Si function on Gibbs phenomenon

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I'm reading Gibbs phenomenon but I'm confused about the statement below.

Since the Dirichlet sum of indicator functions $\chi_{[0,1]}$ is $S_{R} \chi_{[0,1]} (x) = \frac{1}{2}(Si(2R\pi(1-x)) + Si(2R\pi x))$ with $Si(x) = 2/\pi\int_{0}^x \frac{\sin(t)}{t} dt$ then for a decreasing sequence ${x_n}$, we have $\limsup_{n \rightarrow \infty} S_n\chi_{[0,1]} (x_n) \leq \frac{1}{2}(1+Si(\pi)).$

I know the properties of the $Si$ function that $0 \leq Si(x) \leq Si(\pi)$ for all $x>0$ but it can't imply the less equal sign.

Thank you!