A equality in the proof of $T$ compact $\iff T^*$ compact

Question

Problem: I am reading through Rudin's text on functional analysis and came across the proof for the following theorem: Suppose $X, Y$ are Banach spaces, and $T$ is a bounded linear operator from $X$ to $Y$. Then $T$ is compact $\iff$ $T^*$ is compact. In his proof of the theorem (picture below)

He has the equality $$\|T^{\ast}y_{n_i}^* - T^{\ast}y_{n_{j}}^* \| = \sup_{x \in U} \| \langle Tx, y_{n_i}^* - y_{n_j}^* \rangle \|$$

Where did this equality come from? Is it part of some elementary results covered in the text before which I missed? Any help is appreciated!

Answer 1

$\|T^{\ast}y_{n_i}^* - T^{\ast}y_{n_{j}}^* \| = \sup_{x \in U} |((T^* (y_{n_i})-T^*(y_{n_j}))(x)|=Sup_{x\in U}|\langle y_{n_i}-y_{n_j},T(x)\rangle|$ adjunction formula.

Answer 2

This is an ordinary definition of the operator norm - a supremum of its values on a unit ball.