Ivan Shishkin, Rye (1878)

Problems/Functional analysisExerciseUnreviewed

Ma première application du théorème de convergence monotone

by SalixBabylonica·
40
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Dans l’espace mesuré ([0,1],B([0,1]),λ)([0,1],\mathcal B([0,1]),\lambda), calculer
lim⁡n→+∞∫0111+xn dx.\lim_{n\to+\infty} \int_0^1 \frac{1}{1+x^n}\,dx.

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Solution by SalixBabylonica

Discussions0 useful votes

On pose fn(x)=11+xn.f_{n}(x)=\frac{1}{1+x^{n}}.
Alors fnf_{n} est continue sur [0,1][0,1] donc mesurable sur cet intervalle.
Soit x∈[0,1],x \in [0,1],

  • fn(x)≥0f_n(x)\ge0,
  • Comme xn+1≤xnx^{n+1}\le x^n, alors fn+1(x)≥fn(x)f_{n+1}(x)\ge f_{n}(x).

On calcule également la limite de fnf_{n} :
lim⁡n→+∞fn(x)={1si x∈[0,1[,12si x=1.=1 p.p sur [0,1].\lim_{n \to +\infty}f_n(x)=\begin{cases} 1 & \text{si } x\in[0,1[,\\ \dfrac12 & \text{si } x=1. \end{cases} = 1 \text{ p.p sur } [0,1].On peut donc appliquer le théorème de convergence monotone :
lim⁡n→+∞∫[0,1]fn(x)dx=∫[0,1]lim⁡n→+∞fn(x)dx=∫[0,1]1dx=1.\lim_{n\to + \infty} \int_{[0,1]} f_{n}(x) dx = \int_{[0,1]} \lim_{n \to + \infty}f_{n}(x) dx = \int_{[0,1]} 1 dx = 1.

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