[proofplan]
Set $A=B^{-1}E$, so the hypothesis says that $A$ has operator norm strictly smaller than $1$. We prove directly, by the Neumann series, that $I+A$ is invertible in $\mathcal{L}(X)$. Then we factor $B+E$ as $B(I+A)$ and invert the product by reversing the order of the inverse factors. The final sufficient condition follows from [submultiplicativity of the operator norm](/theorems/1054).
[/proofplan]
[step:Invert $I+B^{-1}E$ by the Neumann series]
Define the bounded operator $A\in\mathcal{L}(X)$ by
\begin{align*}
A:=B^{-1}E.
\end{align*}
By hypothesis,
\begin{align*}
\|A\|_{\mathcal{L}(X)}<1.
\end{align*}
Since $X$ is Banach, $\mathcal{L}(X)$ is complete in the operator norm. For every $n\in\mathbb{N}\cup\{0\}$, define $A^n\in\mathcal{L}(X)$ by $A^0:=I$ and $A^{n+1}:=A^nA$. The submultiplicativity of the operator norm gives
\begin{align*}
\|A^n\|_{\mathcal{L}(X)}\leq \|A\|_{\mathcal{L}(X)}^n.
\end{align*}
Because the geometric series $\sum_{n=0}^{\infty}\|A\|_{\mathcal{L}(X)}^n$ converges, the series
\begin{align*}
\sum_{n=0}^{\infty}(-A)^n
\end{align*}
converges absolutely in $\mathcal{L}(X)$. Define $S\in\mathcal{L}(X)$ by
\begin{align*}
S:=\sum_{n=0}^{\infty}(-A)^n.
\end{align*}
For $N\in\mathbb{N}\cup\{0\}$, define the partial sum $S_N\in\mathcal{L}(X)$ by
\begin{align*}
S_N:=\sum_{n=0}^{N}(-A)^n.
\end{align*}
Using the finite telescoping identity in the algebra $\mathcal{L}(X)$, we have
\begin{align*}
(I+A)S_N=I+(-1)^N A^{N+1}.
\end{align*}
Also,
\begin{align*}
S_N(I+A)=I+(-1)^N A^{N+1}.
\end{align*}
Since
\begin{align*}
\|A^{N+1}\|_{\mathcal{L}(X)}\leq \|A\|_{\mathcal{L}(X)}^{N+1}\to 0,
\end{align*}
passing to the limit in operator norm gives
\begin{align*}
(I+A)S=I.
\end{align*}
Similarly,
\begin{align*}
S(I+A)=I.
\end{align*}
Thus $I+A$ is invertible in $\mathcal{L}(X)$, with inverse $S$.
[guided]
The first task is to invert the operator that is close to the identity. Define $A\in\mathcal{L}(X)$ by
\begin{align*}
A:=B^{-1}E.
\end{align*}
The assumption of the theorem is exactly
\begin{align*}
\|A\|_{\mathcal{L}(X)}<1.
\end{align*}
This strict inequality is what makes the Neumann series converge.
For every $n\in\mathbb{N}\cup\{0\}$, define the power $A^n\in\mathcal{L}(X)$ by $A^0:=I$ and $A^{n+1}:=A^nA$. The operator norm is submultiplicative, so induction gives
\begin{align*}
\|A^n\|_{\mathcal{L}(X)}\leq \|A\|_{\mathcal{L}(X)}^n.
\end{align*}
Because $\|A\|_{\mathcal{L}(X)}<1$, the scalar geometric series $\sum_{n=0}^{\infty}\|A\|_{\mathcal{L}(X)}^n$ converges. Hence the operator series
\begin{align*}
\sum_{n=0}^{\infty}(-A)^n
\end{align*}
is absolutely convergent in the [Banach space](/page/Banach%20Space) $\mathcal{L}(X)$. Completeness of $\mathcal{L}(X)$ is used here: it ensures that the [Cauchy sequence](/page/Cauchy%20Sequence) of partial sums has a limit in $\mathcal{L}(X)$. Define that limit by
\begin{align*}
S:=\sum_{n=0}^{\infty}(-A)^n.
