Return to Answer

I think that it is possible with a large enough vector space $V$. I first misread the question, and constructed something where the inner product depends on $f$ while the mapping $F$ does not. The construction can be adapted to the true question as stated, so I'll still give it first as a warmup.

I'll construct $F$ and $V$ together, and then construct the bilinear pairing last. Let $V_0 = \mathbb{R}$ with its basis vector $1$. Then let $V_{n+1}$ be the direct sum of $V_n$ and the vector space $W_n$ of formal linear combinations of elements of $V_n \setminus V_{n-1}$, where in this formula $V_{-1} = \emptyset$. If $x \in V_n \setminus V_{n-1}$, let $[x]$ denote the corresponding element in $W_n \subset V_{n+1}$. Let $V$ be the union of all $V_n$, and let $F(x) = [x]$. Note that every $x \in V$ has a degree $d(x)$, by definition the first $n$ such that $x \in V_n$.

To construct the pairing, let $\langle 1,1 \rangle = 1$. We need to choose values of $\langle e,f \rangle$ for every other unordered pair of basis vectors $e,f$. I claim that your constraints are triangular with respect to degree, in other words that the values can be constructed by induction. Also the diagonal values $\langle e, e \rangle$ are unrestricted. To see this, consider your equation $$\langle F(x), F(x) \rangle + \langle F(y),F(y) \rangle - 2\langle F(x), F(y) \rangle = \langle F(x) - F(y), F(x) - F(y) \rangle = f(\langle x-y, x-y \rangle)$$ with $x \ne y$. By construction, the arguments of the cross-term $\langle F(x), F(y) \rangle$ are both basis vectors, and only occur once for any given $x$ and $y$. Let's say that $\max(d(x), d(y)) = n$. Then $d(x-y) \le n$. In defining the inner product on $V_{n+1}$, the right side of your equation is already chosen, two terms on the left are unrestricted, and the third term can be chosen to satisfy the equality.

Suppose instead that the inner product is to be fixed and instead $F$ can change with $f$. In this case, let $W_n$ be the vector space of formal linear combinations of elements of $(V_n \setminus V_{n-1}) \times \mathbb{R}^\mathbb{R}$, and as before let $V_{n+1} = V_n \oplus W_n$. In this case, $W_n$ has a basis vector $[x,f]$ for every $f$ and every suitable $v$. For any fixed $f$, define $F(x) = [x,f]$.

As before, say that $\langle 1,1 \rangle = 1$ and that $\langle [x,f], [x,f] \rangle$ is unrestricted. Also $\langle [x,f], [y,g] \rangle$ is unrestricted when $f \ne g$, for all $x$ and $y$. Finally, as before, $$\langle F(x), F(y) \rangle = \langle [x,f], [y,f] \rangle$$ with $x \ne y$ is uniquely determined by induction on $\max(d(x),d(y))$.

I think that it is possible with a large enough vector space $V$. I first misread the question, and constructed something where the inner product depends on $f$ while the mapping $F$ does not. The construction can be adapted to the true question as stated, so I'll still give it first as a warmup.

Version 1

I'll construct $F$ and $V$ together, and then construct the bilinear pairing last. Let $V_0 = \mathbb{R}$ with its basis vector $1$. Then let $V_{n+1}$ be the direct sum of $V_n$ and the vector space $W_n$ of formal linear combinations of elements of $V_n \setminus V_{n-1}$, where in this formula $V_{-1} = \emptyset$. If $x \in V_n \setminus V_{n-1}$, let $[x]$ denote the corresponding element in $W_n \subset V_{n+1}$. Let $V$ be the union of all $V_n$, and let $F(x) = [x]$. Note that every $x \in V$ has a degree $d(x)$, by definition the first $n$ such that $x \in V_n$.

