[proofplan] We construct $T^*$ pointwise using the [Riesz Representation Theorem](/theorems/221): for each $y \in K$, the map $x \mapsto (Tx, y)_K$ is a bounded linear functional on $H$, which Riesz identifies with a unique element $T^*y \in H$. We then verify linearity, boundedness, and the norm identity $\|T^*\| = \|T\|$ via a double supremum argument. [/proofplan] [step:Construct $T^*y$ for each $y \in K$ via the Riesz Representation Theorem] [claim:Existence Of Adjoint Image] For each $y \in K$, there exists a unique $z_y \in H$ with $(Tx, y)_K = (x, z_y)_H$ for all $x \in H$, and $\|z_y\|_H \leq \|T\| \|y\|_K$. [/claim] [proof] Define $\varphi_y: H \to \mathbb{R}$ by $\varphi_y(x) = (Tx, y)_K$. This is linear and bounded: \begin{align*} |\varphi_y(x)| = |(Tx, y)_K| \leq \|Tx\|_K \|y\|_K \leq \|T\| \|y\|_K \|x\|_H, \end{align*} where the first inequality is Cauchy-Schwarz in $K$. By the [Riesz Representation Theorem](/theorems/221), there exists a unique $z_y \in H$ with $\varphi_y(x) = (x, z_y)_H$ and $\|z_y\|_H = \|\varphi_y\|_{H^*} \leq \|T\| \|y\|_K$. [/proof] [/step] [step:Define $T^*$ and verify linearity] Define $T^*: K \to H$ by $T^*y = z_y$. For $y_1, y_2 \in K$ and $\lambda \in \mathbb{R}$, and every $x \in H$: \begin{align*} (x, T^*(\lambda y_1 + y_2))_H &= (Tx, \lambda y_1 + y_2)_K = \lambda(Tx, y_1)_K + (Tx, y_2)_K \\ &= \lambda(x, T^*y_1)_H + (x, T^*y_2)_H = (x, \lambda T^*y_1 + T^*y_2)_H. \end{align*} By non-degeneracy of the inner product, $T^*(\lambda y_1 + y_2) = \lambda T^*y_1 + T^*y_2$. [/step] [step:Prove the norm identity $\|T^*\| = \|T\|$] [claim:Norm Identity] $\|T^*\|_{\mathcal{L}(K,H)} = \|T\|_{\mathcal{L}(H,K)}$. [/claim] [proof] The bound $\|T^*y\|_H \leq \|T\| \|y\|_K$ from the first step gives $\|T^*\| \leq \|T\|$. For the reverse: for every $x \in H$, \begin{align*} \|Tx\|_K = \sup_{\|y\|_K \leq 1} |(Tx, y)_K| = \sup_{\|y\|_K \leq 1} |(x, T^*y)_H| \leq \|x\|_H \|T^*\|, \end{align*} where the first equality uses $\|v\|_K = \sup_{\|y\| \leq 1} |(v, y)_K|$ (Cauchy-Schwarz, with equality at $y = v/\|v\|$). Dividing by $\|x\|_H$ and taking the supremum: $\|T\| \leq \|T^*\|$. [/proof] [/step] [step:Verify uniqueness of the adjoint] If $S \in \mathcal{L}(K, H)$ also satisfies $(Tx, y)_K = (x, Sy)_H$ for all $x, y$, then $(x, T^*y - Sy)_H = 0$ for all $x$. Taking $x = T^*y - Sy$ gives $\|T^*y - Sy\|_H = 0$, so $S = T^*$. [/step]