{s_n +s_q+ s_l +1},\frac{nql}{2^{s_q +s_n + s_l}}\right) }{2^{s_n +s_q+ s_l +1}} +{\mathbb {Z}}\\&\quad = (mk2^{s_q}\frac{q}{2^{s_q}}) \frac{\varphi \left( 2^{s_n +s_q+ s_l +1},\frac{nl}{2^{s_n+ s_l}}\cdot \frac{q}{2^{s_q}}\right) }{2^{s_n +s_q +s_l +1}} \\&\qquad + \left( pk2^{s_n}\frac{n}{2^{s_n}}\right) \frac{\varphi \left( 2^{s_n +s_q+ s_l +1},\frac{ql}{2^{s_q+ s_l}}\cdot \frac{n}{2^{s_n}}\right) }{2^{s_n +s_q+ s_l +1}} +{\mathbb {Z}}\\&\quad = (mk) \frac{\varphi \left( 2^{s_n +s_q+ s_l +1},\frac{nl}{2^{s_n+ s_l}}\right) }{2^{s_n+ s_l +1}} + (pk) \frac{\varphi \left( 2^{s_n +s_q+ s_l +1},\frac{ql}{2^{s_q+ s_l}}\right) }{2^{s_q+ s_l +1}} +{\mathbb {Z}}\\&\quad = (mk) \frac{\varphi \left( 2^{s_n+ s_l +1},\frac{nl}{2^{s_n+ s_l}}\right) }{2^{s_n+ s_l +1}} +{\mathbb {Z}} + (pk) \frac{\varphi \left( 2^{s_q+ s_l +1},\frac{ql}{2^{s_q+ s_l}}\right) }{2^{s_q+ s_l +1}} +{\mathbb {Z}}\\&\quad =C\left( \frac{m}{n},\frac{k}{l}\right) +C\left( \frac{p}{q},\frac{k}{l}\right) . \end{aligned}$$

The proof is complete. \(\square \)

Now we introduce some theory of the adjoint operator on groups.

Definition 5

Let S, H, X be groups, \(A:S^2\rightarrow X\), \(B:H^2\rightarrow X\) be biadditive functions. Let further \(T:S\rightarrow H\) and

$$\begin{aligned} D(T^*)=\{ v\in H:\, \exists _{y\in S} \forall _{x\in S}\, B(T(x),v)=A(x,y) \}. \end{aligned}$$

A function \(T^*:D(T^*)\rightarrow S\) is called a (B, A)-adjoint operator (to T) if and only if

$$\begin{aligned} B(T(x),v)=A(x,T^*(v)),\ x\in S,\ v\in D(T^*). \end{aligned}$$

Lemma 5

Let S, H, X be groups, \(A:S^2\rightarrow X\), \(B:H^2\rightarrow X\) be biadditive functions. Let further \(T:S\rightarrow H\) and \(T^*:D(T^*)\rightarrow S\) be a (B, A)-adjoint operator to T,

$$\begin{aligned} S_{AR}&:= \{ y\in S:\, \forall _{x\in S}\, A(x,y)=0 \}, \end{aligned}$$
(3)
$$\begin{aligned} S_{ALT^*}&:= \{ x\in S:\, \forall _{y\in \text {im }T^*}\, A(x,y)=0 \}, \end{aligned}$$
(4)
$$\begin{aligned} H_{BTR}&:= 3 \{ v\in H:\, \forall _{u\in \text {im }T}\, B(u,v)=0 \}, \end{aligned}$$
(5)
$$\begin{aligned} H_{BLD^*}&:= \{ u\in H:\, \forall _{v\in D(T^*)}\, B(u,v)=0 \}. \end{aligned}$$
(6)

Then

  1. 1.

    \(D(T^*)\) is a group, \(S_{AR}\), \(S_{ALT^*}\) are normal subgroups of S, \(H_{BTR}\), \(H_{BLD^*}\) are normal subgroups of H. Moreover in the case when X is torsion-free, if H is divisible, then \(H_{BTR}\), \(H_{BLD^*}\) are divisible, if S is divisible, then \(S_{AR}\), \(S_{ALT^*}\) are divisible, if S, H are divisible, then \(D(T^*)\) is divisible;

  2. 2.

    \(\forall _{x,y\in S}\, T(x+y)-T(y)-T(x)\in H_{BLD^*}\);

  3. 3.

    \(\forall _{x,y\in S}\, x-y\in S_{ALT^*} \Leftrightarrow T(x)-T(y)\in H_{BLD^*}\);

  4. 4.

    \(\forall _{u,v\in D(T^*)}\, T^*(u+v)-T^*(v)-T^*(u)\in S_{AR}\);

  5. 5.

    \(\forall _{u,v\in D(T^*)}\, u-v\in H_{BTR} \Leftrightarrow T^*(u)-T^*(v)\in S_{AR}\);

  6. 6.

    \(H_{BTR}\subset D(T^*)\);

  7. 7.

    Assume that H is abelian and divisible. Let K be a subgroup of H such that \(H=K \oplus H_{BTR}\), \(\varkappa :S\rightarrow S/S_{AR}\) be a canonical homomorphism. Then \(D(T^*)\cap K\) is a group and \({\widetilde{T}}^*{:}{=}\varkappa \circ T^* :D(T^*)\cap K \rightarrow \text {im }T^*/S_{AR}\) is an isomorphism.

Proof

  1. 1.

    Since kernel of any homomorphism is a normal subgroup, then \(S_{AR}\), \(S_{ALT^*}\) are normal subgroups of S, \(H_{BTR}\), \(H_{BLD^*}\) are normal subgroups of H.

