[proofplan]
The proof uses the defining adjoint identity and the [Cauchy-Schwarz inequality](/theorems/432) for Hilbert spaces. First we compare $\|T^*\|$ and $\|T\|$ in both directions by testing the adjoint identity on the special vectors $T^*y$ and $Tx$, with zero-vector cases separated before division. Then we apply the [basic operator norm inequality](/theorems/9321) to obtain the upper bound for $\|T^*T\|$, and use the identity $(T^*Tx,x)_H=(Tx,Tx)_K$ to get the matching lower bound by taking the supremum over unit vectors.
[/proofplan]
[step:Use the adjoint identity to prove $\|T^*\|_{\mathcal L(K,H)}\le \|T\|_{\mathcal L(H,K)}$]
By definition of the [Hilbert space](/page/Hilbert%20Space) adjoint, $T^*:K\to H$ is the unique bounded [linear map](/page/Linear%20Map) satisfying
\begin{align*}
(Tx,y)_K=(x,T^*y)_H
\end{align*}
for every $x\in H$ and every $y\in K$.
Fix $y\in K$. If $T^*y=0$, then
\begin{align*}
\|T^*y\|_H=0\le \|T\|_{\mathcal L(H,K)}\|y\|_K.
\end{align*}
If $T^*y\ne 0$, apply the adjoint identity with $x=T^*y$. Since the [inner product](/page/Inner%20Product) on Androma is linear in the first variable,
\begin{align*}
\|T^*y\|_H^2=(T^*y,T^*y)_H=(T(T^*y),y)_K.
\end{align*}
Taking absolute values and applying the Cauchy-Schwarz inequality in $K$ gives
\begin{align*}
\|T^*y\|_H^2\le \|T(T^*y)\|_K\|y\|_K.
\end{align*}
By the basic [operator norm](/page/Operator%20Norm) inequality [citetheorem:9321] applied to $T\in\mathcal L(H,K)$ and the vector $T^*y\in H$,
\begin{align*}
\|T(T^*y)\|_K\le \|T\|_{\mathcal L(H,K)}\|T^*y\|_H.
\end{align*}
Therefore
\begin{align*}
\|T^*y\|_H^2\le \|T\|_{\mathcal L(H,K)}\|T^*y\|_H\|y\|_K.
\end{align*}
Since $\|T^*y\|_H>0$, division by $\|T^*y\|_H$ yields
\begin{align*}
\|T^*y\|_H\le \|T\|_{\mathcal L(H,K)}\|y\|_K.
\end{align*}
Thus the same inequality holds for every $y\in K$. Taking the infimum over admissible operator bounds, equivalently using [citetheorem:9320], gives
\begin{align*}
\|T^*\|_{\mathcal L(K,H)}\le \|T\|_{\mathcal L(H,K)}.
\end{align*}
[guided]
We want to bound $T^*$ by the known bound for $T$. The adjoint identity is the bridge between them: for every $x\in H$ and $y\in K$,
\begin{align*}
(Tx,y)_K=(x,T^*y)_H.
\end{align*}
To estimate $\|T^*y\|_H$, we choose the input $x$ in this identity to be the vector whose norm we are trying to control, namely $x=T^*y$.
Fix $y\in K$. If $T^*y=0$, then the desired estimate is immediate:
\begin{align*}
\|T^*y\|_H=0\le \|T\|_{\mathcal L(H,K)}\|y\|_K.
\end{align*}
Now suppose $T^*y\ne 0$. Substituting $x=T^*y$ into the adjoint identity gives
\begin{align*}
\|T^*y\|_H^2=(T^*y,T^*y)_H=(T(T^*y),y)_K.
\end{align*}
The right-hand side is an inner product in $K$, so the Cauchy-Schwarz inequality in $K$ gives
\begin{align*}
\|T^*y\|_H^2\le \|T(T^*y)\|_K\|y\|_K.
\end{align*}
Now $T^*y$ is a vector in $H$, so the basic operator norm inequality [citetheorem:9321] applies to $T\in\mathcal L(H,K)$ with input $T^*y$. Hence
\begin{align*}
\|T(T^*y)\|_K\le \|T\|_{\mathcal L(H,K)}\|T^*y\|_H.
\end{align*}
Combining the two estimates gives
\begin{align*}
\|T^*y\|_H^2\le \|T\|_{\mathcal L(H,K)}\|T^*y\|_H\|y\|_K.
\end{align*}
Because we are in the case $T^*y\ne 0$, the number $\|T^*y\|_H$ is positive, so division is valid and yields
\begin{align*}
\|T^*y\|_H\le \|T\|_{\mathcal L(H,K)}\|y\|_K.
\end{align*}
Together with the zero case, this holds for every $y\in K$. Therefore $\|T\|_{\mathcal L(H,K)}$ is an admissible bound for the operator $T^*:K\to H$, and the equivalent formula for the operator norm [citetheorem:9320] gives
\begin{align*}
\|T^*\|_{\mathcal L(K,H)}\le \|T\|_{\mathcal L(H,K)}.
\end{align*}
[/guided]
[/step]
[step:Use the adjoint identity again to prove $\|T\|_{\mathcal L(H,K)}\le \|T^*\|_{\mathcal L(K,H)}$]
Fix $x\in H$. If $Tx=0$, then
\begin{align*}
\|Tx\|_K=0\le \|T^*\|_{\mathcal L(K,H)}\|x\|_H.
