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Let $ f $ be an increasing function on $ [0,1] $ such that $ 0\leq f(x)\leq 1 $ and $ \int_0^1(f(x)-x)dx=0 $. Show that \begin{align} \int_0^1|f(x)-x|dx\leq 1/4. \end{align}

Here is my try. I find that the equality is satisfied for $ f\equiv 1/2 $. Moreover, I want to use the Poincaré inequality but I cannot bound $ \int_0^1|f'(x)-1|dx $ by $ 1/4 $. Can you give me some references or hints?

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    $\begingroup$ Is $f$ continuous ? Moreover, why should $f$ be derivable ? $\endgroup$
    – Surb
    Commented Apr 26, 2022 at 12:11
  • $\begingroup$ Is the upper bound $1/4$ correct? I just wonder because here is the same question (with answers), only with $1/2$ instead of $1/4$: math.stackexchange.com/q/3188689/42969. $\endgroup$
    – Martin R
    Commented Apr 26, 2022 at 12:14
  • $\begingroup$ @ Surb, As $ f $ is increasing, $ f' $ exists a.e.. Though it maybe inacurate, it is only my attempt, I assume that the smoothness of $ f $ and want to get some observation about it. $\endgroup$ Commented Apr 26, 2022 at 12:15
  • $\begingroup$ @MartinR, I do not find counterexample and in the question you post, I do not find the condition for the equality. $\endgroup$ Commented Apr 26, 2022 at 12:21
  • $\begingroup$ The upper bound $1/4$ is also claimed here, but without proof: math.stackexchange.com/q/3190661/42969. $\endgroup$
    – Martin R
    Commented Apr 26, 2022 at 12:35

1 Answer 1

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Step 1. (Proof under extra assumptions) Assume that

  • $f : [0, 1] \to [0, 1]$ is continuous and non-decreasing;
  • $\int_{0}^{1} (f(x) - x) \, \mathrm{d}x = 0$;
  • $f(0) = 0$, and $f(1) = 1$.

Then the set $U_+ = \{ x \in [0, 1] : f(x) > x \}$ is open, hence it is written as the union of at most countably many disjoint open intervals $(a_i, b_i)$, $i = 1, 2, \ldots $ Also, the continuity of $f$ forces that $f(b_i) = b_i$. So,

\begin{align*} I_+ &:= \int_{U_+} |f(x) - x| \, \mathrm{d}x = \sum_{i} \int_{(a_i, b_i)} (f(x) - x) \, \mathrm{d}x \\ &\leq \sum_{i} \int_{(a_i, b_i)} (b_i - x) \, \mathrm{d}x = \sum_{i} \frac{(b_i - a_i)^2}{2} \\ &\leq \frac{1}{2}\biggl( \sum_{i} (b_i - a_i) \biggr)^2 = \frac{1}{2} \operatorname{Leb}(U_+)^2. \end{align*}

A similar argument shows that, for $U_- = \{ x \in [0, 1] : f(x) < x \}$ we have

\begin{align*} I_- := \int_{U_-} |f(x) - x| \, \mathrm{d}x \leq \frac{1}{2} \operatorname{Leb}(U_-)^2. \end{align*}

Moreover, from $\int_{0}^{1} (f(x) - x) \, \mathrm{d}x = 0$ we get $I_+ = I_-$. Therefore, together with the observations$ \int_{0}^{1} |f(x) - x| \, \mathrm{d}x = 2I_+ = 2I_-$ and $\operatorname{Leb}(U_+) + \operatorname{Leb}(U_-) \leq 1$, we conclude that

$$ \int_{0}^{1} |f(x) - x| \, \mathrm{d}x \leq \min\{ \operatorname{Leb}(U_+), \operatorname{Leb}(U_-) \}^2 \leq \frac{1}{4}. $$

These inequalities have a nice interpretation in terms of areas:

figure

Step 2. (General case by approximation) Now suppose $f : [0, 1] \to [0, 1]$ is non-decreasing and satisfies $\int_{0}^{1} (f(x) - x) \, \mathrm{d}x = 0$. Then it is not hard to find a sequence $f_n(x)$ satisfying the conditions in Step 1 and $f_n(x) \to f(x)$ for almost every $x$. So, by the dominated convergence theorem,

$$ \int_{0}^{1} |f(x) - x| \, \mathrm{d}x = \lim_{n\to\infty} \int_{0}^{1} |f_n(x) - x| \, \mathrm{d}x \leq \frac{1}{4}. $$

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    $\begingroup$ @MartinR, I revamped my answer. I am not fully satisfied with this as I was looking for a more elegant solution, but at least the current version faithfully reflects our geometric intuition about this problem. $\endgroup$ Commented Apr 26, 2022 at 20:44
  • $\begingroup$ This is really nice, and the image makes it clear why the inequality holds! – I wonder if $\int_{U_+} ( f(x) - x) \, dx \le \frac 12 \lambda(U_+)^2$ can somehow be proven “directly,” without resorting to continuous functions and sums first. $\endgroup$
    – Martin R
    Commented Apr 27, 2022 at 1:36
  • $\begingroup$ Is $f(0) = 0$, $f(1) = 1$ really needed in the first part of your proof? $\endgroup$
    – Martin R
    Commented Apr 27, 2022 at 1:43
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    $\begingroup$ @MartinR, Not quite, but they make both $U_{\pm}$ open sets of $\mathbb{R}$. $\endgroup$ Commented Apr 27, 2022 at 1:56

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