I am reading a paper where at some point they analyse the following linear operator:

$$L_\lambda(\phi)= - \Delta \phi - C_\lambda(x) \phi$$

where $ C_\lambda(x)>0$ (smooth) in $ \Omega$ a bounded domain in $ \mathbb{R}^N$ containing the origin and where the function $ C_\lambda$ concentrates at the origin as $ \lambda$ gets big. They analyse the operator in some weighted $L^\infty$ spaces, where the weights depends on the parameter $ \lambda$. Lets denote the spaces by $ X_\lambda$ and $Y_\lambda$. For each fixed $ \lambda$ the weights are bounded above by one and bounded away from zero by a constant which depends on $ \lambda$. As $ \lambda$ gets big both the weights concerge to zero at the origin. Here is their result: there is some explicit function $ v_\lambda $ which is smooth and zero on the boundary of $ \Omega$, and there is some $ C>0$ and $ \lambda_0 >0$ such that for all $ \lambda > \lambda_0$ and all $ f \in L^\infty$ there is some $ \phi_\lambda $ such that

$$L_\lambda(\phi_\lambda)=f \text{ in } \Omega \text{ with } \phi_\lambda=0 \text{ on } \partial \Omega.$$

Moreover $ \phi_\lambda$ is perpendicular to $ v_\lambda$ in $H_0^1$. Moreover one has the following estimate:

$$ \| \phi_\lambda\|_{X_\lambda} \le C \| f\|_{Y_\lambda}$$

So here is my question. Lets fix $ \lambda$ and analyse the operator. Since $L_\lambda$ is self adjoint there should be a basis of orthogonal eigenfunctions for $L^2$ or $H_0^1$. If one of the eigenvalues was zero then there would be some $f \in L^\infty$ such that $L_\lambda(\phi)=f$ would not be solvable. Hence all the eigenvalues must be non zero. But then there is no way one can hope to find a solution $ \phi$ for each $f$ which is perpendicular to $ v_\lambda$.

I am missing something? Any comments would be greatly appreciated.

thanks Craig

  • $\begingroup$ $\pOm$ should really be $ \partial \Omega$. $\endgroup$
    – Craig
    Jun 18, 2013 at 4:28
  • $\begingroup$ I do not see it an the moment (a reference to the paper would be a help), but your argument works only for $f\in L^2$. It is a big restriction to have $f\in L^\infty$. $\endgroup$ Jul 1, 2013 at 10:12
  • $\begingroup$ Andras. If you are interested I asked a very related question with regards to this material and would greatly appreciate your comments. thanks $\endgroup$
    – Craig
    Jul 19, 2013 at 3:26


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