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?

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    $\begingroup$ Perhaps it would be more profitable to ask why the nonsurjective operators form a closed set. $\endgroup$ – Cheerful Parsnip Jan 11 '11 at 14:07
  • $\begingroup$ @Jim: No. The solution is a corollary of the open mapping theorem and can be for instance found in Lang's book on functional analysis $\endgroup$ – OrbiculaR Jan 11 '11 at 16:59
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    $\begingroup$ If you know where to find the solution what's your motivation for asking? $\endgroup$ – Rasmus Jan 11 '11 at 17:21
  • $\begingroup$ Well, I found the reference afterwards... Sorry! $\endgroup$ – OrbiculaR Jan 11 '11 at 20:06
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    $\begingroup$ If 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). $\endgroup$ – Asaf Karagila Jan 11 '11 at 20:36

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.

  • $\begingroup$ Among the many Banach-Schauder theorems the one I have in mind is the following: Let $p: E \to F$ have norm at most one. If $p(B_{\leq 1}E)$ is dense in $B_{\leq 1}F$ then $p$ is onto and the map $E/\text{Ker}\,p \to F$ is isometric. $\endgroup$ – t.b. Jan 12 '11 at 0:33
  • $\begingroup$ This is essentially the proof given by Lang. The proof reminds me of the proof that the space of right- or left-invertible operators is open (usually the proof uses a series, but one can also write down a sequential versions via partial sums - yielding the proof you just gave) $\endgroup$ – OrbiculaR Jan 12 '11 at 7:21
  • $\begingroup$ @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. $\endgroup$ – t.b. Jan 12 '11 at 10:33

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 $$\|B^* \phi\| \ge \|A^* \phi\| - \|A^* - B^*\| \|\phi\| \ge (c - \|A-B\|) \|\phi\|$$

  • $\begingroup$ That'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. $\endgroup$ – t.b. Apr 6 '12 at 23:09

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