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Assume $f(x)>0, x\in[0,1]$. Prove that $$\iint_{\left[ 0,1 \right] \times \left[ 0,1 \right]}{\frac{f\left( x \right)}{f\left( y \right)}\text{d}x\text{d}y\ge 1} $$

It gives me the sense that the integration of $\dfrac{f(x)}{f(y)}$ will be small in one part of $[0,1]\times [0,1]$, but large in another part of $[0,1]\times [0,1]$ and finally their sum will be more than $1$. But how to formulate this sense formally? I guess we need to make some inequality scalings. Can anyone help?

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  • $\begingroup$ is $f$ continuous ? $\endgroup$
    – Surb
    Dec 15, 2020 at 12:36
  • $\begingroup$ yeah $f$ is continuous $\endgroup$
    – FFjet
    Dec 15, 2020 at 12:41
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    $\begingroup$ 1. For $A>0$, $A + \frac{1}{A} \ge 2$; 2. $\iint \frac{f(x)}{f(y)} dx dy = \iint \frac{f(y)}{f(x)} dx dy$. $\endgroup$
    – Fnacool
    Dec 15, 2020 at 12:48

1 Answer 1

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Suppose $\int_{0}^{1} f$ and $\int_{0}^{1} \frac{1}{f}$ exists and let $I$ be the given integral,

Observe that the given integral can be written as $$ I = \left(\int_{0}^{1} f(x) \, dx \right)\left(\int_0^1\frac{1}{f(x)} \, dx\right) $$

Then by Cauchy Schwartz inequality for integrals we have $$ I \geq \left(\int_0^1 1\, dx\right)^2 = 1 $$

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