Show $\int_0^a f(x)\,dx + \int_0^b f^{-1}(x)\,dx \ge ab$ for strictly increasing function $f(x)$

Question

Let $f: [0, \infty) \to [0, \infty)$ be continuous and strictly increasing and with $f(0) = 0$. Prove that $$\int_0^a f(x)\,dx + \int_0^b f^{-1}(x)\,dx \ge ab$$ for any $a, b > 0$, and give a condition for equality to hold.

Answer 1

We remark that the integrals are well-defined by Theorem 6.9 of Rudin's Principles of Mathematical Analysis.

Theorem 6.9 (Rudin). If $f$ is monotonic on $[a, b]$, and if $\alpha$ is continuous on $[a, b]$, then $f \in \mathscr{R}(\alpha)$. (We still assume, of course, that $\alpha$ is monotonic.)

By definition, there exists a sequence of partitions $(P_n)$ of $[0, a]$ such that$$U(P_n, f) \to \int_0^a f(x)\,dx$$and sequence of partitions $(Q_n)$ of $[0, b]$ such that$$L(Q_n, f^{-1}) \to \int_0^b f^{-1}(x)\,dx.$$Let $P_n' = P_n \cap f^{-1}(Q_n)$, $\tilde{P}_n = P_n' \cap [0, a]$, $Q_n' = Q_n \cap f(P_n)$, and $\tilde{Q}_n = Q_n' \cap [0, b]$. Then $U(\tilde{P}_n, f) \le U(P_n, f)$, $L(\tilde{Q}_n, f^{-1}) \ge L(Q_n, f^{-1})$ and so we have$$U(\tilde{P}_n, f) \to \int_0^a f(x)\,dx,\text{ }L(\tilde{Q}_n, f^{-1}) \to \int_0^b f^{-1}(x)\,dx.$$Replace $P_n$ by $\tilde{P}_n$ and $Q_n$ with $\tilde{Q}_n$. Because $f$ is strictly increasing,$$U(P_n, f) = \sum_{i=1}^p f(x_i)(x_i - x_{i-1}),$$where $P_n = \{x_0, \dots, x_p\}$, and similarly,$$L(Q_n, f^{-1}) = \sum_{j=1}^q f^{-1}(y_{j-1})(y_j - y_{j-1}),$$where $Q_n = \{y_0, \dots, y_q\}$. Also,$$ab = \left(\sum_{i=1}^p (x_i - x_{i-1})\right)\left( \sum_{j=1}^q(y_j - y_{j-1})\right).$$Let $P_n' = \{x_0, \dots, x_p, \dots, x_{p'}\}$ and $Q_n' = \{y_0, \dots, y_q, \dots, y_{q'}\}$. Realizing that, we can write$$U(P_n, f) = \sum_{i=1}^p \sum_{y_j \le f(x_i)} (y_j - y_{j-1})(x_i - x_{i-1}) = \sum_{i=1}^p \sum_{y_j \le f(x_i)} (x_i - x_{i-1})(y_j - y_{j-1})$$and$$L(Q_n, f^{-1}) = \sum_{j=1}^q \sum_{x_i \le f^{-1}(y_{j-1})}(x_i - x_{i-1})(y_j - y_{j-1}) = \sum_{j=1}^q \sum_{f(x_i) < y_j} (x_i - x_{i-1})(y_j - y_{j-1}),$$we obtain $ab \le U(P_n, f) + L(Q_n, f^{-1})$ since the left-hand side involves summing over fewer terms. Taking limits gives the required identity.

When $f(a) = b$, we obtain equality because the sums are equal. To see the converse, we use an argument like the one in Exercise 6.2 of Rudin's Principles of Mathematical Analysis.

Exercise 6.2 (Rudin). Suppose $f \ge 0$, $f$ is continuous on $[a, b]$, and $\int_a^b f(x)\,dx = 0$. Then $f(x) = 0$ for all $x \in [a, b]$. (See here for a solution.)

If $f(a) > b$, find a neighborhood $N$ of which this is true; one of the sums in the above argument will consist of more terms and the extra contribution will not converge to zero as this contribution is giving $\int_N (f(x) - b)\,dx > 0$. Similarly, if $f(a) < b$.

Answer 2

Note that $$\int_0^cf(x)\,dx+\int_0^{f(c)}f^{-1}(x)\,dx=cf(c)$$

You can think of this as the area of the rectangle with $(0,0)$ for one corner and $(c,f(c))$ for the opposite corner.

Without loss of generality, let $f(a)\le b$. Then $$\int_0^af(x)\,dx+\int_0^bf^{-1}(x)\,dx=af(a)+\int_{f(a)}^bf^{-1}(x)\,dx$$

Now since $f(x)$ is strictly increasing, we have $f^{-1}(x)>a$ for $x>f(a)$. Hence $$\int_{f(a)}^bf^{-1}(x)\,dx\ge a\left(b-f(a)\right)$$ with equality if and only if $b=f(a)$.

Putting this together, we get $$\int_0^af(x)\,dx+\int_0^bf^{-1}(x)\,dx\ge af(a)+a(b-f(a))=ab$$ with equality if and only if $b=f(a)$.