Let $H$ be a Hilbert space and $U$ a unitary operator on $H$, that is, $U^\ast U=UU^\ast =I$ where $U^*$ denotes the adjoint of $U$. Then it is immediately clear that for $f,g\in H$ we have $$ \langle Uf,Ug\rangle = \langle f,U^*Ug\rangle = \langle f,g\rangle. $$ However in the physics text I am reading, the proof is as follows:
Let $\vert V_1'\rangle=U\vert V_1\rangle$ and $\vert V_2'\rangle=U\vert V_2\rangle$ be kets. Then $$ \langle V_2'\vert V_1'\rangle = \langle UV_2\vert UV_1\rangle = \langle V_2\vert U^*U\vert V_1\rangle = \langle V_2\vert V_1\rangle. $$
I am confused by this notation. Wouldn't $\langle V_2'\vert V_1'\rangle$ denote a $(\mathrm{bra}, \mathrm{ket})$ pair? Or in this case is it simply the inner product of $V_2'$ with $V_1'$? Moreover, what is the meaning of $\langle V_2\vert U^*U\vert V_1\rangle$? I get that of course $U^*U=I$ but the double $\langle\cdot\vert\cdot\vert\cdot\rangle$ notation is throwing me off.
Another example of the double $\vert$ notation:
$\delta_{ij} = \langle i\vert U^*U\vert j\rangle$,
where $\delta_{ij} = \mathsf 1_{\{i=j\}}(i,j)$ is the Kronecker delta. What is the meaning of $\langle i\vert U^*U\vert j\rangle$?
In the bra-ket notation, $\langle x | y\rangle$ means the scalar product $\langle x,y\rangle$ of $x$ and $y$, and $\langle x | A| y\rangle$ means the scalar product $\langle x,Ay\rangle$ (here it's convenient to use the convention that the scalar product is linear in the second and antilinear in the first variable). Thus, the meaning of $\langle V_2|U^*U|V_1\rangle$ is just $\langle V_2,U^*UV_1\rangle$ which is equal to $\langle V_2,V_1\rangle$ as $U^*U=\operatorname{id}$.