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Let $f: [0,1] \to \mathbb{R}$ be a Lebesgue measureable function and let $1<c<\infty$ , so $$c^{-1}\le f(x)\le c$$ for every $x \in [0,1]$ .

Let: $$\alpha=\int_0^1\frac{1}{f(x)}dx$$ $$\beta=\bigg(\int_0^1f(x)dx\bigg)^{-1}$$

Which is greater $\alpha$ or $\beta$ ?

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After setting some specific simple functions, it seems that $\alpha$ is greater.

So I first tried to prove it for some simple function $f=\sum_{j=1}^N\alpha_j\chi_{E_j}$ , where $\{E_j\}$ are pairwise disjoint, $[0,1]=\bigcup_{j=1}^{N} E_j$ and $c^{-1}\le \alpha_j \le c$ for every $j$.

Also $\frac{1}{f}=\sum_{j=1}^N\frac{1}{\alpha_j}\chi_{E_j}$, and $c^{-1}\le \frac{1}{\alpha_j} \le c$ for every $j$.

However, from this point any inequation I tried led me to the obvious result which is $c^{-1}\le \alpha, \beta \le c$ .

Thanks!

Z17Math
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1 Answers1

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Hint: as $f(x) \ge c^{-1}>0$, just apply the Cauchy–Schwarz_inequality:

$$\frac{\alpha}{\beta} = \bigg(\int_0^1f(x)dx\bigg)\bigg(\int_0^1\frac{1}{f(x)}dx\bigg) \ge \left(\int_0^1\sqrt{f(x)}\cdot \sqrt{\frac{1}{f(x)}}dx \right)^2 = 1$$

NN2
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