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Let $X_0, X_1, \dots, X_k$ be smooth vector fields over ${\mathbb R}^n$, and let us consider the operator $$ L = \sum_{i=1}^k X_i^2 + X_0~. $$ Here, I assume that Hörmander's bracket condition is satisfied, that is to say that the Lie algebra generated by $X_0, X_1, \dots, X_k$ has full rank at every point in ${\mathbb R}^n$. This implies that $L$ is hypoelliptic. Now the question is: what are the conditions (if any) for the formal adjoint $L^\dagger$ of $L$ to be also hypoelliptic?

Judging by the answer to this question it appears that this is trivially true since Hörmander's condition "does not change by taking adjoints". What this means is unclear to me. We have indeed that $$ L^\dagger = \sum_{i=1}^k X_i^2 - X_0 + f~, $$ where $f$ is a smooth scalar function, right? If there were no $f$, then this would indeed be trivial, since we find again the same Lie algebra. Since all the treatments of Hörmander's theorem that I have seen discuss operators without such an $f$ (except Hormander's paper of 1967), I guess there is an easy way to get rid of this $f$. Do you know how? Or do you have another argument to show that $L^\dagger$ is also hypoelliptic? Thanks a lot.

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Not related to the question, but to Hormander: ams.org/news?news_id=1710 –  Alain Valette Dec 4 '12 at 22:01
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1 Answer

up vote 4 down vote accepted

The hypoellipticity result is more precise: you have $$ Lu \in H^s_{loc}\Longrightarrow u\in H^{s+2-\delta}_{loc}\quad\text{ for some $\delta\in [0,2)$,} $$ and that $\delta$ is linked to the number of brackets of the $X_j$ needed to generate the full tangent space. For instance, in the elliptic case, where $X_0=0$ and the $(X_j)_{1\le j\le n}$ span the tangent space, we have $\delta=0$. Now if $f\in C^\infty$,

$$ u\in H^s_{loc},\quad Lu+fu \in H^s_{loc}\Longrightarrow Lu \in H^s_{loc}\Longrightarrow u\in H^{\epsilon+s}_{loc} $$ with $\epsilon=2-\delta>0$.

We assume now $Lu+fu\in C^\infty$ and we want to prove $u\in C^\infty$: since it is a local result, we may assume for some $s$ that $ u\in H^s_{loc} $.

Since $Lu+fu\in H^s_{loc}$, we get from above $u\in H^{s+\epsilon}_{loc}$.

Since $u, Lu+fu\in H^{s+\epsilon}_{loc}$ we get

$u\in H^{s+2\epsilon}_{loc}$

and so on: $u$ is smooth.

P.S. The weird typesetting is due to MathJax.

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Thank you very much. Do you happen to have a good reference to recommend for this more precise notion of hypoellipticity? –  Nown Dec 4 '12 at 21:34
Yes, Hörmander's ALPDO, Chapter 22. –  Bazin Dec 4 '12 at 21:38
Ok, I'll have a look at it. Thank you very much, again! –  Nown Dec 4 '12 at 21:44
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