To construct the pairing, let $\langle 1,1 \rangle = 1$. We need to choose values of $\langle e,f \rangle$ for every other unordered pair of basis vectors $e,f$. I claim that your constraints are triangular with respect to degree, in other words that the values can be constructed by induction. Also the diagonal values $\langle e, e \rangle$ are unrestricted. To see this, consider your equation $$\langle F(x), F(x) \rangle + \langle F(y),F(y) \rangle - 2\langle F(x), F(y) \rangle = \langle F(x) - F(y), F(x) - F(y) \rangle = f(\langle x-y, x-y \rangle)$$ with $x \ne y$. By construction, the arguments of the cross-term $\langle F(x), F(y) \rangle$ are both basis vectors, and only occur once for any given $x$ and $y$. Let's say that $\max(d(x), d(y)) = n$. Then $d(x-y) \le n$. In defining the inner product on $V_{n+1}$, the right side of your equation is already chosen, two terms on the left are unrestricted, and the third term can be chosen to satisfy the equality.

Version 2

Suppose instead that the inner product is to be fixed and instead $F$ can change with $f$. In this case, let $W_n$ be the vector space of formal linear combinations of elements of $(V_n \setminus V_{n-1}) \times \mathbb{R}^\mathbb{R}$, and as before let $V_{n+1} = V_n \oplus W_n$. In this case, $W_n$ has a basis vector $[x,f]$ for every $f$ and every suitable $v$. For any fixed $f$, define $F(x) = [x,f]$.

As before, say that $\langle 1,1 \rangle = 1$ and that $\langle [x,f], [x,f] \rangle$ is unrestricted. Also $\langle [x,f], [y,g] \rangle$ is unrestricted when $f \ne g$, for all $x$ and $y$. Finally, as before, $$\langle F(x), F(y) \rangle = \langle [x,f], [y,f] \rangle$$ with $x \ne y$ is uniquely determined by induction on $\max(d(x),d(y))$.

Version 3

Ady reminds me that the second version still misses the condition that the bilinear form on $V$ should be non-degenerate. I think that the same trick works a third time: We can just enlarge $V$ to also guarantee this condition. This time let $W_n$ be as in the second version, and let $$V_{n+1} = V_n \oplus W_n \oplus V_n^*,$$ where $V_n^*$ is the (algebraic) dual vector space to $V_n$. Define the bilinear form on $V_n \oplus W_n$ as in version 2, and define $F$ as in version 2. The bilinear form on $V_n^*$ is unrestricted, and so is the bilinear pairing between $V_n^*$ and $W_n$. Finally the bilinear pairing between $V_n^*$ and $V_n$ should be the canonical pairing $\langle \phi, x \rangle = \phi(x)$. This guarantees that for every vector $x \in V_n$, there exists $y \in V_{n+1}$ such that $\langle y,x \rangle = 1$.

Every version of the construction is cheap in the sense that the image of $F$ is a linearly independent set. Moreover, in the second and third versions, the image of $F$ is far from a basis. My feeling is that it is difficult to ask for much better than that in a universal construction.

If, as you say, you have complete freedom to pick an $F$ yourself, then I think that it is possible with a large enough vector space $V$. Edit: I misinterpted the question and the construction is something weaker: The inner product depends on $f$.

I'll construct $F$ and $V$ together, and then construct the bilinear pairing last. Let $V_0 = \mathbb{R}$ with its basis vector $1$. Then let $V_{n+1}$ be the direct sum of $V_n$ and the vector space $W_n$ of formal linear combinations of elements of $V_n \setminus V_{n-1}$, where in this formula $V_{-1} = \emptyset$. If $v \in V_n \setminus V_{n-1}$, let $[v]$ denote the corresponding element in $W_n \subset V_{n+1}$. Let $V$ be the union of all $V_n$, and let $F(v) = [v]$. Note that every $v \in V$ has a degree $d(v)$, by definition the first $n$ such that $v \in V_n$.