    Moreover, if S is divisible and X is torsion-free, then for \(x\in S_{ALT^*}\) and \(n\in {\mathbb {N}}\) there exists \(z\in S\) such that \(nz=x\). We have

    $$\begin{aligned}&nA(z,T^*(u))=A(nz,T^*(u))=A(x,T^*(u))=0,\ u\in D(T^*). \end{aligned}$$

    Since X is torsion-free, then \(z\in S_{ALT^*}\).

  2. 2.

    Let \(x,y\in S\), \(v\in D(T^*)\). Then

    $$\begin{aligned}&B(T(x+y)-T(y)-T(x),v)\\&\quad =B(T(x+y),v)-B(T(y),v)-B(T(x),v)\\&\quad =A(x+y,T^*(v))-A(y,T^* (v))-A(x,T^*(v))\\&\quad =A(x+y-y-x,T^*(v))=A(0,T^*(v))=0, \end{aligned}$$

    which shows that \(T(x+y)-T(y)-T(x)\in H_{BLD^*}\).

  3. 3.

    Let \(x,y\in S\), \(v\in D(T^*)\). Then

    $$\begin{aligned} B(T(x)-T(y),v)&=B(T(x),v)-B(T(y),v)\\&=A(x,T^* (v))-A(y,T^*(v))=A(x-y,T^*(v)), \end{aligned}$$

    which shows that \(x-y\in S_{ALT^*} \Leftrightarrow T(x)-T(y)\in H_{BLD^*}\).

  4. 4.

    Let \(u,v\in D(T^*)\), \(x\in S\).

    $$\begin{aligned}&A(x,T^*(u+v)-T^*(v)-T^*(u))\\&\quad =A(x,T^*(u+v))-A(x,T^*(v))-A(x,T^*(u))\\&\quad =B(T(x),u+v)-B(T(x),v)-B(T(x),u)\\&\quad =B(T(x),u+v-v-u)=B(T(x),0)=0, \end{aligned}$$

    which shows that \(T^*(u+v)-T^*(v)-T^*(u)\in S_{AR}\).

  5. 5.

    Let \(u,v\in D(T^*)\), \(x\in S\). Then

    $$\begin{aligned} B(T(x),u-v)&=B(T(x),u)-B(T(x),v)=A(x,T^* (u))-A(x,T^*(v))\\&=A(x,T^*(u)-T^*(v)), \end{aligned}$$

    which shows that \(u-v\in H_{BTR} \Leftrightarrow T^*(u)-T^*(v)\in S_{AR}\).

  6. 6.

    Let \(u\in H_{BTR}\) and \(y\in S_{AR}\). Then

    $$\begin{aligned}&B(T(x),u)=0=A(x,y),\ x\in S, \end{aligned}$$

    which shows that \(u\in D(T^*)\).

  7. 7.

    Let \(u,v\in D(T^*)\). Then using property 4 we obtain

    $$\begin{aligned}&(T^*(u)+S_{AR}) + (T^*(v)+S_{AR})=T^*(u+v)+S_{AR},\\&(T^*(u)+S_{AR}) + (T^*(-u)+S_{AR})=T^*(0)+S_{AR}=S_{AR}, \end{aligned}$$

    so \(\text {im }T^*/S_{AR}\) is a group. Using property 4 we obtain that \({\widetilde{T}}^*\) is a homomorphism, from 5 we get that \({\widetilde{T}}^*\) is injective. Let \(y=T^* (u)\) for some \(u\in D(T^*)\). Since \(H=K\oplus H_{BTR}\), then \(u=u_1+u_2\), where \(u_1\in K\), \(u_2\in H_{BTR}\). From 6 we have \(u_1=u-u_2\in D(T^*)\). Using property 5 we get \(T^* (u)-T^* (u_1)\in S_{AR}\), so

    $$\begin{aligned}&{\widetilde{T}}^* (u_1) =\varkappa (T^* (u_1))=\varkappa (T^* (u))=\varkappa (y), \end{aligned}$$

    which shows that \({\widetilde{T}}^*\) is surjective.

\(\square \)

Using property 7 from Lemma 5 we can accept the following

Definition 6

Let S, H, X be groups, H be abelian and divisible, \(A:S^2\rightarrow X\), \(B:H^2\rightarrow X\) be biadditive functions. Let further \(T:S\rightarrow H\) and \(T^*:D(T^*)\rightarrow S\) be a (B, A)-adjoint operator to T, \(\text {im }T^*/S_{AR}=S/S_{AR}\), K be a subgroup of H such that \(H=K \oplus H_{BTR}\). We define the function \((T^*)^{-1}:S\rightarrow D(T^*)\cap K\) by the formula

$$\begin{aligned} (T^*)^{-1} (x)= ({\widetilde{T}}^*)^{-1} (\varkappa (x)),\ x\in S. \end{aligned}$$
(7)

Remark 3

The function \((T^*)^{-1}\) from the above definition is additive and \(\text {im }(T^*)^{-1}=D(T^*)\cap K\).

3 Main results

Assume that \((S,+)\) is a semigroup, \((H,+)\) is a divisible abelian group, \((X,+)\) is a torsion-free group, \(A:S^2 \rightarrow X\), \(B:H^2 \rightarrow X\) are biadditive functions.

