[proofplan] We prove the two parts separately. For Part 1, we show that the adjoint $i^*: Y^* \to X^*$ is bounded (with $\|i^*\| \le \|i\|$) by a direct operator-norm estimate, and injective by using the density of $i(X)$ in $Y$. For Part 2, we assume $i$ is compact and show $i^*$ is compact by applying the Arzela-Ascoli theorem to the sequence of functionals restricted to the compact set $\overline{i(B_X)}$, extracting a uniformly convergent subsequence that yields Cauchy convergence in $X^*$. [/proofplan] [step:Define the adjoint operator $i^*$ and verify continuity] For $y^* \in Y^*$, the adjoint $i^*: Y^* \to X^*$ is defined by \begin{align*} (i^*(y^*))(x) = y^*(i(x)) \quad \text{for all } x \in X. \end{align*} Since $i$ is bounded, there exists $C > 0$ with $\|i(x)\|_Y \le C\|x\|_X$ for all $x \in X$. Estimating the operator norm: \begin{align*} \|i^*(y^*)\|_{X^*} &= \sup_{\substack{x \in X \\ \|x\|_X \le 1}} |y^*(i(x))| \\ &\le \sup_{\substack{x \in X \\ \|x\|_X \le 1}} \|y^*\|_{Y^*} \|i(x)\|_Y \\ &\le \|y^*\|_{Y^*} \cdot \|i\|_{\mathcal{L}(X,Y)}. \end{align*} Therefore $i^*$ is bounded with $\|i^*\|_{\mathcal{L}(Y^*, X^*)} \le \|i\|_{\mathcal{L}(X,Y)}$. [/step] [step:Prove injectivity of $i^*$ using density of $i(X)$ in $Y$] Suppose $y^* \in Y^*$ satisfies $i^*(y^*) = 0$ in $X^*$. Then \begin{align*} y^*(i(x)) = (i^*(y^*))(x) = 0 \quad \text{for all } x \in X. \end{align*} This means $y^*$ vanishes on the subspace $i(X) \subset Y$. By hypothesis, $i(X)$ is [dense](/page/Dense%20Subset) in $Y$. Since $y^*$ is [continuous](/page/Continuity) and vanishes on a dense subspace, it vanishes on all of $Y$, so $y^* = 0$. Therefore $\ker(i^*) = \{0\}$, and $i^*$ is injective. [guided] We need to show that $i^*(y^*) = 0$ forces $y^* = 0$. The condition $i^*(y^*) = 0$ says that the functional $y^*$ vanishes on every element of the form $i(x)$ for $x \in X$, i.e., $y^*$ vanishes on $i(X)$. Now $i(X)$ is [dense](/page/Dense%20Subset) in $Y$ by hypothesis. Take any $y \in Y$. By density, there exists a sequence $(x_k)_{k=1}^\infty \subset X$ with $i(x_k) \to y$ in $Y$. By continuity of $y^*$: \begin{align*} y^*(y) = y^*\left(\lim_{k \to \infty} i(x_k)\right) = \lim_{k \to \infty} y^*(i(x_k)) = \lim_{k \to \infty} 0 = 0. \end{align*} Since $y \in Y$ was arbitrary, $y^* = 0$. Therefore $\ker(i^*) = \{0\}$ and $i^*$ is injective. [/guided] [/step] [step:Assume $i$ is compact and restrict functionals to $\overline{i(B_X)}$] Now assume $i: X \to Y$ is compact. Let $B_X = \{x \in X : \|x\|_X \le 1\}$ denote the closed unit ball. Since $i$ is compact, the set $K := \overline{i(B_X)}$ is a compact metric space in $Y$. Let $(y_k^*)_{k=1}^\infty$ be a bounded sequence in $Y^*$ with $\|y_k^*\|_{Y^*} \le M$ for all $k$. Define \begin{align*} f_k: K &\to \mathbb{R}, \\ y &\mapsto y_k^*(y). \end{align*} [/step] [step:Verify uniform boundedness and equicontinuity, then apply Arzela-Ascoli] **Uniform boundedness:** For any $y \in K$, \begin{align*} |f_k(y)| \le \|y_k^*\|_{Y^*} \|y\|_Y \le M \sup_{z \in K} \|z\|_Y < \infty. \end{align*} **Equicontinuity:** For any $y, z \in K$, \begin{align*} |f_k(y) - f_k(z)| = |y_k^*(y - z)| \le M\|y - z\|_Y. \end{align*} The Lipschitz constant $M$ is independent of $k$, so $(f_k)$ is equicontinuous. By the Arzela-Ascoli theorem (applicable because $K$ is a compact metric space, and $(f_k)$ is uniformly bounded and equicontinuous), there exists a subsequence $(f_{k_j})$ converging uniformly on $K$. [guided] The Arzela-Ascoli theorem states that on a compact metric space, a uniformly bounded and equicontinuous family of real-valued functions is relatively compact in $C(K)$ (the space of continuous functions on $K$ with the supremum norm). We verify both conditions. **Uniform boundedness:** Every $y \in K$ satisfies $\|y\|_Y \le \sup_{z \in K}\|z\|_Y =: R < \infty$ (since $K$ is compact, hence bounded). Then \begin{align*} |f_k(y)| = |y_k^*(y)| \le \|y_k^*\|_{Y^*}\|y\|_Y \le MR. \end{align*} This bound is independent of both $k$ and $y$. **Equicontinuity:** For any $y, z \in K$: \begin{align*} |f_k(y) - f_k(z)| = |y_k^*(y - z)| \le \|y_k^*\|_{Y^*}\|y - z\|_Y \le M\|y - z\|_Y. \end{align*} Given $\varepsilon > 0$, choosing $\delta = \varepsilon/M$ makes $|f_k(y) - f_k(z)| < \varepsilon$ whenever $\|y - z\|_Y < \delta$, uniformly in $k$. By Arzela-Ascoli, there exists a subsequence $(f_{k_j})$ and a continuous function $f: K \to \mathbb{R}$ such that $\sup_{y \in K} |f_{k_j}(y) - f(y)| \to 0$. [/guided] [/step] [step:Convert uniform convergence on $K$ to norm convergence in $X^*$] We show that $(i^*(y_{k_j}^*))$ is Cauchy in $X^*$. For any indices $k_j, k_l$ in the subsequence: \begin{align*} \|i^*(y_{k_j}^*) - i^*(y_{k_l}^*)\|_{X^*} &= \sup_{\substack{x \in X \\ \|x\|_X \le 1}} |(y_{k_j}^* - y_{k_l}^*)(i(x))| \\ &= \sup_{y \in i(B_X)} |f_{k_j}(y) - f_{k_l}(y)| \\ &\le \sup_{y \in K} |f_{k_j}(y) - f_{k_l}(y)|. \end{align*} Since $(f_{k_j})$ converges uniformly on $K$, it is uniformly Cauchy on $K$. Therefore $(i^*(y_{k_j}^*))$ is Cauchy in the [Banach space](/page/Banach%20Space) $X^*$, and hence converges. Since every bounded sequence in $Y^*$ has a subsequence whose image under $i^*$ converges in $X^*$, the operator $i^*: Y^* \to X^*$ is compact. [/step]