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I am looking at a proof of Liouville's Theorem, which states that for $F, G \in C_0^\infty$ and a Hamiltonian $H$, the operator $$D_H = \sum_{i=1}^n\Big(\frac{\partial H}{\partial p_i} \frac{\partial}{\partial q_i} - \frac{\partial H}{\partial q_i} \frac{\partial}{\partial p_i}\Big)$$ is formally skew adjoint, i.e. $$\langle D_HF, G \rangle = -\langle F, D_H G\rangle.$$

The proof supposedly follows from integration by parts. That is, let $\mu$ be the Liouville measure $$d\mu = \prod_{i=1}^n dp_i dq_i.$$ then $$\int F^- (D_HG)\,d\mu = - \Big[\sum_{i=1}^n(F^-G) \Big(\frac{\partial^2 H}{\partial q_i \partial p_i} - \frac{\partial^2 H}{\partial p_i \partial q_i}\Big) + \int(D_H F)^- G \,d\mu\Big].$$

I have not seen such integration by parts done before. Can anyone walk through how this calculation was done, and also why we are considering $F^-$?

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    $\begingroup$ what is $F^{-}$ here? $\endgroup$
    – peek-a-boo
    Jun 18 at 4:18
  • $\begingroup$ @peek-a-boo The text I am using does not specify. My intuition is that it is the limit as an increasing sequence approaches $F$. $\endgroup$
    – CBBAM
    Jun 18 at 4:22
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    $\begingroup$ still, that makes no sense, and I don't see its purpose. could you quote your reference? Anyway, the equality $\langle D_HF,G\rangle=-\langle F,D_HG\rangle$ (assuming $\langle\cdot,\cdot\rangle$ denotes the $L^2$ inner product) does indeed follow from integration by parts and compact support of $F,G$. $\endgroup$
    – peek-a-boo
    Jun 18 at 4:28
  • $\begingroup$ @peek-a-boo Other than the superscript, does the rest of the integration by parts make sense? $\endgroup$
    – CBBAM
    Jun 18 at 4:30
  • $\begingroup$ sure, though I'm not sure why they carry around the first term (the mixed partials of $H$ are equal so those terms cancel). $\endgroup$
    – peek-a-boo
    Jun 18 at 4:30

1 Answer 1

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The first thing to note, which I leave to you to verify, is that $D_H$ is a derivation on smooth functions, meaning it is linear and satisfies a product rule: \begin{align} D_H(FG)=(D_HF)\cdot G+F\cdot (D_HG). \end{align} In fact this is one of the nice properties of the Poisson bracket $\{\cdot,\cdot\}$; it is $\Bbb{R}$-bilinear and satisfies the product rule with respect to each slot. The map $D_H$ is just $\{H,\cdot\}$ (or maybe its the negative of this... sign conventions here are a pain).

So, we have $F(D_HG)=D_H(FG)- (D_HF)G$. Integrating both sides over $\Bbb{R}^{2n}$, we get \begin{align} \langle F,D_HG\rangle=\int_{\Bbb{R}^{2n}}D_H(FG)\,d\mu - \langle D_HF, G\rangle.\tag{$*$} \end{align} So, all that remains is showing why the first term vanishes. For this, it suffices to show that for every $f\in C^{\infty}_c(\Bbb{R}^{2n})$, we have $\int_{\Bbb{R}^{2n}}D_Hf\,d\mu=0$. We now calculate: \begin{align} D_Hf&:=\sum_{i=1}^n\frac{\partial H}{\partial p_i}\frac{\partial f}{\partial q_i}- \frac{\partial H}{\partial q_i}\frac{\partial f}{\partial p_i}\\ &=\sum_{i=1}^n\left[\frac{\partial}{\partial q_i}\left(\frac{\partial H}{\partial p_i}f\right)-\frac{\partial^2H}{\partial q_i\partial p_i}f\right] - \left[\frac{\partial}{\partial p_i}\left(\frac{\partial H}{\partial q_i}f\right)-\frac{\partial^2H}{\partial p_i\partial q_i}f\right]\\ &=\sum_{i=1}^n\frac{\partial}{\partial q_i}\left(\frac{\partial H}{\partial p_i}f\right) - \frac{\partial}{\partial p_i}\left(\frac{\partial H}{\partial q_i}f\right), \end{align} where the second mixed partials of $H$ are equal, so those terms cancel out. Finally, recall the following basic lemma from analysis: if $\phi:\Bbb{R}^k\to\Bbb{R}$ is $C^1$ and has compact support, then for any $1\leq i \leq k$, we have $\int_{\Bbb{R}^k}\frac{\partial\phi}{\partial x^i}\,dV=0$. You can view this as a special case of the divergence theorem if you want (though really this fact is used in the proof of the divergence theorem). The idea is you use Fubini to integrate over $x^i$ first, then use the fundamental theorem of calculus; but then since $\phi$ has compact support, the values at the 'endpoints' are just $0$, so the whole integral is $0$. So, applying this result yields the fact $\int_{\Bbb{R}^{2n}}D_Hf\,d\mu=0$, and hence that the integral in $(*)$ vanishes. This completes the proof of skew-adjointness of $D_H$ with respect to the $L^2$ inner product.

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  • $\begingroup$ Thank you very much, this was very helpful and clear! $\endgroup$
    – CBBAM
    Jun 18 at 4:58

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