On the Spectral Theorem 
Let $H$ be a Hilbert space, $T\in B(H)$ be normal and $E$ its spectral measure.
a- Let $\delta >0$ , and let $M_{\delta}$ = $\left\{\lambda\in \sigma(T): |\lambda|\geq \delta\right\}$. Prove that $Range(E(M_{\delta}))\subseteq T(H)$. 
b- Let $\sigma\subseteq \sigma(T)$ be open then $E(\sigma)\neq 0$. 
c- Use the spectral theorem to prove that $T$ is positive $\Longleftrightarrow$ $\langle Tx,x\rangle \geq0$ , for all $ x\in H$. 

Note we say $T$ is positive if there exists $S\in B(H)$ self-adjoint such that $T=S^2$.
Any idea how to start?
Thank you .
 A: a- Note that $1_{M_\delta}(x)=x\cdot 1_{M_\delta}(x)\cdot \left( \frac{1}{x}\right)1_{M_\delta}(x) =f_1(x)f_2(x)f_3(x).$ Applying the$L^\infty$ functional calculus of $T$, this yields
$$
E(M_\delta)=1_{M_\delta}(T)=f_1(T)f_2(T)f_3(T)=TE(M_\delta)f_3(T).
$$
Now clearly the range of the latter is contained in the range of $T$.
b- This is false, as $\sigma=\emptyset $ yields $E(\sigma)=0$. But it is true of $\sigma $ is a nonempty open set. Indeed, by the spectral theorem 
$$\|E(\sigma)\|=\|1_\sigma(T)\|=\|1_\sigma\|_\infty=1$$
in this case, hence $E(\sigma)\neq 0$.
c- If $T$ is positive, i.e. $T=S^2$ for some self-adjoint $S$, then $(Tx,x)=(S^2x,x)=(Sx,Sx)=\|Sx\|^2\geq 0$ for every $x$. Now assume $(Tx,x)\geq 0$ for every $x$. First show that $T$ is self-adjoint using polarization. Then check the spectrum of $T$ is nonnegative. By functional calculus, we can then set $S:=\sqrt{T}$, so that $S$ is self-adjoint and $S^2=T$.
Note: the if direction of c is false in the real case. Consider
$$
\left( \matrix{0&1\\-1&0}\right)
$$
for example.
