i got the following operator. Let H be a Hilbert space, $(\lambda_n)_{n \in \mathbb{N}}$ a bounded sequence in $\mathbb{K}$ and $T^: H \to H, x \to \sum_{n\in\mathbb{N}} \lambda_n \left<x, e_n\right> e_n$. for a given orthonormal system $(e_n)_{n \in \mathbb{N}}$ I have shown quite a bit about this operator, but i got some open questions.
Is there an easy way to show that $T$ is well-defined? Is it sufficient to show that $\Vert Tx \Vert$ is bounded for a $x \in H$? Because that is pretty easy.
Im supposed to show that $T$ is compact if and only if $\lambda_n \to 0$ for $n \to \infty$. For the "$\Rightarrow$" direction I have the following: Let $T$ compact, and $(x_n)_n$ a bounded sequence in $H$. Suppose $\lambda_n \not\to 0$ for $n \to \infty$. Since $T$ is compact, $(Tx_n)_n$ should have a convergent subsequence $(Tx_{n_k})_k$. My Idea is to use the Assumption $\lambda_n \not\to 0$ to show, that $\Vert Tx_{n_k} - Tx_{n_l} \Vert > C$ for a constant $C$. But i dont know how to continue the proof. For the "$\Leftarrow$" direction i got no idea how to start.
- Has this operator any eigenvalues beside the $\lambda_n$'s? And if so, how can i show this?
Showing that $T$ is well-defined amounts to observing that, for any orthonormal set $\{ e_n \}_{n=1}^{\infty}$ in a Hilbert space, the sum $\sum_{n=1}^{\infty}\alpha_n e_n$ converges in $H$ to a vector $y$ iff $\sum_{n=1}^{\infty}|\alpha_n|^2 < \infty$, and, in that case, $\|y\|^2 = \sum_{n=1}^{\infty}|\alpha_n|^2$. Because $\sum_{n=1}^{\infty}|(x,e_n)|^2 \le \|x\|^2$ for any $x \in H$, and because the $\lambda_n$ are uniformly bounded, then $Tx=\sum_{n=1}^{\infty}\lambda_n(x,e_n)e_n$ always converges. If $M$ is a uniform bound for $|\lambda_n|$, then $\|Tx\| \le M\|x\|$ easily follows.
For the second part, if $\lambda_n\rightarrow 0$, then, for every $\epsilon > 0$, there exists $N$ such that $|\lambda_n| < \epsilon/2$ for all $n > N$, which gives \begin{align} \left\|Tx -\sum_{n=1}^{N}\lambda_n(x,e_n)e_n\right\|^2 & = \sum_{n=N+1}^{\infty}|\lambda_n|^2|(x,e_n)|^2 \\ & \le \frac{\epsilon^2}{4}\sum_{n=N+1}^{\infty}|(x,e_n)|^2 \\ & \le \frac{\epsilon^2}{4}\|x\|^2 \end{align} If $T_N$ denotes the finite-rank operator $T_N x = \sum_{n=1}^{N}\lambda_n(x,e_n)e_n$, then the above shows that $\|T-T_k\| < \epsilon$ for all $k > N$. That's enough to prove that $T$ is compact. If $\lambda_n$ does not converge to $0$, there exists $\delta > 0$ and a subsequence $\{ \lambda_{n_{k}} \}$ such that $|\lambda_{n_{k}}| \ge \delta$. Then $\{ Te_{n_{k}} =\lambda_{n_{k}}e_{n_{k}} \}_{k=1}^{\infty}$ has no convergent subsequence (why?), which proves that $T$ is not compact.
Assume that none of the $\lambda_n$ are $0$; you can do this by omitting such terms from the sum that defines $T$. Then $T$ has eigenvalues $\lambda_n$ because $Te_n=\lambda_n e_n$. If the remaining $\{ e_n \}_{n=1}^{\infty}$ is not a complete orthonormal set, then $0$ is an eigenvector of $T$ as well because $Tx=0$ if $x\ne 0$ is orthogonal to every $e_n$. If $Tx=\lambda x$ for some $x\ne 0$, then \begin{align} 0=(Tx-\lambda x,e_k)&=\left(\sum_{n=1}^{\infty}\lambda_n(x,e_n)e_n-\lambda x,e_k\right) \\ &= \lambda_k(x,e_k)-\lambda(x,e_k) \\ &=(\lambda_k-\lambda)(x,e_k),\;\;\; k=1,2,3,\cdots. \end{align} There are two choices: either
$(x,e_k)=0$ for all $k$, which means $Tx=0$ and, hence $\lambda=0$, or
$(x,e_k)\ne 0$ for some $k$ and $\lambda=\lambda_k$ for that $k$.
Hence, the only possible eigenvalues are $\{ 0,\lambda_1,\lambda_2,\cdots\}$. And $0$ is an eigenvalue iff $\{ e_1,e_2,e_3,\cdots\}$ is not a complete orthonormal set.