How to find the image of an arbitrary element under this operator? Let $(e_n)$ be a total orthonormal sequence in a separable Hilbert space $H$ and define the right shift operator to be the linear operator $T \colon H \to H$ such that $T e_n = e_{n+1}$ for $n = 1, 2, 3, \ldots$. Then how to find $Tx$ for an arbitrary element $x \in H$? 
By total, we mean that the span of $(e_n)$ is dense in $H$. In this case, every $x \in H$ can be written as 
$$x = \sum_{n=1}^\infty \langle x, e_n \rangle e_n$$ 
because the $e_n$ are orthonormal. 
What next? 
If we could show that $T$ is bounded (and hence continuous), then we can show that 
$$Tx = \sum_{n=1}^\infty \langle x, e_n \rangle T e_n = \sum_{n=1}^\infty \langle x, e_n \rangle  e_{n+1}.$$ 
But how to show that $T$ is bounded? 
 A: Let $D \subset H$ be dense subspace and assume $T: D \to H$ is bounded on $D$ by say $M \in \mathbb R^+$. This means 
$$
 \sup_{ x \in D \setminus \{0\}} \frac{\Vert Tx \Vert} {\Vert x \Vert } = M.
$$
We claim that there exists a linear map  $\tilde T: H \to H$ which extends $T$ and has norm $M$. 
Now, let $y \in H$ be arbitrary. Approximate $y$ by elements of $D$, say $x_n \to y$ in norm. Then $T(x_n)$ converges since 
$$
 \Vert Tx_n - Tx_m \Vert = \Vert {T(x_n-x_m)}\Vert \leq M \Vert x_n-x_m\Vert 
$$ (Use that $(x_n)$ is Cauchy and that $H$ is complete.) 
Define $\tilde T(y) := \lim_n T(x_n)$. Note also, that if $x_n' \to y$ then $\lim_n T(x_n) = \lim_n T(x_n')$, which proves that $\tilde T$ is well defined. That $\tilde T$ is linear, you can also check by hand using limit properties.
To see that $\tilde T$ has norm less or equal to $M$, note that if $x_n \to x$ then 
$$
 \Vert \tilde Tx \Vert = \Vert \lim_n T(x_n) \Vert = \lim_n \Vert T(x_n) \Vert \leq M.
$$ For equality we know that there exists a  sequence $x_n$ in $D$ with $\lim_n \Vert T(x_n) \Vert = M$.
