Some context: I am going through some literature on empirical risk minimization for bipartite ranking [1] that shows how certain "low-noise" conditions lead to fast rates of convergence of the excess risk to $0$ of the empirical risk minimizer. The low noise condition at hand is:


There exists constants $c>0$ and $\alpha \in [0, 1]$ such that for all $x \in \mathcal{X}$:


$$\mathbb{E}\_X \frac{1}{\left| \eta(x) - \eta(X) \right|^\alpha} \leq c$$


where $\eta(x) = \mathbb{P}(Y = 1 | X=x)$ where $Y$ is supported on $\{0, 1\}$


The discussion following the statement of this condition in the paper mentions that $\alpha = 0$ is the least restrictive and $\alpha=1$ is the most restrictive. It is clear that $\alpha = 0$ imposes no constraint. They also mention that in the case where $X \sim uniform([0,1])$ then $\eta$ must be nondifferentiable to satisfy the condition. Why is this the case?


**What I've tried**


The problem seems equivalent to:


Show that if there exists some constant $c>0$ such that the following condition holds for a non-negative $f : [0,1] \rightarrow \mathbb{R}$ with positive and finite integral then $f$ cannot be differentiable:


$$\int\_0^1 \frac{1}{\left| f(x) - f(z) \right |}dz \leq c$$


This looks like an irregularity condition to me, but I'm having trouble getting a clearer idea on how to show the claim. It's clear that a constant function does not satisfy this condition since in that case, the integrand is nowhere finite. I can think of examples that do not satisfy this condition, for example: $f(x) = x$ does not satisfy this condition. However I cannot find an example that does satisfy this condition.


Any intuition would also be helpful.


[1] Clémençon, Stéphan; Lugosi, Gábor; Vayatis, Nicolas. Ranking and Empirical Minimization of U -statistics. Ann. Statist. 36 (2008), no. 2, 844--874. doi:10.1214/009052607000000910. <https://projecteuclid.org/euclid.aos/1205420521>



Fast rates in ERM: Extreme case of low-noise assumption implies non-differentiability
This can be shown using Jensen's inequality. For all fixed $x$:


$$\mathbb{E}\_X \frac{1}{\left| \eta(x) - \eta(X) \right|}
\\= \mathbb{E}\_X \left [ \frac{1}{ \eta(x) - \eta(X) } \mid \eta(x) > \eta(X) \right]\mathbb P(\eta(x) > \eta(X)) 
\\+ \mathbb{E}\_X \left [ \frac{1}{ \eta(X) - \eta(x) } \mid \eta(x) \leq \eta(X) \right]\mathbb P(\eta(x) \leq \eta(X)) 
\\\geq \frac{1}{ \eta(x) - a } \mathbb P(\eta(x) > \eta(X)) 
+ \frac{1}{ b - \eta(x) }\mathbb P(\eta(x) \leq \eta(X)) $$


Where $a, b$ are the corresponding conditional expectations after applying jensen's inequality. If $\eta$ is continuous then by the intermediate value theorem, there is some $y$ such that $\eta(y)=a$ so that the above expression is unbounded. So $\eta$ cannot be continuous.


