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I've a question which is mentioned below.

If $f(f(x)) = x$ where $x \in [0, 1]$ and $f(0) = 1$, then find the value of $\displaystyle\int_0^1 (x - f(x))^{2n} dx$ where $n\in \mathbb{N}.$

I tried it and I think I've solved it. I wish if someone could check my work or provide any alternative or easy way to do the same.

Here's my work.

Given that $f(f(x)) = x$ so $f(x) = f^{-1}(x)$ $\forall\ x\in [0, 1] $ assuming inverse exists.

And since $f(0) = 1$, we have $f^{-1}(0) = 1$.

Now, from the definition of inverse function, graph of $f(x)$ and $f^{-1}(x)$ are mirror images of each other about the line $y = x$. Since $f(0) = 1, f^{-1}(1) = 0$. This means that $f(x)$ and $f^{-1}(x)$ are coincident lines with equation $x + y = 1$ or $f(x) = 1 -x$.

Graph of the same is attached below. enter image description here Now
Since, $f(x) = 1-x$, our next task is to find $\displaystyle \int_0^1 (x - f(x))^{2n} dx$ i.e. $\displaystyle\int_0^1 (2x - 1)^{2n} dx$

which is not a hard nut to crack.

We have, $$\int_0^1 (2x - 1)^{2n} dx = \frac{(2x-1)^{2n+1}}{2(2n+1)}\bigg|^1_0 = \frac{1}{2(2n+1)} + \frac{1}{2(2n+1)} = \boxed{\frac{1}{2n+1}}.$$

I'm unsure if assuming inverse of the function exists is right.

Can anyone please check my work and let me know if I can improve it?

Utkarsh
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    The community appreciates your efficient use of mathjax as a new user. – insipidintegrator Sep 28 '22 at 11:05
  • "-2" on my post within seconds. Is it how community appreciates? – Utkarsh Sep 28 '22 at 11:06
  • Yes, I can explain that $f(x)$ can't be $e^x$ because I've already proved that $f(x) = f^{-1}(x)$. So I think $f(x)$ is surely a line. – Utkarsh Sep 28 '22 at 11:08
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    I agree. Nice . Upvoted because I think the soln is correct assuming inverse exists, and also because this is an exemplary post for a new user to write, as compared to what we generally receive daily. – insipidintegrator Sep 28 '22 at 11:11
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    There are a lot of possibilities for $f(x)$. For instance, $f(x) = \sqrt{1 - x^2}$ would work also. Your attempt is at best an instance of "if the answer is well-defined, then it is $\frac1{2n+1}$." – Brian Moehring Sep 28 '22 at 11:17
  • @BrianMoehring no, that is $f(f(x)) = |x|$ – Exodd Sep 28 '22 at 11:19
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    @Exodd and since $x \in [0,1]$? – Brian Moehring Sep 28 '22 at 11:19
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    You conclusion $f(x) = 1-x$ is wrong, I think. See https://math.stackexchange.com/q/3422890/42969 or https://math.stackexchange.com/q/1776769/42969 for some correct solutions. – Martin R Sep 28 '22 at 11:20
  • ok, I admit I missed the domain – Exodd Sep 28 '22 at 11:20
  • Thanks @MartinR – Utkarsh Sep 28 '22 at 11:20
  • If you also know that $f$ maps $[0,1]$ into itself then $f$ is necessarily bijective. – geetha290krm Sep 28 '22 at 11:27
  • $f$ is bijective if you assume that $f$ has domain $[0,1]$, because $f(f(x))$ is bijective. However, the substitution $x \to f(x)$ which was used in linked questions needs some justification. Indeed, one cannot "naively" perform the substitution by saying $y = f(x)$, then $dy = f'(x)dx$. However, hopefully $f$ is of bounded variation, so writing $f = h_1 - h_2$ where $h_1,h_2$ are monotone increasing functions allows the use of the Riemann-Stieltjes integral where one can somehow work out the rigorous details. What's here isn't "right", but is good for a multiple-choice exam. – Sarvesh Ravichandran Iyer Sep 28 '22 at 11:35
  • I assume that, were this to be given as an answer with complete details, $f$ has to be differentiable (or continuous , I believe one can get by with that much) but without any regularity, I find it difficult to believe that the integral can even be performed (measurability issues for $f$). – Sarvesh Ravichandran Iyer Sep 28 '22 at 11:41
  • Reminds me of my (incorrect) work in this question-https://math.stackexchange.com/questions/4302061/if-fffxfx-2-for-all-0%e2%89%a4x%e2%89%a42-then-find-int-02-fx-dx – Amadeus Sep 28 '22 at 11:49
  • Searching on this site before posting is really useful https://approach0.xyz/search/?q=OR%20content%3A%24%5Cint_%7B%5C%200%7D%5E1%5Cleft(x%5C%20-%5C%20f%5Cleft(x%5Cright)%5Cright)%5E%7B2n%7Ddx%24&p=1 – Amadeus Sep 28 '22 at 11:56
  • @Amadeus That is really nice. I learnt something from your post. – Utkarsh Sep 29 '22 at 16:06
  • There should be some more assumptions on $f$. For example, if we allow discontinuity $f(x) =\begin{cases} 1-x & x \in [0,\alpha)\cup(1-\alpha,1]\ x & \text{otherwise} \end{cases}$ – Dark Malthorp Oct 26 '22 at 15:40
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    If we do not restrict $f(x)$ to map $[0,1]$ to itself, I think you can get any real number as the result. – Dark Malthorp Oct 26 '22 at 15:51

1 Answers1

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I assume that $f(x)$ is monotonic, continuous and differentiable over $(0,1)$.

Let$$t=f(x)$$ then $$f(t)=x$$ $$f'(t)\mathbb dt=\mathbb dx$$ Therefore, \begin{align} I=\int_0^1\left(x-f(x)\right)^{2n}\mathbb dx&=\int_1^0\left(f(t)-t\right)^{2n}f'(t)\mathbb dt\\ &=-\int_0^1\left(x-f(x)\right)^{2n}f'(x)\mathbb dx\\\\\\ \therefore\,2I&=\int_0^1\left(x-f(x)\right)^{2n}(1-f'(x))\mathbb dx\\ &=\frac{1}{2n+1}\left[\left(x-f(x)\right)^{2n+1}\right]_0^1\\ &=\frac{2}{2n+1}\\\\ I&=\boxed{\frac{1}{2n+1}} \end{align}

Note: $f(x)=1-x$ is not the only $f(x)$ that meets the condition. One example is $f(x)=\dfrac{1-x}{1+x}$. I presume there are infinitely many $f(x)$'s.

Kay K.
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