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Let $ f : [0,1] \to \mathbb{R} $ be a function satisfying: 1.) $ |f(x)| \leqslant a $ for some $ a < 1 $, and 2.) $ \int_0^1 f(x) dx = 0 $. I would like to know whether the following inequality holds: $$ \int_0^1 \frac{1}{1+f(x)} dx \stackrel{?}{\leqslant} \frac{1}{1-a^2} . $$ I've confirmed it numerically in some special cases, e.g., $ f(x) = a \cos(2\pi n x) $ for various $ a, n $. It is also easy to see that it holds (with equality) when $ f(x) = \pm a $, each on one half of the domain.

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  • $\begingroup$ There is a problem since $a^2 < a$ $\endgroup$ Jun 26, 2023 at 12:03
  • $\begingroup$ did you add the property $\int f = 0$? $\endgroup$ Jun 26, 2023 at 12:18

1 Answer 1

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Since $|f(x)| \leq a$, then one has $$f(x)^2 \leq a^2, \quad \quad \text{so }\quad\quad 1-f(x)^2 \geq 1-a^2$$

Dividing by $(1+f(x))(1+a)$, one gets $$\dfrac{1-f(x)}{1+a} \geq \dfrac{1-a}{1+f(x)}$$

Now integrate between $0$ and $1$ : using the assumption $\displaystyle \int_0^1 f(x) dx = 0$, you get $$(1-a)\int_0^1 \dfrac{dx}{1+f(x)} \leq \dfrac{1}{1+a}\int_0^1 (1-f(x)) dx = \dfrac{1}{1+a}$$

so finally $$\boxed{\int_0^1 \dfrac{dx}{1+f(x)} \leq \dfrac{1}{1-a^2}}$$

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    $\begingroup$ Very nice! A minor remark: If you divide by $(1+f)(1-a^2)$ then you get $\frac{1}{1+f} \le \frac{1-f}{1-a^2}$ and the result follows even faster. $\endgroup$
    – Martin R
    Jun 26, 2023 at 12:51
  • $\begingroup$ @MartinR Yes, indeed, thanks ! Sometimes I like to complicate my thinking... $\endgroup$ Jun 26, 2023 at 12:53
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    $\begingroup$ Smart trick! well done $\endgroup$ Jun 26, 2023 at 12:57
  • $\begingroup$ @MartinR is $1+f$ shorthand for $1+f(x)$? This is the first time I'm experiencing notation like this. $\endgroup$ Jun 27, 2023 at 0:32
  • $\begingroup$ @stickynotememo: If you interpret $1$ as the constant function $x \mapsto 1$ then $1+f$ is the function $x \mapsto 1 + f(x)$. But here I was just sloppy to save time and space. $\endgroup$
    – Martin R
    Jun 27, 2023 at 8:45

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