Let $(X,\langle\cdot ,\cdot\rangle)$ an inner product space and $A\in\mathcal L(X)$. I have to show that $$\|A\|=\sup\{|\langle Ax,x\rangle|\mid x\in X, \|x\|\leq 1\}.$$ The fact that $\|A\|\geq \sup\{|\langle Ax,x\rangle|\mid x\in X, \|x\|\leq 1\}$ is a consequence of Cauchy-Schwarz. For the other inequality, we set $m=\inf\{\langle Ax,x\rangle\mid x\in X, \|x\|=1\}$ and $M=\sup\{\langle Ax,x\rangle\mid x\in X, \|x\|=1\}$. So we want to prove that $\|A\|\leq \max\{-m,M\}$.
Q1) Why $m\leq 0$ ?
Moreover, in the proof, we have $$\max\{-m,M\}=\sup\{|\langle Ax,x\rangle|\mid x\in X,\|x\|=1\}\underset{(*)}{=}\sup\{|\langle Ax,x\rangle|\mid x\in X,\|x\|\leq 1\}.$$
Q2) Why $(*)$ is an equality ? Shouldn't it be a $\leq$ ? I asked this question to my teacher, and he told me that it was correct but with no explanation.
Q1) This is false, take $A=Id$, then $\langle x, x\rangle =\|x\|^2$ for all $x$. So $m=\inf\{\langle Ax,x\rangle\mid x\in X, \|x\|=1\}$ can be positive.
Q2) If $\|x\|\leq 1$, then we can write $\langle Ax, x\rangle =\|x\| ^2\langle \frac{Ax}{\|x\|}, \frac{x}{\|x \|}\rangle \leq \sup\{|<Ay,y>|\mid y\in X, \|y\|=1 \}$ since $\|x\|^2\leq 1$. The converse inequality is obvious.