Let $V$ be Hilbert space, $A:V\rightarrow V$ be bounded, linear and self-adjoint operator. Show (without using general spectral theory) that $$\|A\|:= \sup\limits_{v\in V,v\neq 0}\frac{\|Av\|_V}{\|v\|_V}=\sup\limits_{v\in V,v\neq 0}\frac{|(Av,v)_V|}{\|v\|^2_V}$$ where $(\cdot,\cdot)_V$ and $\|\cdot\|_V$ is inner product and norm on $V$.
I would really appreciate it if anyone could give me a hint on how to start this proof. Thanks!
Hints: Let $M=\sup \{|\langle Av, v \rangle|\}: \|v\| \leq 1\}$. Then $$\Re \langle Ax, y \rangle =\langle A(\frac {x+y} 2), \frac {x+y} 2 \rangle-\langle A(\frac {x-y} 2), \frac {x-y} 2 \rangle$$ $$ \leq M (\|\frac {x+y} 2\|^{2}+\|\frac {x-y} 2\|^{2}) \leq M$$ if $\|x\| \leq 1$ and $\|y\|\leq 1$. This shows that $\|Ax\| \leq M$ for $\|x\| \leq 1$ so $\|A\|\leq M$. Reverse inequality is obvious.