I consider the following scenario. Let $I$ be a compact interval in space and $f$ a nice function in the space $C^{\infty}(I)$. In the following we consider a self-adjoint realization of our operators on said interval.


We can consider the perturbed heat semigroup $T=e^{(\Delta+f)}$ at fixed time $1$. The heat semigroup, as we all know is smoothing. The unbounded operator $S=e^{-\Delta }$ corresponding to the inverse heat semigroup is also well-defined by the functional calculus for self-adjoint operators.


Now one could be tempted to think that the smoothing by the heat semigroup was enough that $ST$ was bounded as an operator on $L^2$ and this is true if there was no $f$. Is it still true with having the $f$ around?



Backward heat equation and forward perturbed heat equation well posed?
No, this cannot be true if $f$ is just $C^\infty$. Let $u=e^{(\Delta+f)t}u\_0$.
At $t=1$, $u=e^{\Delta+f}u\_0=e^\Delta v\_0$ for some $v\_0$. Then, by well known properties of the heat equation, $u$ is spatially analytic. Moreover, $u\_t=e^{\Delta+f}(\Delta+f)u\_0$. If $u\_0$ is sufficiently smooth, then $(
\Delta+f)u\_0$ is in $L^2$, so $u\_t$ would have to equal $e^\Delta w\_0$ for some $w\_0$. This implies that both $u$ and $u\_t$ are analytic, which makes $f$ analytic wherever $u\neq 0$.


