Proving that the set of continuous linear maps (From a Banach space $X$ to another Banach space $Y$) is an open subset of $L(X,Y)$

Question

I am a university undergraduate student, and I am reading up on the German Functional Analysis textbook "Introduction to Functional Analysis" by Friedrich Hirzebruch/Winfried Scharlau.

My question goes in this manner:

Let $X$ and $Y$ be Banach spaces. Prove that the set of continuous bijective linear maps from $X\to Y$ is an open subset of $L(X,Y)$ (with respect to the sup norm, defined by $||T||=\sup_{x\neq0}\frac{||Tx||}{||x||}$).

As for the theorems that are involved, I think the "Open mapping theorem" and its corollary, "The theorem of the inverse operator" might be useful, but I could not think of even a way to even start with tackling the problem, not even with the hint:

Consider $T$ $\in$ $L(X)$ with $||T||$ sufficiently small. Does the identity $(\operatorname{Id}-T)^{-1}=T^0+T^1+T^2+\dots$ holds?

And extensive searches with the internet/google along with various textbooks doesn't help for me.

Therefore, I humbly ask for help in solving this question. Further hints to point me out in the right way would be much appreciated.

Answer 1

I denote with $\operatorname{GL}(X,Y)$ the set of continuous (and therefore bounded) invertible linear operators from $X$ to $Y$. We first show the following Lemma:

If $A \in L(X)$ such that $\| A \| < 1$, then $I - A \in \operatorname{GL}(X)$ (where $I$ denotes the identity on $X$) and $(I - A)^{-1} = \sum_{k = 0}^\infty A^k$.

Proof therefore. By the submultiplicativity of the operator norm we have $\| A^k \| \leq \|A\|^k$ and therefore

$$\sum_{k = 0}^\infty \| A \|^k \stackrel{\|A\| < 1}{=} \frac{1}{1 - \|A\|}$$

using geometric series. Using Weierstraß-M-test it follows that $\sum_{k = 0}^\infty A_k$ is an absolutely convergent series in the Banach space $L(X)$ and in particular, $B := \sum_{k = 0}^\infty A^k$ is well-defined in $L(X)$. One easily sees $AB = BA = \sum_{k = 1}^\infty A^k$, therefore we also have $(I - A)B = B(I - A) = I$ and this is equivalent to $(I - A)^{-1} = B$.

Now we prove our desired result:

The set $\operatorname{GL}(X, Y)$ is open in $L(X, Y)$.

Proof. Let $T_0 \in \operatorname{GL}(X, Y)$ and $T \in L(X, Y)$ such that $\| T_0 - T \| < \|T_0^{-1}\|^{-1}$. Also,

$$T = T_0 + T - T_0 = T_0 [I + T_0^{-1} (T - T_0)].$$

We want to show that $I + T_0^{-1}(T- T_0) \in \operatorname{GL}(X)$. Since

$$\| - T_0^{-1}(T - T_0)\| \leq \|T_0^{-1}\| \| T - T_0 \| <1$$

this follows from our preceding Lemma. All in all, we have shown that the open ball in $L(X, Y)$ with center $T_0$ and radius $\|T_0^{-1}\|^{-1}$ is an element of $\operatorname{GL}(X,Y)$, so the proof is complete.