For an orthonormal basis and a linear transformation $T$ which is not possible Let, $H$ be a Hilbert space and let, $\{e_n:n \ge 1\}$ be an orthonormal basis of $H$. Suppose that $T:H\to H$ be a bounded linear operator. Then which of the following CANNOT be true:
(A) $T(e_n)=e_1$.
(B) $T(e_n)=e_{n+1}$
(C) $T(e_n)=\sqrt\frac{n+1}{n}e_n$
(D) $T(e_n)=e_{n-1}$ for all $n \ge 2$ and $T(e_1)=0$
How to start? Any hint?
 A: The following proposition is relevant:

Let $H$ be a Hilbert space and $(e_n)_{n=1}^\infty$ an orthonormal sequence in $H$. Then for a sequence of scalars $(\alpha_n)_{n=1}^\infty$ we have:
$$\sum_{n=1}^\infty \alpha_ne_n \in H\iff (\alpha_n)_{n=1}^\infty \in \ell^2$$
That is, the series $\sum_{n=1}^\infty \alpha_ne_n$ converges if and only if $\sum_{n=1}^\infty |\alpha_n|^2 < \infty$.
Proof:
For every $M, N\in\mathbb{N}$, $M \ge N$ we have:
$$\left\|\sum_{n=1}^M\alpha_ne_n - \sum_{n=1}^N\alpha_ne_n\right\|^2 = \left\|\sum_{n=N+1}^M\alpha_ne_n\right\|^2 = \sum_{n=N+1}^M|\alpha_n|^2 = \left|\sum_{n=1}^M|\alpha_n|^2 - \sum_{n=1}^N|\alpha_n|^2\right|$$
Thus $\left(\sum_{n=1}^N\alpha_ne_n\right)_{N=1}^\infty$ is Cauchy iff $\left(\sum_{n=1}^N|\alpha_n|^2\right)_{N=1}^\infty$ is Cauchy. Since both $H$ and $\mathbb{C}$ are Banach spaces, one series converges iff the other converges.

Let $x = \sum_{n=1}^\infty \alpha_ne_n \in H$ be arbitrary.
For $(A)$: Assume such bounded map $T$ exists:
$$Tx = T\left(\sum_{n=1}^\infty \alpha_ne_n\right) = \sum_{n=1}^\infty \alpha_nTe_n = \sum_{n=1}^\infty \alpha_ne_1 = \left(\sum_{n=1}^\infty\alpha_i\right)e_1$$
However, by the proposition we have that $\sum_{n=1}^\infty \frac1ne_n \in H$, so:
$$T\left(\sum_{n=1}^\infty \frac1ne_n\right) = \left(\sum_{n=1}^\infty\frac1n\right)e_1$$
This is a contradiction since $\sum_{n=1}^\infty\frac1n$ diverges. Thus, such $T$ does not exist.

For $(B)$: Assume such bounded map $T$ exists:

$$Tx = T\left(\sum_{n=1}^\infty \alpha_ne_n\right) = \sum_{n=1}^\infty \alpha_nTe_n = \sum_{n=1}^\infty \alpha_ne_{n+1} = \sum_{n=2}^\infty \alpha_{n-1}e_n$$
Since we have $(\alpha_n)_{n=1}^\infty \in \ell^2$ then certainly also $(\alpha_{n-1})_{n=2}^\infty \in \ell^2$ so the linear map $Tx = \sum_{n=2}^\infty \alpha_{n-1}e_n$ is well-defined.
We have to check that it's bounded:
$$\|Tx\|^2 = \left\|\sum_{n=2}^\infty \alpha_{n-1}e_n\right\|^2 = \sum_{n=1}^\infty |\alpha_n|^2 = \|x\|^2$$
Thus, such $T$ exists, and is given by $Tx = \sum_{n=2}^\infty \alpha_{n-1}e_n$.

For $(C)$: Assume such bounded map $T$ exists:
$$Tx = T\left(\sum_{n=1}^\infty \alpha_ne_n\right) = \sum_{n=1}^\infty \alpha_nTe_n = \sum_{n=1}^\infty \alpha_n\sqrt{\frac{n+1}n}e_n$$
We have $(\alpha_n)_{n=1}^\infty \in \ell^2$. Let's check that $\left(\alpha_n\sqrt{\frac{n+1}n}\right)_{n=1}^\infty \in \ell^2$:
$$\sum_{n=1}^\infty |\alpha_n|^2\underbrace{\frac{n+1}n}_{\le 2} \le 2\sum_{n=1}^\infty |\alpha_n|^2 < +\infty$$
Thus, the map $Tx = \sum_{n=1}^\infty \alpha_n\sqrt{\frac{n+1}n}$ is well-defined.
We have to check that it's bounded:
$$\|Tx\|^2 = \left\|\sum_{n=1}^\infty \alpha_n\sqrt{\frac{n+1}n}e_n\right\|^2 = \sum_{n=1}^\infty |\alpha_n|^2\frac{n+1}n \le 2\sum_{n=1}^\infty |\alpha_n|^2 = 2\|x\|^2$$
Thus, such $T$ exists, and is given by $Tx = \sum_{n=1}^\infty \alpha_n\sqrt{\frac{n+1}n}$.

For $(D)$: Assume such bounded map $T$ exists:
$$Tx = T\left(\sum_{n=1}^\infty \alpha_ne_n\right) = \sum_{n=1}^\infty \alpha_nTe_n = \sum_{n=2}^\infty \alpha_ne_{n-1} = \sum_{n=1}^\infty \alpha_{n+1}e_{n}$$
Since we have $(\alpha_n)_{n=1}^\infty \in \ell^2$ then certainly also $(\alpha_{n+1})_{n=1}^\infty \in \ell^2$ so the linear map $Tx = \sum_{n=1}^\infty \alpha_{n+1}e_{n}$ is well-defined.
We have to check that it's bounded:
$$\|Tx\|^2 = \left\|\sum_{n=1}^\infty \alpha_{n+1}e_n\right\|^2 = \sum_{n=2}^\infty |\alpha_n|^2\le \sum_{n=1}^\infty |\alpha_n|^2 = \|x\|^2$$
Thus, such $T$ exists, and is given by $Tx = \sum_{n=1}^\infty \alpha_{n+1}e_{n}$.
