Let $H$ be a separable Hilbert space with two orthonormal bases $e_{n}$ and $f_{n}$, and let $s_{n}>0$ be a strictly decreasing sequence s.t. $s_{n} \to 0$. Let $A: H \to H$ be an operator defined by $Ax = \sum_{n=1}^{\infty} s_{n} \langle x, e_{n} \rangle f_{n}$.
I wish to prove several properties of this operator:
1) $A$ is well-defined, bounded and compact, and $A^{*}x = \sum_{n=1}^{\infty} s_{n} \langle x, f_{n} \rangle e_{n},$ where $A^{*}$ is the adjoint operator.
2) If the property $||A^{*}x|| \leq ||Ax||$ holds for all $x \in H$, then $e_{n} = f_{n}$ for all $n \in \mathbb{N}$.
I was able to prove 1) fairly easily. The only part that required a little bit of effort was compactness, where I used the fact that $A$ is compact iff $$x_{n} \rightharpoonup 0 \implies Ax_{n} \to 0.$$
The arrow on the left denoted weak convergence, and the arrow on the right strong convergence.
However, I can only prove 2) partially. Here's what I have. We have $\langle Ax, Ax \rangle \leq \langle A^{*}x, A^{*}x \rangle$, i.e. $$\sum_{n=1}^{\infty}s_{n}^{2} | \langle x, f_{n} \rangle |^2 \leq \sum_{n=1}^{\infty} s_{n}^2 |\langle x, e_{n} \rangle |^2,$$ i.e. $$\sum_{n=1}^{\infty} s_{n}^2(|\langle x, f_{n}\rangle |^2 - |\langle x,e_{n} \rangle|^2) \leq 0,$$ i.e. $$|\langle x, f_{n} \rangle| = |\langle x, e_{n} \rangle|$$ for all $n$.
By plugging in $x=f_{m}$, I get $f_{m} \bot e_{n}$ for all $m \neq n$ and $|\langle f_{m}, e_{m} \rangle| = 1$, i.e. $f_{m} = e^{i\theta_{m}} e_{m}$ for all $m$, where $e^{i\theta_{m}} = \langle f_{m}, e_{m} \rangle$.
Furthermore, $||f_{n} - e_{n}||^{2} = 2(1-\text{Re}\langle f_{n}, e_{n} \rangle)$, so if I prove that $\langle f_{n}, e_{n} \rangle = 1$, I'm done. However, I can't seem to prove this.
Does anybody know a neat trick or a hint that could solve this?