# Derivatives of quantile loss function [duplicate]

I'm reading a text - Roger Koenker (2005) Quantile Regression [page 8] - that goes like this:

Consider the function $$R(\xi) = \sum_{i=1}^n \rho_\tau(y_i-\xi)$$ where $$\rho_\tau(y_i-\xi) =(y_i-\xi)(\tau - I[y_i-\xi<0])$$ If $$\hat \xi$$ solves $$\min_\xi R(\xi)$$ then the objective function must be increasing as one moves away from $$\hat \xi$$. This requires that the left and right derivatives are both non-negative at $$\hat \xi$$. Thus

$$R'(\xi+) := \lim_{h \rightarrow 0} \frac{R(\xi + h) - R(\xi)}{h} = \sum_i( I[y_i-\xi<0]- \tau )$$ with $$R'(\hat \xi+) \geq 0$$ and

$$R'(\xi-) := \lim_{h \rightarrow 0} \frac{R(\xi - h) - R(\xi)}{h} = \sum_i (\tau - I[y_i-\xi<0] )$$

with $$R'(\hat \xi- )\geq 0$$.

My question is: Shouldn't $$R'(\xi+) = \sum_i( I[y_i-\xi\leq 0]- \tau )$$ with a weak inequality in the indicator function rather than the strict?

Because for $$y_i - \xi = 0$$ the limit

$$\lim_{h \rightarrow 0^+} \frac{(y_i-\xi-h)(\tau - I[y_i-\xi-h<0]) - (y_i-\xi)(\tau - I[y_i-\xi<0])}{h} = \\[8pt] \lim_{h \rightarrow 0^+} \frac{-h(\tau - I[-h<0]) }{h} = -(\tau - \lim_{h \rightarrow 0^+} I[-h < 0]) = 1-\tau$$ so the i'th summand in $$R'(\xi+)$$ should be $$I[y_i-\xi\leq 0] - \tau$$.