Convergence of self-adjoint operators

Question

Let $\left(L_n\right)$ be a sequence of self-adjoint operators on a Hilbert space and let $L$ be a self-adjoint operator. Assume that for a family $\left(\Phi_\alpha\right)_{\alpha \in I}$ of measurable bounded functions from $\mathbb{R}$ to $\mathbb{R}$ and some index set $I$ we have $$ \lim _{n \rightarrow \infty} \Phi_\alpha\left(L_n\right) f=\Phi_\alpha(L) f $$ for all $f$ in the Hilbert space and for all $\alpha \in I$. Let $\mathcal{A}$ be the closure of $\left\{\Phi_\alpha \mid \alpha \in I\right\}$ with respect to the supremum norm. Show that $$ \lim _{n \rightarrow \infty} \Phi\left(L_n\right) f=\Phi(L) f $$ for all $\Phi \in \mathcal{A}$ and $f$ in the Hilbert space.

My thoughts so far:

1. Let $\Phi\in \overline{\{\Phi_{\alpha}: \alpha \in I \}}$ and $\varepsilon >0$. Then we can find and $\alpha \in I$ with $||\Phi_{\alpha}-\Phi||_{\infty}=\sup\limits_{x \in \mathbb{R}}|\Phi_{\alpha}(x)-\Phi(x)| < \varepsilon$ Furthermore $\Phi$ is also bounded.
2. By the spectral theorem for a self-adjoint operator $A$ there exist a $\sigma-$finite measure space $(X,\mu)$, a measurable function $u:X\rightarrow \mathbb{R}$ and a unitary map $U:L^{2}(X,\mu)\rightarrow H$ with $A=UM_{u}U^{-1}$ where $M_{u}$ is the multiplication operator.
3. By assumption for $\varepsilon>0$ we can find an $n_{0}$ such that for all $n>n_{0}$ $$ \begin{align} \varepsilon > ||\Phi_{\alpha}(L_{n})f-\Phi_{\alpha}(L_{n})f||_{H} &= ||\left(\Phi_{\alpha}(L_{n})-\Phi_{\alpha}(L)\right)f||_{H}\\ &= ||U\left(M_{\Phi_{\alpha}\circ u_{n}}-M_{\Phi\circ u}\right)U^{-1}f||_{H}\\ &= ||\left(M_{\Phi_{\alpha}\circ u_{n}}-M_{\Phi\circ u}\right)\psi||_{L^{2}}\\ &= ||\left(\Phi_{\alpha}\circ (u_{n}-u)\right)\psi||_{L^{2}}\\ &= ||\left(\Phi_{\alpha}(u_{n})-\Phi_{\alpha}(u)\right)\psi||_{L^{2}} \end{align} $$

where I've used that $U$ is an isometry and $\psi=U^{-1}f$. However I've assumed that $U$ is identical for all $L_{n}$ and $L$ which I'm not sure about.

Using the spectral theorem I think this should yield an expression similar to the one below which one can bound by some $\varepsilon$ which would proof the claim.

$$ \begin{align} ||\Phi(L_{n})f-\Phi_{\alpha}(L_{n})f||_{H} &= \ldots = ||\left(\Phi\circ(u_{n}-u)\right)\psi||_{L^{2}}\\ &\leq ||\psi||_{L^{2}} ||\Phi\circ (u_{n}-u)||_{L^{\infty}}\\ &\leq ||\psi||_{L^{2}}\left( ||\Phi\circ (u_{n}-u)-\Phi_{alpha}(u_{n}-u)||_{L^{\infty}}+||Phi_{\alpha}(u_{n}-u)||_{L^{\infty}}\right)\\ &\leq ||\psi||_{L^{2}}\left( ||\Phi-\Phi_{\alpha}||_{\infty} ||u_{n}-u||_{L^{\infty}}+||\Phi_{\alpha}(u_{n}-u)||_{L^{\infty}}\right)\\ \end{align} $$

Is my approach correct? Or do I need to think in another direction?

Answer 1

You have all the ingredients, they just need to be put together.

Since $\{L_n\}$ is convergent, it is bounded. That is, there exists $c$ with $\|L_n\|≤c$ for all $n$. You also have $\|L\|≤c$.

$\def\e{\varepsilon}$ Fix $\e>0$. Then there exists $\alpha$ such that $$\|\Phi_\alpha-\Phi\|_\infty<\frac\e{3c}.%,\qquad\qquad \alpha\geq\alpha_0. $$ You also have an $n_0$ such that $$ \|\Phi_\alpha(L_n)f-\Phi_\alpha(L)f\|<\frac\e{3c} $$ for all $n\geq n_0$.

The final ingredient is more or less done in your question. A unitary is an isometry. So $$ \|g(A)\|=\|M_g\,u\|≤\|M_g\|\,\|u\|=\|g\|_\infty\,\|A\|. $$ In our case, this gives \begin{align} \|\Phi_\alpha(A)-\Phi(A)\|=\|(\Phi_\alpha-\Phi)A\|≤\|\Phi_\alpha-\Phi\|_\infty\,\|A\|. \end{align}

Then, assuming without loss of generality that $\|f\|=1$, \begin{align} \|\Phi(L_n)f-\Phi(L)f\| &\leq\|\Phi(L_n)f-\Phi_\alpha(L_n)f\| + \|\Phi_\alpha(L_n)f-\Phi_\alpha(L)f\|\\[0.3cm] &\qquad+ \|\Phi_\alpha(L)f-\Phi(L)f\|\\[0.3cm] &\leq\|\Phi_\alpha-\Phi\|_\infty\,\|L_n\| + \frac\e3 +\|\Phi_\alpha-\Phi\|_\infty\,\|L\|\\[0.3cm] &\leq\frac\e3+\frac\e3+\frac\e3=\e \end{align} for all $n\geq n_0$.