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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?

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    $\begingroup$ This reminds me the following book: R. Lattes & J.-L. Lions, Méthode de quasi-réversibilité et applications. Dunod (1967). Perhaps there was an english translation in 1969. $\endgroup$ Mar 9, 2021 at 20:45
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    $\begingroup$ @Is there a particular result in that book that you feel is related to my question? $\endgroup$
    – Sascha
    Mar 9, 2021 at 23:22
  • $\begingroup$ The authors compose the backward heat equation with a forward modified heat equation. This is the relation with your question. I don't pretend that they solve exactly this question. $\endgroup$ Mar 10, 2021 at 6:40

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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 $u\neq 0$.

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  • $\begingroup$ I have difficulties to follow your argument for the existence of $v_0$ and $w_0$: the operator $e^{\Delta}$ is not surjective on $L^2$ (or, say, on similar function spaces). $\endgroup$ Mar 10, 2021 at 19:46
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    $\begingroup$ @Jochen Glueck: The question was whether $S^{-1}T$ is a bounded operator. $\endgroup$ Mar 10, 2021 at 22:21
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    $\begingroup$ @MichaelRenardy nice argument. I am curious, do you have then any intuition if it holds for analytic $f$? $\endgroup$
    – Sascha
    Mar 11, 2021 at 16:20
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    $\begingroup$ No, it cannot be expected to hold for analytic $f$ either. Solutions to the heat equation are in a sense smoother than just analytic. $\endgroup$ Mar 11, 2021 at 16:47
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    $\begingroup$ @MichaelRenardy I am wondering actually what the take-home message from your answer is. I understand it is a counterexample to what I thought was true, I understand it means that if true $f$ would have to be extremely regular. But what I am after are mapping properties of $S^{-1}T:X \rightarrow Y$ and I wonder what are the right space to consider for $X,Y$? Do you have any thoughts (you may speculate of course)? $\endgroup$
    – Sascha
    Mar 11, 2021 at 22:37
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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.

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    $\begingroup$ thank you for the many insights, this gives a good idea what to expect. One comment: since we are on a bounded interval, having f convex and $e^{tf}$ bounded might be okay. $\endgroup$
    – Sascha
    Mar 10, 2021 at 14:00
  • $\begingroup$ @Sascha. When you speak of a bounded interval, you mean in space ? Your question was not clear about that. It could have a time interval. $\endgroup$ Mar 10, 2021 at 14:26
  • $\begingroup$ indeed, I meant in space, sorry for the confusion $\endgroup$
    – Sascha
    Mar 10, 2021 at 14:50

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