If $A:X\rightarrow Y$ is an invertible bounded linear operator between Banach spaces $X$ and $Y$, I want to show that there is an $\epsilon>0$ such that whenever $|B|<\epsilon$ , $A+B$ is invertible. Any hints on how to start this? What theorems do I need here? (this is not homework, just practice)
There exists an $\epsilon>0$ such that $A+B$ is invertible whenever $|B|<\epsilon$
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functional-analysis
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0Step 1: Write $A + B = A(I + A^{-1}B)$. Step 2: Think about $(I + C)^{-1}$. – 2017-01-11
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0I think I see what your saying. I can expand $1/(1+A^{-1}B)$ using geometric series? – 2017-01-11
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0That depends. Under what conditions can you guarantee the convergence? – 2017-01-11
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0I need for $|B|<|A|$ right? So $\epsilon=|A|$? – 2017-01-11
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0No, that will usually be too large. Consider $A$ given by $\begin{pmatrix} 1 & 0 \\ 0 & \delta\end{pmatrix}$ with small $\delta > 0$. – 2017-01-11
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0I don't think you need to consider expansions. I did it by straight computation. It's more or less an adaptation of the general result that the set of linear operators is open. – 2017-01-12
1 Answers
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Use $|\cdot | \ $for the norms on $X$ and $Y$, and $\left \| \cdot \right \|$ for the operator norm.
Now, choose $\alpha>0$ such that $\alpha \left \| A^{-1} \right \|=1$, and suppose $B:X\to Y$ satisfies $\left \| B \right \| \left \| A^{-1} \right \|<1$.
Then, $\left \| B \right \| <\alpha$ and for $0\neq x\in X$, we compute:
$\alpha \left | x \right |=\alpha \left | Ix \right |=\alpha \left | A^{-1}Ax \right |\le \alpha \left \| A^{-1} \right \|\cdot \left | Ax \right |\le \left | Ax \right |=\left | (B+A)x-Bx \right |\le \left | (B+A)x \right |+\left | Bx \right |\le \left | (B+A)x \right |+\left \| B \right \|\cdot \left | x \right |\Rightarrow \left | (B+A)x \right |\ge (\alpha-\left \| B \right \|)\left | x \right |>0.$
The foregoing calculation shows that, as soon as $ \left \| B \right \|<1/\left \| A^{-1} \right \|,\ $ if $x\neq 0$ then neither is $(B+A)x.\ $i.e.$\ B+A$ is injective, hence invertible.