The plan is to express $\|T^*\|$ as a double supremum over both spaces, use the definition of $T^*$, and bound the result using the operator norm of $T$. **Step 1: Express the adjoint norm as a double supremum.** \begin{align*} \|T^*\|_{\mathcal{L}(Y^*, X^*)} &= \sup_{\substack{g \in Y^* \setminus \{0\}}} \frac{\|T^*(g)\|_{X^*}}{\|g\|_{Y^*}} = \sup_{\substack{g \in Y^* \setminus \{0\}}} \frac{1}{\|g\|_{Y^*}} \sup_{\substack{x \in X \setminus \{0\}}} \frac{|(T^*(g))(x)|}{\|x\|_X}. \end{align*} **Step 2: Substitute the definition of $T^*$.** By definition, $(T^*(g))(x) = g(T(x))$ for all $x \in X$ and $g \in Y^*$. Therefore: \begin{align*} \|T^*\|_{\mathcal{L}(Y^*, X^*)} &= \sup_{\substack{g \in Y^* \setminus \{0\}}} \sup_{\substack{x \in X \setminus \{0\}}} \frac{|g(T(x))|}{\|g\|_{Y^*} \cdot \|x\|_X}. \end{align*} **Step 3: Bound using the operator norm of $g$.** Since $g \in Y^*$, the definition of the operator norm gives $|g(T(x))| \le \|g\|_{Y^*} \cdot \|T(x)\|_Y$. Substituting: \begin{align*} \|T^*\|_{\mathcal{L}(Y^*, X^*)} &\le \sup_{\substack{g \in Y^* \setminus \{0\}}} \sup_{\substack{x \in X \setminus \{0\}}} \frac{\|g\|_{Y^*} \cdot \|T(x)\|_Y}{\|g\|_{Y^*} \cdot \|x\|_X} = \sup_{\substack{x \in X \setminus \{0\}}} \frac{\|T(x)\|_Y}{\|x\|_X} = \|T\|_{\mathcal{L}(X,Y)}. \end{align*}