Suppose $E$ and $F$ are given Banach spaces. Let $A$ be a continuous surjective map. Why is there a small ball around $A$ in the operator topology, such that all elements in this ball are surjective?
Why is the space of surjective operators open?
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2If you have a solution, for sake of completeness I think it'd be wise if you post an answer, so the next time someone comes to ask this question there can be a reference to direct them to (or they could find it on their own). – 2011-01-11
2 Answers
Since this question bugged me, I decided to write down the proof (I don't have access to Lang's book, so I hope my argument is not much more complicated than necessary). The idea is the same as in the proof of the Banach-Schauder theorem.
By the open mapping theorem we may scale the norm on $E$ in such a way that $A$ maps the unit ball of $E$ onto the unit ball of $F$, that is $B_{\leq 1} F \subset A(B_{\leq 1}E)$. Since $A$ is linear we have $B_{\leq r} F \subset A(B_{\leq r}F)$ for all $r > 0$.
Claim. If $B: E \to F$ is such that $\alpha := \|A - B\| < 1$ then $B$ is onto.
Proof. Let $f \in F$. We want to show that there is $e$ such that $f = Be$. For convenience, we put $f_{0} = f$ and assume $\|f_{0}\| \leq 1$.
Choose $e_{0}$ with $\|e_{0}\| \leq 1$ such that $Ae_{0} = f_{0}$. Define $f_{1} = f_{0} - Be_{0}$ and observe $\|f_{1}\| = \|(A - B) e_{0}\| \leq \alpha$, so we may choose $e_{1}$ with $\|e_{1}\| \leq \alpha$ such that $Ae_{1} = f_{1}$. Now $f_{2} = f_{1} - Be_1$ has norm $\|f_{2}\| = \|(A - B)e_{1}\| \leq \alpha^{2}$, so we obtain by induction two sequences $\{f_{n}\}_{n=0}^{\infty}$ and $\{e_{n}\}_{n=0}^{\infty}$ having the following properties:
- $\|e_{n}\|, \|f_{n}\| \leq \alpha^{n}$ for all $n$.
- $e_{n}$ is such that $f_{n} = A(e_{n})$,
- $f_{n+1} = f_{n} - B(e_{n}) = (A-B)(e_{n})$.
Finally $e = \sum_{n = 0}^{\infty} e_{n}$ has norm $\|e\| \leq \sum_{n=0}^{\infty} \alpha^{n} = \frac{1}{1-\alpha}$ and, moreover, \[ B(e) = \sum_{n=0}^{\infty} B(e_{n}) = \sum_{n=0}^{\infty} (f_{n} - f_{n+1}) = f_{0} = f, \] as we wanted to show.
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0@OrbiculaR: Thanks for this info. Yes, there is an entire family of results whose proof follows this `telescopic' pattern, but I'm unable to put my finger on exactly what their statements have in common (up to the fact that it always has to do with special classes of surjective maps). It definitely is a nice argument and probably worth remembering. – 2011-01-12
A bounded linear operator $A: E \to F$, where $E$ and $F$ are Banach spaces, is surjective if and only if there is $c > 0$ such that for all $\phi \in F^*$, $\|A^* \phi\| \ge c \|\phi\|$ (see e.g. Rudin, "Functional Analysis", Theorem 4.15). Now if $A$ is such an operator, so is $B$ for $\|A-B\| < c$, since $\|B^* \phi\| \ge \|A^* \phi\| - \|A^* - B^*\| \|\phi\| \ge (c - \|A-B\|) \|\phi\|$
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0That's very nice. I had a feeling that there must be a slicker way than the proof I gave... On the other hand, it uses much less machinery than yours. – 2012-04-06