I have some questions about the following exercise (Stein's Real Analysis / Measure Theory)
Exercise $33$ Let $\mathcal H$ be a Hilbert space with 'basis' $\{\varphi_k\}_k$. Verify that the operator $T$ defined by
$$T(\varphi_k) = \frac{1}{k}\varphi_{k+1}$$
is compact, but has no eigenvector.
1st
I think the 'basis' should be 'orthonormal' to solve the compactness. As written in this page, we have to define a compact operator (finite rank) $T_n$, and obtain an inequality $\Vert (T-T_n)x \Vert ^2 \le \frac{1}{(n+1)^2}\Vert x \Vert ^2$ to induce $\Vert T-T_n \Vert \le \frac{1}{n+1}$. But this process is estimated that the basis has to be orthonormal, which is not written in the exercise. Can I regard the text 'basis' in the exercise as an omitted one of 'orthonormal' basis? Or, is there any other way to show the compactness without the orthonormality?
2nd
I'd like to check my approach regarding the existence of eigenvector.
Let $f$ be an eigenvector of $\mathcal H$ : $T(f) = \lambda f$ where $f = \sum_{k=1}^{\infty}a_k \varphi_k$, $a_k \in \mathbb C$.
$T(f) = \sum_{k=1}^{\infty}a_k T(\varphi_k) = \sum_{k=1}^{\infty}a_k \frac{1}{k}\varphi_{k+1}$, which is equal to $ \sum_{k=1}^{\infty}\lambda a_k \varphi_k$
Then, $\lambda a_1 \varphi_1 + \lambda a_2 \varphi_2 + \lambda a_3 \varphi_3 + ... = 0\varphi_1 + a_1\frac{1}{1}\varphi_2 + a_2\frac{1}{2}\varphi_3 + ...$
Inducively, we can obtain $a_1 = a_2 = a_3 = ... = 0$ (whether the $\lambda = 0$ or not).
So $f = 0$, which is a contradiction.
Any comments about this would be grateful.