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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,

2 Answers 2

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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 $\left\|Q^{-1}y\right\|\le{1\over m}\left\|y\right\|$.

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    Thank you very much. I have only one question, Why can we assert that $$\vert \vert Q \vert \vert=\sup_{x\in H}\frac{(x,Qx)_{H}}{(x,x)_H}$$ – 2017-02-02
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    @DiegoVargas Good question! That is so because $Q$ is a self-adjoint operator...in general we have $\left\|Q\right\|=\sup_{x\in H}{\left\|Qx\right\|\over\left\|x\right\|}=\sup_{x,y\in H}\frac{(x,Qy)_H}{(x,y)_H}\ne\sup_{x\in H}\frac{(x,Qx)_H}{(x,x)_H}$, but if $Q= Q^*$ the last equality holds. – 2017-02-02
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    @DiegoVargas I can add the demonstration of that statement later – 2017-02-02
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    Thanks @MattG88, I didn´t know that fact. – 2017-02-02
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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, $$ \frac{(x,Q^{-1}x)}{(x,x)} = \frac{(Q^{-1}x,x)}{(x,x)} \le \|Q^{-1}\| \le \frac{1}{m},\;\;\; x\ne 0. $$