Norm of an inverse operator: $\|T^{-1}\|=\|T\|^{-1}$?

Question

I am a beginner of functional analysis. I have a simple question when I study this subject.

Let $L(X)$ denote the Banach algebra of all bounded linear operators on Banach space $X$, $T\in X$ is invertible, then $||T^{-1}||=||T||^{-1}$? Is this result correct?

Answer 1

In general you have $\|T^{-1}\|\geq \dfrac{1}{\|T\|}$ and cannot say much more. If $x_1$ and $x_2$ are nonzero with $Tx_1=y_1$ and $Tx_2=y_2$, then $\|T\|\geq\max\left\{\dfrac{\|y_1\|}{\|x_1\|},\dfrac{\|y_2\|}{\|x_2\|}\right\}$, while $\|T^{-1}\|\geq\max\left\{\dfrac{\|x_1\|}{\|y_1\|},\dfrac{\|x_2\|}{\|y_2\|}\right\}$. The only way it is possible for these lower bounds to be reciprocals is if $\dfrac{\|y_1\|}{\|x_1\|}=\dfrac{\|y_2\|}{\|x_2\|}$, which will typically not happen. If this happens for all $x_1,x_2$, then $T$ must be a scalar multiple of an isometric isomorphism.

To see this another way, suppose that $\|T^{-1}\|=\dfrac{1}{\|T\|}$, and let $S=\dfrac{1}{\|T\|}T$. Then $\|S\|=1$ and $\|S^{-1}\|=1$. Thus for all $x$, $\|Sx\|\leq \|x\|$ and $\|x\|=\|S^{-1}(Sx)\|\leq \|Sx\|$, which implies that $\|Sx\|=\|x\|$ for all $x$, and $T=\|T\|S$ is a scalar multiple of the isometric isomorphism $S$.

---

A useful upper bound for $\|T^{-1}\|$ can be given in one particular context. If $\|T-I\|<1$, then $T^{-1}=\sum_{k=0}^\infty(I-T)^k$, which implies that $\|T^{-1}\|\leq \dfrac{1}{1-\|I-T\|}$.

Answer 2

Consider operator given by matrix $$ M= \begin{pmatrix} 0 & 2^{-1}\\ 2 & 0\\ \end{pmatrix} $$ then $M^{-1}=M$. Note that $\Vert M\Vert\geq \Vert Me_1\Vert/\Vert e_1\Vert=2$. Hence $\Vert M^{-1}\Vert=\Vert M\Vert\geq 2$ and equality $\Vert M^{-1}\Vert=\Vert M\Vert^{-1}$ doesn't hold in general.