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Fall 2009 - Problem 3

decomposition of scales, measure theory

Let f ⁣:[0,1]→R\func{f}{\br{0,1}}{\R} be continuous with

min⁡0≤x≤1f(x)=0.\min_{0 \leq x \leq 1} f\p{x} = 0.

Assume that for all 0≤a<b≤10 \leq a < b \leq 1 we have

∫abf(x)−min⁡a≤y≤bf(y) dx≤12∣b−a∣.\int_a^b f\p{x} - \min_{a \leq y \leq b} f\p{y} \,\diff{x} \leq \frac{1}{2} \abs{b - a}.

  1. Prove that for all λ≥0\lambda \geq 0:

    ∣{x∣f(x)>λ+1}∣≤12∣{x∣f(x)>λ}∣\abs{\set{x \mid f\p{x} > \lambda + 1}} \leq \frac{1}{2} \abs{\set{x \mid f\p{x} > \lambda}}

    Here ∣S∣\abs{S} denotes the Lebesgue measure of the set SS.

  2. Prove that for all 1≤c<21 \leq c < 2,

    ∫01cf(x) dx≤1002−c.\int_0^1 c^{f\p{x}} \,\diff{x} \leq \frac{100}{2 - c}.
Solution.

  1. Set Eλ={x∣f(x)>λ}E_\lambda = \set{x \mid f\p{x} > \lambda}. By continuity of ff, Eλ=f−1((λ,∞))E_\lambda = f^{-1}\p{\p{\lambda, \infty}} is open, so we can write

    Eλ=⋃n=1∞(an,bn),E_\lambda = \bigcup_{n=1}^\infty \p{a_n, b_n},

    where the (an,bn)\p{a_n, b_n} are pairwise disjoint. On [an,bn]\br{a_n, b_n}, ff attains its minimum by continuity and satisfies

    λ≤min⁡x∈Eλf(x)≤min⁡x∈[an,bn]f(x),\lambda \leq \min_{x \in E_\lambda} f\p{x} \leq \min_{x \in \br{a_n, b_n}} f\p{x},

    so we have for all n≥1n \geq 1 that

    ∫anbnf(x)−λ dx≤12∣bn−an∣.\int_{a_n}^{b_n} f\p{x} - \lambda \,\diff{x} \leq \frac{1}{2}\abs{b_n - a_n}.

    By countable additivity of the Lebesgue measure, adding these up gives

    ∫Eλf(x)−λ dx≤12∣Eλ∣.\int_{E_\lambda} f\p{x} - \lambda \,\diff{x} \leq \frac{1}{2}\abs{E_\lambda}.

    Notice that

    ∫Eλf(x)−λ dx=∫Eλ+1f(x)−λ dx+∫{x∣λ<f(x)≤λ+1}f(x)−λ dx≥∫Eλ+1(λ+1)−λ dx+∫x∣λ<f(x)≤λ+1λ−λ dx=∣Eλ+1∣.\begin{aligned} \int_{E_\lambda} f\p{x} - \lambda \,\diff{x} &= \int_{E_{\lambda+1}} f\p{x} - \lambda \,\diff{x} + \int_{\set{x \mid \lambda < f\p{x} \leq \lambda + 1}} f\p{x} - \lambda \,\diff{x} \\ &\geq \int_{E_{\lambda+1}} \p{\lambda + 1} - \lambda \,\diff{x} + \int_{x \mid \lambda < f\p{x} \leq \lambda + 1} \lambda - \lambda \,\diff{x} \\ &= \abs{E_{\lambda+1}}. \end{aligned}

    Thus,

    ∣Eλ+1∣≤12∣Eλ∣,\abs{E_{\lambda+1}} \leq \frac{1}{2} \abs{E_\lambda},

    which was what we wanted to show.

  2. Notice that because f(x)≥0f\p{x} \geq 0 and bounded,

    [0,1]={x∣f(x)=0}∪⋃n=0∞(En∖En+1).\br{0, 1} = \set{x \mid f\p{x} = 0} \cup \bigcup_{n=0}^\infty \p{E_n \setminus E_{n+1}}.

    From the previous part, we have for all n≥0n \geq 0 that

    ∣En+1∣≤12∣En∣  ⟹  ∣En+1∣≤12n+1∣E0∣≤12n+1.\abs{E_{n+1}} \leq \frac{1}{2} \abs{E_n} \implies \abs{E_{n+1}} \leq \frac{1}{2^{n+1}} \abs{E_0} \leq \frac{1}{2^{n+1}}.

    So,

    ∫01cf(x) dx≤∫{x∣f(x)=0}cf(x) dx+∑n=0∞∫En∖En+1cf(x) dx≤∫{x∣f(x)=0}c0 dx+∑n=0∞∫En∖En+1cn+1 dx≤∫[0,1] dx+∑n=0∞∫Encn+1 dx=1+∑n=0∞cn+1∣En∣≤1+∑n=0∞cn+12n=1+c1−c2=2+c2−c≤1002−c.\begin{aligned} \int_0^1 c^{f\p{x}} \,\diff{x} &\leq \int_{\set{x \mid f\p{x} = 0}} c^{f\p{x}} \,\diff{x} + \sum_{n=0}^\infty \int_{E_n \setminus E_{n+1}} c^{f\p{x}} \,\diff{x} \\ &\leq \int_{\set{x \mid f\p{x} = 0}} c^0 \,\diff{x} + \sum_{n=0}^\infty \int_{E_n \setminus E_{n+1}} c^{n+1} \,\diff{x} \\ &\leq \int_{\br{0,1}} \,\diff{x} + \sum_{n=0}^\infty \int_{E_n} c^{n+1} \,\diff{x} \\ &= 1 + \sum_{n=0}^\infty c^{n+1} \abs{E_n} \\ &\leq 1 + \sum_{n=0}^\infty \frac{c^{n+1}}{2^n} \\ &= 1 + \frac{c}{1 - \frac{c}{2}} \\ &= \frac{2 + c}{2 - c} \\ &\leq \frac{100}{2 - c}. \end{aligned}