Let $X$ be a separable Hilbert space, and let $(e_i)_{i=1}^\infty$ be an orthonormal basis of $X$. For each $n\in \mathbb{N}$, let $X_n$ be the subspace spanned by $(e_i)_{i=1}^n$, and consider the operator $I_n:\mathbb{R}^n\to X$ be such that $I_n u=\sum_{i=1}^n u_i e_i$ for all $u=(u_i)_{i=1}^n$. The adjoint $I_n^*:X\to \mathbb{R}^n$ is then given by $I^*_n x= (\langle x, e_i\rangle_X)_{i=1}^n$.

Let $M$ be a symmetric positive semidefinite operator on $X$ (we can assume $M$ is of trace class if necessary). Given $\lambda>0$ and $U\in X$. Is it true that $$ \lim_{n\to \infty} \langle I_n(I_n^* MI_n+\lambda \operatorname{id}_{\mathbb{R}^n})^{-1} I_n^* U, U\rangle_X = \langle ( M +\lambda \operatorname{id}_{X})^{-1} U,U\rangle_X. $$

The claim seems to be related to the convergence of projection operator in the weak operator topology. The matrix $I_n^* MI_n$ has a matrix representation $(\langle Me_i,e_j\rangle_X)_{ij}$. By using $I_n^*\operatorname{id}_{X} I_n=\operatorname{id}_{\mathbb{R}^n}$, we can rewrite $I_n^* MI_n+\lambda \operatorname{id}_{\mathbb{R}^n}$ as $I_n^* (M+\lambda\operatorname{id}_{X})I_n $. But I don't know how to handle the inverse operator, as $I_n:\mathbb{R}^n\to X$ is not invertible.