 Lower norm of an operator

For a linear operator $T:D(T)\subseteq H\to H$ on a nonzero Hilbert space, its lower norm isIt 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 $\mu (T)=\min (\nu (T),\nu (T^{\ast }))$. Then $\mu (T)>0$ exactly when $T$ is bijective, and in that caseIndeed, $\nu (T)>0$ gives injectivity and, by closedness, closed range. The same bound for the adjoint operator gives $\ker T^{\ast }=0$, so the range of $T$ is dense and therefore all of $H$. For an invertible $T$, applying the definition to its bounded inverse gives $\nu (T)=\Vert T^{-1}{\Vert }^{-1}$, 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.