Linear operator and extension of its inverse Let $K:H_1 \to H_2$ be a linear operator between Hilbert spaces that may not be bounded. $K$ is bounded below. So $K$ has an inverse $K^{-1}:\text{Range}(K) \to H_1$.

$K^{-1}$ extends by continuity to an operator defined on the completion of $\text{Range}(K)$ that maps into the completion of $H_1$.

Why this is true? How does continuity work -- $K$ is not even continuous so why would $K^{-1}$ be (maybe the bounded below bit makes this work). How does it map into the completion of $H_1$?
 A: Since $K$ is bounded below,
$$
\|Kx\|\geq c\|x\|
$$
for all $x\in H_1$. If $y\in\mbox{Range}(K)$, then $y=Kx$ for some $x\in H_1$. Then
$$
\|K^{-1}y\|=\|x\|\leq\frac1c\,\|Kx\|=\frac1c\,\|y\|.
$$
So $K^{-1}$ is bounded. As such, it can be extended by continuity to the closure of its domain. This is standard: if $y_n\to y$ with $y_n=Kx_n$, then 
$$
\|K^{-1}y_n-K^{-1}y_m\|=\|K^{-1}(y_n-y_m)\|\leq\frac1c\,\|y_n-y_m\|,
$$
which shows that the sequence $\{K^{-1}y_n\}$ is Cauchy, and has a limit that can be defined as $K^{-1}y$.
A: Since $K$ is bounded below, there is $c>0$ so that $c\|Kx\|\ge\|x\|$ for all $x$ in the domain of $K$. In particular, $Kx=0\implies x=0$, so $K$ has trivial kernel, hence is injective, so $K^{-1}$ exists as a linear map $\mbox{Range}(K)\to H_1$. For $y\in \mbox{Range}(K)$, say $y=Kx$, we have $\|K^{-1}y\|=\|x\|\le c\|Kx\|=c\|y\|$, so $K$ is bounded, i.e. continuous.
In general, if $T_0:X_0\to Y_0$ is a continuous linear map between normed vector spaces, then $T_0$ extends uniquely to a continuous linear map $T:X\to Y$, where $X,Y$ are the completions of $X_0,Y_0$ respectively. To see this, you define $Tx=\lim T_0x_n$ where $x_n\in X_0$ with $x_n\to x$ as $n\to \infty$ and check that everything works out.
So the statement in the gray box holds. (BTW, $H_1$ is a Hilbert space, so is already complete).
