Show that there exists a unique continuous linear map Let H be a separable Hilbert space with orthonormal basis {$e_n$}.Suppose
that elements $h_{n} \  \ (n = 1, 2, . . .)$ of H are given such that $\sum\limits_{i=1}^{\infty} ||h_{n}||^2 < \infty$.
Show that there exists a unique continuous linear map $T : H → H$ such that
$T e_{n} = h_{n} \ (n = 1, 2, . . .)$.
I have no idea how to tackle this problem. Maybe something like $Tx = \sum\limits_{i=1}^{\infty} h_{i}(x, e_{i})e_{i}$. Also how to use the separability of H?
All help is much appreciated.
 A: Suppose there are two such maps $T,T'$. Then there exists an $x$ in your space such that $Tx\neq T'x$. Write $x=\sum_{i=1}^\infty \langle e_i,x\rangle e_i$. Show that you can exchange $T$ with the sum, which gives: $Tx=\sum_{i=1}^\infty \langle e_i,x\rangle h_i$. Consider the finite sum $\sum_{i=1}^N \langle e_i,x\rangle h_i$ and show that it converges to the infinite sum. This will allow you to conclude that $Tx=T'x$ which implies $T$ must be unique.
A: Since $\mathbb{H}$ is separable, the Schauder basis is countable. This means that for any $x \in \mathbb{H}$, there is a sequence $x_n \to x$ where each $x_n$ lies in a finite dimensional subspace spanned by a finite number of basis elements.
Since the basis is orthonormal, we can give a convenient, explicit
expression for such an $x_n$, in particular
$x_n = \sum_{k \le n} \langle e_k , x \rangle e_k$.
Uniqueness follows from the following: If a continuous linear operator $L$ satisfies $L e_k = 0$ on a Schauder basis $e_k$ then $L=0$. To see this choose some $x$ and let $x_n$ be the sequence
above, we see that $Lx_n = 0 $ and hence $Lx = 0$ by continuity.
Hence if $T,T'$ are continuous linear operators that agree on $e_k$ then we must have $T=T'$ (choose $L=T-T'$).
For convenience, let me use the notation $y(k) = \langle e_k , y \rangle$.
Existence needs a little more work. Define
$Tx = \sum_k x(k) h_k$, then
$\|Tx\|^2 = \sum_n \langle e_n , Tx \rangle^2 = \sum_n \langle e_n , \sum_k x(k) h_k \rangle^2 = \sum_n (\sum_k x(k) h_k(n))^2$.
Note that $(\sum_k x(k) h_k(n) )^2 \le (\sum_k |x(k)|^2) \sum_k (h_k(n))^2$ from which we get
$\|Tx\|^2 \le (\sum_n \sum_k (h_k(n))^2) \|x\|^2 = (\sum_k \sum_n (h_k(n))^2) \|x\|^2 =(\sum_k \|h_k\|^2 ) \|x\|^2$ and so
$\|T\| \le \sqrt{\sum_k \|h_k\|^2}$. Hence $T$ is continuous and
$T e_n = h_n$.
