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4 changes: 2 additions & 2 deletions Analysis/Section_10_3.lean
Original file line number Diff line number Diff line change
Expand Up @@ -28,12 +28,12 @@ theorem derivative_of_antitone (X:Set ℝ) {x₀:ℝ} (hx₀: ClusterPt x₀ (.p
sorry

/-- Proposition 10.3.3 / Exercise 10.3.4 -/
theorem strictMono_of_positive_derivative {a b:ℝ} (hab: a < b) {f:ℝ → ℝ}
theorem strictMono_of_positive_derivative {a b:ℝ} {f:ℝ → ℝ}
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vacuously true even if a < b is not true.

(hderiv: DifferentiableOn ℝ f (.Icc a b)) (hpos: ∀ x ∈ Set.Ioo a b, derivWithin f (.Icc a b) x > 0) :
StrictMonoOn f (.Icc a b) := by
sorry

theorem strictAnti_of_negative_derivative {a b:ℝ} (hab: a < b) {f:ℝ → ℝ}
theorem strictAnti_of_negative_derivative {a b:ℝ} {f:ℝ → ℝ}
(hderiv: DifferentiableOn ℝ f (.Icc a b)) (hneg: ∀ x ∈ Set.Ioo a b, derivWithin f (.Icc a b) x < 0) :
StrictAntiOn f (.Icc a b) := by
sorry
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17 changes: 8 additions & 9 deletions Analysis/Section_10_4.lean
Original file line number Diff line number Diff line change
Expand Up @@ -57,7 +57,6 @@ theorem inverse_function_theorem {X Y: Set ℝ} {f: ℝ → ℝ} {g:ℝ → ℝ}
(hfXY: ∀ x ∈ X, f x ∈ Y) (hgYX: ∀ y ∈ Y, g y ∈ X)
(hgf: ∀ x ∈ X, g (f x) = x) (hfg: ∀ y ∈ Y, f (g y) = y)
{x₀ y₀ f'x₀: ℝ} (hx₀: x₀ ∈ X) (hfx₀: f x₀ = y₀) (hne : f'x₀ ≠ 0)
(hcluster: ClusterPt x₀ (.principal (X \ {x₀})))
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This assumption is not needed. Full proof without it - https://github.com/rkirov/analysis/blob/main/Analysis/Section_10_4.lean#L81-L113

(hf: HasDerivWithinAt f f'x₀ X x₀) (hg: ContinuousWithinAt g Y y₀) :
HasDerivWithinAt g (1/f'x₀) Y y₀ := by
-- This proof is written to follow the structure of the original text.
Expand All @@ -74,27 +73,27 @@ theorem inverse_function_theorem {X Y: Set ℝ} {f: ℝ → ℝ} {g:ℝ → ℝ}
convert (hf _ hx _).inv₀ _ using 2 with n <;> grind

/-- Exercise 10.4.1(a) -/
example {n:ℕ} (hn: n > 0) : ContinuousOn (fun x:ℝ ↦ x^(1/n:ℝ)) (.Ici 0) := by sorry
example {n:ℕ} : ContinuousOn (fun x:ℝ ↦ x^(1/n:ℝ)) (.Ioi 0) := by sorry
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n = 0 is true due to garbage value of 1 / n but I can see an argument to keeping it even if not used in proof.

The Ici to Ioi change is just fixing a typo.


/-- Exercise 10.4.1(b) -/
example {n:ℕ} (hn: n > 0) {x:ℝ} (hx: x ∈ Set.Ici 0) : HasDerivWithinAt (fun x:ℝ ↦ x^(1/n:ℝ))
((n:ℝ)⁻¹ * x^((n:ℝ)⁻¹-1)) (.Ici 0) x := by sorry
example {n:ℕ} {x:ℝ} (hx: x ∈ Set.Ioi 0) : HasDerivWithinAt (fun x:ℝ ↦ x^(1/n:ℝ))
((n:ℝ)⁻¹ * x^((n:ℝ)⁻¹-1)) (.Ioi 0) x := by sorry

/-- Exercise 10.4.2(a) -/
example (q:ℚ) {x:ℝ} (hx: x ∈ Set.Ici 0) :
HasDerivWithinAt (fun x:ℝ ↦ x^(q:ℝ)) (q * x^(q-1:ℝ)) (.Ici 0) x := by
example (q:ℚ) {x:ℝ} (hx: x ∈ Set.Ioi 0) :
HasDerivWithinAt (fun x:ℝ ↦ x^(q:ℝ)) (q * x^(q-1:ℝ)) (.Ioi 0) x := by
sorry

/-- Exercise 10.4.2(b) -/
example (q:ℚ) : (nhdsWithin 1 (.Ici 0 \ {1})).Tendsto (fun x:ℝ ↦ (x^(q:ℝ)-1)/(x-1)) (nhds q) := by
example (q:ℚ) : (nhdsWithin 1 (.Ioi 0 \ {1})).Tendsto (fun x:ℝ ↦ (x^(q:ℝ)-1)/(x-1)) (nhds q) := by
sorry

/-- Exercise 10.4.3(a) -/
example (α:ℝ) : (nhdsWithin 1 (.Ici 0 \ {1})).Tendsto (fun x:ℝ ↦ (x^α-1^α)/(x-1)) (nhds α) := by
example (α:ℝ) : (nhdsWithin 1 (.Ioi 0 \ {1})).Tendsto (fun x:ℝ ↦ (x^α-1^α)/(x-1)) (nhds α) := by
sorry

/-- Exercise 10.4.3(b) -/
example (α:ℝ) {x:ℝ} (hx: x ∈ Set.Ici 0) : HasDerivWithinAt (fun x:ℝ ↦ x^α) (α * x^(α-1)) (.Ici 0) x := by
example (α:ℝ) {x:ℝ} (hx: x ∈ Set.Ioi 0) : HasDerivWithinAt (fun x:ℝ ↦ x^α) (α * x^(α-1)) (.Ioi 0) x := by
sorry

end Chapter10
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