Theorem 3

Let \(f,g:S\rightarrow H\). Then (f, g) satisfies

$$\begin{aligned} B(f(x),g(y))=A(x,y),\ x,y\in S, \end{aligned}$$
(8)

if and only if there exist divisible groups \(H_0, H_1, H_2, H_3\), additive functions \(f_a:S\rightarrow H_2 \oplus H_3\), \(g_a:S\rightarrow H_1 \oplus H_3\) and functions \(f_r :S\rightarrow H_0 \oplus H_1\), \(g_r :S\rightarrow H_0 \oplus H_2\) such that

$$\begin{aligned}&H=\bigoplus \limits _{i=0}^3 H_i \text { and } H_1,H_2,H_3 \text { are torsion-free}, \end{aligned}$$
(9)
$$\begin{aligned}&f=f_a +f_r ,\ g=g_a +g_r , \end{aligned}$$
(10)
$$\begin{aligned}&(H_0 \oplus H_1 )\times (H_0 \oplus H_2 )\subset B^{-1}(\{0\}), \end{aligned}$$
(11)
$$\begin{aligned}&\text {im }f_a \times (H_0 \oplus H_2 )\subset B^{-1}(\{0\}), \end{aligned}$$
(12)
$$\begin{aligned}&(H_0 \oplus H_1 )\times \text {im }g_a \subset B^{-1}(\{0\}), \end{aligned}$$
(13)
$$\begin{aligned}&B(f_a(x),g_a(y))=A(x,y),\ x,y\in S. \end{aligned}$$
(14)

Moreover, we can assume that \(H_0\oplus H_2=\{ v\in H:\,\forall _{u\in \text {im }f}\, B(u,v)=0 \}\).

Proof

(\(\Rightarrow \)) Let

$$\begin{aligned}&D_1 := \{ v\in H:\ \forall _{u\in \text {im }f}\ B(u,v)=0 \},\\&D_2 := \{ u\in H:\ \forall _{v\in \text {im }g + D_1}\ B(u,v)=0 \}. \end{aligned}$$

It is easy to see that above sets are groups. We show that \(D_1, D_2, D_1 \cap D_2\) are divisible.

Let \(v\in D_1\) and \(n\in {\mathbb {N}}\). Then there exists \(w\in H\) such that \(v=nw\). For every \(u\in \text {im }f\) we have

$$\begin{aligned}&nB(u,w)=B(u,nw)=B(u,v)=0, \end{aligned}$$

and since X is torsion-free, then \(w\in D_1\).

Let \(u\in D_2\) and \(n\in {\mathbb {N}}\). Then there exists \(w\in H\) such that \(u=nw\). For every \(v\in \text {im }g+D_1\) we have

$$\begin{aligned}&nB(w,v)=B(nw,v)=B(u,v)=0, \end{aligned}$$

and since X is torsion-free, then \(w\in D_2\).

Let \(x\in D_1 \cap D_2\) and \(n\in {\mathbb {N}}\). Then there exists \(z\in H\) such that \(x=nz\). Let \(u\in \text {im }f\) and \(v\in \text {im }g +D_1\). We have

$$\begin{aligned}&nB(u,z)=B(u,nz)=B(u,x)=0,\\&nB(z,v)=B(nz,v)=B(x,v)=0, \end{aligned}$$

and since X is torsion-free, then \(z\in D_1 \cap D_2\).

In view of Theorem 1 there exist divisible groups \(H_0, H_1, H_2, H_3\) such that \(D_2=H_0 \oplus H_1\), \(D_1=H_0 \oplus H_2\) and \(H=\bigoplus \limits _{i=0}^3 H_i\). In view of Lemma 2 every element of H of the finite order belongs to \(D_1 \cap D_2 = H_0\), so \(H_1, H_2, H_3\) are torsion-free. Let \(f=f_0 +f_1 +f_2 +f_3\), \(g=g_0 +g_1 +g_2 +g_3\), where \(f_i ,g_i :S\rightarrow H_i\) for \(i\in \{ 0,1,2,3\}\). Let further \(f_a {:}{=}f_2 +f_3\), \(g_a {:}{=} g_1 +g_3\). Hence \(f_r{:}{=}(f-f_a) :S\rightarrow H_0 \oplus H_1\) and \(g_r{:}{=}(g-g_a) :S\rightarrow H_0 \oplus H_2\).

We observe also that

$$\begin{aligned}&(H_0 \oplus H_1 )\times (H_0 \oplus H_2 )=D_2 \times D_1 \subset B^{-1}(\{0\}),\\&\text {im }f_a \times (H_0 \oplus H_2 )\subset (\text {im }f +D_2)\times D_1\subset B^{-1}(\{0\}),\\&(H_0 \oplus H_1 )\times \text {im }g_a \subset D_2 \times (\text {im }g +D_1) \subset B^{-1}(\{0\}). \end{aligned}$$

Now we show that \(f_a\) and \(g_a\) are additive. Let \(x,y\in S\), \(v\in D_1\). Then

$$\begin{aligned}&B(f_a (x+y)-f_a (y)-f_a (x),g(z)+v)= B(f(x+y)-f(y)-f(x),g(z))\\&\quad =B(f(x+y),g(z))-B(f(y),g(z))-B(f(x),g(z))\\&\quad =A(x+y,z)-A(y,z)-A(x,z)=0,\quad z\in S, \end{aligned}$$

which means that \(f_a (x+y)-f_a (y)-f_a (x)\in D_2\), so \(f_a (x+y)=f_a (x)+f_a (y)\). Similarly for \(g_a\) we have

$$\begin{aligned}&B(f(z),g_a (x+y)-g_a (y)-g_a (x))= B(f(z),g(x+y)-g(y)-g(x))\\&\quad =B(f(z),g(x+y))-B(f(z),g(y))-B(f(z),g(x))\\&\quad =A(z,x+y)-A(z,y)-A(z,x)=0,\quad z\in S, \end{aligned}$$

which means that \(g_a (x+y)-g_a (x)-g_a (y)\in D_1\), so \(g_a (x+y)=g_a (x)+g_a (y)\).

