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).



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> Is there an example of a continuous, bilinear and surjective map between Banach spaces that is *open* but *not uniformly open*?
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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**](http://dx.doi.org/10.1215/17358787-3599741) (2016), no. 3, 482-494).



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> I would also welcome examples of $n$-linear maps with this property, where $n$ is arbitrary.
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Open bilinear maps that are not uniformly open
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.



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> Stanisław Kowalczyk, Małgorzata Turowska, [Multiplication in the space of functions of bounded variation](https://www.sciencedirect.com/science/article/pii/S0022247X18309910), *Journal of Mathematical Analysis and Applications* **472** (April 2019), 696–704.
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