To construct the pairing, let $\langle 1,1 \rangle = 1$. We need to choose values of $\langle e,f \rangle$ for every other unordered pair of basis vectors $e,f$. I claim that your constraints are triangular with respect to degree, in other words that the values can be constructed by induction. Also the diagonal values $\langle e, e \rangle$ are unrestricted. To see this, consider your equation $$\langle F(x), F(x) \rangle + \langle F(y),F(y) \rangle - 2\langle F(x), F(y) \rangle = \langle F(x) - F(y), F(x) - F(y) \rangle = f(\langle x-y, x-y \rangle)$$ with $x \ne y$. By construction, the arguments of the cross-term $\langle F(x), F(y) \rangle$ are both basis vectors, and only occur once for any given $x$ and $y$. Let's say that $\max(d(x), d(y)) = n$. Then $d(x-y) \le n$. In defining the inner product on $V_{n+1}$, the right side of your equation is already chosen, two terms on the left are unrestricted, and the third term can be chosen to satisfy the equality.

You could economize in the construction a little bit by setting $V_{-1} = \{0\}$ instead. In this case, the diagonal terms $\langle F(x), F(x) \rangle$ are not unrestricted. Instead, you should set $$\langle F(x), F(x) \rangle = f(\langle x, x \rangle),$$ taking $y = 0$ in your equation.

I think that it is possible with a large enough vector space $V$. I first misread the question, and constructed something where the inner product depends on $f$ while the mapping $F$ does not. The construction can be adapted to the true question as stated, so I'll still give it first as a warmup.

I'll construct $F$ and $V$ together, and then construct the bilinear pairing last. Let $V_0 = \mathbb{R}$ with its basis vector $1$. Then let $V_{n+1}$ be the direct sum of $V_n$ and the vector space $W_n$ of formal linear combinations of elements of $V_n \setminus V_{n-1}$, where in this formula $V_{-1} = \emptyset$. If $x \in V_n \setminus V_{n-1}$, let $[x]$ denote the corresponding element in $W_n \subset V_{n+1}$. Let $V$ be the union of all $V_n$, and let $F(x) = [x]$. Note that every $x \in V$ has a degree $d(x)$, by definition the first $n$ such that $x \in V_n$.

To construct the pairing, let $\langle 1,1 \rangle = 1$. We need to choose values of $\langle e,f \rangle$ for every other unordered pair of basis vectors $e,f$. I claim that your constraints are triangular with respect to degree, in other words that the values can be constructed by induction. Also the diagonal values $\langle e, e \rangle$ are unrestricted. To see this, consider your equation $$\langle F(x), F(x) \rangle + \langle F(y),F(y) \rangle - 2\langle F(x), F(y) \rangle = \langle F(x) - F(y), F(x) - F(y) \rangle = f(\langle x-y, x-y \rangle)$$ with $x \ne y$. By construction, the arguments of the cross-term $\langle F(x), F(y) \rangle$ are both basis vectors, and only occur once for any given $x$ and $y$. Let's say that $\max(d(x), d(y)) = n$. Then $d(x-y) \le n$. In defining the inner product on $V_{n+1}$, the right side of your equation is already chosen, two terms on the left are unrestricted, and the third term can be chosen to satisfy the equality.

Suppose instead that the inner product is to be fixed and instead $F$ can change with $f$. In this case, let $W_n$ be the vector space of formal linear combinations of elements of $(V_n \setminus V_{n-1}) \times \mathbb{R}^\mathbb{R}$, and as before let $V_{n+1} = V_n \oplus W_n$. In this case, $W_n$ has a basis vector $[x,f]$ for every $f$ and every suitable $v$. For any fixed $f$, define $F(x) = [x,f]$.

As before, say that $\langle 1,1 \rangle = 1$ and that $\langle [x,f], [x,f] \rangle$ is unrestricted. Also $\langle [x,f], [y,g] \rangle$ is unrestricted when $f \ne g$, for all $x$ and $y$. Finally, as before, $$\langle F(x), F(y) \rangle = \langle [x,f], [y,f] \rangle$$ with $x \ne y$ is uniquely determined by induction on $\max(d(x),d(y))$.