Moreover, using (11)–(13) we have

$$\begin{aligned} B(f_a (x),g_a (y))&=B(f_a (x),g_a (y))+B(f_r (x),g_a (y))\\&\quad +B(f_a (x),g_r (y))+B(f_r (x),g_r (y))\\&=B(f_a(x)+f_r(x),g_a (y)+g_r (y))\\&=B(f(x),g(y))=A(x,y),\quad x,y\in S. \end{aligned}$$

(\(\Leftarrow \)) Assume that there exist divisible groups \(H_0, H_1, H_2, H_3\), additive functions \(f_a:S\rightarrow H_2 \oplus H_3\), \(g_a:S\rightarrow H_1 \oplus H_3\) and functions \(f_r :S\rightarrow H_0 \oplus H_1\), \(g_r :S\rightarrow H_0 \oplus H_2\) such that conditions (9)–(14) holds. Then

$$\begin{aligned} B(f(x),g(y))&=B(f_a (x)+f_r (x),g_a (y)+g_r (y))\\&=B(f_a (x),g_a (y))+B(f_a (x),g_r (y))\\&\quad +B(f_r (x),g_a (y))+B(f_r (x),g_r (y))\\&=B(f_a (x),g_a (y))=A(x,y),\quad x,y\in S. \end{aligned}$$

\(\square \)

The following example shows that we cannot drop the assumption that X is torsion-free in the previous theorem.

Example 1

Let \(S={\mathbb {Z}}^2\), \(H={\mathbb {Q}}^2\), \(X={\mathbb {Q}}\times {\mathbb {Z}}(2^\infty )\), \(f,g:S\rightarrow H\) be functions given by formulas

$$\begin{aligned}&f(n,m)={\left\{ \begin{array}{ll} (n,1) &{}\quad n\in {\mathbb {Z}},\ m\in 2{\mathbb {Z}}+1\\ (n,2^{|m|+1}) &{}\quad n\in {\mathbb {Z}},\ m\in 2{\mathbb {Z}}\\ \end{array}\right. },\\&g(n,m)=(n,m),\ n,m\in {\mathbb {Z}}. \end{aligned}$$

Let further \(B:H^2\rightarrow X\), \(A:S^2 \rightarrow X\) be functions given by formulas

$$\begin{aligned} B\Big ((n,m),(p,q)\Big )&=(np,C(m,q)),\ n,m,p,q\in {\mathbb {Q}},\\ A(x,y)&=B(f(x),g(y)),\ x,y\in S, \end{aligned}$$

where \(C:{\mathbb {Q}}^2 \rightarrow {\mathbb {Z}} (2^\infty )\) is a biadditive and symmetric function such that \(C(1,1)=\frac{1}{2}+{\mathbb {Z}}\) (see Theorem 2).

It is easy to see that g is additive, B is biadditive and symmetric.

Since for all \(x,y\in S\) we have \(f(x+y)-f(x)-f(y)\in \{0\}\times 2{\mathbb {Z}}\), then for every \(z=(z_1,z_2)\in S\) there is an \(n\in {\mathbb {Z}}\) such that

$$\begin{aligned}&A(x+y,z)-A(x,z)-A(y,z)\\&\quad =B(f(x+y),z)-B(f(x),z)-B(f(y),z)\\&\quad =B(f(x+y)-f(x)-f(y),z)=(0\cdot z_1,C(2n,z_2))\\&\quad =(0,2n z_2 C(1,1))=\left( 0 ,2n z_2 \frac{1}{2} +{\mathbb {Z}}\right) =(0,{\mathbb {Z}}). \end{aligned}$$

Hence A is biadditive and (f, g) solves (8).

Suppose that there exist divisible groups \(H_0, H_1, H_2, H_3\), additive functions \(f_a:S\rightarrow H_2 \oplus H_3\), \(g_a:S\rightarrow H_1 \oplus H_3\) and functions \(f_r :S\rightarrow H_0 \oplus H_1\), \(g_r :S\rightarrow H_0 \oplus H_2\) such that conditions (9)–(13) holds. Since

$$\begin{aligned} {\mathbb {Z}}^2 =\text {im }g\subset \text {im }g_a +(H_0 \oplus H_2), \end{aligned}$$

then from (11), (13) we obtain

$$\begin{aligned}&(H_0 \oplus H_1) \times {\mathbb {Z}}^2 \subset B^{-1} (\{(0,{\mathbb {Z}})\}). \end{aligned}$$

Let \((p,q)\in H_0\oplus H_1\). Then there exists \(k\in {\mathbb {N}}\) such that \((kp,kq)\in {\mathbb {Z}}^2\). Hence, since \((kp,kq)\in H_0 \oplus H_1\), we get

$$\begin{aligned} (0,{\mathbb {Z}})=B\Big ((kp,kq),(1,1)\Big )=\bigg (kp,\frac{kq}{2}+{\mathbb {Z}}\bigg ), \end{aligned}$$

so \(p=0\) and \(kq\in 2{\mathbb {Z}}\). On the other hand, if \(q\ne 0\) and \((0,kq)\in H_0 \oplus H_1\), then, by Lemma 1, \((0,1)\in H_0\oplus H_1\). Consequently,

$$\begin{aligned} (0,{\mathbb {Z}})=B\Big ((0,1),(0,1)\Big )=\bigg (0,\frac{1}{2}+{\mathbb {Z}}\bigg ), \end{aligned}$$
(15)

a contradiction. Thus \(H_0 = H_1 = \{ 0\}\) and \(f_a =f\), but f is not additive, which give us a contradiction.

In the theorem below we investigate the preservation of the biadditivity by only one function, namely we solve the following generalization of the orthogonality equation.