\end{align*}
If $Tx\ne 0$, apply the adjoint identity with $y=Tx$. Then
\begin{align*}
\|Tx\|_K^2=(Tx,Tx)_K=(x,T^*Tx)_H.
\end{align*}
Taking absolute values and applying the Cauchy-Schwarz inequality in $H$ gives
\begin{align*}
\|Tx\|_K^2\le \|x\|_H\|T^*(Tx)\|_H.
\end{align*}
By [citetheorem:9321] applied to $T^*\in\mathcal L(K,H)$ and the vector $Tx\in K$,
\begin{align*}
\|T^*(Tx)\|_H\le \|T^*\|_{\mathcal L(K,H)}\|Tx\|_K.
\end{align*}
Thus
\begin{align*}
\|Tx\|_K^2\le \|T^*\|_{\mathcal L(K,H)}\|x\|_H\|Tx\|_K.
\end{align*}
Since $\|Tx\|_K>0$, division by $\|Tx\|_K$ gives
\begin{align*}
\|Tx\|_K\le \|T^*\|_{\mathcal L(K,H)}\|x\|_H.
\end{align*}
The inequality holds for every $x\in H$, so [citetheorem:9320] gives
\begin{align*}
\|T\|_{\mathcal L(H,K)}\le \|T^*\|_{\mathcal L(K,H)}.
\end{align*}
Combining this with the previous step,
\begin{align*}
\|T^*\|_{\mathcal L(K,H)}=\|T\|_{\mathcal L(H,K)}.
\end{align*}
[/step]
[step:Bound $\|T^*T\|_{\mathcal L(H,H)}$ from above by $\|T\|_{\mathcal L(H,K)}^2$]
Define the composition operator
\begin{align*}
A:H&\to H
\end{align*}
by $A(x)=T^*(Tx)$ for every $x\in H$. Since $T\in\mathcal L(H,K)$ and $T^*\in\mathcal L(K,H)$, the map $A=T^*T$ is linear and bounded, hence $A\in\mathcal L(H,H)$.
For every $x\in H$, applying [citetheorem:9321] first to $T^*$ and then to $T$ gives
\begin{align*}
\|T^*Tx\|_H\le \|T^*\|_{\mathcal L(K,H)}\|Tx\|_K.
\end{align*}
Also,
\begin{align*}
\|Tx\|_K\le \|T\|_{\mathcal L(H,K)}\|x\|_H.
\end{align*}
Using the equality of adjoint norms already proved,
\begin{align*}
\|T^*Tx\|_H\le \|T\|_{\mathcal L(H,K)}^2\|x\|_H.
\end{align*}
Therefore [citetheorem:9320] gives
\begin{align*}
\|T^*T\|_{\mathcal L(H,H)}\le \|T\|_{\mathcal L(H,K)}^2.
\end{align*}
[/step]
[step:Bound $\|T^*T\|_{\mathcal L(H,H)}$ from below by $\|T\|_{\mathcal L(H,K)}^2$]
If $H=\{0\}$, then $T$ is the zero map from $H$ to $K$, so $\|T\|_{\mathcal L(H,K)}=0$. The upper bound already proved gives $\|T^*T\|_{\mathcal L(H,H)}\le 0$, while the operator norm is non-negative by [citetheorem:9323]. Hence
\begin{align*}
\|T^*T\|_{\mathcal L(H,H)}=0=\|T\|_{\mathcal L(H,K)}^2.
\end{align*}
It remains to consider the case $H\ne\{0\}$.
Let $x\in H$ satisfy $\|x\|_H=1$. The adjoint identity with $y=Tx$ gives
\begin{align*}
\|Tx\|_K^2=(Tx,Tx)_K=(x,T^*Tx)_H.
\end{align*}
Taking absolute values and applying the Cauchy-Schwarz inequality in $H$,
\begin{align*}
\|Tx\|_K^2\le \|x\|_H\|T^*Tx\|_H.
\end{align*}
Since $\|x\|_H=1$, this becomes
\begin{align*}
\|Tx\|_K^2\le \|T^*Tx\|_H.
\end{align*}
By [citetheorem:9321] applied to $T^*T\in\mathcal L(H,H)$,
\begin{align*}
\|T^*Tx\|_H\le \|T^*T\|_{\mathcal L(H,H)}\|x\|_H=\|T^*T\|_{\mathcal L(H,H)}.
\end{align*}
Hence
\begin{align*}
\|Tx\|_K^2\le \|T^*T\|_{\mathcal L(H,H)}
\end{align*}
for every unit vector $x\in H$. Taking the supremum over all $x\in H$ with $\|x\|_H=1$ and using [citetheorem:9320],
\begin{align*}
\|T\|_{\mathcal L(H,K)}^2\le \|T^*T\|_{\mathcal L(H,H)}.
\end{align*}
Combining this lower bound with the upper bound from the previous step gives
\begin{align*}
\|T^*T\|_{\mathcal L(H,H)}=\|T\|_{\mathcal L(H,K)}^2.
\end{align*}
This proves both asserted identities.
[/step]
