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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.

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In the one-dimensional case, do you require $x\ge y$ in (1)? Otherwise $f(x-y)$ is not defined. So the first order of business is either to figure out what should take place of $x\ge y$ in a vector space, or to generalize the one-dimensional construction so that $x\ge y$ is not required. (BTW equation numbers are great, but not as much when two equations have the same number.) –  user31373 Aug 6 '12 at 22:40
    
@LeonidKovalev In general, I would like the conditions to hold whenever $f(x-y)$ is defined. In the one-dimensional case, $f$ has a unique continuous solution on either signed half ray, both of which can be pasted together for a function with a simple discontinuity at zero. I couldn't discern whether or not it would be useful to consider $f$ as a function on all of $V$ in the general case. –  user17794 Aug 6 '12 at 22:53
    
But note that you ask for continuity on the closed unit ball of a normed space. In the one-dimensional case the closed unit ball is the interval $[-1,1]$, on which you don't have continuity. –  user31373 Aug 6 '12 at 23:05
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I still don't understand the one-dimensional situation. Your example for $[a,b]=[-1,1]$ does not appear to work: it has $f_{\delta}(0)=\delta-1$, which is inconsistent with $0=f_\delta(x)-f_{\delta}(x)=f_\delta(0)-\delta$. –  user31373 Aug 13 '12 at 1:22
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Condition (2) implies that $f_\delta(0)=\delta$, and that $g(u)=f_\delta(u)-\delta$ is a linear map: i.e. $g(u+v)=g(u)+g(v)$ and $g(au)=ag(a)$ for any real $a$. From this it follows that $V$ cannot be finite dimensional. –  Einar Rødland Aug 15 '12 at 18:06

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