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.

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 $$ \lim_{n\to \infty} \langle I_n(I_n^* MI_n+\lambda \operatorname{id}_{\mathbb{R}^n})^{-1} I_n^* U, U\rangle_X = \langle ( M +\lambda \operatorname{id}_{X})^{-1} U,U\rangle_X. $$

Thanks to Prof. Ozawa's comment, here is what I have got so far:

As $B= I_n^* MI_n+\lambda \operatorname{id}_{\mathbb{R}^n}$ is a symmetric positive definite matrix on $\mathbb{R}^n$, $I_nBI_n^*$ preserves the space $X_n$, and hence one can directly verify that $$ I_nB^{-1} I_n^*=P_n (I_nBI_n^*)^{-1} P_n, $$ where $P_n$ is the orthogonal projection onto $X_n$. As $I_nI_n^*=P_n =P_n \operatorname{id}_{X} P_n$, \begin{align*} I_n(I_n^* MI_n+\lambda \operatorname{id}_{\mathbb{R}^n})^{-1} I_n^* &= P_n (I_n(I_n^* MI_n+\lambda \operatorname{id}_{\mathbb{R}^n}) I_n^*)^{-1} P_n \\ &=P_n (P_n( M +\lambda \operatorname{id}_{X}) P_n)^{-1} P_n. \end{align*} Then the desired claim reduces to the convergence of $$ P_n (P_n( M +\lambda \operatorname{id}_{X}) P_n)^{-1} P_n \to ( M +\lambda \operatorname{id}_{X})^{-1} $$ in the weak operator topology. However, I don't know how to handle the inverse operator. In particular, $P_n( M +\lambda \operatorname{id}_{X}) P_n$ is only invertible on $X_n$, $(P_n( M +\lambda \operatorname{id}_{X}) P_n)^{-1}$ is only defined on $X_n$.

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 $u=(u_i)_{i=1}^n$. 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.

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.