The plan is to combine the upper bound $\|T^*\| \le \|T\|$ (already established in the [Adjoint Operator Norm Inequality](/theorems/876)) with a matching lower bound that uses the existence of support functionals. **Step 1: Upper bound.** We have $\|T^*\|_{\mathcal{L}(Y^*, X^*)} \le \|T\|_{\mathcal{L}(X,Y)}$ by the [Adjoint Operator Norm Inequality](/theorems/876). **Step 2: Lower bound.** Let $\varepsilon > 0$. If $\|T\|_{\mathcal{L}(X,Y)} = 0$ then the result is immediate, so assume $\|T\| > 0$. By definition of the operator norm, there exists $v \in X$ with $\|v\|_X = 1$ and $\|T(v)\|_Y \ge \|T\|_{\mathcal{L}(X,Y)} - \varepsilon$. For $\varepsilon$ small enough, $T(v) \ne 0$. By the [Existence of Support Functionals](/theorems/881), there exists $f_{T(v)} \in Y^*$ with $f_{T(v)}(T(v)) = \|T(v)\|_Y$ and $\|f_{T(v)}\|_{Y^*} = 1$. Then: \begin{align*} \|T^*\|_{\mathcal{L}(Y^*, X^*)} \ge \frac{\|T^*(f_{T(v)})\|_{X^*}}{\|f_{T(v)}\|_{Y^*}} = \|T^*(f_{T(v)})\|_{X^*} \ge |(T^*(f_{T(v)}))(v)| = |f_{T(v)}(T(v))| = \|T(v)\|_Y \ge \|T\| - \varepsilon. \end{align*} Since $\varepsilon > 0$ was arbitrary, $\|T^*\| \ge \|T\|$. **Step 3: Combine.** The two inequalities give $\|T^*\|_{\mathcal{L}(Y^*, X^*)} = \|T\|_{\mathcal{L}(X,Y)}$.