Problem 5

Exercise 4.4 (Inversions and Estimates in Banach spaces)   Let $X$ be a Banach space and $A\in L(X)$ be a bounded linear operator. Show that there exists a bounded linear operator $B\in L(X)$ satsifying $AB=BA=I_X$ if and only if there exists a constant $\gamma>0$ such that

$$\Vert x\Vert \leq \gamma\Vert Ax\Vert \quad \Vert\phi\Vert \leq \gamma\Vert A^*\phi\Vert x\in X \phi \in X^*$$ (4.10)

Proof. Suppose a bounded inverse $B$ exists satisfying $AB=BA=I_X$. Set $\gamma=\Vert B\Vert$. Then for any $x\in X$ we have

$$\Vert x\Vert = \Vert I_X x\Vert = \Vert B(Ax)\Vert \leq \gamma\Vert Ax\Vert$$ (4.11)

To prove the other estimate we take adjoints: $(I_X)^* = (AB)^*=B^*A^*$. Recall that $\Vert B^*\Vert=\Vert B\Vert=\gamma$, so for any $\phi \in X^*$ we can estimate directly

$$\Vert\phi\Vert = \Vert I_X^*\phi\Vert = \Vert B^*A^*\phi\Vert \leq \Vert B^*\Vert\Vert A^*\phi\Vert = \gamma\Vert A^*\phi\Vert$$ (4.12)

For the converse, let $\gamma>0$ entail the above estimates. We can see $A$ is injective because if $Ax=Ay$, then setting $z=x-y$ shows

$$\Vert x-y\Vert = \Vert z\Vert \leq \gamma \Vert Az\Vert = \gamma\Vert Ax-Ay\Vert= 0.$$ (4.13)

The second estimate will let us show $A$ is surjective. To apply the open mapping theorem, we verify that $\frac{1}{\gamma}U\subseteq \overline{A(U)}$, where $U=\{x\in X\;\vert\; \Vert x\Vert < 1\}$. Suppose $y\notin \overline{A(U)}$. The set $\overline{A(U)}$ is closed, balanced, and convex, so there exists a linear functional $\phi : X\to\mathbb{C}$ such that $\vert\phi(y)\vert>1$ and $\vert\phi(Ax)\vert\leq 1$ for $\Vert x\Vert\leq 1$. Since $\phi(Ax)=A^*(\phi)(x)$, the second estimate shows $\Vert A^*\phi\Vert\leq 1$. Putting these all together,

$$\frac{1}{\gamma} < \frac{1}{\gamma}\vert\phi(y)\vert \leq \frac{1... ...Vert\phi\Vert\Vert y\Vert \leq \Vert A^*\phi\Vert\Vert y\Vert \leq \Vert y\Vert$$ (4.14)

we see that $\Vert y\Vert\geq 1/\gamma$. Therefore, if $\Vert y\Vert<1/\gamma$, then $y\in\overline{A(X)}$, so that $\frac{1}{\gamma}U \subseteq \overline{A(X)}$ which implies $A$ is surjective. Now that $A$ is a continuous bijection, the inverse mapping theorem implies that $A^{-1}$ is a bounded linear operator. $\qedsymbol$

The interested reader is welcomed to read Proposition 6.8.5 of [4], which outlines a more general case of the surjectivity aspect of this exercise, but not the injectivity. Theorem 4.13 of [11] does the same.