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If $Q$ is a symmetric nonnegative operator on a real separable Hilbert space $H$. We have known that an orthonormal set $\{ e_i \}$ can be chosen such that $Q e_i=\lambda_i e_i, \lambda_i > 0$. And $Q$ can be represented by $Q = \lambda_i e_i \otimes e_i$.

I want to know

(1) When the range of $Q$ is closed? I know when $\lambda_i \geq c >0$, the range of $Q$ is closed. Is it true conversely?

(2) When $\lambda_i \to 0$, the range of $Q$ may be not closed, can an element be given such that it is not in the range but in the closure of the range?(I really want to know such an element.)

(3) The range of $Q$ is $(\ker Q )^{\bot}$, isn't it?

(4) Is there any other theorems to verify the closure of the range of the operator?(If the question is too wide to answer, it can be omitted.)

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migrated from mathoverflow.net May 12 '14 at 20:13

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  1. Suppose the eigenvalues are not bounded below, then we may find a sequence of eigenvalues $\lambda_n$ with $\lambda_n \le \frac{1}{n}$. If $e_n$ are the corresponding orthonormal eigenvectors, show that the sum $\sum_n \lambda_n e_n$ converges to some $y \in H$, and $y$ is not in the range of $Q$ but is in its closure. (Intuitively, if we had $Qx=y$ then we would have to have $x = \sum_n e_n$ but this sum does not converge.)

  2. See 1.

  3. No, $(\ker Q)^\perp$ is necessarily closed but we have just shown the range need not be. However, you can say that $(\ker Q)^\perp$ is the closure of the range of $Q$; this is easy to prove using the eigenvectors and eigenvalues of $Q$ (or directly, using the self-adjointness of $Q$).

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