In the 'interactive' proof of Lomonosov's Theorem about hyperinvariant subspaces (in the book Hilbert Space Operators, A Problem Solving Approach), one is asked to prove the compactness of the following nonlinear operator:

$F(x) = \displaystyle\sum_{A\in\mathscr{F}}\beta_A(Kx)AKx $

where $\mathscr{F}$ is a finite set, K is a linear compact operator, A is a linear bounded operator and

$\beta_A(x)=\displaystyle\frac{\alpha_A(x)}{\displaystyle\sum_{A\in\mathscr{F}}\alpha_A(x)}$

$\alpha_A(x) = max\{0,1-\|Ax-x_0\| \}$

As $\beta_A$ maps into the compact set $[0,1]$ and $AK$ being a compact operator, is it not possible to immediately say that $F$ is a (nonlinear) compact operator? This would heavily rely on $\beta_A\cdot AK$ and the finite sum of (nonlinear) compact operators being compact though.

In the 'interactive' proof of Lomonosov's Theorem about hyperinvariant subspaces (in the book Hilbert Space Operators, A Problem Solving Approach), one is asked to prove the compactness of the following nonlinear operator:

$F(x) = \displaystyle\sum_{A\in\mathscr{F}}\beta_A(Kx)AKx $

where $\mathscr{F}$ is a finite set, K is a linear compact operator, A is a linear bounded operator and

$\beta_A(x)=\displaystyle\frac{\alpha_A(x)}{\displaystyle\sum_{A\in\mathscr{F}}\alpha_A(x)}$

$\alpha_A(x) = max\{0,1-\|Ax-x_0\| \}$

As $\beta_A$ maps into the compact set $[0,1]$ and $AK$ being a compact operator, is it not possible to immediately say that $F$ is a (nonlinear) compact operator? This would heavily rely on $\beta_A\cdot AK$ and the finite sum of (nonlinear) compact operators being compact though.

Edit: This would skip the use of Mazur's theorem and be a nice 'parallel' to the linear case. I've found some other posts but they don't cover the sum part. Shouldn't the same proof as in the linear case work though?