\end{align*}
We now prove that $S$ is the two-sided inverse of $I+A$. For $N\in\mathbb{N}\cup\{0\}$, define the partial sum $S_N\in\mathcal{L}(X)$ by
\begin{align*}
S_N:=\sum_{n=0}^{N}(-A)^n.
\end{align*}
Multiplying the finite sum on the left by $I+A$ gives a telescoping cancellation:
\begin{align*}
(I+A)S_N=I+(-1)^N A^{N+1}.
\end{align*}
The same cancellation occurs on the right:
\begin{align*}
S_N(I+A)=I+(-1)^N A^{N+1}.
\end{align*}
The remainder tends to zero in operator norm, because
\begin{align*}
\|A^{N+1}\|_{\mathcal{L}(X)}\leq \|A\|_{\mathcal{L}(X)}^{N+1}\to 0.
\end{align*}
Since multiplication in $\mathcal{L}(X)$ is continuous with respect to the operator norm, passing to the limit as $N\to\infty$ gives
\begin{align*}
(I+A)S=I.
\end{align*}
Likewise,
\begin{align*}
S(I+A)=I.
\end{align*}
Thus $S$ is a two-sided inverse for $I+A$, so $I+A$ is invertible in $\mathcal{L}(X)$.
[/guided]
[/step]
[step:Factor $B+E$ through the invertible operator $I+B^{-1}E$]
With $A=B^{-1}E$, we compute in $\mathcal{L}(X)$:
\begin{align*}
B(I+A)=B+BA.
\end{align*}
Since $A=B^{-1}E$, we have
\begin{align*}
BA=B(B^{-1}E)=E.
\end{align*}
Therefore
\begin{align*}
B(I+A)=B+E.
\end{align*}
The operator $B$ is invertible by hypothesis, and $I+A$ is invertible by the previous step. Hence their product $B(I+A)=B+E$ is invertible. For invertible operators $S,T\in\mathcal{L}(X)$, the inverse of $ST$ is $T^{-1}S^{-1}$, because
\begin{align*}
(ST)(T^{-1}S^{-1})=I
\end{align*}
and
\begin{align*}
(T^{-1}S^{-1})(ST)=I.
\end{align*}
Applying this with $S=B$ and $T=I+A$ gives
\begin{align*}
(B+E)^{-1}=(I+A)^{-1}B^{-1}.
\end{align*}
Substituting $A=B^{-1}E$ yields
\begin{align*}
(B+E)^{-1}=(I+B^{-1}E)^{-1}B^{-1}.
\end{align*}
[/step]
[step:Derive the sufficient norm condition on $E$]
Assume now that $X$ is nonzero and that
\begin{align*}
\|E\|_{\mathcal{L}(X)}<\|B^{-1}\|_{\mathcal{L}(X)}^{-1}.
\end{align*}
Since $B$ is invertible on a nonzero Banach space, $B^{-1}\neq 0$, so
\begin{align*}
\|B^{-1}\|_{\mathcal{L}(X)}>0.
\end{align*}
By submultiplicativity of the operator norm,
\begin{align*}
\|B^{-1}E\|_{\mathcal{L}(X)}\leq \|B^{-1}\|_{\mathcal{L}(X)}\|E\|_{\mathcal{L}(X)}.
\end{align*}
The assumed strict inequality for $\|E\|_{\mathcal{L}(X)}$ gives
\begin{align*}
\|B^{-1}\|_{\mathcal{L}(X)}\|E\|_{\mathcal{L}(X)}<1.
\end{align*}
Consequently,
\begin{align*}
\|B^{-1}E\|_{\mathcal{L}(X)}<1.
\end{align*}
The first part of the proof therefore applies, and $B+E$ is invertible. This proves the stated sufficient condition.
[/step]
