When can one write $f(x,y)$ as $|g(x)-g(y)|_2^2$ for some $g(\cdot)$?

Question

Suppose I have a function $f:\mathbb{R}^n \times \mathbb{R}^n \rightarrow \mathbb{R}$. Fix a natural number $k$. Under what conditions can I write $$ f(x,y) = [g(x) - g(y)]^{T}[g(x) - g(y)] $$ for some appropriate function $g:\mathbb{R}^n \rightarrow \mathbb{R}^k$?

Necessary conditions are obviously that $f(x,y) \geq 0$, $f(x,x) = 0$ and that $f(x,y) = f(y,x)$, but are these sufficient?

Answer 1

No, there is one more condition: $f(x,y)=f(x,0)+f(y,0)-2\sqrt {f(x,0)f(y,0)}$. This condition and the three conditions you have are necessary and sufficient.

Answer 2

Assuming you wrote:

$f(x,y)= |g(x)-g(y)|^2$, we can find a $k\in \mathbb{R^n}$, such that

$|g(x)-g(y)|^2 \leq |kx-ky|^4 \implies |g(x)-g(y)| \leq |kx-ky|^2 \implies \frac{ |g(x)-g(y)|}{|kx-ky|} \leq |kx-ky| \implies -|kx-ky| \leq \frac{ g(x)-g(y)}{kx-ky} \leq |kx-ky| \implies -k|kx-ky| \leq \frac{ g(x)-g(y)}{x-y} \leq k|kx-ky|$ As $x \rightarrow y$, $|kx-ky| \rightarrow 0$. Hence, by Sandwich Theorem, we can see that $ lim_(x \rightarrow y) [\frac{ g(x)-g(y)}{x-y}]=0 \implies g(x)=g(y)=constant$ Hence $f(x,y)=0$