$\underset{_{a\rightarrow 0}}{\lim }L_{a}^{\ast }\left( L_{a}L_{a}^{\ast }\right) ^{-1}L_{a}=?$

Question

Let $\varepsilon >0$ and let $L$ be a bounded operator acting on a Hibert space such that $L_{\lambda }=L-\lambda I$ is surjective of every $\lambda $ such that $\varepsilon >\left\vert \lambda \right\vert >0$.

I'm trying, by applying the Cauchy criterion, to show that the limit

$\underset{_{\lambda \rightarrow 0}}{\lim }L_{\lambda }^{\ast }\left( L_{\lambda }L_{\lambda }^{\ast }\right) ^{-1}L_{\lambda }$

exists, where $L_{\lambda }^{\ast }=L^{\ast }-\overline{\lambda }I.$ If this is not true, can you give me some counter-example ?

Thank you !

Answer 1

Setting $L=0$ shows $L_\lambda = -\lambda I$, and the operator $L_\lambda^*(L_\lambda L^*_\lambda)^{-1}L_\lambda $ is equal to the identity. So it seems to disprove convergence is going to be hard.

It is quite easy to show that $L_\lambda^*(L_\lambda L^*_\lambda)^{-1}L_\lambda =I$ holds for finite-dimensional $H$ or normal $L$.

Here is a partial answer for the general case: Using singular value decomposition of $L=UT_fV^*$, where $U$ is partial isometry ($U:N(U)^\perp\to R(U)$ is isometry), $V$ isometry, $T_f$ multiplication operator on some $L^2(\mu)$.

Then $$ L_\lambda^*(L_\lambda L^*_\lambda)^{-1}L_\lambda = VT_{\bar f-\bar\lambda}U^* (UT_{f-\lambda}V^* VT_{\bar f-\bar\lambda}U^*)^{-1}UT_{f-\lambda}V^* =VT_{\bar f-\bar\lambda}U^* (UT_{|f|^2+|\lambda|^2}U^*)^{-1}UT_{f-\lambda}V^*. $$ If $U$ would be a (full) isometry, then clearly this operator reduces to the identity.

I do not know how to show this for the general case. Maybe someone else has an idea?