A map $f\colon X\to Y$ between metric spaces is uniformly open whenever for each $\varepsilon >0$ there is $\delta >0$ such that for any $x\in X$ one has $$B_Y\big(f(x),\delta\big)\subseteq f\big(B_X(x,\varepsilon)\big), $$ where $B_X, B_Y$ denote open balls in the respective spaces.

Uniformly open maps have the property that they are surjective as long as the codomain is connected. The exponential function is an example of an open map that is not uniformly open.

A continuous, surjective bilnear map $A\colon X\times X\to Y$ between Banach spaces need not be open, contrary to the linear case. (Already matrix multiplication is an example; another example is multiplication in $C[0,1]$ in the case of real scalars; this is due to Fremlin).

Is there an example of a continuous, bilinear and surjective map between Banach spaces that is open but not uniformly open?

I have a suspicion that it could be already convolution in $\ell_1(\mathbb Z)$ but I am not sure how to approach this (in the latter case I know that it is not uniformly open as $\ell_1(\mathbb{Z}/n\mathbb{Z})$ do not have equi-uniformly open convolutions).

Edit (27.06.2017). This question was also asked in a paper by Balcerzak, Behrends, and Strobin (Banach J. Math. Anal. 10 (2016), no. 3, 482-494).

I would also welcome examples of $n$-linear maps with this property, where $n$ is arbitrary.

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    $\begingroup$ Note that due to the homogeneity of $f$, it is equivalent to just ask: there is $\delta_1 >0$ such that for any $x\in X$ one has $$B_Y\big(f(x),\delta_1\big)\subseteq f\big(B_X(x,1)\big), $$ because then for any $\varepsilon$ one can take $\delta:=\delta_1\varepsilon^2 $. $\endgroup$ Jun 3 '17 at 13:31
  • $\begingroup$ @TomekKania It looks like the Banach algebra of functions of bounded variation is a counterexample. Are you (or anybody else) still interested? $\endgroup$
    – fedja
    May 27 '18 at 1:31
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    $\begingroup$ @TomekKania That was the hardest part of my argument. I proved that for every $f,g\in BV$, you can find arbitrarily close $F,G$ such that $fg=FG$ but $|F^2|+|G|^2\ge\delta(f,g)>0$. If that is how they did it, the only (trivial) thing I have to add is that $0f=0$, the available perturbations are $\psi(f+\varphi)$, $\|\varphi\|,\|\psi|\le 1$ and if $f$ goes over an $1/N$ net in the disk of radius $2$ on $N$ disjoint intervals, then $f+\varphi$ goes within $2/N$ from $0$ on one of those intervals, so the perturbation is below $2/N$ somewhere and cannot be identically $4/N$, say. $\endgroup$
    – fedja
    May 27 '18 at 11:37
  • $\begingroup$ @TomekKania On the other hand, if their argument is drastically different, then it still may make some sense for me to post mine. $\endgroup$
    – fedja
    May 27 '18 at 11:39
  • $\begingroup$ @TomekKania Done. $\endgroup$
    – fedja
    May 27 '18 at 14:16

Recently it was proved that multiplication in the space of functions of bounded variation is open but not uniformly so it is a counterexample to the above question. This also reflects the above comments, which predicted that.

Stanisław Kowalczyk, Małgorzata Turowska, Multiplication in the space of functions of bounded variation, Journal of Mathematical Analysis and Applications 472 (April 2019), 696–704.


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