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Consider an operator $T: L^2 \mapsto L^2$ of the form $TA = g (h \ast A)$ where $g$ is and $h$ are bounded $C^\infty$ functions.

This operator $T$ can be shown to be Hilbert-Schmidt, hence compact. The function $h$ is given by a series $$h(x, y) = \sum_m h_m(x, y).$$ As a first step in solving the eigenvalue problem associated to $A$, we approximate the operator $T$ by $T_n$ where we replace $h$ by a truncated series up to $n$. That is, $$h_n(x, y) = \sum_{m \leq n} h_m(x, y).$$

So, a question we now might ask is: if we approximate the eigenvalue problem associated to $T$, namely -$TA = \lambda A$- by $T_n A_n = \lambda_n A_n$ can anything be said about the difference between the maximal eigenvalues $\lambda^*$ and $\lambda_n^*$?

For instance, is there a relationship between $|\lambda^* - \lambda_n^*|$ and $\|T - T_n\|$ as the latter can be bound above by the $L^2$ norm of $h - h_n$.

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  • $\begingroup$ Not an answer, but: perhaps you should ask the moderators to merge this profile with your older one? mathoverflow.net/users/5295/j $\endgroup$
    – Yemon Choi
    Commented May 11, 2016 at 16:20
  • $\begingroup$ He's alive! ALIVE!!!! :-) $\endgroup$
    – Asaf Karagila
    Commented May 11, 2016 at 16:43
  • $\begingroup$ @YemonChoi Thanks, I will do that. Asaf: Yes. $\endgroup$
    – Jonas T
    Commented May 11, 2016 at 17:45

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