English
The smoothing function ε is strictly decreasing on (1, ∞). Consequently, the map x ↦ (1 + ε(x)) is strictly decreasing on (1, ∞).
Русский
Функция сглаживания ε строго убывает на (1, ∞). Следовательно, отображение x ↦ 1 + ε(x) строго убывает на (1, ∞).
LaTeX
$$$$\text{StrictAntiOn}(\varepsilon, (1,\infty))\quad\text{and}\quad \text{StrictAntiOn}(1+\varepsilon(x), (1,\infty)).$$$$
Lean4
theorem strictAntiOn_smoothingFn : StrictAntiOn ε (Set.Ioi 1) :=
by
change StrictAntiOn (fun x => 1 / log x) (Set.Ioi 1)
simp_rw [one_div]
refine StrictAntiOn.comp_strictMonoOn inv_strictAntiOn ?log fun _ hx => log_pos hx
refine StrictMonoOn.mono strictMonoOn_log (fun x hx => ?_)
exact Set.Ioi_subset_Ioi zero_le_one hx