TITLE: Backward heat equation and forward perturbed heat equation well posed?
QUESTION [5 upvotes]: 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?

REPLY [4 votes]: 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 $u\neq 0$.