Backward heat equation and forward perturbed heat equation well posed? 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?
 A: Here is a hint. Let me begin with a formal calculus. The Baker-Campbell-Hausdorff formula tells you that
$$e^{-Y}e^{-X}e^{X+Y}\sim e^{\frac12[Y,X]},$$
where $[\cdot,\cdot]$ is the commutator. Applying this to
$$X=t\Delta,\quad Y=tf,$$
we find that, for small $t>0$, $e^{-tf}ST$ should behave as $e^{\frac12 t^2L}$ where
$$L=[f,\Delta]=-\Delta f-2\nabla f\cdot\nabla.$$
Notice that $L$ is anti-Hermitian, as the commutator of two Hermitian operators. Thus it generates a group of unitary operators. So far so good, this is good news, but it postpones the question of well-posedness to the next corrector.
Thus let us consider the composition
$$e^{-\frac{t^2}2L}e^{-tf}e^{-t\Delta}e^{t(\Delta+f)}$$
for small $t>0$. Zassenhaus' formula gives us the next term. This product should behave as $e^{\frac16R}$ where
$$R=2[Y,[X,Y]]+[X,[X,Y]]=t^3([f,[\Delta,f]]+[\Delta,[\Delta,f]])=:t^3Q.$$
Remark that $Q$ is a second-order operator.
My guess, or my advice, is that for the product $ST$ to be well-defined, it is necessary that $Q$ generates a semi-group of bounded operators. In particular, the principal symbol of $Q$ should be negative semi-definite. Since the principal part is
$$Q_p=4\nabla^2f:\nabla^2,$$
this necessary condition is that $f$ be convex.
Remark that this convexity may be annoying, because we have to postmultiply by $e^{tf}$, which will not be a bounded operator over $L^2$, unless $f$ is constant. But you could have a good result by replacing $L^2$ by a weighted $L^2$-space in your analysis.
A: 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 $u\neq 0$.
