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About the notation: $E$ and $F$ are normed vector spaces. The space $\mathcal L(E,F)$ is the normed algebra of "bounded" linear maps. Bounded here means that if $A\in\mathcal L(E,F)$ then exists some $\alpha>0$ such that $\|Ax\|_F\le\alpha \|x\|_E$ for all $x\in E$.

(Trivially $\mathcal L(E,F)$ is a space of Lipschitz functions.)

The norm on $\mathcal L(E,F)$ is defined as

$$\|A\|_{\mathcal L(E,F)}:=\inf\{\alpha\ge 0:\|Ax\|\le\alpha \|x\|,\forall x\in E\}$$

and from a previous exercise I know that if $A^{-1}\in\mathcal L(F,E)$ then $\|A^{-1}\|_{\mathcal L(F,E)}\ge \|A\|_{\mathcal L(E,F)}^{-1}$.

Suppose $E$ and $F$ are finite dimensional and $A\in\mathcal L(E,F)$ is bijective with continuous inverse $A^{-1}\in\mathcal L(F,E)$. Show that if $B\in\mathcal L(E,F)$ satisfies $$\|A-B\|_{\mathcal L(E,F)}<\|A^{-1}\|_{\mathcal L(F,E)}^{-1}\tag{1}$$ then $B$ is invertible.

I was trying to prove that $B$ is injective, what is equivalent to say that the kernel of $B$ is trivial, that is $\ker(B)=\{0_E\}$ but I dont found something useful.

From (1), and the notes in the top, is clear that

$$\|A-B\|<\|A^{-1}\|^{-1}\le\|A\|\implies 0<\|B\|<2\|A\|$$

but this doesnt help me to show that $B$ is injective. Some help will be appreciated.

I think I solved it. I was overlooking some theoretic statements over the $\mathcal L$ spaces, by example that

$$\|AB\|\le \|A\|\|B\|\tag{2}$$

where $AB$ is a well-defined composition of linear functions from spaces $\mathcal L$. We must show that $B$ is injective or, in other words, that $\ker(B)=\{0_E\}$. Now observe that from (2) and (1)

$$\|A-B\|<\|A^{-1}\|^{-1}\implies \|A^{-1}\|\|A-B\|<1\implies \|I-A^{-1}B\|<1\tag{3}$$

Now I define $C:=A^{-1}B\in\mathcal L(E)$. Now observe that $C$ is injective if and only if $B$ is injective. Suppose that $C$ is not injective, then exists some $x\neq 0$ such that $Cx=0$, and observe that the RHS of (3), from the norm definition on $\mathcal L$ spaces, would imply that

$$\|x-Cx\|<\|x\|\implies \|x-0_E\|<\|x\|$$

what cannot be possible, hence $B$ is injective.

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    Write $A - B = A(I - A^{-1}B)$. Do you know something about the invertibility of $I - C$ for $C \in \mathcal{L}(E,E)$? – 2017-02-10
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    @DanielFischer I dont see before something related to the invertibility of $I-C$. But I can try to see where it goes. – 2017-02-10
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    @DanielFischer I solved it, thank you! – 2017-02-10
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    And now let $E,F$ be Banach spaces and get the same conclusion. The injectivity doesn't automatically imply that $B$ is an isomorphism in the infinite-dimensional case. Have you an idea how to argue in that case? – 2017-02-10
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    @DanielFischer but I was trying to prove that $B$ is invertible, if it is an isomorphism is a different problem, right? By now I dont have an idea if $B^{-1}$ is isomorphic. To be isomorphic we must ensure that $B$ is also surjective... – 2017-02-10
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    An isomorphism is (here) an invertible (continuous) linear map (in the finite-dimensional case, continuity is automatic, in the Banach space case [which includes the finite-dimensional case] the continuity of the inverse then follows from the open mapping theorem). In the finite-dimensional case, since $A$ is assumed invertible, we have $\dim E = \dim F$ and the rank formula shows that $L \in \mathcal{L}(F,E)$ is bijective if and only if it is injective, if and only if it is surjective. So in the case considered, showing the injectivity of $B$ suffices to conclude invertibility. – 2017-02-10
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    In the infinite-dimensional case, let $D = I - A^{-1}B$. As you showed, we have $\lVert D\rVert < 1$. It follows that $A^{-1}B = I - D$ is invertible, and hence, since $A$ is invertible, so is $B = AA^{-1}B$. To see $\lVert D\rVert < 1 \implies I - D\text{ is invertible}$, look at the Neumann (Carl, not Janos von) series, $$\sum_{k = 0}^{\infty} D^k$$ is absolutely convergent, hence, since $\mathcal{L}(E,E)$ is Banach, it is convergent. One verifies $$(I - D)^{-1} = \sum_{k = 0}^{\infty} D^k$$ pretty much the same way as the geometric series for $(1-z)^{-1}$ with complex $z$. – 2017-02-10

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