Integral of an unknown $f$

Real Analysis
Post Reply
User avatar
Riemann
Posts: 176
Joined: Sat Nov 14, 2015 6:32 am
Location: Melbourne, Australia

Integral of an unknown $f$

#1

Post by Riemann »

Let $f:\mathbb{R} \rightarrow \mathbb{R}$ be a differentiable function in $[0, 1]$ , strictly monotonic , $f(0)=1$ as well as $f(f(x))=x$ forall $x \in \mathbb{R}$. Evaluate:

$$\mathcal{J} = \left(\int_0^1 \left(x - f(x) \right)^{2016} \, {\rm d}x\right)^{-1}$$
(Exam Question at ISI)
Message
Welcome to $2017$.
$\displaystyle \sum_{n=1}^{\infty}\frac{1}{n^s}= \prod_{p \; \text{prime}}\frac{1}{1-p^{-s}}$
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Integral of an unknown $f$

#2

Post by Papapetros Vaggelis »

Hi Riemann. We have that

\(\displaystyle{f(f(0))=0\implies f(1)=0}\) and since \(\displaystyle{f}\) is continuous and strictly decreasing at \(\displaystyle{\left[0,1\right]}\)

we get \(\displaystyle{f(\left[0,1\right])=\left[f(1),f(0)\right]=\left[0,1\right]}\). Also, according to the

relation \(\displaystyle{f(f(x))=x\,,\,\forall\,x\in\mathbb{R}}\), we have that \(\displaystyle{f^{-1}=f}\)

at \(\displaystyle{\left[0,1\right]}\). There exists \(\displaystyle{a\in\left[0,1\right]}\) such that \(\displaystyle{f(a)\neq a}\).

Then, \(\displaystyle{\int_{0}^{1}(x-f(x))^{2016}\,\mathrm{d}x>0}\) and

\(\displaystyle{\begin{aligned} J^{-1}&=\int_{0}^{1}(x-f(x))^{2016}\,\mathrm{d}x\\&=\int_{0}^{1}(1-f^\prime(x))\,(x-f(x))^{2016}\,\mathrm{d}x+\int_{0}^{1}f^\prime(x)\,(x-f(x))^{2016}\,\mathrm{d}x\,\,(I)\end{aligned}}\)

where

\(\displaystyle{\int_{0}^{1}(1-f^\prime(x))\,(x-f(x))^{2016}\,\mathrm{d}x=\left[\dfrac{(x-f(x))^{2017}}{2017}\right]_{0}^{1}=\dfrac{2}{2017}}\)

and in order to calculate the integral \(\displaystyle{\int_{0}^{1}f^\prime(x)\,(x-f(x))^{2016}\,\mathrm{d}x}\)

we make the sub \(\displaystyle{u=f(x)\,,u\in\left[0,1\right]}\) and now \(\displaystyle{x=f^{-1}(u)=f(u)}\)

so,

\(\displaystyle{\int_{0}^{1}f^\prime(x)\,(x-f(x))^{2016}\,\mathrm{d}x=-\int_{0}^{1}(f(u)-u)^{2016}\,\mathrm{d}u=-J}\)

Finally, the relation \(\displaystyle{(I)}\) gives us

\(\displaystyle{J=\dfrac{2}{2017}-J\iff J=\dfrac{1}{2017}\iff J^{-1}=2017}\).
Papapetros Vaggelis
Community Team
Posts: 426
Joined: Mon Nov 09, 2015 1:52 pm

Re: Integral of an unknown $f$

#3

Post by Papapetros Vaggelis »

A function which satisfies all the conditions of the above exercise is \(\displaystyle{f(x)=1-x\,,x\in\mathbb{R}}\).
Post Reply

Create an account or sign in to join the discussion

You need to be a member in order to post a reply

Create an account

Not a member? register to join our community
Members can start their own topics & subscribe to topics
It’s free and only takes a minute

Register

Sign in

Who is online

Users browsing this forum: Ahrefs [Bot] and 13 guests