[proofplan]
We prove the forward implication by testing the vector $T^*y$ against arbitrary unit vectors $x \in H$. The adjoint identity transfers the [inner product](/page/Inner%20Product) from $T^*y$ to $Tx$, and the assumed bound for $T$ controls the resulting expression. A finite-dimensional Hilbert-space norm is recovered by taking the supremum of inner products against unit vectors; applying the same argument to $T^*$ and using $T^{**} = T$ gives the converse.
[/proofplan]
[step:Recover a vector norm from inner products with unit vectors]
For a finite-dimensional [Hilbert space](/page/Hilbert%20Space) $E$ and a vector $v \in E$, define its unit sphere $S_E \subset E$ by
\begin{align*}
S_E := \{w \in E : \|w\|_E = 1\}.
\end{align*}
We will use the identity
\begin{align*}
\|v\|_E = \sup_{w \in S_E} |(v,w)_E|
\end{align*}
whenever $E \neq \{0\}$.
Indeed, for every $w \in S_E$, positivity of the inner product gives the elementary Cauchy-Schwarz estimate
\begin{align*}
|(v,w)_E| \leq \|v\|_E\|w\|_E = \|v\|_E.
\end{align*}
Thus the supremum is at most $\|v\|_E$. If $v \neq 0$, choosing $w := v/\|v\|_E \in S_E$ gives
\begin{align*}
|(v,w)_E| = \left|\left(v,\frac{v}{\|v\|_E}\right)_E\right| = \|v\|_E.
\end{align*}
If $v = 0$, both sides are $0$. This proves the identity.
[guided]
We first isolate the finite-dimensional Hilbert-space fact that lets us convert bounds on all scalar products into a norm bound. Let $E$ be a finite-dimensional Hilbert space and let $v \in E$. Define the unit sphere $S_E \subset E$ by
\begin{align*}
S_E := \{w \in E : \|w\|_E = 1\}.
\end{align*}
We claim that, when $E \neq \{0\}$,
\begin{align*}
\|v\|_E = \sup_{w \in S_E} |(v,w)_E|.
\end{align*}
The upper bound follows from the positivity of the inner product. For each $w \in S_E$, the elementary Cauchy-Schwarz estimate gives
\begin{align*}
|(v,w)_E| \leq \|v\|_E\|w\|_E.
\end{align*}
Since $w \in S_E$, we have $\|w\|_E = 1$, and therefore
\begin{align*}
|(v,w)_E| \leq \|v\|_E.
\end{align*}
Taking the supremum over $w \in S_E$ gives
\begin{align*}
\sup_{w \in S_E} |(v,w)_E| \leq \|v\|_E.
\end{align*}
For the reverse inequality, there are two cases. If $v = 0$, then $|(v,w)_E| = 0$ for every $w \in S_E$, so the supremum is $0 = \|v\|_E$. If $v \neq 0$, define $w := v/\|v\|_E$. Then $w \in S_E$, and
\begin{align*}
|(v,w)_E| = \left|\left(v,\frac{v}{\|v\|_E}\right)_E\right| = \frac{(v,v)_E}{\|v\|_E} = \|v\|_E.
\end{align*}
Thus the supremum is at least $\|v\|_E$, and the two inequalities prove the identity.
[/guided]
[/step]
[step:Use the bound for $T$ to bound $T^*$]
Assume
\begin{align*}
\|Tx\|_K^2 \leq C\|x\|_H^2 \quad \text{for every } x \in H.
\end{align*}
If $H = \{0\}$, then $T^*y = 0$ for every $y \in K$, so the desired estimate for $T^*$ holds. Suppose $H \neq \{0\}$.
Fix $y \in K$. For every $x \in S_H$, the adjoint identity gives
\begin{align*}
|(T^*y,x)_H| = |(y,Tx)_K|.
\end{align*}
Using the inner-product estimate and then the assumed bound for $T$, we obtain
\begin{align*}
|(y,Tx)_K| \leq \|y\|_K\|Tx\|_K \leq \sqrt{C}\,\|y\|_K\|x\|_H = \sqrt{C}\,\|y\|_K.
\end{align*}
Taking the supremum over $x \in S_H$ and using the norm-recovery identity with $E=H$ and $v=T^*y$ yields
\begin{align*}
\|T^*y\|_H \leq \sqrt{C}\,\|y\|_K.
\end{align*}
Squaring both sides gives
\begin{align*}
\|T^*y\|_H^2 \leq C\|y\|_K^2.
\end{align*}
Since $y \in K$ was arbitrary, the [second inequality](/theorems/2136) holds for every $y \in K$.
[/step]
[step:Apply the same argument to the adjoint to obtain the converse]
Assume
\begin{align*}
\|T^*y\|_H^2 \leq C\|y\|_K^2 \quad \text{for every } y \in K.
\end{align*}
Apply the previous implication to the [linear map](/page/Linear%20Map) $T^*: K \to H$. Its Hilbert-space adjoint is $T^{**}: H \to K$, and [finite dimensionality](/theorems/1534) gives $T^{**} = T$. Therefore, for every $x \in H$,
\begin{align*}
\|Tx\|_K^2 = \|T^{**}x\|_K^2 \leq C\|x\|_H^2.
\end{align*}
This is exactly the [first inequality](/theorems/2897), completing the equivalence.
[/step]
