It's known that for a matrix, $\max|λ|≤\sqrt{tr(A^*A)}=\sqrt{∑_{i,j=1}^n|A_{i,j}|^2}$ where $\lambda$ denotes its eigenvalue. I'm wondering whether there's an analog of this inequality for a general bounded operator.

For example an integral operator, $T(f(x))=\int k(x,y)f(y)dy$, can we say its spectrum is bounded by something like $\sqrt{\int k(x,y)^2dydx}$, it looks similar to the property that $\rho (T)\leq ||T||=\underset {||f||\leq 1}\sup ||Tf||$ if we use the norm of $L_2 R$ space.


Yes there is. If $A$ is an operator in a Hilbert space, $A^*A$ is a trace class operator, and $\lambda$ is an eigenvalue of $A$ with the corresponding unit eigenvector $u$, then $|\lambda|^2=(Au,Au)=(A^*Au,u)\le\operatorname{tr}A^*A$, because $\operatorname{tr}A^*A=\sum (A^*Ae_j,e_j)=\sum||Ae_j||^2$ for any orthonormal basis $\{e_j\}$, and we can always arrange that $e_1=u$. Moreover, $$ ||A||^2=\sup_{||u||=1}(Au,Au)\le \operatorname{tr}A^*A. $$ But note that not all operators in an infinite-dimensional Hilbert space are trace class...

  • $\begingroup$ Thank you very much! Could you please tell me the name of this theorem if there is one? I'd like to read about more details about it. $\endgroup$ – iridium Jun 18 '14 at 19:07
  • $\begingroup$ Apparently there is no special name. You might find this link en.wikipedia.org/wiki/Trace_class useful, just for starters. $\endgroup$ – Vladimir Jun 18 '14 at 19:11

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