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Schauder's Lemma in functional analysis states the following:

Let $E$ and $F$ be metrizable locally convex topological vector spaces, and let $E$ be Fréchet. Then if the linear continuous map $A:E\to F$ is nearly open, that is to say, for any neighborhood of the origin $U\subset E$ we have that $\emptyset\ne\operatorname{Int}(\overline{A(U)})\subset F$, then $A$ is surjective and open.

However, I've been unable to find a reference for a proof of this, and I haven't managed to figure out why it is true either. Either a proof or a reference to a proof would be much appreciated. The most I've been able to show is that $\overline{A(E)}=F$.

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  • $\begingroup$ If the space $E$ is Banach, then there is a simple proof, which exploits the notion of an $\infty$-convex set, see matstud.org.ua/texts/1999/11_1/11_1_083-084.pdf $\endgroup$
    – Taras Banakh
    Commented Jul 17, 2017 at 19:31
  • $\begingroup$ I don't see a simple way to adapt that argument though. $\endgroup$
    – Dominic Wynter
    Commented Jul 17, 2017 at 20:34
  • $\begingroup$ If $E$ and $F$ are Banach, then its unit ball $U$ is $\infty$-convex, and so is its image $A(U)$. By the assumption, the closure of $A(U)$ in $F$ contain some open ball $V$ of $F$. Then $V\cap A(U)$ is dense $\infty$-convex set in $V$ and hence coincides with $V$ by the Static Property proved in that paper. Therefore, $A(U)$ contains the ball $V$ and $A(U-U)$ contains the ball $V-V$ centered at zero, which means that the operator $A$ is open. This gives the proof of the "Banach" case. I do not know if this argument can be adapted to the "Frechet" case. $\endgroup$
    – Taras Banakh
    Commented Jul 18, 2017 at 1:50

3 Answers 3

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The result has not much to do with the linear structure. In the book Introduction to Functional Analysis of Meise and Vogt you find a version for metric spaces (Lemma 3.9):

Let $X$ and $Y$ be metric spaces; $X$ be complete. Let $f:X\to Y$ be continuous and assume that for every $\varepsilon>0$ there exists a $\delta>0$, such that for all $x\in X$, $\overline{f(U_\varepsilon(x))} \supset U_\delta(f(x))$. Then the map $f$ is open.

As far as I remember this can also be found in Bourbaki.

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It follows from

I. M. Dektjarev, A closed graph theorem for ultracomplete spaces, Dokl. Akad. Nauk. Sov., 154 (1964), 771–773 (in Russian).

See also discussion on p. 5 here.

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Below you can find some references. They do not give the result directly in the form you want but instead give more general results where $E$ is assumed to be a Pták space (= B-complete = fully complete) which then gives the result you want by the implication Fréchet $\Rightarrow$ Pták.

[1] J. Horváth: Topological Vector Spaces and Distributions, Theorem 3.17.2, p. 296.

[2] H. Jarchow: Locally Convex Spaces, Theorem 9.7.1, p. 186.

[3] H.H. Schaefer: Topological Vector Spaces, Theorem 8.3, p. 163.

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