The sum of integrals of a function and its inverse: $\int_{0}^{a}f+\int_{f(0)}^{f(a)}f^{-1}=af(a)$

Question

Regarding real numbers, the following appears to be true, or at least true with some modifications. Could you help me for the proof? $$\int_0^af(x)dx+\int_{f(0)}^{f(a)}f^{-1}(x)dx=af(a)$$

Answer 1

Geometrically, this equality represents:

The area of the rectangle with side lengths $a\times f(a)$ is equal to the sum of the area below the graph of $f(x),\,x\in[0,a]$ and the area above the graph of $f(x)$, which completes the rectangle. The later area is in fact the area below the inverse of $f$ in the interval $[0,f(a)]$.

Addition: The minimal requirements on the function $f$ are to be continuous and injective (differentiability is not needed), so that $f^{-1}$ makes sense. The condition of continuity is necessary, because otherwise it might happen that the range of $f$ is a strict subset of $[f(0),f(a)]$ and that $f^{-1}$ is not defined on a whole subinterval of $[f(0),f(a)]$ and so the integral $\int\limits_{f(0)}^{f(a)}{f^{-1}(y)dy}$ would not make sense. Now, because $f$ is continuous and a bijection from $[0,a]$ to $[f(0),f(a)]$ it follows that $f$ is strictly monotone.

Case 1 ($f$ is increasing) (look at the picture of Nikolaos Skout):

For each point $M(x_M,y_M)$ in the rectangle $a\times f(a)$ we have that either $y_M>f(x_M)$ ($M$ is above the graph of $f$, i.e $M\in B$) or $y_M\leq f(x_M)$ ($M$ is below the graph of $f$, i.e $M\in A$). Therefore $A\cap B=\emptyset$ and $A\cup B=$ the rectangle $\{a\times f(a)\}$ $\quad\quad(*)$.

Also we have that according to the Rieman integral ($f$ and $f^{-1}$ are integrable as continuous functions defined on fintie intervals) that $\mu(A)=\int\limits_{0}^{a}{f(x)dx}$ and $\mu(B)=\int\limits_{f(0)}^{f(a)}{f^{-1}(y)dy}$, where $\mu$ denotes area. Now according to $(*)$ it follows that $\mu(\{a\times f(a)\})=\mu(A)+\mu(B)$.

Case 2 ($f$ is decreasing) is done analogously, having in mind that $\int\limits_{f(0)}^{f(a)}{f^{-1}(y)dy}$ is a negative value, because $f(0)>f(a)$.

Answer 2

A geometric approach is as follows: the integral $\displaystyle \int_{0}^{a}f(x)dx$ is represented by the region A and the
$\displaystyle \int_{f(0)}^{f(a)}f^{-1}(x)dx$ is represented by the region B. Of course the sum of the areas of the above regions equals the whole area, that is af(a), as desired.

Answer 3

We assume $f$ has an inverse over the interval $(0,a)$ and both $f$ and $f^{-1}$ are smooth. Letting $x = f^{-1}(y) \iff y = f(x)$, by substitution we have

$$\int_{f(0)}^{f(a)} f^{-1}(y) \, dy = \int_0^a x f'(x) dx$$

Using integration by parts, we have $$\int_0^a x f'(x) dx = \int_0^a x df(x) = xf(x) \Big\vert_{x=0}^{x=a} - \int_0^a f(x) dx = af(a) - \int_0^a f(x) dx $$

Answer 4

Hint: (Assuming $f$ is differentiable). Differentiate the function $g$ defined by $$ g(a) = \int_0^a f + \int_{f(0)}^{f(a)} f^{-1} - af(a) $$ (which is indeed differentiable as $f$ is).

In detail (place your mouse over the gray area to reveal its contents)

$$g^\prime(a) = f(a) + f^\prime(a)\cdot f^{-1}(f(a)) - (af^\prime(a) + f(a)) = f(a) + af^\prime(a) - af^\prime(a) - f(a) = 0$$ using the chain rule to derive the second term. To conclude, observe that $g(0)=0$.

