$\newcommand{\norm}[1]{\|#1\|}$ $\newcommand{\ab}[1]{\langle #1\rangle}$ $\newcommand{\C}{\mathbf C}$
Questions
Let $I$ be the unit interval. Let $H=L^2(I)$ and $T:H\to H$ be a compact self-adjoint operator. Let $f_n:I\to I$ be a sequence of function in $L^\infty(I)$ such that $f_n\to \mathbf 1$ in $L^\infty$ where $\mathbf 1$ denotes the function $I\to \C$ which takes the value $1$ everywhere. Define $T_n=f_n T$, that is $T_n:H\to H$ takes $\xi$ to $(T\xi)f _n$, that is, the pointwise product of $T\xi$ and $f_n$.
Question 1. I am trying to see if $$ \lambda_{\text{max}}(T_n) \to \lambda_{\text{max}}(T) $$ where $\lambda_\text{max}(T_n)$ denotes the largest eigenvalue of $T_n$ and similarly for $T$.
(We know that since each $T_n$ is compact and has real spectra as argued in the Appendix).
Progress on the Question 1. If $\rho$ denotes the spectral radius, then by the spectral radius formula, we have $$ \rho(T_n) = \limsup_{k\to \infty} \norm{(T_n)^k}^{1/k} = \limsup_{k\to \infty} \norm{(f_n T)^k}^{1/k} $$ If $n$ is large enough then we have $\norm{f_n-\mathbf 1}\leq \varepsilon$, which given from above that $$ \rho(T_n) \leq \lim\sup_{k\to \infty} \norm{f_n^k T^k}^{1/k}\leq (1+\varepsilon)\limsup_{k\to \infty} \norm{T^k}^{1/k} = (1+\varepsilon) \rho(T) $$ Similarly we can show that for large enough $n$ we have $$ (1-\varepsilon)\rho(T) \leq \rho(T_n) $$ giving that $\rho(T_n)\to \rho(T)$ as $n\to \infty$. Since the spectrum may have negative values, this does not necessarily give $\lambda_{\text{max}}(T_n)\to \lambda_{\text{max}}(T)$ but is a step in that direction.
Also, if $\lambda_{\text{penmax}}(T)$ denotes the second to largest eigenvalue of $T$, then,
Question 2. Is it also true that $$ \lambda_{\text{penmax}}(T_n) \to \lambda_{\text{penmax}}(T) $$ as $n\to \infty$.
If there is something more general known about convergence of spectra (of compact operators or otherwise) then please feel free to share.
Appendix
Let $\varepsilon>0$ be a small positive number and $f:I\to I$ be such that $\norm{f-\mathbf 1}_\infty<\varepsilon$, where $\mathbf 1$ denotes the constant function $1$. Let $H=L^2(I)$. Let $T:H\to H$ be a compact operator and define $fT: H\to H$ as $(fT)\xi = f\ T\xi$ for all $\xi\in H$.
We will show that if $T$ is a compact self-adjoint operator then $fT$ is compact with all its eigenvalues real.
If $T$ is compact then there is a sequence $(T_n)$ of bounded linear operators on $H$ such that each $T_n$ has finite rank and $T_n\to T$ in operator norm. Clearly, $fT_n\to fT$ in operator norm. Since each $fT_n$ is also of finite rank, we see that $fT$ is indeed compact.
For any $g\in L^\infty(I)$ close to $\mathbf 1$ in the $L^\infty$ norm, define an inner product $\ab{\cdot, \cdot}_g$ on $L^2(I)$ as $\ab{\xi, \zeta}_g = \int_I \xi\bar \zeta g\ d\mu$. It is easy to check that this is indeed an inner product. Let us write $H_g$ to denote $L^2(I)$ equipped with the inner product $\ab{\cdot, \cdot}_f$. Then we have $$ \ab{(fT)\xi, \zeta}_{1/f} = \ab{f\ T\xi, \zeta}_{1/f} = \ab{T\xi, \zeta} = \ab{\xi, T \zeta} = \ab{\xi, f\ T\zeta}_{1/f} = \ab{\xi, (fT) \zeta}_{1/f} $$ This shows that $fT$ is self-adjoint with respect to the inner product $\ab{\cdot, \cdot}_{1/f}$.
Cross Posted on MSE.
