let $T:X \to Y$ be a bounded linear map. $X$ and $Y$ are two normed vector space. Each norm is denoted by $\| \cdot \|_X$ and $\| \cdot \|_Y$
define its operator norm by $$\|T\|_\text{op}=\sup_{x \ne 0} \frac{\|T_x\|_Y}{\|x\|_X}$$ The book said that
$$\|T\|_\text{op}=\sup_{\|x\|_X=1} \|T_x\|_Y = \sup_{\|x\|_X \le 1}\|T_x\|_Y $$
I can understand the first equation, but have no idea about the second equation.
The inequality $\sup_{\|x\|=1} \|Tx\| \le \sup_{\|x\| \le 1} \|Tx\|$ holds immediately due to the definition of supremum. So it suffices to show the reverse inequality.
Fix $\epsilon > 0$ such that $\epsilon < \sup_{\|x\| \le 1} \|Tx\|$. Then there exists $x_\epsilon$ satisfying $\|x_\epsilon\| \le 1$ such that $\|Tx_\epsilon\| \ge \sup_{\|x\| \le 1} \|Tx\| - \epsilon > 0$. We cannot have $x_\epsilon = 0$ else $\|Tx_\epsilon\| = 0$; thus $x_\epsilon \ne 0$. Then defining $\tilde{x}_\epsilon := x_\epsilon / \|x_\epsilon\|$ we have $$\|T\tilde{x}_\epsilon\| = \|Tx_\epsilon\| / \|x_\epsilon\| \ge \|Tx_\epsilon\| \ge \sup_{\|x\| \le 1} \|Tx\| - \epsilon.$$ Taking the supremum over the left-hand side over $\|x\|=1$ we obtain $$\sup_{\|x\|=1} \|Tx\| \ge \sup_{\|x\| \le 1} \|Tx\| - \epsilon.$$ The above argument holds for all sufficiently small $\epsilon$, so we have $$\sup_{\|x\|=1} \|Tx\| \ge \sup_{\|x\| \le 1} \|Tx\|.$$