Answer 5

Here is another solution based on some elements of Lebesgue-Stieltjes integration.

Proposition: If $f$ is strictly monotone increasing and continuous over the interval $[0,c]$, then for any $0\leq a\leq c$ \begin{align}\int^a_0 f(x)\,dx + \int^{f(a)}_{f(0)} f^{-1}(y)\,dy = a f(a)\tag{0}\label{zero}\end{align}

Proof: As $f$ is strictly monotone increasing continuous on $[0,c]$, $f$ is invertible on $[f(0),f(c)]$. There are unique Borel finite continuous measures $\mu_f$ and $\mu_{f^{-1}}$ on $(0,c]$ and $(f(0), f(c)]$ respectively such that $f(s)-f(t)=\mu_f((s,t])$ and $f^{-1}(v)-f^{-1}(u)=\mu_{f^{-1}}((u,v])$ for all $0\leq t<s\leq c$ and $f(0)<u<v\leq f(c)$. Let $0<a\leq c$. Notice that

$$\{(x,y): 0<x\leq a, f(0)<y\leq f(x)\}=\{(x,y):f(0)<y\leq f(a), f^{-1}(y)\leq x\leq a\}$$

An application of Lebesgue integration by parts and Fubini-Tonelli's there yield

\begin{align} \int_{(0,a]} f(x)\,dx &= af(a)-\int_{(0,a]} x\mu_f(dx)\\ &=af(a)-\int_{(0,a]}\int_{(f(0),f(x)]}\mu_{f^{-1}}(dy)\,\mu_f(dx)\\ &=a f(a)-\int_{(f(0),f(a)]}\int_{[f^{-1}(y),a]}\mu_f(dx)\,\mu_{f^{-1}}(dy)\\ &=af(a)-\int_{(f(0),f(a)]}(f(a)-y)\mu_{f^{-1}}(dy)\\ &=af(a)-f(a)a+\int_{(f(0),f(a)]}y\mu_{f^{-1}}(dy)\\ &=f(a)a-\int_{(f(0),f(a)]} f^{-1}(y)\,dy \end{align}

---

Remark: With very little additional effort, from \eqref{zero} we obtain the following well known result:

(Young's inequality) If $f$ is strictly monotone continuous over $[0,c]$, then for any $0\leq a\leq c$ and $f(c)\leq b\leq f(c)$ $$\int^a_0f(x)\,dx+\int^b_{f(0)}f^{-1}(y)\,dy\geq ab$$ with equality when $b=f(a)$.

If $f(0)\leq b\leq f(a)$, then \begin{align} \int^{f^{-1}(b)}_0 f(x)\,dx+\int^b_{f(0)} f^{-1}(y)\,dy&=bf^{-1}(b)=ba+b(f^{-1}(b)-a) \end{align} Hence \begin{align} ab&=\int^{f^{-1}(b)}_0 f(x)\,dx+b(a-f^{-1}(b)) + \int^b_{f(0)} f^{-1}(y)\,dy\\ &< \int^{f^{-1}(b)}_0 f(x)\,dx + \int^a_{f^{-1}(b)}f(x)\,dx+\int^b_{f(0)} f^{-1}(y)\,dy\\ &=\int^a_0 f(x)\,dx + \int^b_{f(0)} f^{-1}(y)\,dy \end{align}

Similarly, if $f(a)<b\leq f(c)$, then \begin{align} \int^a_0 f(x)\,dx+\int^{f(a)}_{f(0)}f^{-1}(y)\,dy=af(a)=ab+a(f(a)-b) \end{align} Hence \begin{align} ab& =\int^a_0 f(x)\,dx+\int^{f(a)}_{f(0)}f^{-1}(y)\,dy+a(b-f(a))\\ &< \int^a_0 f(x)\,dx+\int^{f(a)}_{f(0)}f^{-1}(y)\,dy+\int^b_{f(a)}f^{-1}(y)\,dy\\ &=\int^a_0f(x)\,dx+\int^b_{f(0)}f^{-1}(y)\,dy \end{align}