Conductivity is not automatically a smooth quantum potential

For smooth positive scalar conductivity $\sigma$ on a Euclidean domain,

Thus the scalar conductivity energy is the corresponding Riemannian Dirichlet energy. Smoothness and uniform positivity make this a smooth metric comparison.

If only $\sigma \in L^{\infty }$ is assumed, that formula gives only rough coefficients. It does not produce an admissible smooth metric in the earlier class.

Similarly, putting $v=\sqrt{\sigma }u$ gives, in the smooth case,

At bounded-measurable regularity, $\Delta \sqrt{\sigma }/\sqrt{\sigma }$ need not be an ordinary function or even a canonically defined product of distributions.

Consequently, even if the counterexample asserted in 365-03 is verified, it does not automatically provide two conventional smooth quantum potentials with equal measurements. Nor does it refute a smooth inverse theorem on a genuinely narrower class. The distinction between smooth scalar uniqueness and lower-regularity questions is already essential in the classical inverse literature SylvesterUhlmann,OAI365c.