Let $(M,g)$ be a compact Riemannian manifold without boundary of dimension $2m$. Let
$$T:W^{2,2}(M)\rightarrow L^{2}(T^{*}M\otimes TM)$$ be a second order, linear, differential operator (coefficients in local coordinates are bounded smooth functions).

Now consider the fourth order linear operator $$\Lambda:= T^{*}T: W^{4,2}(M)\rightarrow L^{2}(M)$$

Suppose we know that: $\Lambda$ is Fredholm (finite dimensional kernel and cokernel), is elliptic with index 0 and its principal symbol is equal to that of $\Delta_g^2$. Let $$\ker(\Lambda):=span\left[\psi_1,\ldots, \psi_N \right]$$ with $\psi_i$ a $L^{2}$-orthonormal basis of $\ker(\Lambda)$. Now let $$\pi_{D}:W^{2,2}(M)\rightarrow W^{2,2}(M)$$ be the continuous projection $$\pi_{D}(f)=f-\sum_{i=1}^{N}(f,\psi_i)_{L^{2}(M)}\psi_i$$ and define $$D=\pi_{D}(W^{2,2}(M))$$

Under the above assumptions does exist $C>0$ s.t. $\forall f\in D$ we have

$$\left\| \left|T f\right|_{g} \right\|_{L^{2}(M)}\geq C\left\|f\right\|_{W^{2,2}(M)}$$

In other words, is $\left\| \left|T f\right|_{g} \right\|_{L^{2}(M)}^2$

a coercive functional on $D$?


1 Answer 1


I could not quite understand your question due to TeX problems and here is my best guess. There exists $C>0$ such that for any $f\in W^{4,2}$ orthogonal to $\ker \Lambda$ we have

$$C \Vert f\Vert_{L^2} \leq \Vert \Lambda f\Vert_{L^2}= \Vert Tf\Vert_{L^2}^2.\tag{1}\label{1} $$

(For a proof of (\ref{1}) check Lemma 10.4.9 of these notes.) Using elliptic estimates we deduce

$$ \Vert f\Vert_{W^{4,2}} \leq C'(\Vert f\Vert_{L^2}+ \Vert\Lambda f\Vert_{L^2} ). $$

Using (\ref{1}) we deduce

$$ \Vert f\Vert_{W^{4,2}}\leq C_1 \Vert Tf\Vert_{L^2}^2$$

Now let

$$\mu =\inf\bigl\lbrace \Vert Tf\Vert_{L^2};\;\; \Vert f\Vert_{4,2}=1,\;\; f\perp \ker \Lambda\;\bigr\rbrace\geq \frac{1}{\sqrt{C_1}}. $$

We deduce that

$$ \mu \Vert f\Vert_{W^{2,2}}\leq \mu \Vert f\Vert_{W^{4,2} } \leq \Vert Tf\Vert_{L^2},\;\;\forall f \in W^{4,2},\;\;f\perp \ker \Lambda. $$

Since $W^{4,2}$ is dense in $W^{2,2}$ we obtain the coercivity you were seeking.

  • $\begingroup$ Thank you very much, it's exactly what i was looking for! $\endgroup$
    – Italo
    Dec 9, 2012 at 17:12

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