# Does there exist a scalar function $g({\bf{x}})$ that satisfies $g({\bf{x}} +\,{\bf{f}}({\bf{x}}))= g ({\bf{x}})\det(I+\,{\bf{f}}'({\bf{x}}))$?

Given a vector valued function $$\bf{f}:\mathbb{R}^n\rightarrow\mathbb{R}^n$$, what is a scalar function $$g:\mathbb{R}^n\rightarrow\mathbb{R}$$ that satisfies the following $$g({\bf x} +\mathbf{f}(\mathbf{x}))=g(\mathbf{x})\det(I+\mathbf{f}'(\mathbf{x})),$$ where $$\bf{f}'(\bf{x})$$ is the Jacobian matrix of $$\bf{f}(\bf{x})$$ and $$I$$ is the $$n\times n$$ identity matrix.

If a solution can't be found for arbitrary $$\bf{f}$$, what structure can one impose on $$\bf{f}$$ for there to exist a particular function $$g$$ that satisfies the above condition.

One particular example of a function $$g$$ that satisfies the above is all I'm after (i.e., I don't need the most general solution).

Alternatively: Can one prove that there does not exist a function $$g$$ that satisfies the above?

• $g\equiv 0$ works. – zhw. Sep 24 '19 at 19:51

A couple of observations in 1D that may be helpful (but I have not tried to generalize),

Let, $$g(x) = \exp{\int_0^x h(y) dy}$$

Then, $$g(x+f(x))=\exp{\int_0^{x+f(x)} h(y) dy} = g(x) \exp{\int_x^{f(x)}h(y)dy}$$

If we specialize to $$h(y) = 1/y$$, and $$f(x) = -\int\log(x)$$, the last exponential becomes

$$\exp{\log(\frac{-\int \log(x)}{x})} = 1 - \log(x) = 1 + f'(x)$$

So in 1D we can write, $$g(x+f(x)) = g(x)(1+f'(x))$$

• That is very nice solution to the problem in 1D! Looks like the solution relies on $f(x)$ having a particular form so I guess it is quite hard to do solve the problem 1D while keeping $f$ arbitrary, which probably means that in higher dimensions it's even more difficult as we have to deal with the determinant. Perhaps a general $n$-D solution is a bit ambitious... – Ben Tapley Sep 25 '19 at 13:34