Exercise :
Let $T$ be a bounded linear operator such that $T : X \to Y$. Show that : $$\|T\| = \sup\{\|Tx\| : \|x\| \leq 1 \} $$ $$\quad\quad = \sup\{\|Tx\| : \|x\| = 1 \}$$
Attempt :
First of all, we know that the operator norm is defined as :
$$\|T\| = \sup\bigg\{\frac{\|Tx\|}{\|x\|} : x \in X, \forall x \neq 0\bigg\}$$
Since the operator $T$ is bounded, it holds that : $\|Tx\| \leq \|T\|\cdot \|x\|$ .
Thus, for the first set we can see that it it is : $\|Tx \| \leq \|T\| \cdot \|x\|$, where $\|x\| \leq 1$ and for the second set, it simply is : $\|Tx \| \leq \|T\|$. Thus, as sets, it is :
$$\sup\{\|Tx\| : \|x\| = 1 \} \leq \sup\{\|Tx\| : \|x\| \leq 1 \}$$
But, it is :
$$\frac{\|Tx\|}{\|x\|} = \bigg\|\frac{Tx}{\|x\|}\bigg\| = \bigg\|T\bigg(\frac{x}{\|x\|}\bigg)\bigg\| \leq\|T\|\cdot \bigg\|\frac{x}{\|x\|}\bigg\| = \|T\|$$
Thus, it is : $\sup\bigg\{\frac{\|Tx\|}{\|x\|} : x \in X, \forall x \neq 0\bigg\} \leq \sup\{\|Tx\| : \|x\| = 1 \}$.
Finally, if $\|x\| \leq 1$, it would be :
$$\|Tx\| \leq \frac{\|Tx\|}{\|x\|}$$
Τhus it is : $\sup\{\|Tx\| : \|x\| \leq 1 \} \leq \sup\bigg\{\frac{\|Tx\|}{\|x\|} : x \in X, \forall x \neq 0\bigg\} $.
But, this leads to :
$$\boxed{\|T\| = \sup\bigg\{\frac{\|Tx\|}{\|x\|} : x \in X, \forall x \neq 0\bigg\} = \sup\{\|Tx\| : \|x\| \leq 1 \} = \sup\{\|Tx\| : \|x\| = 1 \}}$$
Question : Is my approach rigorous enough and correct ?