Unbounded operator $T $ is bounded below when $\overline T$ is bounded How to prove the following?
A densely defined symmetric operator $T$ in Hilbert space $H$ has a closure $\overline T$ which is bounded iff both $T,-T$ are bounded below (there exist  constants $c,c' \in \mathbb{R}$ with $(Tx,x)\geq c||x||^2$ and $(-Tx,x)\geq c'||x||^2$).
 A: In Perdersen's book, Analysis Now, where this observation is made, bounded below is also called semibounded. I'll go with the latter, as I've spent too many years giving bounded below the more restrictive meaning it has in (bounded) operator theory.
Essentially, this result boils down to: $T$ symmetric is bounded if and only if $AI\leq T\leq BI$ on its domain for some $A,B\in \mathbb{R}$ . 

Lemma: a symmetric operator $S:H\longrightarrow H$ is bounded if and only if $S$ and $-S$ are semibounded.

Proof: For a symmetric operator $S:H\longrightarrow H$, we have
$$
\|S\|=\sup_{\|x\|=1}|(x,Sx)|
$$
where each $(x,Sx)$ is real. So if $S$ is bounded, we have
$$
-\|S\|\leq (x,Sx)\leq \|S\|\qquad\forall \|x\|=1
$$
hence
$$
-\|S\|\|x\|^2\leq (x,Sx)\leq \|S\|\|x\|^2\qquad\forall x\in H.
$$
Thus $S$ and $-S$ are semibounded. 
Conversely, if $S$ and $-S$ are semibounded, we get $A,B$ real numbers such that
$$
A\|x\|^2\leq (x,Sx)\leq B\|x\|^2\qquad\forall x\in H.
$$
Setting $M:=\max\{|A|,|B|\}$, we obtain
$$
-M\|x\|^2\leq (x,Sx)\leq M\|x\|^2\iff|(x,Sx)|\leq M\|x\|^2\qquad\forall x\in H.
$$
Hence $\|S\|\leq M$ and $S$ is bounded. QED.
Direction $\Rightarrow$: if $T$ is closable with bounded closure, then  the domain of $\overline{T}$ is $H$ and $\overline{T}$ is still symmetric. So the above entails that $\pm \overline{T}$, a fortiori $\pm T$, are semibounded.
Direction $\Leftarrow$: if $\pm T$ are semibounded, then the argument in the proof above applies and yields $\|T\|\leq M<\infty$ where $\|T\|$ is the sup of $|(x,Tx)|$ over the unit sphere of the domain of $T$. It follows that $T$ is bounded and extends continuously to a bounded operator on the closure of its domain, i.e. $H$. In other words, $T$ is closable with bounded closure.
Note: the crucial formula $\|T\|=\sup_{\|x\|=1}|(x,Tx)|$ follows from Cauchy-Schwarz in one direction, and from polarization+parallelogram law in the other direction. See here, p.198, for a proof in the bounded case. The general symmetric case follows by the same argument, essentially.
A: I suppose the question ask on the page "
On boundedly invertible
" is almost similar to the one above.
