When does $\| T^{-1} \| = \inf_{x \neq 0} \frac{\| x \|}{\| T x \|}$ hold?

Question

Let $T: X \to Y$ be a linear operator between normed Banach spaces $X$ and $Y$. The definition of the operator norm $$ \| T \| := \sup_{x \neq 0} \frac{\| Tx \|}{\| x \|} $$ is well known.

Now, let $T$ be bijective. When can we say that $$ \| T^{-1} \| = \inf_{x \neq 0} \frac{\| x \|}{\| T x \|} $$ holds?

Answer 1

First of all, you need to assume that $T^{-1}$ is bounded. If $X$ and $Y$ are Banach spaces, this holds automatically, but not for normed spaces. For example, consider $X=Y=c_{00}(\mathbb N)$, the space of sequences with finite support, and let $T$ be defined by $(Tx)(n)=\frac{1}{n}x(n)$. Then $T$ a bounded linear bijection (under any $p$-norm), but it's inverse is not bounded.

Under this additional hypothesis, your question is equivalent to asking when $\|T^{-1}\|=\|T\|^{-1}$. I am not aware of any well-known conditions which guarantees this, but it doesn't happen a lot. In fact, if wikipedia is to be believed, this only happens in finite-dimensional spaces when $T$ is a scalar multiple of an isometry.

Answer 2

This fails in finite dimensions.

Take $$ T=\begin{pmatrix} 2&0\\0&3 \end{pmatrix} $$ So $||T||=3$ and $$ T^{-1}=\begin{pmatrix} 1/2&0\\0&1/3 \end{pmatrix} $$ and $||T^{-1}||=1/2$.

But, $$ \inf_{x\ne0}\frac{||x||}{||Tx||}\geq\frac{1}{||T||}=\frac13 $$ while $x=(0,1)$ achieves this lower bound, showing that $$ \inf_{x\ne0}\frac{||x||}{||Tx||}=\min_{x\ne 0}\frac{||x||}{||Tx||}=\frac13\ne \frac12 $$