$\int_0^1\bigl( f(x)+f^{-1}(x )\bigr)\>dx=1$

Question

Given any strictly-increasing continuous function $f$ such that $f(0) = 0$ and $f(1) = 1$, show that $$\int_0^1 [f(x)+f^{-1}(x) ]dx= 1.$$

I tried using mean value theorem but it won't work. I tried few examples it is correct.

Edit:

If I draw graph of picture, and seeing integral as area, we can see the integral is just area of the square $\{(0,0),(0,1),(1,1),(1,0)\}$.

Please give me a hint to attack this problem analytically.

The inverse is the function $f$ but reflected at $y=x$, if I recall correctly. I think it may be of help. — I was suspended for talking, Jun 23 '19 at 12:17
@xbh,@sisyphus,thanks, i got geometric idea , I'm looking for analytical ways . — Cloud JR K, Jun 23 '19 at 12:20
If you are allowed to use conclusions from Riemann-Stieltjes integration, then the theorem of change variables and integration by parts would work. — xbh, Jun 23 '19 at 12:32
Possible duplicate of Show: $f(a) = a,f(b) = b \implies \int_a^b \left[ f(x) + f^{-1}(x) \right] , \mathrm{d}x = b^2 - a^2$ — Martin R, Jun 23 '19 at 13:00

score 4 · Accepted Answer · answered Jun 23 '19 at 12:43

4

By the change of variables theorem, if $f:[0,1]\to[0,1]$ is differentiable then $$\int_0^1f^{-1}(x)\,dx=\int_{f(0)}^{f(1)}f^{-1}(x)\,dx=\int_{0}^{1}f^{-1}(f(x))f'(x)\,dx=$$ $$=\int_0^1 xf'(x)\,dx=[xf(x)]_0^1-\int_0^1 f(x)\,dx=1-\int_0^1 f(x)\,dx$$ using integration by parts

answered Jun 23 '19 at 12:43

augustoperez

3,216

1

If $f$ isn't absolutely continuous this won't work. – Umberto P. Jun 23 '19 at 13:55
@UmbertoP. thanks for the note! Could differentiable + strictly increasing imply absolute continuity? – augustoperez Jun 23 '19 at 16:20
1

Unfortunately not. If $g$ is any nondecreasing non-AC function, then $x \mapsto x+g(x)$ is strictly increasing but not AC. – Umberto P. Jun 23 '19 at 18:53

score 1 · Answer 2 · answered Jun 23 '19 at 12:22

1

If $\mu$ is two dimensional Lebesgue measure then $1=\mu (0,1) \times (0,1))=\mu \{(x,y): y\leq f(x)\} + \mu \{(x,y): y> f(x)\}=\mu \{(x,y): y\leq f(x)\} + \mu \{(x,y): x <f^{-1}(y)\}$.

Use Fubini's Theorem to see that the first term is $\int_0^{1} f(x)dx$ and the second term is $\int_0^{1}f^{-1}(y)dy$.

answered Jun 23 '19 at 12:22

Kavi Rama Murthy

311,013

Didn't expect ISI professor here . Thanks sir – Cloud JR K Jun 23 '19 at 12:25
1

This proof is analytical as asked by OP. Why the downvote? – Kavi Rama Murthy Jun 23 '19 at 12:25

$\int_0^1\bigl( f(x)+f^{-1}(x )\bigr)\>dx=1$

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