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The text book says that the following inequality "follows quite easily" but I don't see why.

Given $\alpha \lt 1 \lt \beta$ and $\psi : \mathbb{R}_{\ge 0} \rightarrow \mathbb{R}_{\ge 0}$ increasing, show that that:

$$\frac{1}{(1-\alpha)x}\int_{\alpha x}^x\psi(u)du \le \psi(x) \le \frac{1}{(\beta-1)x}\int_x^{\beta x} \psi(u)du$$

I think it may be assumed that $\alpha$ is non-negative.

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3 Answers 3

up vote 3 down vote accepted

You have $\psi(u) \leq \psi(x)$ for $u \in [\alpha x,x]$. Integrating with respect to $u$ gives $\int_{\alpha x}^x \psi(u) du \leq \psi(x)(x - \alpha x) = \psi(x) (1-\alpha) x$, or equivalently, $\frac{1}{1-\alpha} \int_{\alpha x}^x \psi(u) du \leq \psi(x)$. The other side of the inequality follows the same line of reasoning.

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Oh of course. It didn't click that the denominator of the fraction is the length of the interval being integrated over! –  Mark Feb 5 '13 at 3:05

Hint: Sice it is assumed that $\psi$ is a continous function on $[\alpha,\beta]$, so using The First Mean Value Theorem is useful.

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+1 for the helpful hint! –  amWhy Feb 6 '13 at 0:08

Let $\Phi$ be the primitive of $\phi$. We have $$\frac{\Phi(x)-\Phi(\alpha x)}{x-\alpha x}\le \phi(x)$$ Because this is the average of $\Phi' = \phi$ over $(\alpha x,x)$ which is less than $\phi (x)$ by monotonicity. The same thing applies for $\beta$ in the opposite direction.

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