Today I thought of a question. Which normed vector spaces (over $\mathbb{R}$) have the following property? For all $\epsilon >0,$ there exist nonzero $\delta \in V$ and continuous $f_{\delta }$ mapping the closed unit ball of $V$ into the open unit ball of $V$ satisfying
(1) $ \parallel \delta \parallel < \epsilon $
(2) $f_{\delta}(x) - f_{\delta}(y) = f_{\delta}(x-y) -\delta $ (where defined)
(3) $\parallel f_{\delta }(u) \parallel = 1 - \epsilon $ for all $\parallel u \parallel = 1.$
(4) $\parallel f_{\delta }(x) - x\parallel < \epsilon $ (where defined)
The motivation here comes from the one-dimensional case. I had practical occasion to construct a map $f:[0,1] \rightarrow (0,1)$ satisying
(1) $f(x) - f(y) = f(x-y) - \delta $ (where defined)
(2) $f(1) = 1 - \delta $
Intuitively, $f$ shrinks $[0,1]$ in such a way that is which slightly perturbs the domain, is sensitive to differences, and removes boundary points. Using the same methods as typically applied to the Cauchy Functional Equation, we can show that $f(x) = (1- 2\delta ) x + \delta $ is the unique continuous solution. In practice, $\delta $ should be small (much smaller than $\frac{1}{2}$) so that $f$ is strictly increasing. The same approach can be modified to the case when $f$'s domain is any closed interval $[a,b],$ giving $f_{\delta } (x) = = (b- a - 2\delta ) x + a + \delta .$ In the case of $a=-1, b=1,$ we have $f_{\delta } (x) = 2(1 -\delta) x + \delta - 1.$
Even though I used this approach to solve a data-related problem, I was still curious as to how it might generalize. I decided that normed spaces would be the most appropriate setting given the goals (not being familiar with any notion of size for topological groups.) The original problem seems so simple that I can't help but think I've somehow reinvented the wheel. I couldn't think of a good name for such $f$ ("contraction mapping" was already taken.) Let me know if any of this makes no sense.