Norm of operator vs. norm of its inverse

[AxiomOfChoice]

Are there any circumstances under which we can conclude that, for an invertible, bounded linear operator T,


\| T^{-1} \| = \frac{1}{\| T \|} ?


E.g., does this always hold if we know the inverse is bounded?

[micromass]

No, this doesn't even hold for finite-dimensional spaces! (i.e. for matrices).

Consider the matrix

\left(\begin{array}{cc} 2 & 0\\ 0 & 1\end{array}\right).

The norm of this operator is 2. However, the inverse operator is

\left(\begin{array}{cc} 1/2 & 0\\ 0 & 1\end{array}\right)

and this has norm 1.

However, you do have an inequality (for bounded operators of course): Since 1=\|id\|=\|TT^{-1}\|\leq \|T\|\|T^{-1}\|, it follows that \frac{1}{\|T\|}\leq \|T^{-1}\|.

[Landau]

Or simpler, the 1x1-matrix (a) has inverse (1/a), and these have norms a and 1/a, respectively :p

In general, it's good advice to test statements in functional analysis in the easy case of finite dimensions first.

[AxiomOfChoice]

Or simpler, the 1x1-matrix (a) has inverse (1/a), and these have norms a and 1/a, respectively :p

In general, it's good advice to test statements in functional analysis in the easy case of finite dimensions first.

Good advice. Thanks to all of you :biggrin: