L
Lv 7
L 發問時間： 科學數學 · 1 0 年前

# 高微證明題 (其實也可算微積分的東西) ~ 20點

1) 證明 ∫_[a,b] |sin(nx)| dx -> (2/π)(b-a) as n -> oo

2) 若 f 在 [a,b] 上黎曼可積, 證明

∫_[a,b] f(x)|sin(nx)| dx -> (2/π)∫_[a,b] f(x) dx as n -> oo

(好像要用到實變的M.C.T.)

### 1 個解答

• Eric
Lv 6
1 0 年前
最佳解答

1)Proof. This is trivial if a = b, so assume a < b. For each n ∈ N, let K(n) = ⌊(b - a)/(π/n)⌋ (⌊x⌋ is the greatest integer smaller than x); then clearlyhn = (b - a)/(π/n) - K(n) has 0 ≤ hn < 1. Consequently,K(n)/n = (b - a)/π - hn/n → (b - a)/π.For each n, ∫ab |sin(nx)| dx = ∑k=1K(n) ∫a+(k-1)(π/n)a+k(π/n) |sin(nx)| dx + ∫a+K(n)(π/n)b |sin(nx)| dx= ∑k=1K(n) ∫0π/n sin(nx) dx + ∫a+K(n)(π/n)b |sin(nx)| dx= K(n) [-(1/n)cos(nx)]0π/n dx + ∫a+K(n)(π/n)b |sin(nx)| dx= 2K(n)/n + ∫a+K(n)(π/n)b |sin(nx)| dx,which converges to (2/π)(b-a) since|∫a+K(n)(π/n)b |sin(nx)| dx| ≤ |∫a+K(n)(π/n)b dx|= b - a - K(n)(π/n)→ 0. ∎2) 此法可行。Proof.  The result is trivial if a = b, so assume a < b.Simple functions. By (1), the proposition is true for indicator functions on subintervals of [a,b], and hence true for all simple functions.Nonnegative functions. Say f ≥ 0 is integrable, and let ε > 0. We can find a simple function φ with ∫|f - g| < ε. Then|∫ab f(x)(|sin(nx)| - 2/π) dx| ≤ |∫ab (f(x) - φ(x))(|sin(nx)| - (2/π))| + |∫ab φ(x)(|sin(nx)| - 2/π) dx| ≤ (1+2/π)|∫ab (f(x) - φ(x)) dx| + |∫ab φ(x)(|sin(nx)| - 2/π) dx| < (1+2/π)ε + |∫ab φ(x)(|sin(nx)| - 2/π) dx|.In light of the result for simple functions, we havelim supn |∫ab f(x)(|sin(nx)| - 2/π) dx| ≤ (1+2/π)ε.Since ε was arbitrary, it follows thatlim supn |∫ab f(x)(|sin(nx)| - 2/π) dx| = 0.The result follows.Integrable functions.Say f is integrable, and decompose f by sign to get f = f+ - f-, where f+ ≥ 0 and f- ≥ 0. The result holds for f+ and f-, and by linearity of the integral, it also holds for f. ∎

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