Theorem 4

Let \(f:S\rightarrow H\). Then f satisfies

$$\begin{aligned} B(f(x),f(y))=A(x,y),\ x,y\in S, \end{aligned}$$
(16)

if and only if there exist divisible groups \(H_0, H_1\), an additive function \(F_a:S\rightarrow H_1\), and a function \(F_r :S\rightarrow H_0\) such that

$$\begin{aligned}&H=H_0\oplus H_1 \text { and } H_1 \text { is torsion-free}, \end{aligned}$$
(17)
$$\begin{aligned}&f=F_a +F_r , \end{aligned}$$
(18)
$$\begin{aligned}&H_0\times (H_0 \oplus \text {im }F_a) \subset B^{-1}(\{0\}), \end{aligned}$$
(19)
$$\begin{aligned}&(H_0 \oplus \text {im }F_a)\times H_0 \subset B^{-1}(\{0\}), \end{aligned}$$
(20)
$$\begin{aligned}&B(F_a(x),F_a(y))=A(x,y),\ x,y\in S. \end{aligned}$$
(21)

Moreover, we can assume that \(H_0 \subset \{ v\in H:\ \forall _{u\in \text {im }f}\, B(u,v)=0 \}\).

Proof

(\(\Rightarrow \)) In view of Theorem 3 there exist divisible groups \(K_0, K_1, K_2, K_3\), additive functions \(f_a:S\rightarrow K_2 \oplus K_3\), \({\widetilde{f}}_a:S\rightarrow K_1 \oplus K_3\) and functions \(f_r :S\rightarrow K_0 \oplus K_1\), \({\widetilde{f}}_r :S\rightarrow K_0 \oplus K_2\) such that

$$\begin{aligned}&H=\bigoplus \limits _{i=0}^3 K_i \text { and } K_1,K_2,K_3 \text { are torsion-free},\\&f=f_a +f_r ={\widetilde{f}}_a +{\widetilde{f}}_r ,\\&(K_0 \oplus K_1 )\times (K_0 \oplus K_2 )\subset B^{-1}(\{0\}),\\&\text {im }f_a \times (K_0 \oplus K_2 )\subset B^{-1}(\{0\}),\\&(K_0 \oplus K_1 )\times \text {im }{\widetilde{f}}_a \subset B^{-1}(\{0\}),\\&B(f_a(x),{\widetilde{f}}_a(y))=A(x,y),\ x,y\in S. \end{aligned}$$

Let \(f=f_0 +f_1 +f_2 +f_3\), where \(f_i :S\rightarrow K_i\) for \(i\in \{0,1,2,3\}\). Then \(f_a =f_2+f_3\) and \({\widetilde{f}}_a =f_1 +f_3\). Hence \(f_1,f_2,f_3\) are additive. Let \(H_0=K_0\), \(H_1=\bigoplus \limits _{i=1}^3 K_i\), \(F_a=f_1+f_2+f_3\), \(F_r =f_0\). Then \(F_a :S\rightarrow H_1\) is additive. We have also

$$\begin{aligned}&(H_0 \oplus K_1) \times H_0 \subset (K_0 \oplus K_1 )\times (K_0 \oplus K_2 )\subset B^{-1}(\{0\}),\\&\text {im }f_a \times H_0 \subset \text {im }f_a \times (K_0 \oplus K_2 )\subset B^{-1}(\{0\}),\\&H_0 \times (H_0 \oplus K_2 ) \subset (K_0 \oplus K_1 )\times (K_0 \oplus K_2 )\subset B^{-1}(\{0\}),\\&H_0 \times \text {im }{\widetilde{f}}_a \subset (K_0 \oplus K_1 )\times \text {im }{\widetilde{f}}_a \subset B^{-1}(\{0\}), \end{aligned}$$

and since B is biadditive we obtain that

$$\begin{aligned}&(H_0 \oplus \text {im }F_a )\times H_0 \subset (\text {im }f_a \oplus H_0 \oplus K_1) \times H_0 \subset B^{-1}(\{0\}),\\&H_0 \times (H_0 \oplus \text {im }F_a ) \subset H_0\times (\text {im }{\widetilde{f}}_a \oplus H_0 \oplus K_2) \subset B^{-1}(\{0\}). \end{aligned}$$

Consequently

$$\begin{aligned} B(F_a (x),F_a (y))&= B(F_a (x),F_a (y))+ B(F_a (x),F_r (y))+ B(F_r (x),F_a (y))\\&\quad + B(F_r (x),F_r (y))\\&=B(F_a (x)+F_r (x),F_a (y)+F_r (y))=A(x,y),\quad x,y\in S. \end{aligned}$$

(\(\Leftarrow \)) Assume that there exist divisible groups \(H_0, H_1\), an additive function \(F_a:S\rightarrow H_1\), and a function \(F_r :S\rightarrow H_0\) such that conditions (17)–(21) holds. Then

$$\begin{aligned} B(f(x),f(y))&=B(F_a (x)+F_r (x),F_a (y)+F_r (y))\\&=B(F_a (x),F_a (y))+ B(F_a (x),F_r (y))+B(F_r (x),F_a (y)) \\&\quad + B(F_r(x),F_r (y))\\&=B(F_a (x),F_a (y))=A(x,y),\quad x,y\in S. \end{aligned}$$

\(\square \)

It is a natural question whether given a function f there exists a function g such that (f, g) satisfies equation (8). The theorem below give us an answer for this question.

