1
$\begingroup$

Hi!

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 $D$?

$\endgroup$

1 Answer 1

0
$\begingroup$

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.

$\endgroup$
1
  • $\begingroup$ Thank you very much, it's exactly what i was looking for! $\endgroup$
    – Italo
    Dec 9, 2012 at 17:12

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.