Norm of an inverse operator: $\|T^{-1}\|=\|T\|^{-1}$? I am a beginner of funcional 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?
 A: 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 $\|T^{-1}\|\leq \dfrac{1}{1-\|I-T\|}$.
A: The following formula for $\|T^{-1}\|$ is relevant for the question posted. 
Let $(\mathcal E, \|\cdot\|_{\mathcal E})$ and $(\mathcal F, \|\cdot\|_{\mathcal F})$ be Banach spaces and let $\mathcal L(\mathcal E,\mathcal F)$ be the space of all bounded operators from $\mathcal E$ into $\mathcal F$. Let $T \in \mathcal L(\mathcal E,\mathcal F)$. The following two statements are equivalent.


*

*$T$ is an injection and the range of $T$, $\operatorname{ran} T$, is a closed subspace of $\mathcal F$. 

*$\inf \bigl\{\|Tu\|_{\mathcal F}\, :\, u\in\mathcal E \quad \text{and} \quad \|u\|_{\mathcal E} = 1\bigr\} \gt 0$. 


If either of the equivalent statements above is satisfied, then $T^{-1} \in \mathcal L(\operatorname{ran} T,\mathcal E)$ and 
\begin{equation} \tag{*} 
\inf \bigl\{\|Tu\|_{\mathcal F}\, :\, u\in\mathcal E \quad \text{and} \quad \|u\|_{\mathcal E} = 1\bigr\} = \frac{1}{\|T^{-1}\|}, 
\end{equation}
where $\|T^{-1}\|$ denotes the norm of $T^{-1}$ in the Banach space $\mathcal L(\operatorname{ran} T,\mathcal E)$. 
Here are the proofs. Assume that $T$ is an injection and that $\operatorname{ran} T$ is closed in $\mathcal F$. Then $T: \operatorname{ran} T \to \mathcal E$ is a linear operator with a closed graph defined on a Banach space. By the Closed Graph Theorem the operator $T^{-1}$ is bounded. That is $T^{-1} \in \mathcal L(\operatorname{ran} T,\mathcal E)$. 
Notice that $T$ is a bijection between the sets $\mathcal E\setminus\{0\}$ and $(\operatorname{ran} T)\setminus\{0\}$. We use this fact to calculate
\begin{align*}
\|T^{-1}\| & = \sup \bigl\{ \|T^{-1} x \|_{\mathcal E} : x \in \operatorname{ran} T \quad \text{and} \quad \|x\|_{\mathcal F} = 1 \bigr\} \\
& = \sup \left\{ \frac{\|T^{-1} x \|_{\mathcal E}}{\|x\|_{\mathcal F}} : x\in (\operatorname{ran} T) \setminus\{0\} \right\} \\
& = \sup \left\{ \frac{\|T^{-1} T v \|_{\mathcal E}}{\|T v\|_{\mathcal F}} : v\in {\mathcal E} \setminus\{0\} \right\} \\
& = \sup \left\{ \frac{\|v \|_{\mathcal E}}{\|T v\|_{\mathcal F}} : v\in {\mathcal E} \setminus\{0\} \right\} \\
& = \sup \left\{ \frac{1}{\|T u\|_{\mathcal F}} : u\in {\mathcal E} \quad \text{and} \quad \|u \|_{\mathcal E} = 1 \right\}.
\end{align*}
This proves 2. and (*). 
Now assume 2. and denote by $m$ the infimum in there. Then for every $v \in \mathcal E$ we have $\|Tv\|_{\mathcal F} \geq m \|v\|_{\mathcal E}$. Therefore $T$ is injective. From the last inequality in a straightforward manner it also follows that $\operatorname{ran} T$ is closed. 
Notice that the formula at the bottom of the first answer is a consequence of the triangle inequality and (*). 
To see this, let $T \in \mathcal L(\mathcal E, \mathcal E)$ be invertible and $\|I - T\| \lt 1$. Let $x \in \mathcal E$ be such that $\|x\|_{\mathcal E} = 1$. Then by the triangle inequality 
$$
\|Tx\|_{\mathcal E} = \|x - (I - T)x\|_{\mathcal E} \geq 1 - \|(I - T)x\|_{\mathcal E} \geq 1 - \|I - T\|.
$$
Thus 
$$
\inf \bigl\{\|Tx\|_{\mathcal E}\, :\, x\in\mathcal E \quad \text{and} \quad \|x\|_{\mathcal E} = 1 \bigl\} \geq 1 - \|I - T\|
$$
and therefore by (*)
$$
\frac{1}{\|T^{-1}\|} \geq 1 - \|I - T\|.
$$
A: 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.