Theorem 5

Assume that S is a group, \(f,g:S\rightarrow H\). Then (f, g) satisfies equation (8) if and only if there exist divisible groups \(H_0, H_1, H_2, H_3\), an additive function \(T:S\rightarrow H_2 \oplus H_3\), functions \(f_r :S\rightarrow H_0 \oplus H_1\), \(g_r :S\rightarrow H_0 \oplus H_2\) such that

$$\begin{aligned}&H=\bigoplus \limits _{i=0}^3 H_i \text { and } H_1,H_2,H_3 \text { are torsion-free}, \end{aligned}$$
(22)
$$\begin{aligned}&\text {im }T^* /S_{AR}=S/S_{AR}, \end{aligned}$$
(23)
$$\begin{aligned}&f=T+f_r,\, g=(T^*)^{-1} +g_r, \end{aligned}$$
(24)
$$\begin{aligned}&(H_0 \oplus H_1 )\times (H_0 \oplus H_2 )\subset B^{-1}(\{0\}), \end{aligned}$$
(25)
$$\begin{aligned}&\text {im }T \times (H_0 \oplus H_2 )\subset B^{-1}(\{0\}), \end{aligned}$$
(26)
$$\begin{aligned}&(H_0 \oplus H_1 )\times (D(T^*)\cap K) \subset B^{-1}(\{0\}), \end{aligned}$$
(27)

where \(T^*:D(T^*)\rightarrow S\) is a (B, A)-adjoint operator to T, \(S_{AR}\) is given by (3), \((T^*)^{-1}\) is defined by the formula (7) and K is a subgroup of H such that \(H_{BTR}\oplus K = H\), where \(H_{BTR}\) is given by (5).

Proof

(\(\Rightarrow \)) Assume that (f, g) satisfies equation (8). Then in view of Theorem 3 there exist divisible groups \(H_0, H_1, H_2, H_3\), additive functions \(f_a:S\rightarrow H_2 \oplus H_3\), \(g_a:S\rightarrow H_1 \oplus H_3\) and functions \(f_r :S\rightarrow H_0 \oplus H_1\), \(g_r :S\rightarrow H_0 \oplus H_2\) which satisfy conditions (9)–(14). Let \(T=f_a\). In view of (14) \(\text {im }g_a \subset D(T^*)\). Let \(y\in S\). We have

$$\begin{aligned}&A(x,y)=B(T(x),g_a (y))=A(x,T^* (g_a (y))),\ x\in S, \end{aligned}$$

so \(y-T^* (g_a(y))\in S_{AR}\) and \(\varkappa (y)={\widetilde{T}}^* (g_a(y))\). Hence \(S/S_{AR}=\text {im }T^* /S_{AR}\) and

$$\begin{aligned}&(T^*)^{-1} (y)=({\widetilde{T}}^*)^{-1} (\varkappa (y)) =({\widetilde{T}}^*)^{-1} ({\widetilde{T}}^* (g_a (y))) =g_a (y). \end{aligned}$$

In view of Remark 3 and (13) we get

$$\begin{aligned}&(H_0 \oplus H_1)\times (D(T^*) \cap K)= (H_0 \oplus H_1)\times \text {im }(T^*)^{-1}\\&\quad = (H_0 \oplus H_1)\times \text {im }g_a \subset B^{-1} (\{0\}). \end{aligned}$$

Conditions (25), (26) are exactly the same as (11) and (12).

(\(\Leftarrow \)) Assume that there exist divisible groups \(H_0, H_1, H_2, H_3\), an additive function \(T:S\rightarrow H_2 \oplus H_3\), functions \(f_r :S\rightarrow H_0 \oplus H_1\), \(g_r :S\rightarrow H_0 \oplus H_2\) which satisfy conditions (22)–(27).

For \(y\in S\) we have

$$\begin{aligned}&\varkappa (T^* ((T^*)^{-1}(y)))= {\widetilde{T}}^* (({\widetilde{T}}^*)^{-1}(\varkappa (y)))=\varkappa (y), \end{aligned}$$

which means that \(y-T^* ((T^*)^{-1}(y))\in S_{AR}\). From Remark 3 we get

$$\begin{aligned}&(H_0 \oplus H_1)\times \text {im }(T^*)^{-1} =(H_0 \oplus H_1)\times (D(T^*) \cap K) \subset B^{-1} (\{0\}). \end{aligned}$$

We have

$$\begin{aligned} A(x,y)&=A(x,y-T^* ( (T^*)^{-1} (y))) +A(x,T^* ( (T^*)^{-1} (y)))\\&=0+B(T(x),(T^*)^{-1} (y))=B(T(x),(T^*)^{-1} (y))\\&\quad + B(T(x),g_r (y)) + B(f_r (x),(T^*)^{-1} (y)) +B(f_r (x),g_r (y))\\&=B(T(x)+f_r(x),(T^*)^{-1} (y)+g_r (y))=B(f(x),g(y)),\quad x,y\in S. \end{aligned}$$

\(\square \)

The following result shows us for which f defined on a group (16) holds.

Theorem 6

Assume that S is a group, \(f:S\rightarrow H\). Then f satisfies (16) if and only if there exist divisible groups \(H_0, H_1\), an additive function \(T:S\rightarrow H_1\), and a function \(F_r :S\rightarrow H_0\) such that

$$\begin{aligned}&H=H_0 \oplus H_1 \text { and } H_1 \text { is torsion-free}, \end{aligned}$$
(28)
$$\begin{aligned}&\text {im }T\subset D(T^*),\, \forall _{y\in S}\, (T^* \circ T)(y)-y\in S_{AR}, \end{aligned}$$
(29)
$$\begin{aligned}&f=T+F_r, \end{aligned}$$
(30)
$$\begin{aligned}&H_0 \times (H_0 \oplus \text {im }T) \subset B^{-1}(\{0\}), \end{aligned}$$
(31)
$$\begin{aligned}&(H_0 \oplus \text {im }T) \times H_0 \subset B^{-1}(\{0\}), \end{aligned}$$
(32)

where \(T^*:D(T^*)\rightarrow S\) is a (B, A)-adjoint operator to T, \(S_{AR}\) is given by (3).

Proof

(\(\Rightarrow \)) Assume that f satisfies (16). In view of Theorem 4 there exist divisible groups \(H_0, H_1\), an additive function \(F_a:S\rightarrow H_1\), a function \(F_r :S\rightarrow H_0\) which satisfy conditions (17)–(21). Let \(T=F_a\). We notice that conditions (28), (30)–(32) hold. From (21) we obtain that \(\text {im }T\subset D(T^*)\) and for \(y\in S\) we have

$$\begin{aligned} A(x,T^* (T(y))-y)&=A(x,T^* (T(y)))-A(x,y)\\&=B(T(x),T(y))-B(T(x),T(y))=0,\ x\in S, \end{aligned}$$

which means that \(T^* (T(y))-y\in S_{AR}\).

