Hilbert space and orthonormal basis.

Question

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.

Answer 1

$(a)$ 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.

Answer 2

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.

Answer 3

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 $\|ne_1\|=n$.

Answer 4

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

$$T_L( (a_1, a_2, ... a_n , ... )) = (a_2, a_3, ... a_n,\ ... ) \\ T_R( (a_1, a_2, ... a_n , ... )) = (0, a_1,\ ... \ a_n,\ ... ) $$