If, as you say, you have complete freedom to pick an $F$ yourself, then I think that it is possible with a large enough vector space $V$. I'll construct $F$ and $V$ together, and then construct the bilinear pairing last. Let $V_0 = \mathbb{R}$ with its basis vector $1$. Then let $V_{n+1}$ be the direct sum of $V_n$ and the vector space $W_n$ of formal linear combinations of elements of $V_n \setminus V_{n-1}$, where in this formula $V_{-1} = \emptyset$. If $v \in V_n \setminus V_{n-1}$, let $[v]$ denote the corresponding element in $W_n \subset V_{n+1}$. Let $V$ be the union of all $V_n$, and let $F(v) = [v]$. Note that every $v \in V$ has a degree $d(v)$, by definition the first $n$ such that $v \in V_n$.

To construct the pairing, let $\langle 1,1 \rangle = 1$. We need to choose values of $\langle e,f \rangle$ for every other unordered pair of basis vectors $e,f$. I claim that your constraints are triangular with respect to degree, in other words that the values can be constructed by induction. Also the diagonal values $\langle e, e \rangle$ are unrestricted. To see this, consider your equation $$\langle F(x), F(x) \rangle + \langle F(y),F(y) \rangle - 2\langle F(x), F(y) \rangle = \langle F(x) - F(y), F(x) - F(y) \rangle = f(\langle x-y, x-y \rangle)$$ with $x \ne y$. By construction, the arguments of the cross-term $\langle F(x), F(y) \rangle$ are both basis vectors, and only occur once for any given $x$ and $y$. Let's say that $\max(d(x), d(y)) = n$. Then $d(x-y) \le n$. In defining the inner product on $V_{n+1}$, the right side of your equation is already chosen, two terms on the left are unrestricted, and the third term can be chosen to satisfy the equality.

You could economize in the construction a little bit by setting $V_{-1} = \{0\}$ instead. In this case, the diagonal terms $\langle F(x), F(x) \rangle$ are not unrestricted. Instead, you should set $$\langle F(x), F(x) \rangle = f(\langle x, x \rangle),$$ taking $y = 0$ in your equation.

If, as you say, you have complete freedom to pick an $F$ yourself, then I think that it is possible with a large enough vector space $V$. Edit: I misinterpted the question and the construction is something weaker: The inner product depends on $f$.

I'll construct $F$ and $V$ together, and then construct the bilinear pairing last. Let $V_0 = \mathbb{R}$ with its basis vector $1$. Then let $V_{n+1}$ be the direct sum of $V_n$ and the vector space $W_n$ of formal linear combinations of elements of $V_n \setminus V_{n-1}$, where in this formula $V_{-1} = \emptyset$. If $v \in V_n \setminus V_{n-1}$, let $[v]$ denote the corresponding element in $W_n \subset V_{n+1}$. Let $V$ be the union of all $V_n$, and let $F(v) = [v]$. Note that every $v \in V$ has a degree $d(v)$, by definition the first $n$ such that $v \in V_n$.

To construct the pairing, let $\langle 1,1 \rangle = 1$. We need to choose values of $\langle e,f \rangle$ for every other unordered pair of basis vectors $e,f$. I claim that your constraints are triangular with respect to degree, in other words that the values can be constructed by induction. Also the diagonal values $\langle e, e \rangle$ are unrestricted. To see this, consider your equation $$\langle F(x), F(x) \rangle + \langle F(y),F(y) \rangle - 2\langle F(x), F(y) \rangle = \langle F(x) - F(y), F(x) - F(y) \rangle = f(\langle x-y, x-y \rangle)$$ with $x \ne y$. By construction, the arguments of the cross-term $\langle F(x), F(y) \rangle$ are both basis vectors, and only occur once for any given $x$ and $y$. Let's say that $\max(d(x), d(y)) = n$. Then $d(x-y) \le n$. In defining the inner product on $V_{n+1}$, the right side of your equation is already chosen, two terms on the left are unrestricted, and the third term can be chosen to satisfy the equality.

You could economize in the construction a little bit by setting $V_{-1} = \{0\}$ instead. In this case, the diagonal terms $\langle F(x), F(x) \rangle$ are not unrestricted. Instead, you should set $$\langle F(x), F(x) \rangle = f(\langle x, x \rangle),$$ taking $y = 0$ in your equation.