Convergence of inverse operator with projections

Question

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

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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 $X_n$.

Answer 1

For simplicity denote the operator $M + \lambda I$ by $A$ then the problem is to prove $$ P_n(P_nAP_n)^{-1}P_n\stackrel{w}{\longrightarrow} A^{-1}, \qquad n\to\infty. $$ Let us prove that the convergence holds even in the strong operator topology. Below we will use the following simple fact: if $B_n$ converges to $0$ in the strong operator topology and $D_n$ is a uniformly bounded sequence of operators then $D_nB_n$ also converges strongly to $0$ (to prove this simply write: $\|D_nB_nx\|\le \|D_n\|\cdot \|B_nx\|$).

Lemma. Let $B$ be a bounded operator on $X$ then in the strong operator topology

• $P_nBP_n$ converge to $B$;
• $(1 - P_n)BP_n$ converge to $0$.

Proof of the lemma: Indeed, we have $$ P_nBP_n - B = P_nBP_n - P_nB - (I - P_n)B = P_nB(I - P_n) + (I - P_n)B, \\ (1 - P_n)BP_n = BP_n - B + B - P_nBP_n = B(P_n - I) + \bigg(B - P_nBP_n\bigg). $$ Since $P_n$ converge strongly to $I$ these two equations imply first and the second parts of the lemma respectively.

In our situation we have (below $P$ is a short notation for $P_n$) $$ P(PAP)^{-1}P - A^{-1} = \bigg(P(PAP)^{-1}P - PA^{-1}P\bigg) + \bigg(PA^{-1}P - A^{-1}\bigg) \\ = P\left((PAP)^{-1} - A^{-1}\right)P + \bigg(PA^{-1}P - A^{-1}\bigg). $$ The first part of the lemma for $B = A^{-1}$ implies the strong convergence to $0$ for the second difference. To deal with the first term write $$ P\left((PAP)^{-1} - A^{-1}\right)P = P\left((PAP)^{-1}[A - PAP]A^{-1}\right)P \\ = P(PAP)^{-1}\bigg[A - PAP\bigg]A^{-1}P \\ = P(PAP)^{-1}\bigg[PA(I - P) + (I - P)A\bigg]A^{-1}P \\ = P(PAP)^{-1}\bigg[PA(I - P)\bigg]A^{-1}P + P(PAP)^{-1}\bigg[(I - P)A\bigg]A^{-1}P \\ = P(PAP)^{-1}\bigg[PA(I - P)\bigg]A^{-1}P = P(PAP)^{-1}PA\bigg[(I - P)A^{-1}P\bigg]. $$ The operator in the brackets converge strongly ro $0$ by the second part of the lemma. Hence the whole expression also converges strongly to $0$.