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

Solution 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 $\|T^{-1}\|\leq \dfrac{1}{1-\|I-T\|}$.

Solution 2:

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.

1. $T$ is an injection and the range of $T$, $\operatorname{ran} T$, is a closed subspace of $\mathcal F$.
2. $\inf \bigl\{\|Tu\|_{\mathcal F}\, :\, u\in\mathcal E \quad \text{and} \quad \|u\|_{\mathcal E} = 1\bigr\} \gt 0$.

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 (*) $$ \frac{1}{\|T^{-1}\|} \geq 1 - \|I - T\|. $$

Solution 3:

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.