Hilbert space and orthonormal basis. Let $H$ be a Hilbert space and let ${e_n} ,\   n=1,2,3,\ldots$ be an orthonormal basis of $H$. Suppose $T$ is a bounded linear oprator on $H$. Then which of the following can not be true?
$$(a)\quad T(e_n)=e_1,  n=1,2,3,\ldots$$ 
$$(b)\quad T(e_n)=e_{n+1}, n=1,2,3,\ldots$$
$$(c)\quad T(e_n)=e_{n-1} , n=2,3,4,\ldots , \,\,T(e_1)=0$$
I think $(a)$ is not true because $e_1$ can not span the range space. I really don't know how to approach to this problem. Could you please give me some hints? Thank you very much.
 A: Let's look at (a).
If $T$ satisfies $T(e_n)=e_1$ for all $n$, then
$$T(e_1+e_2+\cdots+e_n)=ne_1.$$
But $\|e_1+e_2+\cdots+e_n\|=\sqrt n$ and $\|ne_1\|=n$.
A: A bounded operator is completely determined by where it sends each $e_n$.  Specifically, every element of $H$ has the form $\sum a_ne_n$ for some scalars $a_n$ (with $\sum |a_n|^2<\infty$), and then we must have $T(\sum a_ne_n)=\sum a_nT(e_n)$.
So, your task is to determine, in each of the three cases, whether the formula $$T\left(\sum a_ne_n\right)=\sum a_nT(e_n)$$ actually gives a well-defined bounded operator.  This means that you need there exist a constant $C$ such that for all $(a_n)\in \ell^2$, $$\left\|\sum a_nT(e_n)\right\|\leq C\left(\sum |a_n|^2\right)^{1/2}$$ (and in particular, the infinite sum $\sum a_nT(e_n)$ needs to actually converge).  Given the very simple definitions of $T(e_n)$ in each of your three cases, you should be able to write down a simple formula for $\left\|\sum a_nT(e_n)\right\|$ in each case which you can use to try to determine whether such a $C$ exists.
A: $(a)$ cannot be true. Assume that such $T$ exists. Then recall that $\sum_{n=1}^\infty \frac1n e_n \in H$ so 
$$T\left(\sum_{n=1}^\infty \frac1n e_n\right) = \sum_{n=1}^\infty \frac1n Te_n = \left(\sum_{n=1}^\infty \frac1n\right)e_1$$
but the latter sum doesn't converge. This is a contradiction.
$(b)$ and $(c)$ can be true. Consider
$$Sx = \sum_{n=2}^\infty \langle x, e_{n-1}\rangle e_{n}$$
and
$$Tx = \sum_{n=1}^\infty \langle x, e_{n+1}\rangle e_{n}$$
Both are contractive and hence bounded.
A: You're questions been answered, but if you're curious about examples of bounded linear operators that satisfy the last two resp.  Look no further than $l^2(\mathbb{N})$ and consider the left and right shift operators on the standard basis.  That is 
$$T_L( (a_1, a_2, ... a_n , ... )) = (a_2, a_3, ... a_n,\   ... ) \\
T_R( (a_1, a_2, ... a_n , ... )) = (0, a_1,\  ... \ a_n,\  ... ) $$
