Inverse of a bounded self-adjoint operator Let be $Q$ a bounded self-adjoint operator defined on a Hilbert space $H$ that satisfies:
$$\inf_{x\in H}\frac{(x,Qx)_H}{(x,x)_H}=m>0$$
Show that $Q$ is invertible and satisfies:
$$(x,Q^{-1}x)_H \leq \frac{1}{m} (x,x)_H$$
where $(\cdot,\cdot)_H$ represents the inner product in $H$.
Any help with this result would be very appreciated,
 A: By assumption,
$$
         m\|x\|^2 = m(x,x) \le (Qx,x) \le \|Qx\|\|x\| \\
                   m\|x\| \le \|Qx\|.
$$
$Q$ is injective because $m > 0$. $Q$ has a dense range because $\mathcal{R}(Q)^{\perp}=\mathcal{N}(Q)=\{0\}$. In fact $Q$ has a closed range because, if $x$ is in the closure of $\mathcal{R}(Q)$, then there exists $\{ y_n \}$ such that $\{ Qy_n \}$ converges to $x$; then $\|Qy_n - Qy_m\| \ge m\|y_n-y_m\|$ implies that $\{ y_n \}$ is a Cauchy sequence because $\{ Qy_n \}$ is a Cauchy sequence. So $\{ y_n \}$ converges to some $y$, which leads to $Qy = \lim_n Qy_n = x$, which proves that $Q$ is surjective. Hence, $Q$ is a bijection on $H$ and $Q^{-1}$ is bounded because
$$
                 m\|x\| \le \|Qx\| \\
                 m\|Q^{-1}y\| \le \|y\| \\
                  \|Q^{-1}\| \le \frac{1}{m}.
$$
Therefore,
$$
            \frac{(x,Q^{-1}x)}{(x,x)} = \frac{(Q^{-1}x,x)}{(x,x)} \le \|Q^{-1}\| \le \frac{1}{m},\;\;\; x\ne 0.
$$
A: We know that: $$\left\|Q\right\|=\sup_{x\in H}{\left\|Qx\right\|\over\left\|x\right\|}=\sup_{x\in H}\frac{(x,Qx)_H}{(x,x)_H}\ge\inf_{x\in H}\frac{(x,Qx)_H}{(x,x)_H}=m>0$$
Thus, if $Q$ is an operator that satisfies $\left\|Qx\right\|\ge m\left\|x\right\|$, we get that $Q$ is one-to-one ($\ker Q=\{0\}$), in fact:
$$Qx=0\Longrightarrow\left\|Qx\right\|=0\ge m\left\|x\right\|\Longrightarrow\left\|x\right\|=0\Longrightarrow x=0$$
so $Q^{-1}$ exists.
Now let $y=Qx$, then:
$$\left\|y\right\|\ge m\left\|x\right\|=m\left\|Q^{-1}y\right\|$$
hence $Q^{-1}$ is bounded and $\left\|Q^{-1}y\right\|\le{1\over m}\left\|y\right\|$.
