[proofplan]
We use the defining identity of the adjoint: for each $A\in\mathcal L(H)$, $A^*$ is the unique bounded operator satisfying $(Ax,y)_H=(x,A^*y)_H$ for all $x,y\in H$. The algebraic identities are proved by comparing inner products against arbitrary vectors and then using uniqueness of the adjoint. The norm identity follows by estimating matrix coefficients of $T^*$ through those of $T$, and the final identity follows from submultiplicativity for the upper bound and from testing $T^*T$ on unit vectors for the lower bound.
[/proofplan]
[step:Use the adjoint identity to prove conjugate linearity]
Let $A:=\alpha S+\beta T\in\mathcal L(H)$ and let $B:=\overline{\alpha}S^*+\overline{\beta}T^*\in\mathcal L(H)$. For arbitrary $x,y\in H$, linearity of $S,T$ and linearity of the [inner product](/page/Inner%20Product) in the first variable give
\begin{align*}
(Ax,y)_H=\alpha(Sx,y)_H+\beta(Tx,y)_H.
\end{align*}
By the defining identities for $S^*$ and $T^*$,
\begin{align*}
(Ax,y)_H=\alpha(x,S^*y)_H+\beta(x,T^*y)_H.
\end{align*}
Since the inner product is conjugate-linear in the second variable,
\begin{align*}
\alpha(x,S^*y)_H+\beta(x,T^*y)_H=(x,\overline{\alpha}S^*y+\overline{\beta}T^*y)_H=(x,By)_H.
\end{align*}
Thus $(Ax,y)_H=(x,By)_H$ for all $x,y\in H$. By uniqueness in the definition of the adjoint, $A^*=B$, which is the desired identity.
[/step]
[step:Move the factors across the inner product in reverse order]
Let $A:=ST\in\mathcal L(H)$ and let $B:=T^*S^*\in\mathcal L(H)$. For arbitrary $x,y\in H$, the defining identity for $S^*$ gives
\begin{align*}
(Ax,y)_H=(S(Tx),y)_H=(Tx,S^*y)_H.
\end{align*}
Applying the defining identity for $T^*$ to the pair $x,S^*y\in H$, we obtain
\begin{align*}
(Tx,S^*y)_H=(x,T^*S^*y)_H=(x,By)_H.
\end{align*}
Therefore $(Ax,y)_H=(x,By)_H$ for all $x,y\in H$. By uniqueness of the adjoint, $(ST)^*=T^*S^*$.
[/step]
[step:Identify the double adjoint by the same uniqueness argument]
Let $A:=T^*$ and let $B:=T$. For arbitrary $x,y\in H$, the adjoint identity for $T$ gives
\begin{align*}
(Ty,x)_H=(y,T^*x)_H.
\end{align*}
Taking complex conjugates and using conjugate symmetry of the Hilbert inner product, we get
\begin{align*}
(x,Ty)_H=(T^*x,y)_H.
\end{align*}
Equivalently,
\begin{align*}
(Ax,y)_H=(x,By)_H.
\end{align*}
Since this holds for all $x,y\in H$, uniqueness of the adjoint of $A=T^*$ gives $(T^*)^*=T$.
[guided]
We want to prove that the adjoint operation returns to the original operator after being applied twice. The definition of the adjoint says that $(T^*)^*$ is the unique bounded operator $C\in\mathcal L(H)$ satisfying
\begin{align*}
(T^*x,y)_H=(x,Cy)_H
\end{align*}
for every $x,y\in H$. Thus it is enough to show that $C=T$ satisfies this identity.
Fix arbitrary $x,y\in H$. The defining identity for $T^*$, applied to the vectors $y$ and $x$, gives
\begin{align*}
(Ty,x)_H=(y,T^*x)_H.
\end{align*}
Now use conjugate symmetry of the inner product. Taking complex conjugates of both sides yields
\begin{align*}
(x,Ty)_H=(T^*x,y)_H.
\end{align*}
Reordering the equality, we obtain
\begin{align*}
(T^*x,y)_H=(x,Ty)_H.
\end{align*}
This is exactly the defining identity for the adjoint of $T^*$ with candidate adjoint operator $T$. Since the adjoint is unique, $(T^*)^*=T$.
[/guided]
[/step]
[step:Estimate matrix coefficients to prove norm preservation]
First let $u\in H$ with $\|u\|_H\leq 1$. For every $v\in H$ with $\|v\|_H\leq 1$, the adjoint identity and the [Cauchy-Schwarz inequality](/theorems/432) give
\begin{align*}
|(T^*u,v)_H|=|(u,Tv)_H|\leq \|u\|_H\|Tv\|_H\leq \|T\|_{\mathcal L(H)}.
\end{align*}
Taking the supremum over all such $v$ gives
\begin{align*}
\|T^*u\|_H\leq \|T\|_{\mathcal L(H)}.
\end{align*}
Taking the supremum over all $u\in H$ with $\|u\|_H\leq 1$ gives
\begin{align*}
\|T^*\|_{\mathcal L(H)}\leq \|T\|_{\mathcal L(H)}.
\end{align*}
Applying this already proved inequality to the bounded operator $T^*$ and using $(T^*)^*=T$, we get
\begin{align*}
\|T\|_{\mathcal L(H)}=\|(T^*)^*\|_{\mathcal L(H)}\leq \|T^*\|_{\mathcal L(H)}.
\end{align*}
Therefore
\begin{align*}
\|T^*\|_{\mathcal L(H)}=\|T\|_{\mathcal L(H)}.
\end{align*}
[/step]
[step:Combine submultiplicativity and unit-vector testing to prove the final identity]
By [submultiplicativity of the operator norm](/theorems/1054) and the norm preservation just proved,
\begin{align*}
\|T^*T\|_{\mathcal L(H)}\leq \|T^*\|_{\mathcal L(H)}\|T\|_{\mathcal L(H)}=\|T\|_{\mathcal L(H)}^2.
\end{align*}
For the reverse inequality, let $x\in H$ satisfy $\|x\|_H=1$. Using the defining identity for $T^*$ with the pair $Tx,x\in H$, we obtain
\begin{align*}
(T^*Tx,x)_H=(Tx,Tx)_H=\|Tx\|_H^2.
\end{align*}
Also, by Cauchy-Schwarz and the definition of the operator norm,
\begin{align*}
|(T^*Tx,x)_H|\leq \|T^*Tx\|_H\|x\|_H\leq \|T^*T\|_{\mathcal L(H)}.
\end{align*}
Thus, for every unit vector $x\in H$,
\begin{align*}
\|Tx\|_H^2\leq \|T^*T\|_{\mathcal L(H)}.
\end{align*}
Taking the supremum over all $x\in H$ with $\|x\|_H=1$ gives
\begin{align*}
\|T\|_{\mathcal L(H)}^2\leq \|T^*T\|_{\mathcal L(H)}.
\end{align*}
Combining the two inequalities proves
\begin{align*}
\|T^*T\|_{\mathcal L(H)}=\|T\|_{\mathcal L(H)}^2.
\end{align*}
This completes the proof of all stated properties.
[/step]
