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Let $H$ be a separable Hilbert space and K $∈ K(H)$ be a compact, symmetric operator.

Consider the operator $T_μ := μI −K$ for a given $μ$ ∈ ℝ∖{0}.

  1. If $|μ| > ∥K∥_{B(H)}$ , is it true that $T_μ$ is a bijection?

  2. Is this still true if $|μ| = ∥K∥_{B(H)}$ ?

For the first question i would say yes because $∥K∥_{B(H)}$ of a compact, symmetric operator is equal to : $$ max \{|λ| : λ\ is \ eigenvalue \ of \ K\}. $$

So $|μ|$ > $max$ {$|λ|$} then μ cannot be an eigenvalue of $K$. For the Fredholm's alternative theorem this imply that $T_μ$ is injective and surjective.

But if $|μ| = ∥K∥_{B(H)}$? We have $|μ|$ = $max$ {|λ|} but i think is not true that $μ$ is always an eingenvalue of $K$ (for example $μ$ can be equal to -2 and the eigenvalue of $K$ always >0), so in principle $T_μ$ could be bijective. I'm not really sure about the second question. Can i have some hints?

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The spectrum of the 2d Neumann-Poincare operator(which is compact if the curve it is defined is smooth and the operator is symmetrizable) lies inside the interval [-1/2, 1/2] and 1/2 is an eigenvalue.

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  • $\begingroup$ I am Not clearly understand what you would say, in your opinion in the second case cannot be bijective? $\endgroup$
    – maru0032
    Sep 14 at 19:05
  • $\begingroup$ You have to check the second case for a specific operator at hand. The value corresponding to the spectral bounds sometimes contained in the spectrum and sometimes not. $\endgroup$ Sep 14 at 19:17
  • $\begingroup$ Ah ok, so in the second case could be or could Not be bijective, depends on the specific operator. $\endgroup$
    – maru0032
    Sep 14 at 19:22
  • $\begingroup$ Also, symmetric or self-adjoint operators have real spectrum not positive. $\endgroup$ Sep 14 at 19:23
  • $\begingroup$ Yes, it was just an example Not a general result that k>0 $\endgroup$
    – maru0032
    Sep 14 at 19:52

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