Composition of unitary operator with its dual is the identity

Question

Currently I'm self studying functional analysis, namely unitary operators. In the text, the author gives the following proposition without proof:

Proposition: Let $U\colon H\to H$ be an unitary operator, i.e. $\langle Ux, Uy \rangle = \langle x,y \rangle$ for all $x,y\in H$ and $\text{Im}U=H$. Then $UU^*=I=U^*U$; that is, $U$ is invertible and $U^{-1}=U^*$.

I'm unsure how to show that $UU^*=I=U^*U$; however, under the assumption it holds, I believe I can show that $U^{-1}=U^*$ with the following argument:

The dual of $U$, we call it $U^*$, is the unique operator that satisfies $\langle Ux, y \rangle = \langle x, U^*y \rangle$ or all $x,y\in H$. Notice $$ \langle Ux, y \rangle=\langle Ux, Iy \rangle=\langle Ux, UU^{-1}y \rangle=\langle x, U^{-1}y \rangle. $$ Whence $U^*=U^{-1}$.

Answer 1

As mentioned in the comments, you assumed that $U$ is invertible, but this needs to be shown first. Rather, proceed as follows:

• From the assumption: $U$ is isometric. Show that this implies that $U^*U = I$. In particular, $U$ is injective.
• From the assumption: $U$ is surjective. Hence, a left inverse for $U$ is also a right inverse and thus $UU^* = I$.

Answer 2

Here's a hint towards a simpler proof: a bounded operator $A$ is the identity if and only if for all $x,y$ we have $\langle Ax,y\rangle =\langle x,y\rangle$.