# $f$ is monotonically increasing, $0 \le f \le 1$ and $\int_0^1 (f(x) - x) dx = 0$ then $\int_0^1|f(x)-x|dx \le \frac{1}{2}$.

$$f(x)$$ is monotonically increasing in $$[0,1]$$, $$0 \le f \le 1$$ and $$\int_0^1 (f(x) - x) \mathrm{d}x = 0$$. Prove that $$\int_0^1|f(x)-x|\mathrm{d}x \le \frac{1}{2}$$.

It's easy if $$f(x) \ge x$$ in $$[0,1]$$. And even in $$[a,b]$$ we have $$\int_a^b |f(x)-x|\mathrm{d}x \le \frac{(b-a)^2}{2}$$. But the zero points of $$f(x) - x$$ may be infinitely many. This is where difficulty exists.

• Look at the region in $[0,1]^2$ between the graphs of $f$ and $y=x$. Since $f$ is increasing then all pieces of that region that lie below $y=x$ can be reflected along $y=x$ and not overlap those pieces that are above $y=x$. The area of the resulting figure is the integral with the absolute value. At the same time it is contained inside the top triangle in which $y=x$ divides $[0,1]^2$, which has area $1/2$. Apr 15, 2019 at 13:31
• @user647486 Can it be turned to an analytical language? Cuz this function need not to be continuous and its area may not be defined. Apr 15, 2019 at 13:41
• It is monotonic, it can only have countably many jump discontinuities. Therefore, its integral is defined. Apr 15, 2019 at 13:45
• @user647486 I know that integral is defined. But it's not so strict to talk about the area. Apr 15, 2019 at 13:50
• Area is, by definition, the integral. Anywhere where you see the word 'area' and don't like it, replace it with the word integral. Apr 15, 2019 at 13:52

## 3 Answers

Here we present a bit different, calculus-themed approach. In this answer, we will assume that $$f : [0, 1] \to [0, 1]$$ is monotone-increasing. We also write $$I(f) = \int_{0}^{1} |f(x) - x| \, \mathrm{d}x$$ for brevity.

Step 1 - Proof under extra assumptions. Assume further that $$f$$ is piecewise-smooth, $$f(0) = 0$$, and $$f(1) = 1$$. Then by the formula $$\int |x|\,\mathrm{d}x=\frac{1}{2}x|x|+\mathsf{C}$$, we have

$$\int_{0}^{1} |f(x) - x|(f'(x)-1) \, \mathrm{d}x = \left[ \frac{1}{2}|f(x)-x|(f(x)-x) \right]_{0}^{1} = 0.$$

In particular,

$$I(f) = \frac{1}{2}\int_{0}^{1} |f(x) - x|(f'(x)+1) \, \mathrm{d}x.$$

Now pick $$\alpha \in [0, 1]$$ so that $$f(\alpha) + \alpha = 1$$. (This is possible since $$x \mapsto f(x)+x$$ increases from $$0$$ to $$2$$. Then by triangle inequality,

\begin{align*} \int_{0}^{\alpha} |f(x) - x|(f'(x)+1) \, \mathrm{d}x \leq \int_{0}^{\alpha} (f(x)+x)(f'(x)+1) \, \mathrm{d}x = \frac{1}{2}. \end{align*}

Similarly, by writing $$|f(x)-x| = |(1-f(x))-(1-x)| \leq (1-f(x)) + (1-x)$$, we get

\begin{align*} \int_{\alpha}^{1} |f(x) - x|(f'(x)+1) \, \mathrm{d}x \leq \int_{\alpha}^{1} (2-f(x)-x)(f'(x)+1) \, \mathrm{d}x = \frac{1}{2}. \end{align*}

Therefore $$\int_{0}^{1} |f(x)-x| (f'(x)+1) \, \mathrm{d}x \leq 1$$, which in turn implies $$I(f) \leq \frac{1}{2}$$ as required.

Remark. Let $$\gamma(t) = (f(t)+t, f(t)-t)$$. Then $$\int_{0}^{1} |f(t)-t|(f'(t)+1)\,\mathrm{d}t = \int_{\gamma} |y|\,\mathrm{d}x$$ computes the area between the path $$\gamma$$ and the horizontal axis. Note that $$\gamma$$ is essentially the $$-45^\circ$$-rotation of the graph $$y = f(x)$$ up to scaling.

Then the above bounds immediately follow from the fact that the graph of $$\gamma$$ defines a function on $$[0, 2]$$ which is squeezed between lines $$y = \pm x$$ and $$y = \pm (2-x)$$.