(\(\Leftarrow \)) Assume that there exist divisible groups \(H_0, H_1\), an additive function \(T:S\rightarrow H_1\), and a function \(F_r :S\rightarrow H_0\) which satisfy conditions (28)–(32). We have

$$\begin{aligned} B(f(x),f(y))&=B(T(x)+F_r(x),T(y)+F_r (y))\\&=B(T(x),T(y))+B(T(x),F_r (y))+B(F_r(x),T(y)+F_r (y)) \\&=B(T(x),T(y))=A(x,T^* (T(y)))\\&=A(x,T^* (T(y))-y)+A(x,y)=A(x,y),\quad x,y\in S. \end{aligned}$$

\(\square \)

4 Applications

In this part of the article we would like to present some applications of the main results from Section 3 in particular for normed spaces. It is helpful to recall (see [1, Theorem 2.1.1 and Remark 2.1.1]) that the following properties for a real normed space \((X,\Vert \!\cdot \!\Vert )\) are true:

$$\begin{aligned}&\mathrm{If\ } X\ \mathrm{is\ real\ and\ smooth,\ then}\ \rho '_+(x,\cdot )\ \mathrm{is\ linear\ for\ all}\ x\in X. \end{aligned}$$
(33)
$$\begin{aligned}&\mathrm{If\ } X\ \mathrm{is\ real\ and\ smooth,\ then}\ \rho '_+(\cdot ,y)\ \mathrm{is\ homogeneous\ for\ all}\ y\in X. \end{aligned}$$
(34)
$$\begin{aligned}&|\rho '_+(x,y)|\le \Vert x\Vert \!\cdot \!\Vert y\Vert \quad \mathrm{and}\quad \rho '_+(x,x)=\Vert x\Vert ^2. \end{aligned}$$
(35)

Theorem 7

Let X, Y be real and smooth normed spaces, X be reflexive. Let \(f:X\rightarrow Y\) be a mapping satisfying:

$$\begin{aligned} \rho '_+(f(x),f(y))=\rho '_+(x,y), \quad x,y\in X. \end{aligned}$$
(36)

Suppose that \(V\subset \text {im }f\) is a closed subspace of Y such that \(\mathrm{co}\dim V=1\) and \(\mathrm{cl\, span}f^{-1}(V)\ne X\). Then f is a linear isometry.

Before we start the proof, some comments are needed. In the paper [6] this result was proved under the surjectivity assumption (and that X and Y are Banach and Y is separable). However our assumption (that \(\mathrm{cl\, span} f^{-1}(V)\ne X\)) is weaker than the surjectivity. As regards the smoothness, this assumption seems to be reasonable. Indeed (see [6]), there are both smooth and strictly convex normed spaces \(Z_1,Z_2\) and nonlinear mappings \(T:Z_1\rightarrow Z_2\) satisfying (36).

Proof

Let \(W{:}{=}\mathrm{cl\, span}f^{-1}(V)\). By the reflexivity, there is \(x\in X\) such that \(\Vert x\Vert =1\) and \(\Vert x\Vert =\mathrm{dist}(x,W)\). We define two bilinear mappings \(A_x:X^2\rightarrow {\mathbb {R}}\), \(B_{f(x)}:Y^2\rightarrow {\mathbb {R}}\) by the formulas \(A_x(u,w){:}{=}\rho '_+(x,u)\!\cdot \!\rho '_+(x,w)\), \(B_{f(x)}(z,v){:}{=}\rho '_+(f(x),z)\!\cdot \!\rho '_+(f(x),v)\). It follows from (36) that

$$\begin{aligned} A_x(u,w)=B_{f(x)}(f(u),f(w)), \quad u,w\in X. \end{aligned}$$
(37)

Put \(D_1{:}{=}\{ z\in Y:\ \forall _{u\in X}\, B_{f(x)}(z,f(u))=0 \}\). From this we get

$$\begin{aligned} D_1&=\{ z\in Y:\ \forall _{u\in X}\, \rho '_+(f(x),z)\!\cdot \!\rho '_+(f(x),f(u))=0 \}\\&=\{ z\in Y:\ \forall _{u\in X}\, \rho '_+(f(x),z)\!\cdot \!\rho '_+(x,u)=0 \}\\&=\{ z\in Y:\ \rho '_+(f(x),z)=0 \}. \end{aligned}$$

Thus \(D_1\) is a closed linear subspace. In particular, \(D_1\) is a divisible abelian group. We have also

$$\begin{aligned} B_{f(x)}(u,v)=B_{f(x)}(v,u)=0,\, u\in D_1, v\in Y. \end{aligned}$$
(38)

Moreover \(Y=\mathrm{span}\{f(x)\}\oplus D_1\) and so \(f=f_a +f_r\), where \(f_a :X\rightarrow \mathrm{span}\{f(x)\}\), \(f_r :X\rightarrow D_1\). From Theorem 4 there exist divisible groups \(H_0, H_1\), an additive function \(F_a:X\rightarrow H_1\), and a function \(F_r :X\rightarrow H_0 \) such that

$$\begin{aligned}&Y=H_0\oplus H_1 \text { and } H_0\subset \{v\in Y:\, \forall _{u\in \text {im }f}\, B_{f(x)} (u,v)=0 \}=D_1,\\&f=F_a +F_r . \end{aligned}$$

We observe that for \(y,z\in X\) we have

$$\begin{aligned}&f_a (y+z) -f_a (y) -f_a (z) =f(y+z)-f(y)-f(z) \\&\qquad -f_r(y+z)+f_r (y) +f_r (z) =F_a (y+z) -F_a (y) -F_a (z) \\&\qquad +F_r (y+z) -F_r (y) -F_r (z) -f_r(y+z) +f_r (y) +f_r (z)\\&\quad =F_r (y+z) -F_r (y) -F_r (z) -f_r(y+z) +f_r (y) +f_r (z) \in H_0 +D_1\subset D_1 , \end{aligned}$$

which means that \(f_a\) is additive.

