Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Assume that $f$ is a measurable function on the interval $[0,1]$ such that $0<f(x)<\infty$ for $x \in [0,1]$. Then, how can I prove the inequality below? $$\int^1_0 f(x)dx \int^1_0 {1 \over {f(x)}} dx \ge 1$$

share|improve this question
3  
I think your title should be more telling: something like "Integral inequality for measurable function on $[0,1]$". –  Quinn Culver Jun 12 '12 at 17:54

4 Answers 4

up vote 4 down vote accepted

Here is another way. What you have written is nothing but the Arithmetic mean-Harmonic mean inequality for functions. In case, of a finite set of positive numbers $\{a_i\}$'s, the AM-HM inequality reads $$\dfrac{\sum_{i=1}^{n} w_i a_i}{\sum_{i=1}^{n} w_i} \geq \dfrac{\sum_{i=1}^{n} w_i}{\sum_{i=1}^{n} \dfrac{w_i}{a_i}}$$where $w_i \geq 0$ are the weights. The weights can be normalized to $1$ i.e. if we enforce that $\displaystyle \sum_{i=1}^{n} w_i = 1$, then we get that $$\sum_{i=1}^{n} w_i a_i \geq \dfrac1{\displaystyle \sum_{i=1}^{n} \dfrac{w_i}{a_i}}$$ Hence, we get that $$\left(\sum_{i=1}^{n} w_i a_i \right) \left(\sum_{i=1}^{n} \dfrac{w_i}{a_i} \right)\geq 1$$

If you partition the interval $[0,1]$ into $n$ disjoint measurable sets, say $E_1^{(n)},E_2^{(n)},\ldots,E_n^{(n)}$, such that $\displaystyle \bigcup_{k=1}^{n} E_k^{(n)} = [0,1]$, choose $w_i^{(n)} = \mu^{(n)} \left(E_i^{(n)}\right)$. Note that $\sum_i w_i^{(n)} = 1$.

Approximate $f(x)$ from below using step functions on these intervals i.e. $f_{step}^{(n)} = f_i^{(n)}$ on the interval $E_i^{(n)}$ such that $\displaystyle \sum_{i=1}^{n} f_i^{(n)} \mu(E_i^{(n)}) = \int f_{step}^{(n)} \to \int f$.

Since $f>0$ on $[0,1]$, $\dfrac1f$ can be approximated from above using the step function $(1/f)_{step}^{(n)} = 1/f_i^{(n)}$ on the interval $E_i^{(n)}$ such that $\displaystyle \sum_{i=1}^{n} \dfrac1{f_i^{(n)}} \mu(E_i^{(n)}) = \int \dfrac1{f_{step}^{(n)}} \to \int \dfrac1{f}$.

From the AM-HM inequality we have that $\left(\displaystyle \sum_{i=1}^{n} \mu(E_i^{(n)}) f_i^{(n)} \right) \left( \displaystyle \sum_{i=1}^{n} \dfrac{\mu(E_i^{(n)}) }{f_i^{(n)}} \right) \geq 1$.

Take the limit to conclude that $$\left(\int f dx \right) \times \left(\int \dfrac1f dx \right)\geq 1$$

You may need to argue out and justify some parts of the above argument to make it into a rigorous proof but hope the idea is clear.

share|improve this answer
    
It's actually a very rigorous but very nice example for my learning. I'll argue that one with my mates. –  Collins Jun 12 '12 at 17:26
    
@Collins Thanks. –  user17762 Jun 12 '12 at 17:27

We have by Cauchy-Schwarz inequality, $$1=\int_0^1 1dx=\int_0^1\sqrt{f(x)}\frac 1{\sqrt{f(x)}}dx\leq \left(\int_0^1f(x)dx\cdot\int_0^1\frac 1{f(x)}dx\right)^{1/2}.$$ (the result is clear if $f^{1/2}$ or $f^{-1/2}$ is not integrable.

share|improve this answer
    
It's very simple! But I didn't know that theorm yet, Is there any condition for using Cauchy inequality? –  Collins Jun 12 '12 at 16:50
1  
@Collins As long as you know that $L^2$ is an inner product space. You've probably seen a proof of Cauchy-Schwarz before, at least in a special case, and the proof for a general inner product is no harder. Maybe you've shown the Hölder inequality? That's even more general. –  Dylan Moreland Jun 12 '12 at 16:54
    
Oh, I got it. I did not have enough understanding about Hölder inequality and norm space. –  Collins Jun 12 '12 at 17:09
4  
If you want to be very picky, you should also note that technically Cauchy-Schwartz only applies if $f^{1/2}, f^{-1/2} \in L^2(0,1)$, i.e., $f, f^{-1} \in L^1(0,1)$. There should be a special case in which $\int_0^1 \! f(x) \, dx = \infty$ or $\int_0^1 \! f(x) \, dx = \infty$ in which case the inequality is obviously true since both integrals are positive anyway. –  user12014 Jun 12 '12 at 17:38

$$\iint_{[0,1]^2}\frac{(f(x)-f(y))^2}{f(x)f(y)}\mathrm dx\mathrm dy\geqslant0$$ Edit: This is an adaptation of the well-known approach to the inequality $\|f\|_1\leqslant\|f\|_2$ for every probability measure $\mu$ as the expansion of $$ \iint(f(x)-f(y))^2\mathrm d\mu(x)\mathrm d\mu(y)\geqslant0$$ and gives an opportunity to recommend once more the marvelous little book The Cauchy-Schwarz Master Class: An Introduction to the Art of Inequalities by J. Michael Steele.

share|improve this answer
    
That's a nice solution. How did you come up with it? –  Quinn Culver Jun 13 '12 at 3:51
    
@Quinn See Edit. –  Did Jun 13 '12 at 6:51

As @PZZ commented below @David Giraudo's answer, we may assume $f$ is integrable. Then Jensen's inequality, with $\varphi(x)=1/x$ says that $$ \int_{0}^{1} \frac{1}{f(x)}\, dx = \int_{0}^{1}\varphi(f(x))\,dx \geq \varphi\left[\int_{0}^{1} f(x)\, dx\right] = \left[\int_{0}^{1}f(x)\,dx\right]^{-1}. $$

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.