Step 2 - General case. For the general case, let $$f_n$$ be the linear interpolation of the points

$$(0, 0), \quad (\tfrac{1}{n},f(\tfrac{1}{n})), \quad \cdots, \quad (\tfrac{n-1}{n}, f(\tfrac{n-1}{n})), \quad (1, 1).$$

Then by monotonicity,

\begin{align*} |I(f_n) - I(f)| &\leq \int_{0}^{1} |f_n(x) - f(x)| \, \mathrm{d}x = \sum_{k=1}^{n} \int_{\frac{k-1}{n}}^{\frac{k}{n}} |f_n(x) - f(x)| \, \mathrm{d}x \\ &\leq \frac{1}{n}\left( [f(\tfrac{1}{n})-0] + \sum_{k=2}^{n-1} [f(\tfrac{k}{n}) - f(\tfrac{k-1}{n})] + [1-f(\tfrac{n-1}{n})] \right) \\ &= \frac{1}{n}, \end{align*}

hence $$I(f_n) \to I(f)$$ as $$n\to\infty$$ and the desired inequality $$I(f) \leq \frac{1}{2}$$ follows from the previous step.

• Do you know if the bound $1/2$ is sharp? I am asking because the same question, only with an upper bound of $1/4$, has been asked here: math.stackexchange.com/q/4436677/42969. Apr 26 at 12:23
• The upper bound $1/4$ is also claimed in this question math.stackexchange.com/q/3190661/42969, but without proof so far. Apr 26 at 12:37
• @MartinR, I believe that the bound is by no means sharp, not properly utilizing the condition that $\int_{0}^{1}(f(x)-x)\,\mathrm{d}x=0$. Pictures seem to suggest that $\frac{1}{4}$ is indeed the sharp upper bound, attained by $f(x)=\frac{1}{2}$. I will think about it. Apr 26 at 18:01

In case you have some familiarity with measure theory:

Take an increasing sequence of simple functions $$f_k \rightarrow f$$ converging pointwise. By the monotone convergence theorem we have $$\lim_{k \rightarrow \infty} \int_0^1 f_k(x) dx = \int_0^1 f(x) dx$$. So it suffices to show the result for simple functions. The functions $$f_k(x) - x$$ will have only finitely many zeroes, so you know how to do this.

• (+1) Just nitpicking, monotone convergence theorem can be replaced by a more elementary argument if needed, thanks to the monotonicity of $f$. Indeed, we can realize $f_k$'s as $$f_k(x) = \sum_{i=1}^{k} f(\tfrac{i-1}{k})\mathbf{1}[\tfrac{i-1}{k} < x \leq \tfrac{i}{k}]$$ so that $$\left|\int_{0}^{1}|f_k(x)-x|\,\mathrm{d}x-\int_{0}^{1}|f(x)-x|\,\mathrm{d}x\right|\leq\int_{0}^{1}|f_k(x)-f(x)|\,\mathrm{d}x\leq\frac{1}{k}.$$ Apr 15, 2019 at 21:21

Here is a sketch of a different argument. Let $$K$$ be the set of monotone increasing functions $$f:[0,1]\to[0,1]$$ that are right continuous and have limits on the left (the càdlàg functions). One may as well restrict the $$f$$ of the problem statement to be in $$K$$, as any $$f$$ can be modified at its (countably few) discontinuities to make it càdlàg and while preserving all the integrals. Now, $$K$$ is convex, and its extreme points are the cumulative distribution functions of the point masses at points $$a\in[0,1],$$ together with the zero function. In particular, the simple functions constant on $$[0,a)$$ and on $$[a,1]$$. (This claim is, in effect, that the point masses and the zero measure are the extreme points of the sub-probability measures on $$[0,1]$$.) Now, by a theorem of Dubins (see also), the extreme points of $$K\cap L$$, where $$L$$ is the set of $$f$$ for which $$\int_0^1 f = 1/2$$ are all convex combinations of at most two extreme points of $$K$$. Which is to say, simple functions with at most 2 discontinuities. Since $$f\mapsto\int_0^1|f(x)-x|dx$$ is continuous and convex, the maximum is attained at an extreme point of $$K\cap L$$. As the comments indicate, $$\int_0^1|f(x)-x|dx$$ is equal to the area of the union of finitely many triangles, the sum of whose heights is $$\le 1$$ and the sum of whose lengths is $$\le 1$$ and hence have total area $$\le 1/2$$.

• Whoops! I'll edit it. Apr 15, 2019 at 16:01