If $I$ is identity operator on Hilbert space $H$ then show that $I^* =I$
My attempt: For any two bounded linear operators $S$ & $T$ on $H$ we have, $(ST)^∗=T^∗S^∗$ taking $T=I$ & $ S=T^∗$ we obtain, $(T^∗T)^∗=T^∗(T^∗)^∗$ implies $T=T^∗$ as required. is it okay?
According to the definition of adjoint $\langle I^{*}x, y \rangle = \langle x, Iy \rangle =\langle x, y \rangle$ for all $x $ and $y$. Thus $\langle I^{*}x-x, y \rangle =0$ for all $y$ and taking $y=I^{*}x-x$ this gives $\|I^{*}x-x\|^{2}=0$. Thus $I^{*}x=x$ for all $x$ which means $I^{*}=I$.