What is adjoint of extended right shift operator on a Hilbert space? If $H$ is an infinite dimensional (not necessarily separable) Hilbert space and $(e_{n})$ is an orthonormal sequence. One can define $T(e_{n}):=e_{n+1}$, extend linearly to $\operatorname{span}\{e_{n}\}$, extend to a unique bounded operator $\overline{\operatorname{span}\{e_{n}\}}$ and define $T=0$ on $\overline{\operatorname{span}\{e_{n}\}}^{\perp}$. Then how does one compute the adjoint $T^{*}$ of $T\colon H\to H$? If $H$ is separable, then $\overline{\operatorname{span}\{e_{n}\}}^{\perp}=\{0\}$ and it is not hard to see that $T^{*}(e_{1})=0$ and $T^{*}(e_{n})=e_{n-1}$ for $n>1$.
 A: Note that $H$ separable does not guarantee that $\{e_n\}^\perp=0$ for an orthonormal sequence, even if it is infinite. For instance, fix an orthonormal basis $\{f_n\}$ and let $e_n=f_{2n}$. Then $\{e_n\}^\perp$ is infinite-dimensional. 
You can simply see what the adjoint looks like, by decomposing $H=\operatorname{span}\{e_n\}\oplus (\operatorname{span}\{e_n\})^\perp$. Then your extension $\tilde T$ has the form
$$
\tilde T=\begin{bmatrix} T&0\\0&0\end{bmatrix} . 
$$
Then (because the two subspaces are mutually orthogonal) $$\tilde T^*=\begin{bmatrix} T^*&0\\0&0\end{bmatrix}.$$
A: Let $W = \overline{\operatorname{span}\{e_n\}}$. Suppose $x_1, x_2 \in W$ and $y_1, y_2 \in W^\perp$. We have,
$$\langle T^* (x_1 + y_1), x_2 + y_2 \rangle = \langle x_1 + y_1, Tx_2 + Ty_2 \rangle = \langle x_1 + y_1, Tx_2 \rangle = \langle x_1, Tx_2\rangle.$$
If $S$ is the restriction of $T$ to $W$ (i.e. a right shift), then $S^*$ is a left shift, and
$$\langle T^* (x_1 + y_1), x_2 + y_2 \rangle = \langle x_1, Sx_2 \rangle = \langle S^* x_1, x_2\rangle = \langle S^* x_1, x_2\rangle + \langle0y_1, y_2\rangle.$$
Since $S^* x_1 \in W$, perpendicularity implies
$$\langle T^* (x_1 + y_1), x_2 + y_2 \rangle = \langle S^*x_1, x_2 + y_2\rangle.$$
That is, $T^*$ still performs a left shift on $W$, but sends vectors in $W^\perp$ to $0$, like $T$.
