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Ekeland's inverse function theorem gives weak conditions under which a function $f:E\rightarrow F$ between two graded Fréchet-spaces is locally surjective. The theorem requires the codomain $F$ to be standard in the following sense:

Definition. A graded Fréchet-space $(F,\Vert \cdot \Vert_k: k\in \mathbb{N})$ is called standard, if for every $x\in F$ there is an approximating sequence $F\ni x_n\rightarrow x$ and constants $C(x)>0$ and $c(n)>0$ such that $$ \Vert x_n \Vert_k \le C(x) \cdot \Vert x\Vert_k\quad \text{ and} \quad \Vert x_n\Vert_k\le c(n)^k,\qquad (k,n\in\mathbb{N}). $$

Ekeland's main example for a standard Fréchet-space is $C^\infty(\bar \Omega)$ (for a smooth, bounded domain $\Omega\subset \mathbb{R}^d$, graded by either $C^k(\bar \Omega)$ or $H^k(\Omega)$ semi-norms). Here $x_n=S_nx$ for an appropriate smoothing operator $S_n$. While this idea allows to prove standardness of some other function spaces, I wonder whether one can sometimes get away easier:

Question. Suppose $F$ is standard and $F_0\subset F$ is a closed linear subspace. Is $F_0$ also standard?

The obvious problem is that for $x\in F_0$ we only get an approximating sequence in $(x_n)\subset F$, which might not lie in $F_0$.

In some cases life is easier:

  • In the special case that $F_0$ is the image of a projection operator $P:F\rightarrow F_0\subset F$ such that for all $y\in F$ we have $\Vert P y\Vert_k\lesssim_{k} \Vert y \Vert_k$, we can simply put $\tilde x_n = P x_n \in F_0$ to get an approximation from within $F_0$. However, even in Banach spaces we do not always have such projections.
  • If the domain $E$ has the Heine-Borel property, one get's away with non-standard $F$. This was proved here.
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  • $\begingroup$ This does, of course, not answer your question, but my feeling is that a positive answer would be much too good to be true. $\endgroup$
    – Jochen Wengenroth
    Commented Apr 2, 2021 at 11:42
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    $\begingroup$ By the way, the article of Ekeland you refer to has a very bad start: The very first claim that, if the topology of a Fréchet space is given by a sequence of norms (not only semi-norms) then it is an intersection of Banach spaces, is wrong. This is known since a long time and is quite relevant for a problem of Grothendieck about the bounded approximation property. Perhaps the simplest example is in an article of Dietmar Vogt from 2010 in the Proc. Amer. Math. Soc. Hopefully, this is the only mistake in Ekeland's article. $\endgroup$
    – Jochen Wengenroth
    Commented Apr 3, 2021 at 7:41
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    $\begingroup$ You're right. The maps $i_k$ need not be injective as Vogt's example (and others) show. The relation to the bounded approximation property goes back to Pelczynski. This is probably explained in Vogt's article (or in the references cited by Vogt.) $\endgroup$
    – Jochen Wengenroth
    Commented Apr 4, 2021 at 18:12
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    $\begingroup$ Aha, thanks! Just for the record: A Fréchet spaces called countably normed if its topology can be generated by a sequence of norms and if additionally the canonical inclusion maps $i_k$ are injective (p.168 in Dubinsky's The structure of nuclear Fréchet-spaces). Then Vogt's paper, together with reference [3], states: If $E$ admits a continuous norm, then: BAP $\Leftrightarrow$ $E$ countably normed. So the point is: Ekeland's paper only considers countably normed Fréchet-spaces and Vogt's paper gives an example of a Fréchet-space which fails to satisfy this property. $\endgroup$
    – Jan Bohr
    Commented Apr 5, 2021 at 8:35
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    $\begingroup$ The question is a bit inadequate for the following reason: The conditions (3) and especially (5) in Ekeland's results seem to be in such a way restrictive (hence not at all "weak") for the maps to be considered that it is questionable whether these results of Ekeland posses any nontrivial applications. In particular, they do not generalize Nash−Moser. The joint paper with Séré from about 2015 (see arXiv) seems more promising but there the spaces are more restricted sinse they need to allow smoothing operators. $\endgroup$
    – TaQ
    Commented Apr 6, 2021 at 10:20

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The principle for getting examples of nonapplicability of Ekeland's theorem is described by the following simple

Example. With $\mathbb I=[0,1]$ consider the Frechet space $E=F=C^\infty(\mathbb I)$ with the norms $\|x\|_k=\sup\{|{\rm D}^ix\,s|:s\in\mathbb I\text{ and }i\le k\,\}$. Taking $k_0=0$ consider the second order polynomian map $x\mapsto x+x^2$ defined for $\|x\|_0<\frac 14$. Then for $Lxv=\frac v{1+2x}$ Ekeland's condition (5) in his Theorem 3 on page 97 in the Ann. Inst. Henri Poincaré paper [AN 28 (2011) 91–105] requires that $\|\frac v{1+2x}\|_k=\|Lxv\|_k\le m'_k\|v\|_{k+d_2}$ for all $k\in\mathbb N$ and all the appropriate $x$ and all $v$ in $E$. Taking $v:s\mapsto 1$ and $x:s\mapsto\frac 18\sin(2n\pi s)$ and $k=1$ we get $\frac 12n\pi=|{\rm D}(\frac 1{1+2x})0|\le m'_1\|v\|_{1+d_2}=m'_1$ which does not hold if we take $\frac 2\pi m'_1<m'_1\le n$.

The same idea with only more complicated computations can be applied to more general spaces and maps.

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  • $\begingroup$ Sorry, I don't get it. First you considered a fixed polynomial $x(s)=s+s^2$ (since you wrote $\|x\|_0=1/4$,it might rather be $s-s^2$?) and I first thought that the map $L:E\to E$ would be be $v\mapsto v/(1+2x)$, but then you plug in other functions $x$. What is the map $L:E\to E$, then? Do you claim that Nash-Moser applies to $L$ but Ekeland does not? $\endgroup$
    – Jochen Wengenroth
    Commented Sep 21, 2022 at 6:47
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    $\begingroup$ No, I consider the (polynomial) map $f:E\supseteq{\rm dom}\,f\to F=E$ given by $x\mapsto x+x^2$ for which I have not given the specific name "$f$". Here ${\rm dom}\,f$ is given by $\|x\|_0<\frac 14$ and $L:{\rm dom}\,f\to\mathcal L\,(F,E)$ is the family of (right) inverses for the derivative of $f$ occurring in Ekeland's theorem. It it rather "well-known" that this kind of maps (and even much more general ones) satisfy the Nash−Moser conditions. Does this clarify the state of matters? $\endgroup$
    – TaQ
    Commented Sep 21, 2022 at 16:10
  • $\begingroup$ Thank you for this clarification. $\endgroup$
    – Jochen Wengenroth
    Commented Sep 21, 2022 at 17:44

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