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Good morning. I have been thinking about the following question for a while without much success, therefore I'm starting to doubt its validity, although I don't have a clear counterexample in mind.

Let $X$ be a Banach space and $F=(F_1,\dotso, F_m):X\to \mathbb R^m$ be a smooth map such that $F(0)=0$. Assume that $F$ is open at zero, i.e. for every $\epsilon>0$ there exists $\delta=\delta(\epsilon)$ such that $B(0,\delta)\subset F(B_X(0,\epsilon))$.

We consider vectors $\lambda\in S^{m-1}$ and the compositions $\Phi_\lambda:X\to \mathbb R$ defined by $\Phi_\lambda(x):= \langle\lambda, F(x)\rangle=\sum_{i=1}^m \lambda_i F_i(x)$.

If $F$ is open at $0$, $\Phi_\lambda$ is open at zero for every $\lambda\in S^{m-1}$.

I'm now interested in the converse statement, specifically I want to pose these two questions:

  1. Is it true that, under the assumption that $\Phi_\lambda:X\to \mathbb R$ is open at zero for every $\lambda\in S^{m-1}$, then also $F$ is open at zero?
  2. In case point 1. is false, can we strengthen somewhat the hypotheses so that it becomes true, or is it unavoidably false?

Thanks in advance,

Gil

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    $\begingroup$ According to MathOverflow guidelines, there should be only one question per post. Also, your second question is too open and non-specific. So, I suggest it be deleted. $\endgroup$ Commented Sep 12, 2023 at 14:08
  • $\begingroup$ Please feel free to delete the second question if deemed not compliant to the rules. In my opinion though it is so related to the first one that opening a new post would be kind of duplicating the same question. As for the question being too generic, I would be interested in any reference pointing to a result of the type: If a map $F$ as above is open at zero along any scalar projection in $\mathbb R^{m}$ plus some other hypotheses of some kind then $F$ open at zero. I really need a starting point here, and googling has not been of any help unfortunately. $\endgroup$ Commented Sep 13, 2023 at 13:14

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$\newcommand\R{\mathbb R}\newcommand\C{\mathbb C}$The answer to Question 1 is no. Here is a counterexample: $X=\C$, $m=2$, $\R^2$ is identified with $\C$, and $$F(z):=\exp\Big(-\frac1{|z|\,(2\pi-\arg z)\arg z}+i\arg z\Big)$$ for $z\in\C\setminus\{0\}$ with $\arg z\ne0$, with $F(z):=0$ otherwise, and with the values of $\arg z$ assumed to be in the interval $[0,2\pi)$.

Indeed, the function $F$ is smooth. The intersection of the image $F(D)$ under $F$ of any disk $D$ centered at $0$ with any non-horizontal line through $0$ contains an open interval containing $0$, and hence the projection of $F(D)$ on any such line through $0$ contains an open interval containing $0$. It is also easy to that the projection of $F(D)$ on the horizontal line through $0$ contains an open interval containing $0$. However, $F(D)$ does not contain any open set containing $0$.

For an illustration, here is the image $F(D_{1/2})$ under $F$ of the disk $D_{1/2}$ of radius $1/2$ centered at $0$:

enter image description here

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  • $\begingroup$ Thanks for the reply! May I ask you a clarification on the intersection of $F(D)$ with the real axis? Indeed it seems to me that $F(z)$ is real iff $\mathrm{arg}(z)\in {0,\pi}$. If $\mathrm{arg}(z)=0$ then $F(z)=0$ by definition, while if $\mathrm{arg}(z)=\pi$ then $F(z)\to 0$, with $F(z)<0$, as $|z|\to 0$. Even from your picture it is not clear to me how to see that that the projection of $F(D)$ on the horizontal line through $0$ contains an open interval containing $0$. Could you please elaborate a bit on this? $\endgroup$ Commented Sep 13, 2023 at 13:09
  • $\begingroup$ @GilSanders : In particular, the projection of $F(D)$ on the horizontal line through $0$ contains the projection on the horizontal line through $0$ of the intersection (say $I$) of $F(D)$ with an arbitrary line through $0$ that is neither horizontal nor vertical, and the latter projection contains an open interval containing $0$ (because the intersection $I$ contains an open interval containing $0$). $\endgroup$ Commented Sep 13, 2023 at 17:58
  • $\begingroup$ @GilSanders : Also, the picture suggests that the projection of $F(D_{1/2})$ on the horizontal line through $0$ contains the open interval $(-0.8,0.4)$ on the real axis. $\endgroup$ Commented Sep 13, 2023 at 18:03
  • $\begingroup$ Thanks! I was confusing the projection of the image along a line with the intersection of the image with the line. IT's clear now. $\endgroup$ Commented Sep 14, 2023 at 7:39

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