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 $ R^N$ containing the origin and where the function $ C_\lambda$ concentrates at the origin as $ \lambda$ gets big.      They analysis 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  

$L_\lambda(\phi_\lambda)=f$ in $ \Omega$ with $ \phi_\lambda=0$ on $ \pOm$.  Moreover $ \phi_\lambda$ is perpendicular to $ v_\lambda$ in $H_0^1$.  Moreover one has the following estimate: 

$ \| \phi_\lambda\|_X \le C \| f\|_Y$ 

 (these spaces should really be $ X_\lambda$ and $Y_\lambda$ but the latex is not compiling correctly). 

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 or ???  Any comments would be greatly appreciated. 

thanks
Craig