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I'm trying to solve an exercise in Brezis' Functional Analysis, i.e.,

Let $(E, |\cdot|)$ be a real Banach space. Let $T \in \mathcal L(E)$, i.e., $T:E \to E$ is a bounded linear operator. Assume that $\|T\| < 1$. Let $I:E \to E$ be the identity map.

  1. Prove that $(I-T)$ is bijective and that $$ \|(I-T)^{-1}\| \le \frac{1}{1- \|T\|}. $$
  2. Let $S_n := I+T+\cdots+T^{n-1}$. Prove that $$ \|S_n-(I-T)^{-1}\| \le \frac{\|T\|^n}{1- \|T\|}. $$

There are possibly subtle mistakes that I could not recognize in below attempt. Could you please have a check on it?

Assume $(I-T)u=0$. Then $u=Tu$ and thus $u= T^n u$ for $n \in \mathbb N^*$. Then $|u| \le \|T\|^n |u|$ for $n \in \mathbb N^*$. Then $|u| \le |u| \lim_n \|T\|^n=0$ and thus $u=0$. Then $I-T$ is injective.

Let $v \in E$. We want to find $u \in E$ such that $(I-T)u=v$. We define $K:E \to E$ by $K(u)=v+Tu$. Then $|K(u_1) - K(u_2)| = |T(u_1-u_2)| \le \|T\| |u_1-u_2|$. Then $K$ is a strict contraction. By Banach fixed-point theorem, $K$ has a fixed point. This means there is $u\in E$ such that $K(u)=u$. Then $v=(I-T)u$. So $I-T$ is surjective.

We have $\sum_{n=0}^\infty \|T^n\| \le \sum_{n=0}^\infty \|T\|^n = \frac{1}{1-\|T\|}$ because $\|T\|<1$. By characterization of a Banach space by absolute convergence, we get $K:= \sum_{n=0}^\infty T^n \in \mathcal L(E)$. First, $$ K(I-T)= (I-T)K = \sum_{n=0}^\infty T^n- \sum_{n=1}^\infty T^n = I. $$

Hence $K= (I-T)^{-1}$. The claim then follows.

2.

We have $$ S_n-(I-T)^{-1} = S_n-K=\sum_{k=n}^\infty T^k= T^n K. $$

Hence by (1), $$ \|S_n-(I-T)^{-1}\| \le \|T\|^n \cdot \|K\| \le \frac{\|T\|^n}{1- \|T\|}. $$

Akira
  • 17,367

1 Answers1

1

Following the method in your proof you can show a few more things:

  1. That the set of invertible $G_X$ operators in $L(X)$ is an open (with the topology in $L(X)$.
  2. That the map $T\mapsto T^{-1}$ from $G_X$ to itself is differentiable.

Here is a result in the setting of Banach algebras with identity ($L(X)$ is one such thing). As you can see the technique is already in what have in your posting

Theorem: If $A$ is a Banach algebra, then the set $G_A$ of invertible elements in $A$ is open in $(A,\|\;\|)$. Moreover, if $\boldsymbol{x}\in G_A$, then for $\|\boldsymbol{h}\|\|\|\boldsymbol{x}^{-1}\|<1$ \begin{align*} \|(\boldsymbol{x+h})^{-1}\|\leq \frac{\|\boldsymbol{x}^{-1}\|}{1-\|\boldsymbol{x}^{-1}\boldsymbol{h}\|}\leq\frac{\|\boldsymbol{x}^{-1}\|}{1-\|\boldsymbol{x}^{-1}\|\|\boldsymbol{h}\|}\ \end{align*} and \begin{align} (\boldsymbol{x}+\boldsymbol{h})^{-1}-\boldsymbol{x}^{-1}+\boldsymbol{x}^{-1}\boldsymbol{h}\boldsymbol{x}^{-1}=o(\|\boldsymbol{h}\|) \end{align} which implies that the map $\boldsymbol{x}\mapsto \boldsymbol{x}^{-1}$ on $G_A$ is continuous.

Suppose $\boldsymbol{x}\in G_A$. For any $\boldsymbol{y}\in A$, we have \begin{align} \boldsymbol{y}=\boldsymbol{x}-(\boldsymbol{x}-\boldsymbol{y})=\boldsymbol{x}\big(\boldsymbol{e}-\boldsymbol{x}^{-1}(\boldsymbol{x}-\boldsymbol{y})\big). \end{align} If $\|\boldsymbol{x}-\boldsymbol{y}\|<\frac{1}{\|\boldsymbol{x}^{-1}\|}$, then $\|\boldsymbol{x}^{-1}(\boldsymbol{x}-\boldsymbol{y})\|\leq \|\boldsymbol{x}^{-1}\|\|\boldsymbol{x}-\boldsymbol{y}\|<1$ and so, $\big(\boldsymbol{e}-\boldsymbol{x}^{-1}(\boldsymbol{x}-\boldsymbol{y})\big)^{-1}$ exists in $A$. Hence $\boldsymbol{y}$ is invertible and
$\boldsymbol{y}^{-1}=\big(\boldsymbol{e}-\boldsymbol{x}^{-1}(\boldsymbol{x}-\boldsymbol{y})\big)^{-1}\boldsymbol{x}^{-1}$. This shows that the open ball $B(\boldsymbol{x};1/\|\boldsymbol{x}^{-1}\|)$ is fully contained in $G_A$.

Let $f$ be the map $\boldsymbol{x}\mapsto \boldsymbol{x}^{-1}$ on $G_A$. For any $\boldsymbol{h}\in A$ such that $\|\boldsymbol{h}\|<1/\|\boldsymbol{x}^{-1}\|$ we have that \begin{align*} (\boldsymbol{x}+\boldsymbol{h})^{-1}&=(\boldsymbol{e}+\boldsymbol{x}^{-1}\boldsymbol{h})^{-1}\boldsymbol{x}^{-1}=\sum_{n\geq0}(-1)^n(\boldsymbol{x}^{-1}\boldsymbol{h})^n\boldsymbol{x}^{-1}\\ &=\boldsymbol{x}^{-1}-\boldsymbol{x}^{-1}\boldsymbol{h}\boldsymbol{x}^{-1}+\boldsymbol{x}^{-1}(\boldsymbol{h}\boldsymbol{x}^{-1})^2\sum_{n\geq0}(-1)^n(\boldsymbol{h}\boldsymbol{x}^{-1})^n. \end{align*} Hence \begin{align*} \frac{\|(\boldsymbol{x}+\boldsymbol{h})^{-1}-\boldsymbol{x}^{-1} + \boldsymbol{x}^{-1}\boldsymbol{h}\boldsymbol{x}^{-1}\|}{\|\boldsymbol{h}\|}&\leq \frac{\|\boldsymbol{x}^{-1}\|^3\|\boldsymbol{h}\|}{1-\|\boldsymbol{x}^{-1}\|\|\boldsymbol{h}\|}\xrightarrow{\boldsymbol{h}\rightarrow\boldsymbol{0}}0 \end{align*}

Mittens
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