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