Show $\|A\|=\sup_{x\neq 0} \langle Ax,x\rangle/\langle x,x\rangle$ for a positive operator $A$. I have a positive operator $A$ on the Hilbert space $\mathcal{H}$.
I must prove that $\|A\|=\sup_{x \ne 0}\frac{(Ax,x)}{(x,x)}$. I am only able to get one inequality:
Assume $x$ is nonzero:
$$\frac{(Ax,x)}{(x,x)}\le \|Ax\|\|x\|/\|x\|^2\le \|A\|. \text{ So } \|A\|\ge\sup_{x \ne 0}\frac{(Ax,x)}{(x,x)}.$$
Any ideas for the other inequality? I tried something like this:
Let $\epsilon > 0$ be given. Then there is a z, with $\|z\|\le 1$ such that: $\|Az\|>\|A\|-\epsilon$. However, I am not sure how to show that for instance $(Az,z)/(z,z)$ is bigger than something.
Any tips?
 A: I assume that part of your definition of positive operator is that $A^\ast =A$. Otherwise, the claim fails in general. Also, I assume that the Hilbert space I  question is real. Otherwise,  the argument has to be modified slightly.
Let $C $ denote your supremum. Note 
\begin{align*}
0 &\leq \langle A (x+y),x+y\rangle \\ &= \langle Ax,x\rangle + 2\langle Ax,y\rangle +\langle Ay,y\rangle \\ & \leq C (|x|^2+|y|^2) +2\langle Ax,y\rangle.
\end{align*}
Hence, for $|x|=1=|y|$, we get
$$
-\langle Ax,y\rangle \leq C.
$$
But by homogeneity, this yields 
$$
|\langle Ax,y\rangle|\leq C |x||y|
$$
for all $x,y\in H $.
The rest of the proof is easy. 
A: Let $(x,y)_A = (Ax,y)$. Then $(x,y)_A$ is a pseudo inner product, meaning that it satisfies all properties of an inner product except that $(x,x)_A=0$ may occur even though $x \ne 0$. You can replace $A$ by $A+\delta I$ for any positive $\delta$ in order to obtain an actual inner product. Then the Cauchy-Schwarz inequality holds, and you can let $\delta\downarrow 0$ to obtain
$$
                     |(x,y)_A| \le (x,x)_A^{1/2}(y,y)_A^{1/2} \\
           |(Ax,y)| \le (Ax,x)^{1/2}(Ay,y)^{1/2}
$$
Now let $y=Ax$:
$$
              \|Ax\|^2 \le (Ax,x)^{1/2}(AAx,Ax)^{1/2}
$$
If $M=\sup_{\|x\|=1}\frac{(Ax,x)}{(x,x)}$, then
$$
                    \|Ax\|^2 \le \{ M(x,x) \}^{1/2}\{ M(Ax,Ax) \}^{1/2}=M\|x\|\|Ax\| \\
             \implies \|Ax\| \le M\|x\| \\
            \implies \|A\| \le M = \sup_{\|x\|}\frac{(Ax,x)}{(x,x)}.
$$
The other direction is obvious.
