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If $|f(x)-f(y)|\le(x-y)^2$ for all $x,y\in\mathbb R$, then it's easy to show that $f'=0$ everywhere, and the mean value theorem implies that that means $f$ is constant. If there were a gap in the real line and $f=3$ on one side of the gap and $f=4$ on the other side of the gap, then $f'=0$ everywhere but $f$ is not constant. Gaplessness enters via the mean value theorem.

But I wonder if the proposition holds even in a line with gaps. For example if $f:\mathbb Q\to\mathbb Q$ and for all $x,y\in\mathbb Q$ this inequality holds, does that imply $f$ is constant? And what about other gap-filled lines than $\mathbb Q$? Such as $\mathbb R\setminus A$ where $A$ is a finite set, or a countable set, or a set whose complement is dense, or whatever set it might be of interest to ask this question about?

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up vote 6 down vote accepted

You only need the set where $f$ is defined to be dense. The theorem is true in metric spaces, and in that generality, density is more than enough, all you really need is the ability to move from point to point in arbitrarily short steps.

Hopping from $x$ to $y$ in $n$ short steps of size $O(1/n)$ bounds $|f(x)-f(y)|$ as $n O(1/n^2)$, which goes to zero for large $n$.

For the general case in a metric space, define the connected components for "hopping" by $x \sim y$ if for every $e > 0$ there is a finite chain of points starting at $x$ and ending at $y$, with consecutive points all at distance $ < e$ (and a bound, independent of $e$, on the total length of the consecutive jumps). Then

  • a function satisying the inequality is constant on hopping components

  • any two hopping components are at positive finite distance from each other (a "gap"), if the metric space is compact

  • the function can have different values for different components.

The exact amount of freedom to define $f$ differently on the components is an interesting and potentially more difficult question when there are infinitely many components or, in the noncompact case, pairs of components that come arbitrarily close to each other.

The only property of the upper bound $(x-y)^2$ that is used here is that it vanishes to order higher than $1$ for small $|x-y|$.

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Conversely, if the domain of $f$ contains a gap of length $\epsilon$ then one can define $f = 0$ on one side of the gap and $f = \epsilon^2$ on the other. –  Erick Wong Mar 19 '13 at 17:21
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Right. And that is true in any metric space, too. You can define "hopping components" and the function only has to be constant on those. (@ErickWong ) –  zyx Mar 19 '13 at 17:27
    
By hindsight I'm surprised I didn't see this. Haste makes waste. If I'd seen this, maybe I'd have posted both the question and the answer. –  Michael Hardy Mar 20 '13 at 3:34
    
Michael, the exact analysis for the metric space case seems nontrivial. The situation is similar to the 19th century theory point sets on the real line where there is the set, the derived set of limit points, the derived set of that, all transfinitely continued. It is conceivable that there is an example (even in a separable space) with infinitely many hopping components but due to iterated consolidation of the set under the "components are at distance zero" relation, the function has to attain the same value at all points from all components. It would make for a good question of its own. –  zyx Mar 20 '13 at 3:40
    
In the other direction, given finite-distance moats isolating $n$ components, there is a function with at least $n$ independent values (ie, in some open set of that dimension). But this is greatly understating the true dimensionality of the space of solutions when the domain of the function is the union of origin with the reciprocals of all natural numbers. There the solution space contains an open subset of the convergent sequences indexed by the points. So it is potentially an interesting problem from various angles. –  zyx Mar 20 '13 at 3:54
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Let $x = x_0 < x_1 < \dots < x_n = y$ be a family of sequences such that $|x_i - x_{i+1}| \rightarrow 0$ Then use triangle inequality and you can make $|f(x) - f(y)|$ arbitrarily small.

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