Since \(f_a(w)\in \mathrm{span}\{f(x)\}\) for \(w\in X\), there exists a function \(\varphi :X\rightarrow {\mathbb {R}}\) such that \(f_a=\varphi f(x)\). Therefore, by the property of the set \(D_1\) and by (34) we have \(\rho '_+(f_a(w),f_r(y))=0\) for \(w,y\in X\). So, it and (33) and (35) yield

$$\begin{aligned} \Vert f_a(y)\Vert ^2&=\rho '_+(f_a(y),f_a(y))+0=\rho '_+(f_a(y),f_a(y))+\rho '_+(f_a(y),f_r(y))\\&=\rho '_+(f_a(y),f_a(y)+f_r(y))\le \Vert f_a(y)\Vert \!\cdot \!\Vert f_a(y)+f_r(y)\Vert \\&\quad =\Vert f_a(y)\Vert \!\cdot \!\Vert f(y)\Vert . \end{aligned}$$

Since \(\Vert f(y)\Vert =\Vert y\Vert \), it follows from the above inequalities that \(\Vert f_a(y)\Vert \le \Vert y\Vert \) for all \(y\in X\), which implies that \(f_a\) is continuous and linear. Consequently \(f_a(w)=\varphi (w)\!\cdot \!f(x)\) for every \(w\in X\) with some \(\varphi \in X^*\). Next, for all u, w in X we have

$$\begin{aligned} \rho '_+(u,w)&=\rho '_+(f(u),f(w))=\rho '_+(f(u),f_a(w)+f_r(w))\\&=\rho '_+(f(u),\varphi (w)\!\cdot \!f(x)+f_r(w))\\&=\varphi (w)\!\cdot \!\rho '_+(f(u),f(x))+\rho '_+(f(u),f_r(w))\\&=\varphi (w)\!\cdot \!\rho '_+(u,x)+\rho '_+(f(u),f_r(w)). \end{aligned}$$

For given \(u\in X\) we define a \(\gamma _u\in X^*\) by the formula

$$\begin{aligned} \gamma _u(w){:}{=}\rho '_+(u,w)-\varphi (w)\rho '_+(u,x),\quad w\in X. \end{aligned}$$

It follows from the above equalities that \(\gamma _u(w)=\rho '_+(f(u),f_r(w))\). Therefore for fixed \(w,z\in X\) we get

$$\begin{aligned}&\rho '_+(f(u),f_r(\alpha w+\beta z)-\alpha f_r(w)-\beta f_r(z))\\&\quad =\rho '_+(f(u),f_r(\alpha w+\beta z))-\alpha \rho '_+(f(u),f_r(w))-\beta \rho '_+(f(u),f_r(z))\\&\quad =\gamma _u(\alpha w+\beta z)-\alpha \gamma _u(w)-\beta \gamma _u(z)\\&\quad =\gamma _u(\alpha w+\beta z-\alpha w-\beta z)=0. \end{aligned}$$

To summarize, we proved

$$\begin{aligned} \forall _{u\in X} \ \rho '_+(f(u),f_r(\alpha w+\beta z)-\alpha f_r(w)-\beta f_r(z))=0. \end{aligned}$$
(39)

Since \(\Vert x\Vert =\mathrm{dist}(x,W)\), we have the inequality \(\Vert x\Vert \le \Vert x+w\Vert \) for all \(w\in W\). In particular, for all \(t>0\) we obtain \(0\le \Vert x\Vert \!\cdot \!\frac{\Vert x+tw\Vert -\Vert x\Vert }{t}\). Letting \(t\rightarrow 0^+\), we get \(0\le \rho '_+(x,w)\). Putting \(-w\) in place of w (and applying again (33)) we get \(0\ge \rho '_+(x,w)\). So, we proved that \(\rho '_+(x,c)=0\) for all \(c\in W\).

Clearly \(f^{-1}(V)\subset W\). In particular, for all c in \(f^{-1}(V)\) we have \(0=\rho '_+(x,c)=\rho '_+(f(x),f(c))\). Thus \(V\subset D_1\). Since \(\mathrm{co}\dim V=1=\mathrm{co}\dim D_1\), we obtain \(V=D_1\). Since \(f_r(\alpha w+\beta z)-\alpha f_r(w)-\beta f_r(z)\in D_1=V\subset \text {im }f\), there is a \(b_0\in X\) such that \(f(b_0)=f_r(\alpha w+\beta z)-\alpha f_r(w)-\beta f_r(z)\). Hence, applying (39), we get

$$\begin{aligned}&\Vert f_r(\alpha w+\beta z)-\alpha f_r(w)-\beta f_r(z)\Vert ^2=\Vert f(b_0)\Vert ^2=\rho '_+(f(b_0),f(b_0))\\&\quad =\rho '_+(f(b_0),f_r(\alpha w+\beta z)-\alpha f_r(w)-\beta f_r(z))=0. \end{aligned}$$

It holds for all \(w,z\in X\) and \(\alpha ,\beta \in {\mathbb {R}}\), which means that \(f_r\) is linear. Since \(f_a,f_r\) are linear then f also is linear mapping. The equality \(\Vert f(w)\Vert =\Vert w\Vert \) for all w in X implies that f is an isometry. \(\square \)