# Upper bound on an integral

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.

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

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

• 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. Jun 26, 2023 at 12:51
• @MartinR Yes, indeed, thanks ! Sometimes I like to complicate my thinking... Jun 26, 2023 at 12:53
• Smart trick! well done Jun 26, 2023 at 12:57
• @MartinR is $1+f$ shorthand for $1+f(x)$? This is the first time I'm experiencing notation like this. Jun 27, 2023 at 0:32
• @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. Jun 27, 2023 at 8:45