How can we show that $a\leq T\leq b$ if the spectrum $\sigma (T)$ is contained in some interval $[a,b]$? Let $T:V\to V$ be a self-adjoint operator on a Hilbert space $V$. We know that its spectrum $\sigma(T)$ is contained in some finite interval $[a,b]$. How can we show that $a\leq T\leq b$ where scalars refer to scalar operators on $V$? Thanks.
Attempt. For any $v\in V$ we have 
\begin{align}
\langle (b-T)v,v\rangle & \geq b\|v\|^2-\langle Tv,v\rangle 
\\&\geq b\|v\|^2-|\langle Tv,v\rangle|
\\&\geq b\|v\|^2-\|Tv\|\|v\| \qquad \text{by Cauchy-Schwarz inequality}
\\&\geq b\|v\|^2-\|T\|\|v\|^2
\\&\geq \big(b-\|T\|\big)\|v\|^2
\end{align}
So we need to show that $\|T\|\leq b$. I also have difficulty with showing that $\langle (T-a)v,v\rangle\geq 0$ for all $v\in V$. Thanks!
 A: 1. Note that the spectral radius of $T$ equals $\|T\|$ when $T$ is bounded and self-adjoint, i.e., 
$$ \|T\| = \sup\{|\lambda| : \lambda \in \sigma(T) \}. \tag{*} $$
This is a direct consequence of the Gelfand's formula.
2. We consider $r \geq 0$ satisfying $\sigma(T) \subseteq [-r, r]$. Then by $\text{(*)}$, we have $\|T\| \leq r$. Now we fix $v \neq 0$ and consider the function $f_v : \mathbb{R} \to \mathbb{R}$ defined by
$$ f_v(\lambda) = \langle v, (T - \lambda)v \rangle. $$
This is a real polynomial of degree 1 with negative slope. Then by mimicking your computation, we have
$$ |f_v(\lambda)| \geq |\langle v, \lambda v \rangle| - |\langle v, Tv\rangle| \geq (|\lambda| - r) \|v\|^2. $$
It follows that $f_v(\lambda) \neq 0$ whenever $|\lambda| > r$, hence the unique zero $\lambda^*$ of $f_v$ lies in $[-r, r]$. Then
$$ \langle v, (T - r)v \rangle = f_v(r) \leq 0 \leq f_v(-r) = \langle v, (T+r)v \rangle $$
for any $v \neq 0$. This implies that $-r \leq T \leq r$.
3. For general case where $\sigma(T) \subseteq [a, b]$, apply the previous step to $T - \frac{a+b}{2}$ and $r = \frac{b-a}{2}$.
