Infinite-dimensional subspace of a Hilbert space

Let AA be a compact self-adjoint operator on an infinite-dimensional Hilbert space H. Suppose (λn)(\lambda_n) is a sequence of different eigenvalues of AA and (xn)(x_n) is a sequence of corresponding eigenvectors with ‖xn‖=1\| x_n \| = 1.

Is the set S={∑∞i=1αnxn :αn∈C, ∑∞i=1|αn|2<∞}S = \{ \sum_{i=1}^{\infty} \alpha_nx_n\ : \alpha_n \in \mathbb{C}, \ \sum_{i=1}^{\infty} | \alpha_n |^2 < \infty \} closed? I know that self-adjoint operators are closed. If (yn)(y_n) is a sequence in SS converging in HH, say to yy. Then, due to compactness of AA, the sequence (Ay_n)(Ay_n) converges to AyAy. Since AA is closed y \in Sy \in S. But I am not sure if this is a correct proof. =================      I don't understand this step: "Since AA is closed y \in Sy \in S. – GEdgar 2 days ago ================= 1 Answer 1 ================= The reason why S S is closed is that the sequence \{x_n\} \{x_n\} is orthonormal (this follows from A A selfadjoint). In your reasoning, if y_n\to y y_n\to y then Ay_n\to y Ay_n\to y because A A is bounded; compactness has nothing to do with it. And the "A A is closed" bit is nonsense; there are selfadjoint compact operators with dense range (typical example ism A A that maps the canonical basis by e_n\mapsto e_n/n e_n\mapsto e_n/n .      I do not understand your comment about closedness. Every self-adjoint operator is densely defined, closed and symmetric. And closed means, that its graph is closed (in our case in H \timesHH \timesH) – user380460 2 days ago      How do I use the fact, that (x_n)(x_n) is orthonormal to prove that SS is closed? – user380460 2 days ago      Yes, but "closed operator" does not mean "closed range", so the fact is irrelevant to your reasoning. The point is that in general the set SS is strictly larger than the range of AA. – Martin Argerami 2 days ago