question related to the norm of bounded linear operator

Question

If $T\in\mathcal{L}(E)$, we have $$ \|T\|=\sup\left\{|\langle Tx\mid y\rangle|;\;x,y\in E,\;\|x\|\leq1,\|y\|\leq1\right\}. $$

How we can deduce that $$ \|T\|=\sup\left\{|\langle Tx\mid y\rangle|;\;x,y\in E,\;\|x\|= 1,\|y\|=1\right\}? $$

Answer 1

For $x,y$ such that $0<\|x\|,\|y\|<1$, then \begin{align*} \left|\left<T\left(\dfrac{x}{\|x\|}\right),\left(\dfrac{y}{\|y\|}\right)\right>\right|\leq M:=\sup\{\left|\left<Tx,y\right>\right|:x,y\in E,\|x\|=1,\|y\|=1\} \end{align*} by definition, so \begin{align*} |\left<Tx,y\right>|\leq M\cdot\|x\|\cdot\|y\|<M, \end{align*} so \begin{align*} \sup\{\left|\left<Tx,y\right>\right|:x,y\in E,\|x\|\leq 1, \|y\|\leq 1\}\leq M. \end{align*}

Answer 2

For any $x$ with $\|x\|<1$, let $c=1/\|x\|>1$. Then you have $$ |\langle Tx,y\rangle\leq |\langle T(cx),y\rangle|, $$ and $\|cx\|=1$. Similarly with $y$.