Existence of $f$ such that $f(x,|x|^2)f(y,|y|^2)=0$ whenever $x \cdot y=0$

Question

Does there exists a non trivial continuous function (other than $f=0$) with the following :

$f:R^4 \to [0, \infty)$

Let a $x,y \in R^3$ and their respective Euclidean norm squared $|x|^2$ and $|y|^2$ and their dot product $x \cdot y$

$f(x,|x|^2)f(y,|y|^2)=0$ whenever $x \cdot y=0$

$f(x,|x|^2)f(-x,|x|^2)=0$

Answer 1

$f(x_1,x_2,x_3,x_4)=(x_1+|x_1|)(x_2+|x_2|)(x_3+|x_3|)$

or any function which is zero outside of the nonnegative orthant $\mathbb{R}_+^3$.

If $x,y$ are perpendicular then at least one of them has at least one nonpositive coordinate and then $f$ vanishes. One of $x,-x$ also has at least one nonpositive coordinate.

Answer 2

The conditions make no claim about $f(x_1,x_2,x_3,x_4)$ when $x_1^2+x_2^2+x_3^2-x_4\ne0$, hence we can take any continuous $g\colon \Bbb R^4\to[0,\infty)$ and let $$ f(x_1,x_2,x_3,x_4)=\left|x_1^2+x_2^2+x_3^2-x_4\right|g(x_1,x_2,x_3,x_4).$$

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We could get away with $f$ being zero less often, for example as follows: Let $A$ be a closed subset of $S^2$ with the property that of for every $x\notin A$, we have $-x\in A$ and the great circle orthogonal to $x$ is $\subset A$. Then let $f(x_1,x_2,x_3,x_4)$ be something like the distance between $(x_1,x_2,x_3)$ and $\sqrt{|x_4|}\cdot A$ in $\Bbb R^3$.

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Next, we might pick a different $A$ for each $x_4$ in a continous fashion ...