= Lower norm of an operator
{title2=$\nu(T)$}
= Lower norm
{synonym}
For a <linear operator> $T:D(T)\subseteq H\to H$ on a nonzero <Hilbert space>, its <lower norm> is
$$
\nu(T)=\inf\{\|Tx\|:x\in D(T),\ \|x\|=1\}.
$$
It measures how far $T$ is from sending some unit <vector> arbitrarily close to zero. Suppose now that $T$ is closed and densely defined, and set $\mu(T)=\min(\nu(T),\nu(T^*))$. Then $\mu(T)>0$ exactly when $T$ is bijective, and in that case
$$
\mu(T)=\|T^{-1}\|^{-1}.
$$
Indeed, $\nu(T)>0$ gives injectivity and, by closedness, closed range. The same bound for the <adjoint operator> gives $\ker T^*=0$, so the range of $T$ is dense and therefore all of $H$. For an invertible $T$, applying the definition to its bounded inverse gives $\nu(T)=\|T^{-1}\|^{-1}$, and the <adjoint operator> has the same inverse <norm>.
Closedness cannot be dropped. Restrict the identity on $\ell^2$ to the proper dense subspace of finitely supported <sequences>. Both its <lower norm> and its adjoint's <lower norm> equal1, yet the restricted identity is not onto. Its graph is not closed. Thus the symmetric lower-norm criterion concerns closed densely defined <linear operators>, not arbitrary restrictions.
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