# range of compact, self-adjoint operator

Let $A$ be a compact, self-adjoint operator on a Hilbert space $H$. Then, the spectrum theorem says that for every $x\in H$

$$Ax = \sum_{k=1}^{\infty} \lambda_k\langle x,e_k \rangle e_k$$ where $\lambda_k$ are nonzero eigenvalues and $e_k$ are the corresponding orthnormal eigenvectors.

Then, the book that I found ("Applied Analysis" by Hunter) says that assuming $A$ has an infinite sequence of nonzero eigenvalues,

(1) the range of $A$ is

$$\text{ran A} = \big\{ \sum_{k=1}^{\infty} c_ke_k \mid \sum_{k=1}^{\infty} \frac{|c_k|^2}{|\lambda_k|^2} < \infty \}$$

and

(2) the range is not closed since $\lambda_n \to 0$ as $n \to \infty$

But, I do not understand (1) and (2).

For (1), how do I get the condition $\sum_{k=1}^{\infty} \frac{|c_k|^2}{|\lambda_k|^2} < \infty$?

For (2), how does the fact that $\lambda_n \to 0$ as $n \to \infty$ show that the range is not closed?

Any element $Ax = \sum_{k=1}^{\infty} c_k e_k$ of the range has $c_k = \lambda_k \langle x, e_k \rangle$ and therefore $$\sum_{k=1}^{\infty} \frac{|c_k|^2}{|\lambda_k|^2} = \sum_{k=1}^{\infty} |\langle x, e_k \rangle |^2 \le \langle x,x \rangle < \infty$$ by Bessel's inequality. Conversely, if $\sum_{k=1}^{\infty} \frac{|c_k|^2}{|\lambda_k|^2}$ converges then you can let $x = \sum \frac{c_k}{\lambda_k} e_k.$
• For the converse of the first part, how do I know that if I take $u = \sum_{k=1}^{\infty} c_k e_k$ where $\sum_{k=1}^{\infty} \frac{|c_k|^2}{|\lambda_k|^2} < \infty$, $u$ is convergent? Isn't $u$ convergent if and only if $\sum_{k=1}^{\infty} |c_k|^2< \infty$? – nan Jan 9 '17 at 5:57
• @momonao You define $x = \sum \frac{c_k}{\lambda_k} e_k$, not $\sum c_k e_k$. – user6246 Jan 9 '17 at 15:57
• What I mean is that if you want to show the converse, then don't you need to show that the elements in the set$$\big\{ \sum_{k=1}^{\infty} c_ke_k \mid \sum_{k=1}^{\infty} \frac{|c_k|^2}{|\lambda_k|^2} < \infty \}$$ are well-defined? – nan Jan 9 '17 at 20:22
• @momonao $\sum |c_k|^2$ is majorized by $\sum |c_k|^2 / |\lambda_k|^2$ since $\lambda_k$ tends to $0$ – user6246 Jan 10 '17 at 1:11
• Got it Thx for the clarification! As for the second part, could please explain to me how I can see that $\lambda_n \to 0$ implies that the range is not closed (without appealing to the general result)? – nan Jan 10 '17 at 15:32